Algebraic Topology

Wojciech Kryszewski

Algebraic Topology

For Practitioners

Uniwersytet Mikołaja Kopernika Toruń 2009

Contents

Introduction 1 0.1 General notation ...... 1 0.2 General topology ...... 1 0.3 Diﬀerential Calculus ...... 6 0.4 Algebra ...... 9 0.4.A Exterior Algebra ...... 9 0.4.B Complex exterior algebra ...... 12 0.4.C Free R-Modules; Specker modules ...... 15 0.4.D Exact sequences ...... 16 0.4.E Functors ⊗, Hom and Ext ...... 17 0.4.F Direct and inverse systems ...... 19 0.4.G Limits and (co)homology ...... 30 0.5 Functional and nonsmooth analysis ...... 31

1 Categories, Functors and Algebraization of Topological Problems 32 1.1 Categories and Functors ...... 32 1.2 Algebraization ...... 34 1.3 Theory of homotopy ...... 37 1.3.A The space of continuous maps ...... 37 1.3.B Extensions ...... 39 1.3.C Homotopy ...... 41 1.3.D Special Homotopies ...... 45 1.3.E Homotopy Sets ...... 46 1.3.F The Index of a Function relative to a Curve ...... 49 1.4 Absolute Homotopy Groups of a Space ...... 54

2 Differential Forms, de Rham cohomology and singular cubical homology 56 2.1 Gradient Maps Revisited ...... 56 2.2 Differential Forms ...... 60 2.3 De Rham’s Cohomology ...... 65 2.4 Integration of differential forms and smooth singular homology ...... 68 2.4.A Singular homology ...... 71 2.4.B Integration over chains ...... 74 2.5 Orientation of chains ...... 80 2.5.A The Stokes Theorem revisited ...... 82 2.5.B Index of a cycle, topological degree of a continuous map ...... 87

3 General Constructions of (Co)homology of complexes 91 3.1 Homology and cohomology of complexes ...... 91 3.2 General homology and cohomology theories ...... 101 3.2.A Singular homology revisited ...... 102 3.2.B Reduced singular homology theory ...... 112 3.2.C Eilenberg-MacLane theorem ...... 113 3.2.D Various results and consequences of axioms ...... 115 3.2.E Reduced homology ...... 121 3.2.F Mayer-Vietoris sequence ...... 122 3.2.G Compact supports and continuity ...... 124 3.2.H Eilenberg-Steenrod Theorem ...... 127 3.3 Diﬀerentiable singular chains and the De Rham theorem ...... 128 ii Spis treści

4 Various construction of homology and cohomology theories of Čech type 134 4.1 Čech-type homology ...... 134 4.2 Alexander-Spanier cohomology and Massey homology ...... 139 4.2.A Alexander-Spanier cohomology ...... 139 4.2.B Massey homology ...... 144 4.2.C Eilenberg-Steenrod axioms ...... 144 4.2.D Excision axiom ...... 148 4.2.E Continuity of the Massey homology ...... 152 4.3 The Massey homology with compact supports ...... 152 4.4 The Alexander-Spanier cohomology theory with compact supports ...... 152 4.5 The de Rham-Čech Theorem ...... 152 4.6 Final remarks ...... 152

5 Fiber Bundles and Coverings 153

6 Brouwer Theorem 155 6.1 Lemat Spernera ...... 155 Introduction

We shall use diﬀerent techniques in these notes. We assume the basic knowledge of algebra, general topology, functional analysis, diﬀerential and integral calculus in Banach spaces. A good source of necessary material on these subjects are [1], [2], [3] and [4, 5].

0.1 General notation

If X is a metric space, then its metric is denoted by d (or by dX . If A ⊂ X and ε > 0, then B(A, ε) := {x ∈ X | dA(x) = d(x, A) := infa∈A d(a, x) < ε} is the open ball around A of radius ε, D(A, ε) = {x ∈ X | dA(x) 6 ε} is the closed ball and S(A, r) = {x ∈ X | dA(x) = ε}. In n n n−1 n particular D (resp. B ) and S ) is the unit closed (resp. closed) and the unit sphere in R , respectively. ∗ Given a Banach space with a norm k · k, then E is the (Banach) s[ace of continuous linear forms (functionals) over E (in case E is a vector space without topology), then this symbol ∗ denotes the space of all linear forms on E, too). Given p ∈ E and x ∈ E we write hp, xi to ∞ ∗ denote the value p(x). If (pn)n=1 is a sequence from E (sometimes we write, not so formally, ∗ ∗ (pn) ⊂ E ) and p ∈ E , then pn * p means that (pn) converges weakly to p, i.e., for all x ∈ E, limn→∞hpn, xi = hp, xi.

0.2 General topology

Throughout the text by a space we mean a Hausdorff topological space; maps between spaces are always assumed to be continuous, unless otherwise stated (sometimes we explicitly speak of continuity in order to make necessary distinctions). If X is a space and A ⊂ X, then by cl A (or A), int A and ∂A := cl \ int A we denote the closure, the interior and the (topological) boundary of A, respectively. Paracompactness and partitions of unity: Let U, V be families of subsets of a set X. We say that U refines V (or that U is a refinement of V) and write V ≺ U, if, for any U ∈ U, there is V ∈ V such that U ⊂ V . A space X is paracompact if any open covering V of X has an open locally finite refinement U, i.e., V ≺ U and any x ∈ X has a neighborhood W such that #{U ∈ U | U ∩ W 6= ∅} < ∞. For instance any compact space is paracompact and any metrizable space is paracompact (the Stone theorem).

Let X be a space. A family {λs : X → I}s∈S is a partition of unity if: 2 Introduction

1 (i) the family {supp λs}s∈S is a locally finite (closed) covering of X ( ); P 2 (ii) for all x ∈ X, s∈S λs(x) = 1 ( ). Let U be a covering of X. We say that a partition of unity {λs}s∈S is inscribed into U (or that it refines U) if {supp λs}s∈S refines U, i.e., for any s ∈ S, there is Us ∈ U such that supp λs ⊂ Us. We say that {λs}s∈S is subordinate to U = {Us}s∈S if, for any s ∈ S, supp λs ⊂ Us. The following remarkable result is well-known.

0.2.1 Existence of partitions of unity: A space X is paracompact if and only if any open covering U of X admits a partition of unity inscribed into it. To prove it, we need the following lemma.

0.2.2 Lemma: Let {Us}s∈S be an open locally ﬁnite covering of X. There is a an open cover {Vs}s∈S such that cl Vs ⊂ Us for any s ∈ S. Dowód: to be done. The next result, although sometimes very convenient, is less known.

0.2.3 Theorem: Any open covering of a paracompact space admits a subordinate partition of unity.

Proof: Let U = {Us}s∈S. There is a partition of unity {ηa}a∈A inscribed to U. Hence, for any a ∈ A, there is s = r(a) ∈ S such that supp ηa ⊂ Us. Therefore we have deﬁned a function −1 r : A → S. Obviously {r (s)}s∈S is a disjoint cover of A. For s ∈ S, let X λs(x) = ηa(x), x ∈ X, a∈r−1(s)

−1 if r (s) 6= ∅ and λs ≡ 0, otherwise. Clearly λs is well-defined since {supp ηa}a∈A is a locally finite (closed) cover of X. Therefore λs : X → [0, 1] is continuous. Let x ∈ X and consider a neighborhood U of x such that the set Ax := {a ∈ A | U ∩ supp ηa 6= ∅} is finite. We claim that Sx := {s ∈ S | U ∩ supp λs 6= ∅} is also finite. Take s ∈ Sx and observe that the family −1 −1 {ηa ((0, 1])}a∈r−1(s) is locally finite and if λs(y) > 0, then there is a ∈ r (s) such that ηa(y) > 0. Hence −1 [ −1 λs ((0, 1]) ⊂ ηa ((0, 1]). a∈r−1(s) Therefore

−1 [ −1 [ −1 [ supp λs := cl λs ((0, 1]) ⊂ cl ηa ((0, 1]) = cl ηa ((0, 1]) = supp ηa. a∈r−1(s) a∈r−1(s) a∈r−1(s)

Suppose that Sx is inﬁnite; then it contains a countable set {s1, s2, ...} ⊂ Sx. For each i ∈ N, −1 there is ai ∈ r (si) such that U ∩ supp ηai 6= ∅. Of course ai 6= aj if i 6= j, i, j ∈ N. Moreover {a1, a2, ...} ⊂ Ax: contradiction. Hence {supp λs}s∈S is locally ﬁnite. Moreover, for each x ∈ X, X X X X λs(x) = ηa(x) = ηa(x) = 1. s∈S s∈S a∈r−1(s) a∈A

1 Recall that the support of f : X → R is the set supp f := cl {x ∈ X | f(x) 6= 0}. 2Some authors consider a more general notion of a partition of unity – see e.g. [14]. Wstęp 3

0.2.4 Remark: If X is a metric space, then any open covering U admits an inscribed partition of unity {λs}s∈S such that λs is a locally Lipschitz function for all s ∈ S. Indeed, let {Ws}s∈S be a locally ﬁnite open covering reﬁning w U. For s ∈ S and x ∈ X, let 0 gdy x 6∈ Ws; as(x) = d(x, bd Ws) gdy x ∈ Ws.

It is easy to see that as : X → R, s ∈ S, is Lipschitz with constant Ls. Let us deﬁne

−1 λs(x) := as(x)[g(x)] , x ∈ X P where g(x) := s∈S as(x). Since {Ws} is locally finite, g is a correctly defined nonvanishing 0 continuous function being locally Lipschitz with constant Ls. Therefore λs, s ∈ S, is correctly defined and continuous. We claim that λs, ∈ S, is locally Lipschitz. Fix s ∈ S, x ∈ X and s0 ∈ S such that x ∈ Ws0 . Let B(x, 2r) ⊂ Ws0 . For y ∈ B(x, r), g(y) > r. Without loss of generality we may assume that #S0 < ∞ where S0 := {s ∈ S | Ws ∩ B(x, r) 6= ∅}. Then, for y ∈ B(x, r), g(y) = P a (y) and, for y, y0 ∈ B(x, r), s∈S0 s |g(y0)a (y) − g(y)a (y0)| |λ (y) − λ (y0)| = s s s s g(y)g(y0) 6 0 −1 0 0 0 0 (g(y)g(y )) (g(y )|as(y) − as(y )| + as(y )|g(y) − g(y )|) 6 −1 0 0 r (Lsd(y, y ) + |g(y) − g(y )|)

0 0 since as(y ) 6 g(y ). Thus

0 −1 0 0 |λs(y) − λs(y )| 6 r (Ls + Ls)d(y, y ).

Some topological constructions: Let X be a set and let {Xj}j∈J be a family of spaces. Suppose that two families of maps F = {fj : X → Xj}j∈J and G = {gj : Xj → X}j∈J are given.

0.2.5 Weak topology determined by F: There exists the weakest topology in X such that, for each j ∈ J, fj is continuous. This topology is characterized by the following property: given a space Y , a map f : Y → X is continuous if and only if, for any j ∈ J, fj ◦ f : Y → Xj is continuous. Q For instance: (1) let X = j∈J Xj and fj : X → Xj be the projection onto Xj. The weak topology determined by F coincides with the Tychonoﬀ product topology; (2) Let E be a ∗ topological vector space; the weak topology σ(E, E ) in E is the weak topology determined by ∗ E . 0.2.6 Strong topology determined by G: There is the strongest topology in X such that, for all j ∈ J, gj is continuous. This topology is characterized by the following property: given a space Y , a map g : X → Y is continuous if and only if, for each j ∈ J, g ◦ gj : Xj → Y is continuous. W For instance, let X = j∈J Xj (the disjoint union, i.e. the union of disjoint replicas of Xj). The strong topology determined by the inclusions {ij : Xj → X}j∈J coincides with the so-called L topology of the topological union. In this case we usually write X = j∈J Xj. S Let X = j∈J Xj. If we endow X with the strong topology determined by the family of inclusions {ij : Xj → X}j∈J , then A ⊂ X is closed in this topology if and only if, for each j ∈ J, 4 Introduction

the set A ∩ Xj is closed in Xj. S Let X = j∈J Xj have topology. We say that the topology of X is compatible with {Xj} if it coincides with the strong topology determined by inclusions. Suppose that the family {Xj} of subsets of a space X has the following property: for any i, j ∈ J, the set Xi ∩ Xj is closed in Xi and Xj and topologies induced on Xi ∩ Xj by Xi and Xj coincide. The topology of X is compatible with the family {Xj} if and only if A ⊂ X is closed if and only if, for each j ∈ J, the set A ∩ Xj is closed in Xj. For instance: we say that a space X is a k-space (or compactly generated) if its topology is compatible with the family of all compact subsets of X. Exercise: (1) Let X,Y be spaces. Show that a perfect map (i.e. closed and such that the fibers f −1(y), y ∈ Y , are compact) is proper, i.e. f −1(K) is compact for a compact K ⊂ Y . Show that if Y is a k-space, then each proper map is perfect. (2) If a space X is locally compact or satisfies the first axiom of countability, then it is compactly generated.

0.2.7 Quotient and identification topologies: Let X be a space and suppose that R ⊂ X × X is an equivalence relation. In the quotient set X/R the weak topology determined by the quotient projection q : X → X/R is considered. A set U ⊂ X/R is open if and only if q−1(U) is open (in X). This is the so-called quotient topology. By deﬁnition, given a space Y , a map f : X/R → Y is continuous if and only if f ◦q : X → Y is continuous. homeo Exercise: Let R = {Sn−1,Bn} be the decomposition of Dn. Show that Dn/R =∼ Sn.

Note that a decomposition R, i.e., the family R = {Ri}i∈I of pairwise disjoint sets such that S X = i∈I Ri, defines an equivalence relation R in X in the following way: R = {(x, y) ∈ X ×X | x, y ∈ Ri for some i ∈ I}. In this case we write X/R instead of X/R and say that the quotient topology is given by identification (or shrinking) of points from sets Ri, i ∈ I. −1 Conversely, any equivalence relation R defines a decomposition R = {q (y)}y∈X/R. Thus quotient and identification topologies are identical. Quite often we the decomposition of a space is defined by a choice of A ∈ X: it consists of separate points of X \ A and A itself. The quotient space is then denoted by X/A and X/A is said to be obtained by collapsing A to a point. It is also convenient to treat X/A as a pointed space, i.e. the space with a distinguished point ∗. Given a space X, by a suspension SX of X we understand the quotient space of X × I obtained by collapsing X × 0 to one point and X × 1 to another point.

homeo n ∼ n+1 Exercise: Show that SS = S , n > 0. Given topological pairs (X,A) and (Y,B), the smash product (X,A)∧(Y,B) := X ×Y/[X × B ∪ A × Y ].

0.2.8 Fact: Let R be an equivalence relation in a space X. The following conditions are equivalent: (1) the quotient projection q : X → X/R is closed (open); (2) for a closed (open) set A ⊂ X, the set [ {[x]R | [x]R ∩ A 6= ∅} is closed (open); Wstęp 5

(3) for an open (closed) set A ⊂ X, the set [ {[x]R | [x]R ⊂ A} is open (closed). Exercise: Under the above assumptions, let us consider a set-valued map

−1 ϕq : X/R ( X, ϕq(y) := q (y), y ∈ X/R.

Show that the quotient projection q : X → X/R is a closed (open) map if and only if ϕq is upper (lower) semicontinuous.

0.2.9 Corollary: If q : X → X/R is closed, X is paracompact, then X/R is paracompact. If, additionally, for each x ∈ X, the equivalence class [x]R is compact, then the metrizability of X implies the metrizability of X/R. The ﬁrst assertion follows since paracompactness is invariant under closed maps. If the relation R is closing and equivalence classes are compact, then the quotient projection is a 3 perfect map ( ). It is known that metrizability is invariant under perfect maps. 0.2.10 Sewing of spaces and the wedge: We are going now to describe an important example of a certain identiﬁcation topology. Let X, Y be (disjoint) spaces, let A ⊂ X be a closed set and f : A → Y a continuous map. Consider a decomposition R of the topological sum X ⊕ Y comprised of single points of the set X \ A and sets of the form f −1(y) ∪ {y}, y ∈ Y . By the sewing of X and Y along f we understand the quotient space X ∪f Y := X ⊕ Y/R. Let q : X ⊕ Y → X ∪f Y be the quotient projection and let iX : X → X ⊕ Y and iY : Y → X ⊕ Y be the inclusions. Therefore maps

jX := q ◦ iX : X → X ∪f Y, jY : Y → X ∪f Y

−1 are correctly deﬁned and continuous. A set C ⊂ X ∪f Y is open (closed) if and only if jX (C) −1 and jY (C) are open (closed) in X and Y , respectively. Evidently, for B ⊂ X ⊕ Y , q(B) = jX (B ∩ X) ∪ jY (B ∩ Y ). Moreover the following formulae hold:

−1 −1 −1 jX (jY (B)) = f (B), jY (jY (B)) = B, for B ⊂ Y ;

−1 −1 −1 jX (jX (B)) = B ∪ f (f(A ∩ B)), jY (jX (A)) = f(A ∩ B) for B ⊂ X. It follows that: (1)jY is a closed injection, i.e., a homeomorphism onto the closed set jY (Y ); therefore one may identify Y with jY (Y ); (2) jX |X\A is an open injection onto an open set jX (X \ A); moreover X ∪f Y = jX (X \ A) ∪ jY (Y ); these components are disjoint; (3) jX is a closed map if and only if f is a closed map. In particular q is a closed map if and only if f is closed.

0.2.11 Corollary: If X and Y are paracompact, f : A → Y is closed, the sewing X ∪f Y is paracompact. If f is perfect, X, Y are metrizable, then X ⊕ Y and X ∪f Y are metrizable.

Let us remark that the quotient projection q may be treated as a map q :(X,A) → (X ∪f Y,Y ). Moreover q maps homeomorphically X \ A onto (X ∪f Y ) \ Y .

3I.e., a closed map with compact ﬁbers. 6 Introduction

Covering dimension of spaces: Let X be a set and A a family of its subsets. By the order ord (A) of A we mean the largest integers n > −1 such that no subfamily of A consisting of (n + 2)-elements has nonempty intersection (4); if there is no such number, then we put ord (A) = ∞. Now let X be a normal space. We say that X is a space of (covering) dimension 6 n, n > −1, if any ﬁnite open covering of X has an open reﬁnement of order 6 n.

0.2.12 Theorem: Let a space X be paracompact. Then dim X 6 n if and only if any open coverings admits an open reﬁnement of order 6 n.

0.3 Diﬀerential Calculus

A map ϕ : U → F, where U is an open subset of a Banach space E and F is another Banach space, is diﬀerentiable if there is a linear continuous T ∈ L(E, F) such that, for each x ∈ U,

ϕ(x + h) = v(x) + T (h) + o(khk) as h → 0.

r r The map ϕ is of class C (or C -smooth), r > 1, if it is r-times diﬀerentiable and the r-th (r) r derivative ϕ (being a map form U to the Banach space L (E, F) of r-linear continuous maps ∞ ∞ r E × ... × E → F) is continuous. A map is of class C (or C -smooth) if it is of class C for all | {z } r 0 r > 1. We also say that ϕ is of class C when it is continuous. n If ϕ : U → F, where U ⊂ R , is diﬀerentiable and x = (x1, ..., xn), then ∂xi ϕ(x) = ∂iϕ(x) := ∂ ϕ(x) is the partial derivative of f with respect to the i-th variable at x. Sometimes (mainly ∂xi 0 5 in the case ϕ : U → R) we write also ϕxi (x) instead of ∂iϕ(x) ( ). In case U ⊂ R, then we also write ϕ˙(t) = ϕ0(t) to denote the derivative of ϕ at t ∈ U. r n If ϕ is C -smooth, α = (α1, ..., αn) ∈ Z+ (Z+ := {0, 1, 2, ...}) is a multiindex, then |α| := Pn i=1 αi and, if |α| 6 r, then α α ∂ ∂ ϕ(x) := α1 αn ϕ(x). ∂1 ...∂n

r r By a C -diffeomorphism, 1 6 r 6 ∞ we mean we understand a C -smooth and invertible 0 map (i.e., bijection onto its image) ϕ : U → F such that, for each x ∈ U, ϕ (x) is an isomorphism. If ϕ is a Cr-smooth diffeomorphism, then V := ϕ(U) is open and ϕ−1 : V → U is Cr-smooth. n r r In the case E = R , we sometimes say that a C -smooth ϕ : U → F is a C -diffeomorphism if ϕ is invertible and regular, i.e., for all x ∈ U, the vectors ∂iϕ(x), i = 1, ..., n, are linearly independent (6). Remark: In the sequel we shall often speak about Cr-smooth maps. By default any assertion r concerning C -smoothness will be valid for all integers 0 6 r < ∞ and r = ∞ unless stated differently. Differentiability on non-open sets: Sometimes it is convenient to differentiate functions defined on a non-necessarily open set. Suppose that V ⊂ E and ϕ : V → F. We say that r ϕ is differentiable or C -smooth, r > 1 or r = ∞, if there are an open U ⊃ V and a differentiable r (k) (k) or C -smooth map ψ : U → F such that ψ|V = ϕ. For x ∈ V , we put ϕ (x) := ψ (x) for 1 6 k < r + 1 . 4 T I.e., if ord (A) = n, B ⊂ A and A∈B A 6= ∅, then #B 6 n + 1. 5 0 m The symbol ϕi(x) may be confusing since usually for ϕ : E → R , ϕi denotes the i-th coordinate of ϕ. 6Note that a if ϕ is a Cr-diffeomorphism, then ϕ is a homeomorphism onto its image. Wstęp 7

Remark: The problem of the existence of differentiable extensions of a given function ϕ : V → F onto an open neighborhood of V is not well-recognized. One may prove that, for any closed set ∞ A ⊂ E there is a C function ψ : A → F such that ψ|A ≡ 0. The above definition has a disadvantage in that it does not determine the derivatives of ϕ uniquely: usually there are different extensions onto open supersets of V of the same map ϕ (obviously ϕ0(x), as well as higher order derivatives, is determined uniquely if x ∈ int V ); provide a convincing example of such a situation.

On some occasions there is no need to worry. For instance, let ϕ :[a, b] → F, −∞ < a < b < ∞. Then ϕ is (continuously) differentiable if and only if ϕ|(a,b) is (continuously) 0 0 0 0 differentiable and the one-sided derivatives ϕ+(a) and ϕ−(b) exist (and limt→a+ ϕ (t) = ϕ+(a), 0 0 limt→b− ϕ (t) = ϕ−(b)). Indeed if ψ :(a − ε, b + ε) → F is a (continuously) differentiable extension of ϕ, then 0 0 ϕ|(a,b) = ψ|(a,b) is (continuously) differentiable and ψ (a), ψ (b) exist. But

0 ϕ(a + h) − ϕ(a) 0 0 ϕ(a + h) − ϕ(a) 0 ϕ+(a) = lim = ψ (a) and ϕ−(b) = lim = ψ (b). h→0+ h h→0− h Moreover, in case of continuous diﬀerentiability of ψ we see that

0 0 0 0 ϕ+(a) = ψ (a) = lim ψ (t) = lim ϕ (t) t→a+ t→a+ 0 and similarly for ϕ−(b). Conversely, for t ∈ R, set  0  ϕ+(a)(t − a) + ϕ(a) for t 6 a; ψ(t) := ϕ(t) for t ∈ (a, b);  0 ϕ−(b)(t − b) + ϕ(b) for t > b. It is easy to see that ψ is a (continuously) diﬀerentiable extension of ϕ. Of course it the same manner we can discuss the question of Cr-smoothness of maps [a, b] → F. The next example will be of importance. Let U ⊂ E be open and let ϕ : U ×[a, b] → F. Then 1 1 ϕ is C -smooth if and only if ϕ|U×(a,b) is C -smooth and, for each x ∈ U, the partial derivatives 0 0 0 0 ϕx(x, a), ϕx(x, b) and the one-sided partial derivatives ϕt+(x, a), ϕt−(x, b) exist and, moreover, the functions w1, w2 : U × [a, b] → F given by the formulae 0 w1(x, t) := ϕx(x, t), x ∈ U, t ∈ [a, b],  0 ϕt(x, t) for x ∈ U, t ∈ (a, b)  0 w1(x, t) = ϕt+(x, a) for x ∈ U, t = a  0 ϕt−(x, b) for x ∈ U, t = b are continuous, i.e., for all x ∈ U,

0 0 0 0 lim ϕx(y, t) = ϕx(x, a), lim ϕx(y, t) = ϕx(x, b), y→x, t→a+ y→x, t→b− 0 0 0 0 lim ϕt(y, t) = ϕt+(x, a), lim ϕt(y, t) = ϕt−(x, b). y→x, t→a+ y→x, t→b−

1 Indeed: if ψ : W → F is a C -smooth extension of ϕ onto an open superset U × [a, b] ⊂ W ⊂ 0 E × R, then ϕ|U×(a,b) = ψ|U×(a,b) is continuously diﬀerentiable and, for each x ∈ U, ψ (x, a), ψ0(x, b) exist, but (e.g. for t = a)

0 ϕ(x, a + h) − ϕ(x, a) ψ(x, a + h) − ψ(x, a) 0 ϕt+(x, a) = lim = lim = ψt(x, a); h→a+ h h→a+ h 8 Introduction moreover 0 0 0 0 ϕt+(x, a) = ψt(x, a) = lim ψt(y, t) = lim ϕt(y, t). y→x, t→a+ y→x, t→a+ For any x ∈ U and h ∈ E, ϕ(x + th, a) − ϕ(x, a) ψ(x + th) − ψ(x, a) lim = lim = ψ0((x, a); (h, 0)) = ψ0(x, a)(h, 0). t→0 h t→0 t Hence ϕ(·, a) is Gateaux diﬀerentiable at (x, a) and its Gateaux derivative is equal to h 7→ 0 0 ψ (x, a)(h, 0). Clearly it continuously depends on x; hence ϕx(x, a) exists end is equal to ψx(x, a). Therefore 0 0 0 0 ϕx(x, a) = ψx(x, a) = lim ψx(y, t) = lim ϕx(y, t). y→x, t→a+ y→x, t→a+ Conversely, we deﬁne ψ : U × R → F by

 0  ϕt+(x, a)(t − a) + ϕ(x, a) for x ∈ U, t 6 a; ψ(x, t) := ϕ(x, t) for x ∈ U, t ∈ (a, b);  0 ϕt−(x, b)(t − b) + ϕ(x, b) for x ∈ U, t > b. It is clear that ψ is an extension of ϕ. We shall check the continuous differentiability of ψ. It is clear that ψ0(x, t) exists if x ∈ U and t ∈ (a, b) or if t 6∈ [a, b] ; moreover ψ0 is continuous at points of U × (R \{a, b}). Let x ∈ U. In order to check the differentiability of ψ at (x, a) and 0 0 (x, b) it is sufficient that the partial derivatives ψx and ψt exist around (x, a) and (x, b) and are continuous at (x, a) and (x, b). Since these derivatives exist at points of U × (R \{a, b}), we shall check their existence at (x, a) (the same argument will work for (x, b)). To this end observe that, for any h ∈ E,

ψ(x + th, a) − ψ(x, a) ϕ(x + th, a) − ϕ(x) 0 lim = lim = ϕx(x, a)(h). t→0 t t→0 t Thus ψ(·, a) is Gateaux diﬀerentiable with the derivative depending continuously on x; hence 0 ψx(x, a) exists. On the other hand

0 0 ψ(x, a + t) − ψ(x, a) tϕt+(x, a) 0 ψt(x, a) = lim = lim = ϕt+(x, a). t→0 t t→0 t

0 By the very assumption the derivatives ϕx and ϕt are continuous at (x, a). The provided constructions remain valid for derivatives of higher order: we may speak of r (k) C -smoothness of ϕ : U ×[a, b] → F. Arguing as above we see that in this case ϕ , 1 6 k < r+1, is uniquely determined at points of U × [a, b].

Question: Suppose that ϕ : U × [a, b] → F is continuous. Does there exist a continuous extension ψ : U × R → F of ϕ ? The answer is: yes. We shall address this problem in the sequel. There is still another way to define differentiability of functions defined on non-open sets. Suppose that U ⊂ E is open and U ⊂ V ⊂ cl U; we say that a continuous map ϕ : V → F r r is of class C if ϕ|U is of class C and, for any 1 6 k < r + 1, there is a continuous function r (k) (r) vk : cl U → L (E, F) such that vk|U = ϕ |U . Then, for x ∈ V , the k-th derivative ϕ (x), (k) (k) 1 6 k < r + 1, is, by definition, equal to vk(x), i.e., ϕ (x) := limy→x, y∈U ϕ (y). Hence,in this (k) case derivatives ϕ (x), 1 6 k < r + 1 are uniquely determined for any x ∈ cl U. This new definition gives nothing new in case of ϕ :[a, b] → F. For instance, suppose that 1 ϕ is of class C in the sense of this definition: ϕ|(a,b) is continuously differentiable and there is Wstęp 9

∼ 0 a continuous function v :[a, b] → L(R, F) = F such that ϕ (x) = v(x) for x ∈ (a, b). By the Lebesque theorem, for any t ∈ (a, b) and (sufficiently small) h > 0, Z t+h Z t+h ϕ(t + h) − ϕ(t) = ϕ0(s) ds = v(s) ds. t t The continuity of the integral implies that Z t+h Z a+h ϕ(a + h) − ϕ(a) = lim v(s) ds = v(s) ds. t→a+ t a Thus Z a+H 0 ϕ(a + h) − ϕ(a) 1 ϕ+(a) = lim = lim v(s) ds = v(a). h→0+ h h→0+ h a Moreover 0 0 ϕ+(a) = v(a) = lim v(t) = lim ϕ (t). t→a+ t→a+ 0 The same computation is valid for left derivative ϕ−(b). According to the above argument we thus see that ϕ is continuously differentiable in the sense of the first definition. 1 Conversely suppose that there is a C -function ψ :(a − ε, b + ε) → F, ε > 0, such that 0 ψ|[a,b] = ϕ. Then obviously ϕ|(a,b) is continuously differentiable. Moreover ψ |[a,b] is a continuous 0 1 extension of ϕ |(a,b) onto [a, b]. Therefore ϕ is of class C in the sense of the second definition. Analogous statement is true for maps defined on closed cubes, i.e., ϕ : P → F where P = [a1, b1] × ... × [am, bn] where ai, bi ∈ R, ai < bi. In order to speak differentiability or r C -smoothness (1 6 r 6 ∞) we may use both definitions provided above being sure that, for (k) α each t = (t1, ..., tn) ∈ P , derivatives ϕ (t), 1 6 k 6 r and ∂ ϕ(t), |α| < r + 1, are uniquely determined. As before we say that ϕ is regular if ϕ is C1-smooth and, for each t ∈ P , the vectors r ∂iϕ(t), i = 1, ..., n, are linearly independent. We say that ϕ is a C -diffeomorphism, 1 6 r 6 ∞, if ϕ is Cr-smooth, regular and invertible. r r Exercise: Show that a differentiable (C -smooth) ϕ : P → F is regular (a C -diffeomorphism) r r if and only there is a regular (C -smooth) (C -diffeomorphism) map ψ : U → F where U is open, P ⊂ U and ψ|P = ϕ. Some other results concerning differentiability of maps defined on closed cubes may be found in the book of Sikorski [7]. It may be shown that

This new deﬁnition may be applied in the situation of a function ϕ : U × [a, b] → F, too. Again it gives nothing new in comparison to the previous one. We leave to the reader the precise formulation of the statement and its proof.

0.4 Algebra

0.4.A Exterior Algebra

k Let E and F be real Banach spaces and let L (E, F), k > 1, be the space of k-linear continuous k k maps T : E × E × .... × E → F. Recall that L (E, F) is a Banach space. A map T ∈ L (E, F), | {z } k 1 6 k < ∞, is alternating if, for any σ ∈ Sk,

T (eσ(1), ..., eσ(k)) = sgn (σ)T (e1, ..., ek) 10 Introduction

for any e1, ..., ek ∈ E. k Exercise: Show that T ∈ L (E, F) is alternating if and only if T (e1, ..., ek) = 0 for all e1, ..., ek ∈ E such that ei = ej for some 1 6 i 6= j 6 k. k There is a way to alternate a given k-linear map. If T ∈ L (E, F), then 1 X Alt(T )(e , ..., e ) = sgn (σ)T (e , ..., e ) 1 k k! σ(1) σ(k) σ∈Sk where Sk denotes the set of all permutations of the set {1, ..., k} and sgn (σ) is the sign of a permutation σ ∈ Sk. The role of the factor 1/k! is clear. Although the map (e1, ..., ek) 7→ P sgn (σ)T (e , ..., e ) is already alternating, we have added it in order to have a natural σ∈Sk σ(1) σ(k) condition: Alt(Alt(T )) = Alt(T ). k The space A (E, F) of alternating k-linear maps is a closed subspace of the Banach space k k k L (E, F); in particular A (E, F) is a Banach space. If F = R, then we write A (E). Obviously k k Alt : L (E, F) → A (E, F) is a continuous linear map. k m k+m We deﬁne the exterior multiplication ∧ : A (E)×A (E) → A (E), m > 1, by the formula 1 X (T ∧ S)(e , ..., e , e , ..., e ) := sgn (σ)T (e , ..., e )S(e , ..., e ) 1 k k+1 k+m k!m! σ(1) σ(k) σ(k+1) σ(k+m) σ∈Sk+m for e1, ..., ek+m ∈ E. It is clear that equally well we could deﬁne (k + m)! T ∧ S = Alt(T ⊗ S) k!m! where (T ⊗ S)(e1, ..., ek, ek+1, ..., ek+m) := T (e1, ..., ek)S(ek+1, ..., ek+m).

k The multiplication ∧ may be generalized to A (E, H) where H is a Hilbert space. Exercise: Show that; k m k+m (i) ∧ : A (E) × A (E) → A (E) is a bilinear continuous map; k m (ii) For T ∈ A (E) and S ∈ A (E), T ∧ S = (−1)kmS ∧ T ;

k m n (iii) For T ∈ A (E), S ∈ A (E) and U ∈ A (E), T ∧ (S ∧ U) = (T ∧ S) ∧ U;

1 n ∗ (iv) If p , ..., p ∈ E = L(E) and e1, ..., en ∈ E, then 1 2 n i (p ∧ p ∧ ... ∧ p )(e1, ..., en) = det[hp , eji]i,j=1,...n;

1 n ∗ 1 n (v) If p , ..., p ∈ E are linearly dependent, then p ∧ ... ∧ p = 0. 1 n ∗ 0.4.1 The case dim E = n: Let {p , ..., p } be a basis in E and {v1, ..., vn} the basis in E dual i i k to {p }, i.e. hp , vji = δij. Let T ∈ A (E, F). If k > n, then T = 0. Indeed if e1, ..., ek ∈ E, then, Pn j j j for each i = 1, ..., k, ei = j=1 ei vj where ei = hp , eii ∈ R for j = 1, ..., n. Hence n X j1 jk T (e1, ..., ek) = e1 ...ek T (vj1 , ..., vjk ) = 0 j1,j2,..,jk=1 Wstęp 11

since between each k-tuple (vj1 , ...vjk ) at least two vectors are equal. If k > n, then   X X jσ(1) jσ(k) T (e1, ..., ek) =  sgn (σ)e1 ...ek  T (vj1 , ..., vjk ) = 16j1

X jr det[hp , esi]16s,r6kT (vj1 , ..., vjk ) 16j1

i T (e1, ..., en) = det[hp , eji]i,j=1,...,nT (v1, ..., vn), i.e., T = (p1 ∧ p2 ∧ ... ∧ pn) f where f := T (v1, ..., vn).

i1 i2 ik k It follows then {p ∧ p ∧ ... ∧ p }16i1<...

Moreover there is a (linear bijection)

^k ∗ ∗ ∗ ∗ A(E) = E := E ∧ E ... ∧ E . | {z } k

0 0 Remark: It is convenient to set L (E, F) = A (E, F) := F. Vector product: Let n > 3 and let us deﬁne an alternating (n − 1)-linear map T : n n n n R × .... × R → R as follows. If v1, ..., vn−1 ∈ R , then the map | {z } n−1   v1  .  n  .  R 3 w 7→ ϕ(w) := det    vn−1  w is a linear functional; hence, in view of the Riesz theorem, there is a unique vestor T (v1, ..., vn−1) ∈ n R such that hw, T (v1, ..., vn−1)i = ϕ(w).

Instead of T (v1, ..., vn−1) we write v1 × ... × vn−1. This vector is called the vector product of v1, ..., vn−1. It is easy to check that: (i) |v1 × ... × vn−1| is the volume of the prism (pol. graniastosłup) spanned by v1, ..., vn−1; (ii) v1 × ... × vn−1 ⊥ span {v1, ..., vn−1}; (iii) if vectors v1, ..., vn−1 are linearly independent, then the sequence {v1, ..., vn−1, v1 × ... × 12 Introduction

n 7 vn−1} form a base of R equivalent to the canonical base ( ). 2 It is sometimes convenient to introduce a ‘vector product’ in R . Namely repeating the 2 above construction it is natural to put: v× := (−v2, v1) for v = (v1, v2) ∈ R .

0.4.B Complex exterior algebra

Let E, F be complex vector spaces. It is obvious that E, F may also be treated as real vector spaces.

A map T : E → F is C-linear (resp. C-linear) if, for all e1, e2 ∈ E and z1, z2 ∈ C,

T ((z1e1 + z2e2) = z1T (e1) + z2T (e2)(resp. T (z1e1 + z2e2) = z1T (e1) + z2T (e2)) (∗)

We say that T is R-linear if (∗) holds for all z1, z2 ∈ R. Exercise: Check that an R-linear T : E → F is C-linear if and only if T (ie) = iT (e) for all e ∈ E. The sets of continuous -linear, -linear and -linear maps are denoted by L ( , ), L ( , ) C C R C E F C E F and LR(E, F), respectively. Obviously if E and F are complex Banach spaces, then each of this sets has the natural structure of a complex Banach space and L ( , ), L ( , ) are closed sub- C E F C E F spaces of LR(E, F). Observe that any T ∈ LR(E, F) has a unique representation

T = T 0 + T 00 (0.4.1) where T 0 ∈ L ( , ) and T 00 ∈ L ( , ). Indeed, let us put C E F C E F 1 1 T 0(e) := (T (e) − iT (ie)),T 00(e) = (T (e) + iT (ie)), e ∈ . 2 2 E

0 00 It is easy to check that T is C-linear and T is C-linear. Thus, in other words, we have the decomposition L ( , ) = L ( , ) ⊕ L ( , ). R E F C E F C E F 0 Observe that an R-linear T is C-linear if and only if T = T , i.e., T (e) = −iT (ie) for any e ∈ E. k For a positive integer k, by L (E, F) we denote the space of k-R-linear maps E × ... × E → F. R | {z } k As before Ak ( , ) := {T ∈ Lk (E, ) | T is alternating} (8); moreover L0 ( , ) = A0 ( , ) := R E F R F R E F R E F F. For any k 0, Lk (E, ) and Ak ( , ) are complex vector spaces. > R F R E F (p,q) Given p, q > 0 such that p + q = k, we have a (complex vector) subspace L (E, F) of Lk ( , ) consisting of T ∈ L ( , ) such that, for e , .., e ∈ and z ∈ , R E F R E F 1 k E C p q T (ze1, ..., zek) = z z T (e1, ..., ek).

It is easy to see that L(1,0)( , ) = L ( , ) and L(0,1)( , ) = L ( , ). E F C E F E F C E F We put A(p,q)( , ) := L(p,q)( , ) ∩ Ak ( , ) and if T ∈ A(p,q)( , ), then we say that T is E F E F R E F E F (p,q) (p,q) an alternating (p, q)-linear map. Obviously A (E, F) is a closed subspace in L (E, F). 7Two (ordered) bases are equivalent if the determinant of the matrix transferring one of themn onto the other is positive. 8 I.e., for any σ ∈ Sk and e1, ..., ek ∈ E, T (eσ(1), ..., eσ(k)) = sgn (σ)T (e1, ..., ek). Wstęp 13

We have representations

k M (p,q) k M (p,q) LR(E, F) = L (E, F), AR(E, F) = A (E, F). p+q=k p+q=k

As before Lk ( , ), L(p,q)( , ), Ak ( , ) and A(p,q)( , ) are complex Banach spaces if so R E F E F R E F E F are and . Moreover A(p,q)( , ) (resp. L(p,q)( , )) is a closed subspace of Ak ( , ) (resp. E F E F E F R E F Lk ( , ) and Ak ( , ) is a closed subspace of lk ( , ). R E F R E F R E F Exercise: Show that: (p,q) (i) if T ∈ L (E, F), then numbers p, q are uniquely determined; (p,q) (q,p) 9 (ii) if T ∈ L (E) (i.e., F = C), then T ∈ L (E, F) ( ); (p,q) (p0,q0) (p+p0,q+q0) (iii) if T ∈ A (E), S ∈ A (E) , then T ∧ S ∈ A (E). Complex structure and Complexification: Let X be a real vector space and suppose that there is an R-automorphism J : X → X such that J ◦ J = −1X . The pair (X,J) is called a complex structure. Given a complex structure (X,J) let us define the scalar multiplication · : C × X → X by (a + ib)x := ax + bJ(x) for x ∈ X. It is easy to see that X, together with the ordinary addition of vectors and the multiplication defined above, is a complex vector space. The ordinary R-multiplication in X admits an extension to C-multiplication if and only if it admits a complex structure. We have to show the ‘only if’ part: one may ‘multiply’ by i; hence let us define J(x) := ix, x ∈ X. Then J 2(x) = −x for x ∈ X. Exercise: Show that no odd dimensional real vector space admits a complex structure. Consider the (real) vector space X × X and J : X × X → X × X where J(x, y) := (−y, x). It is easy to check that J 2(x, y) = (−x, −y) on X2. Hence X2 admits a complex structure. 2 2 The C-vector space obtained form X (here and above X stands for the real vector space) as above will be denoted by XC or X ⊕ iX and called the complexification of X. Its elements are denoted by x + iy, x, y ∈ X. For (x + iy) ∈ X ⊕ iX and a + ib ∈ C we write (a + ib)(x + iy) := (ax − by) + i(bx + ay) (in other words (a + ib)(x, y) = (ax − by, bx + ay). It is to see that this C-multiplication coincides with the one given by J. Moreover, for x + iy ∈ XC, let x + iy := x − iy. n n ∼ n For instance, we see that (R )C = R ⊕ iR = C . Along with the complexification of a real vector space, on defines the complexification of a linear map f : X → Y where X,Y are real vector spaces. Namely given a linear T : X → Y , let T C”XC → Y C be given by

T C(x + iy) := T (x) + iT (y), ; x + iy ∈ XC.

It is easy to see that T C is C-linear. Fact: Let V be a complex vector space. There is a real vector space X such that V =∼ X ⊕ iX (C-isomorphims). Proof: It is clear that C = R + iR. If B is a basis of V , then V = C(B). Let X := R(B); then X ⊕ iX = V . 9 Here and below, given a C-valued function T : D → C (no matter what domain D), T : D → C denotes the function such that T (x) = T (x) for x ∈ D. 14 Introduction

Let E be a complex vector space and E = X ⊕ iX. If E is a normed (resp. Banach) space, q 2 2 then so is X: for e = x + iy ∈ E where x, y ∈ X, we put kekE := kxkX + kykX . Conversely, if E is a normed (resp. Banach) space, then so is X: it is suﬃcient to put kxk := kx + i0kE for x ∈ X. Let now E = X ⊕ iX, F = Y + iY be complex Banach spaces (and X,Y are real Banach 10 spaces) and let U ⊂ E be open ( ). Given f : U → F we write f(x) = u(x) + iv(x) where u, v : U → Y and say that u, v are the real an imaginary part of f, respectively (11).

Apart from the decomposition of the form (0.4.1), we have that LR(E, F) is the complexiﬁ- cation of the real vector space L(E,Y ), i.e.,

C LR(E, F) = L(E,Y ) = L(E,Y ) ⊕ iL(E,Y ).

More generally we have: Fact: The following equality is true:

k k k AR(E, F) = A (E,Y ) ⊕ iA (E,Y ) (in right-hand side is treated as a real vector space), i.e., Ak ( , ) is a complexiﬁcation of the E R E F k real vector space A (E,Y ). Let A ∈ Ak ( , ) and let T (resp. S) be the real and imaginary part of A, respectively; Proof: R E F hence, for any e1, ..., ek, A(e1, ..., ek) = T (e1, ..., ek) + iS(e1, ..., ek). It is evident that T,S ∈ k k A (E,Y ). Conversely, given T,S ∈ A (E,Y ) we see that A := T + iS, i.e., the map given by

A(e1, ..., ek) := T (e1, ..., ek) + iS(e1, ..., ek), e1, .., ek ∈ E, belongs to Ak ( , ). R E F k If A = T + iS, T,S ∈ A (E,Y ), then obviously T and S are the real and the imaginary part of A, respectively.

Corollary: A ∈ LR(E, F), A = T + iS where T,A ∈ L(E,Y ), is C-linear if and only if T (e) = S(ie) for all e ∈ E. Example: Let U ⊂ E = X ⊕ iX, f : U → C = R ⊕ iR, f = u + iv, where u, v : U → R, 1 2 2 be C -smooth (as a map from X → R ). Hence, for any z ∈ U, z = x + iy, x, y ∈ X, 0 A := f (z) ∈ LR(E, C). The real (resp. imaginary) part of A has the form: for e = e1 + ie2 ∈ E,

0 0 0 0 T (e) = ux(z)(e1) + uy(z)(e2)(resp. S(e) = vx(z)(e1) + vy(z)(e2)).

Consequently 0 A(e) = f (z)(e) = ∂xf(z)(e1) + ∂yf(z)(e2) where ∂xf(z), ∂yf(z) ∈ L(X, C) and

0 0 0 ∂xf(z) = ux(z) + ivx(z) and ∂yf(z) = uy(z) + ivy(z).

0 0 0 0 If we take the complexiﬁcations of ux(z), uy(z), vx(z) and vy(z) we may consider that ∂xf(z) and ∂yf(z) as C-linear maps E → C.

10Evidently U may be treated as on open subset of X2. 11 Generally speaking, given f : D → F, where D is an arbitrary set, f(x) = u(x) + iv(x), where u, v : D → Y are the real and imaginary part of f, respectively. Wstęp 15

Recall (0.4.1). Then it is easy to see that the C-linear part of A and the C-linear one have the form , i.e., 1 1 A0(e) = (A(e) − iA(ie)) = (∂ f(z)(e) − i∂ f(z)(e)) 2 2 x y 1 T 00(e) = (∂ f(z)(e) + i∂ f(z)(e)). 2 x y We put 1 1 ∂ = (∂ − i∂ ), ∂ = (∂ + i∂ ). z 2 x y z 2 x y With this notation 0 f (z)(e) = ∂zf(z)(e) + ∂zf(z)(e).

n n n Remark: If E = C , then, for z ∈ U ⊂ C , ∂xf(z), ∂yf(z), ∂zf(z), ∂zf(z) ∈ L(C , C) and, for n e = (e1, ..., en) ∈ C , n n X X ∂zf(z)(e1, ..., en) = ∂zj f(z)ej, ∂zf(z)(e1, ..., en) = ∂zj f(z)ej, j=1 j=1 where, for j = 1, ..., n, 1 1 ∂ = (∂ − i∂ ), ∂ = (∂ + i∂ ). zj 2 xj yj zj 2 xj yj

0.4.2 Cauchy-Riemann equations: The following conditions are equivalent: (i) A map f is holomorphic; 2 1 0 0 0 (ii) f = (u, v): U → Y is C -smooth, ux = vy and uy = −vx on U (the real form of the Cauchy-Riemann equations; (iii) ∂zf = 0 on U. Proof: is easy if we recall that f is, by deﬁnition, holomorphic if and only if, for each x ∈ U, there is a continuous linear map Tx ∈ LC(E, F) such that f(x + h) − f(x) − T (h) lim x = 0 h→0 h∈E khk 12 and the map U 3 x → Tx ∈ LC(E, F) is continuous ( ).

0.4.C Free R-Modules; Specker modules

Throughout these notes R denote a p.i.d., i.e., a principal ideal domain: a commutative ring with unit, without zero divisors and such that its every ideal is principal (typical examples: R = Z, R = R or C). We consider only left R-modules. (1) For any set A 6= ∅, by R(A) we denote the R-module generated by A, i.e., the R-module of the so-called formal linear combinations of elements of the set A with coeﬃcients in R (13). Precisely speaking the elements in R(A) are functions c : A → R vanishing at all but ﬁnite number of elements of A. Observe that c ∈ R(A) may be written as

n X c = r1a1 + ... + rnan = riai, i=1

12 0 As usual, if f is holomorphic, then Tx is uniquely determined and we write f (x) := Tx, x ∈ U. 13We also put R(∅) := {0}. 16 Introduction

where ai ∈ A, ri ∈ R, i = 1, ..., n, meaning that, for a ∈ A, c(a) = ri if and only if a = ai, i = 1, ..., n. In particular A ⊂ R(A) in the sense that any a ∈ A determines the function 1 · a that takes value 1 at a and 0 elsewhere. (2) We say that an R-module M is free if there is a base i.e., a set A ⊂ M such that the homomorphism R(A) → M (uniquely) determined by the inclusion A,→ M is an isomorphism. Equivalently M is free if there is A ⊂ M such that any x ∈ M can be represented uniquely as Pn the linear combination x = i=1 riai where rißR and ai ∈ A, i = 1, ..., n.In particular R(A) is free for any set A. (3) Each submodule of a free R-module is free (here the assumption that R is a p.i.d. is of importance). (4) Let X be an arbitrary set and let F be the set of all ﬁnite-valued maps fLX → R (R is a p.i.d.). If f ∈ F , then n X f = aiχXi , i=1 where f(X) = {a1, ..., an}, ai ∈ R, ai 6= aj, Xi = {x ∈ X | f(x) = ai}, i, j = 1, ..., n, i 6= j.(χXi denotes the characteristic function of Xi, i = 1, ..., n). Clearly Xi ∩Xj = ∅ if i 6= j, i, j = 1, ..., n. It is clear that F has a structure of an R-module. We say that a submodule S ⊂ F is a Specker module if the condition f ∈ S implies that all characteristic functions χXi taking part in the above representation of f belong to S. It is clear that F itself is a Specker module. We say that a submodule S ⊂ F has a characteristic basis if it is free and has a basis consisting of some characteristic functions.

0.4.3 Theorem: (Nöbeling) Suppose that S, T ⊂ F are Specker submodules and that S ⊂ T . Then there exists a free submodule R ⊂ T such that T = S ⊕ R; moreover R has a characteristic basis. In particular any Specker module T ⊂ F is free and has a characteristic basis; the same holds for T = F . Finally the quotient module T/S is free.

0.4.D Exact sequences

β Let us consider a triple of R-modules and homomorphisms A −→α B −→ C. We say that is exact if Im (α) = Ker (β). More generally a sequence of R-modules and homomorphisms ... → A−2 → A−1 → A0 → A1 → A2 → ... is exact if, for any i, the triple Ai−1 → Ai → Ai+1 is 0 00 exact. An exact sequence of the form 0 → A0 −→α A −→α A00 → 0 is called a short exact sequence. (1) If a sequence β γ A −→α B −→ C −→ D −→δ E is exact, then the sequence

β0 γ 0 −→ Coker α −→ C −→ Ker δ −→ 0 is exact, where Coker α := B/Im α and β0[b] := β(b) for b ∈ B ([a] denotes the class of b in B/Im α); of course β0 is well-deﬁned. Wstęp 17

0 00 (2) We say that a short exact sequence 0 → A0 −→α A −→α A00 → 0 is split if the homomorphism α0 : A0 → A has a left inverse, i.e., there is a homomorphism β0 : A → A0 such that 0 0 β ◦ α = iA0 α0 α00 Exercise: Show that a short exact sequence 0 → A0 −→ A −→ A00 → 0 is split if and only 00 00 00 00 if there exists a homomorphism β : A → A such that α ◦ β = iA00 . In this case the maps (β0, α00): A → A0 ⊕ A00 and (α00, β0): A → A0 ⊕ A00 are isomorphisms.

0 00 (3) If a short exact sequence 0 → A0 −→α A −→α A00 → 0 is split, then the sequence 00 0 00 β β 0 0 0 00 00 0 → A −→ A −→ A → 0 is exact and split and α β + β α = iA. β (4) If A00 is a free R-module, then a short exact sequence 0 → A0 −→α A −→ A00 → 0 is split. (5) (Five Lemma) If a diagram of modules and homomorphisms

ϕ1 ϕ2 ϕ3 ϕ4 M1 / M2 / M3 / M4 / M5

ψ1 ψ2 ψ3 ψ4 ψ5 N1 / N2 / N3 / N4 / N5 β1 β2 β3 β4 consists of exact rows, ψ1, ψ2, ψ4 and ψ5 are isomorphisms, then so is ψ3. In fact one needs that ψ1 is an epimorphism, ψ5 is a monomorphism and ψ2, ψ4 are isomorphisms.

0.4.E Functors ⊗, Hom and Ext

If A, B are R-modules, then A ⊗ B denotes the tensor product of A and B. Recall that A ⊗ B is deﬁned as follows: A ⊗ B is a unique (up to an isomorphism) R-module together with a bilinear map q : A × B → A ⊗ B such that for any R-module C and a bilinear map ϕ : A × B → C, there is a unique homomorphism ϕ : A ⊗ B → C such that the following diagram

A × B FF FF ϕ q FF FF F# A ⊗ B / ϕ C is commutative. This module may be constructed as follows: Let Q be the submodule in R(A × B) generated by elements of the form

(a, b0 + b00) − (a, b0) − (a, b00); (a0 + a00, b) − (a0, b) − (a00, b); (ra, b) − r(a, b); (a, rb) − r(a, b), where a, a0, a00 ∈ A, b, b0, b00 ∈ B and r ∈ R. Then

A ⊗ B := R(A × B)/Q and let q : R(A × B) → A ⊗ B be the quotient homomorphisms. Moreover we put q := q|A×B. It is easy to see that q is bilinear. Given a ∈ A, b ∈ B,

a ⊗ b := q(a, b).

Observe that the set {a ⊗ b | a ∈ A, b ∈ B} generates A ⊗ B, i.e. each element of A × B Pn may be represented as i=1 ri(ai ⊗ bi) where ai ∈ A, bi ∈ B and ri ∈ R, i = 1, ..., n. 18 Introduction

Observe that R ⊗ A =∼ A =∼ A ⊗ R; the (canonical) isomorphism is given by

r ⊗ a 7→ ra, r ∈ R, a ∈ A.

(1) Given homomorphisms ϕ : A → A0, ψ : B → B0, we deﬁne

ϕ ⊗ ψ : A ⊗ B → A0 ⊗ B0 by a ⊗ b 7→ ϕ(a) ⊗ ψ(b) for a ∈ A, b ∈ B. If ϕ, ψ are isomorphisms or epimorphisms, then so is ϕ ⊗ ψ. However if ϕ, ψ are monomorphisms, the ϕ ⊗ ψ may not be a monomorphism. (2) Let us ﬁx an R-module B. The correspondence

A 7→ A ⊗ B that assigns to any R-module A, the tensor product A ⊗ B is a functor; to a homomorphism 0 ϕ : A → A it assigns ϕ ⊗ 1B.

By HomR(A, B) denotes the R-module of all homomorphisms A → B. In case B = R we write sometimes A∗ := Hom(A, R).

(3) Clearly, for any B, Hom(·,B) is a cofunctor in the category ModR of R-modules. Indeed, if α : A → A0 is a homomorphism, then α∗ := Hom(α, B) : Hom(A0,B) → Hom(A, B) is the conjugate of α given by α∗(ϕ)(a) := ϕ(α(a)) for ϕ ∈ Hom(A0,B) and a ∈ A. Similarly, for any R-module A, Hom(A, ·) is a functor in ModR. (4) Given a short exact sequence

β A0 −→α A −→ A00 −→ 0 and an R-module B, the sequences

00 Hom(β,B) Hom(α,B) 0 0 −→ HomR(A ,B) −→ HomR(A, B) −→ Hom(A ,B),

α⊗1 β⊗1 A0 ⊗ B −→B A ⊗ B −→B A00 −→ 0 are exact. (5) If a sequence β 0 −→ A0 −→α A −→ A00 −→ 0 is exact and split, then the sequences

00 Hom(β,B) Hom(α,B) 0 0 −→ HomR(A ,B) −→ HomR(A, B) −→ Hom(A ,B) → 0,

α⊗1 β⊗1 0 −→ A0 ⊗ B −→B A ⊗ B −→B A00 −→ 0 are exact and split.

(6) There is a bifunctor (this will be explained later) Ext on ModR × ModR → ModR, being contravariant with respect to the ﬁrst variable and covariant with respect to the second one, characterized by the following property: given an R-modules A, B, if there is a short exact sequence α β 0 −→ C1 −→ C0 −→ A −→ 0 (∗∗) Wstęp 19

with free R-modules C1,C0, then there is a homomorphism γ : Hom(C1,B) → Ext(A, B) such that the sequence

Hom(β,B) Hom(α,B) γ 0 −→ Hom(A, B) −→ Hom(C0,B) −→ Hom(C1,B) −→ Ext(A, B) → 0 is exact. The deﬁnition is correct since Ext(A, B) does not depend on the sequence (∗∗). More- over it is clear that

∼ Ext(A, B) = Coker Hom(α, B) := Hom(C1,B)/Im Hom(α, B).

If A is free that instead of (∗∗) we may take in particular the sequence 0 → 0 → A → A → 0 which is split and, therefore, Ext(A, B) = 0 for any R-module B. Moreover if R is a ﬁeld, then (∗∗) is split (vector spaces are free R-modules) and again Ext(A, B) = 0 for any vector space B over R.

(7) There is a bifunctor Tor : ModR × ModR → ModR, being covariant with respect to both variables, characterized by the following property: given an R-modules A and B, if there is a short exact sequence α β 0 −→ C1 −→ C0 −→ A −→ 0 (∗∗) with free R-modules C1,C0, then there is a homomorphism γ : Tor(A, B) → C1 ⊗ B such that the sequence γ α⊗1B β⊗1B 0 −→ T or(A, B) −→ C1 ⊗ B −→ C0 ⊗ B −→ A ⊗ B −→ 0 is exact. If A is free, then the sequence 0 → 0 → A → A consists of free modules; hence T or(A, B) = 0. If A or B are torsion free, then Tor(A, B) = 0. Recall that an R-module A is torsion free if

Tor(A) := {a ∈ A | ra = 0 for some 0 6= r ∈ R} = 0.

0.4.F Direct and inverse systems

Recall that a by a directed set we mean a pair (A, 6), where A is a set and 6 is a reﬂexive and transitive relation having the following property: for each α, β ∈ A, there is γ ∈ A such that α, β 6 γ. We say that A0 ⊂ A is coﬁnal if for any α ∈ A there is α0 ∈ A0 such that α 6 α0. 0.4.4 Definition: Let K be an arbitrary category. (1) By a direct system in K we understand a system {Xα, hαβ}α∈A, where for α ∈ A, Xα ∈ OK and, for α, β ∈ A, hα,β ∈ MK(Xα,Xβ) are such that:

(1) hαα = idXα ; (2) if α 6 β 6 γ, then the diagram

hαβ Xα / Xβ BB BB BB hβγ hαγ B B Xγ is commutative. (2) By an inverse system we mean a system {Xα, hαβ}α∈A, where, for α ∈ A, Xα ∈ OK and, 20 Introduction

for α 6 β, hαβ ∈ MK(Xβ,Xα) are such that:

(1) hαα = idXα ; (2) if α 6 β 6 γ, then the diagram

hβγ Xγ / Xβ B BB BB hαβ hαγ BB B! Xα is commutative.

0.4.5 Definition: (1) Let Y ∈ OK and let {pα}α∈A be a family of morphisms pα ∈ MK(Xα,Y ). This family is compatible with the direct system {Xα, hαβ}α∈A if, for any α, β ∈ A, if α 6 β, then the diagram:

hαβ Xα / Xβ C CC CC pβ pα CC C! Y is commutative. (2) Similarly we say that a family {pα}α∈A, where pα ∈ MK(Y,Xα) is compatible with an inverse system {Xα, hαβ}α∈A if for all α, β ∈ A, α 6 β, the diagram

pβ Y / Xβ @ @@ @@ hαβ pα @@ @ Xα is commutative.

0.4.6 Definition: Suppose that {Xα, hαβ} is a direct system. By a direct limit of this system we understand a pair {X, pα}α∈A, where pα ∈ MK(Xα,X) and the family {pα} is compatible with the system such that: for any family {qα : Xα → Y } compatible with the system there is a unique morphism p : X → Y such that, for any α ∈ A, the diagram

pα Xα / X B BB BB p qα BB B! Y is commutative.

0 0.4.7 Lemma: If {X, pα}α∈A and {X , p_α}α∈A are direct limits of a direct system {Xα, kαβ}, then there is an isomorphism g : X → X0 such that, for each α ∈ A, the diagram

pα Xα / X B BB BB g p0 BB α B x0 is commutative. The proof is immediate and follows directly from the very deﬁnition of a direct limit. Wstęp 21

In view of the above lemma, if a direct limit exists, then its object is denoted by

X = lim X . −→ α α∈A

0.4.8 Definition: Given an inverse system {Xαhαβ}α∈A, by an inverse limit we mean a pair {X, pα}, where X ∈ OK and {pα : X → Xα}α∈A is a family of morphism compatible with the system such that: for any Y ∈ OK and a compatible family {qα : Y → Xα}α∈A there is a unique morphism p : Y → X such that the diagram

p Y / X B BB BB pα qα BB B Xα is commutative. Exactly as before one shows that the inverse limit is determined uniquely up to isomorphism. If an inverse limit exists then its object is denoted by

X = lim X . ←− α α∈A

0.4.9 Remark: Sometimes direct (resp. inverse) systems and their direct (resp. inverse) limits are called injective or inductive (resp. projective) systems and limits.

0.4.10 Proposition: Suppose that A is directed and let A0 ⊂ A is coﬁnal. Then A0 is directed. If {X, pα}α∈A is a direct (resp. inverse) limit of a direct (resp. inverse) system

{Xα, hαβ}α∈A,then {X, pα}α∈A0 is a direct (resp. inverse) limit of the system {Xα, hαβ}α∈A0 . We shall deal only with the so-called concrete categories, i.e., such that their objects are sets endowed with some structure. Given a concrete category K, let T be the „forgetting” functor that assigns to any object X ∈ K, the sets T (X) underlying the object X and to any morphism in K, the underlying map of sets. It is easy to see that if a direct (or an inverse system) {Xα, hαβ}α∈A in K admits a direct limit (resp. an inverse limit), then {T (Xα),T (hαβ)}α∈A forms a direct (resp. inverse) system of sets. Since, as we shall see in a moment direct (resp. inverse) systems of sets admit direct (resp. inverse) limits, the following holds ! T lim X = lim T (X ), −→ α −→ α α∈A αA ! T lim X = lim T (X ). ←− α ←− α α∈A α∈A This means that in order to get a direct (resp. inverse) limit in some concrete category on should ﬁnd a direct (resp. inverse) limit of the underlying system of sets and then check whether it carries the ‘compatible’ structure.

Let {Xα, hαβ}α∈A be a direct system of sets. Let _ X := Xα/ ∼ α∈A 22 Introduction

where the relation ∼ is deﬁned as follows: xα ∼ xβ (xα ∈ Xα, xβ ∈ Xβ) if and only if there is α, β 6 γ, such that hαγ(xα) = hβγ(xβ). For each α ∈ A, let pα : Xα → X be given by

pα(x) = [x]∼, x ∈ Xα.

It is immediate to see that {pα}α∈A is compatible with the system and {X, pα}α∈A is a direct limit. Using this we easily check that direct limits exist in the following categories: category of topological spaces, category of topological pairs, category of modules, rings, groups, vector spaces etc. Direct limits do not exists in the category of ﬁnite-dimensional vector spaces. For instance we shall check the construction for the category of modules over a ring R. Let {Xα, hαβ}α∈A be a direct system in ModR. Let X be given as above. In X we introduce the module structure as follows: given [xα]∼, [xβ]∼ ∈ X,

[xα] + [xβ] := [hαγ(xα) + hβγ(xβ)].

If r ∈ R, then r[xα] = [rxα].

These deﬁnitions are correct: they do not depend on the choice of γ > α, β. Indeed let δ > α, β and take κ > γ, δ. Then

hαγ(xα) + hβγ(xβ) ∼ hακ(xα) + hβκ(xβ) ∼ hαδ(xα) + hβδ(xβ).

Similarly we check that the deﬁnition do not depend on the choice of representants of [xα] and [xβ].

It is clear that the map pα : Xα → X is a homomorphism:

pα(rxα) = [rxα] = r[xα] = rpα(xα), xα ∈ Xα,

0 0 0 0 pα(xα + xα) = [xα + xα] = [xα] + [xα ] = pα(xα) + pα(xα).

Now we shall see that also inverse limit exist for sets. Let ( ) Y X := (xα) ∈ Xα | xα = hαβ(xβ), α 6 β α∈A and let pα : X → Xα be the (restriction of the) projection. It is easy to see that {pα}α∈A forms the family compatible with the given (inverse) system and that {X, pα}α∈A is am inverse limit. Inverse limit exists for the above mentioned categories of topological spaces, topological pairs and algebraic categories such as the categories of R-modules etc. We shall need the following abstract result (having its counterparts for topological spaces, pairs of topological spaces and other above mentioned algebraic categories).

0.4.11 Theorem: (1) Let {Xα, hαβ}α∈A be a direct system of sets, let {X, pα}α∈A ne its direct limit and let {qα : Xα → Y }α∈A be a family compatible with the system. Then there is a unique map p : X → Y such that qα = ppα, α ∈ A. The map p is a bijection if and only if the following conditions are met: S (i) Y = α∈A qα(Xα); (ii) for any α, β ∈ A and xα ∈ Xα, xβ ∈ Xβ, qα(xα) = qβ(xβ) if and only if there is γ > α, β Wstęp 23

such that hαγ(xα) = hβγ(xβ). (2) If {Xα, hαβ}α∈A is an inverse system with an inverse limit {X, pα}α∈A and a family {qα : Y → Xα}α∈A is compatible with the system the unique map p : Y → X such that qα = pαp, α ∈ A, is a bijection if and only if the following conditions are satisﬁed: 0 0 0 (i) if for some y, y ∈ Y , qα(y) = qα(y ) for all α ∈ A, then y = y ; (ii) for any x ∈ X there is y ∈ Y such that qα(y) = pα(x) for any α ∈ A.

0.4.12 Remark: Suppose that we deal with a direct system {Xα, hαβ}α∈A in some algebraic category (e.g. ModR). Condition (ii) from (1) above may equivalently read: (ii)’ if for some α ∈ A, xα ∈ Ker qα, then there is β ∈ A, β > α such that x ∈ Ker hαβ. The necessity of (ii)’ is immediate. To see the suﬃciency assume that qα(xα) = qβ(xβ) for some α, β ∈ A and xα ∈ Xα, xβ ∈ Xβ. Take any δ > α, β. Then we have that

qδ(hαδ(xα)) = qα(xα) = qβ(xβ) = qδ(hβδ(xβ), i.e., hαδ(xα) − hβδ(xβ) ∈ Ker qδ. Therefore, for some γ > δ > α, β,

0 = hδγ(hαδ(xα) − hβδ(xβ)) = hαγ(xα) − hβγ(xβ).

Suppose that two direct systems {Xα, hαα0 }α∈A and {Yβ, gββ0 }β∈B in a category K are given 0 0 and there is an increasing map ϕ : A → B, i.e., such that ϕ(α) 6 ϕ(α ) whenever α 6 α in A. By a map of these direct system we mean a family F = {fα : Xα → Yϕ(α)}α∈A of morphisms in K such that if α ≤ α0, then the diagram

fα Xα / Yϕ(α)

g 0 hαα0 ϕ(α)ϕ(α ) Xα0 / Yϕ(α0) fα0 is commutative. Suppose that pairs {X, pα}α∈A and {Y, qβ}β∈B are direct limit of these systems, respectively. Hence, for any α ∈ A, a morphism qϕ(α)fα : Xα → Y is deﬁned. It is easy to see that the family {qϕ(α)fα}α∈A is compatible with the system {Xα, hαα0 }. Therefore there is a unique morphism f : X → Y such that, for any α ∈ A, the diagram

pα Xα / X B BB BB f q fα BB ϕ(α) B! Y is commutative. This morphism is called the direct limit of F and is denoted by f = lim f : lim X → lim Y . −→ α −→ α −→ β α∈A α∈A β∈B

0 0.4.13 Theorem: Under the above assumptions, suppose that two families {sα : Xα → X }α∈A 0 and {tβ : Yβ → Y }β∈B compatible with these systems are given. Then there are (uniquely determined) morphisms s : X → X0 and t : Y → Y 0. If there is a morphism k : X0 → Y 0 such that, for all α ∈ A, the diagram s α 0 Xα / X ,

fα k Yϕ(α) / Y 0 tϕ(α) 24 Introduction then the diagram s X / X0 ,

f∞ k / 0 Y t Y where f := lim f , is commutative. ∞ −→α∈A α 0.4.14 Remark: In practice A = B and then ϕ is the identity.

0.4.15 Theorem: Suppose that three direct systems {Xα}, {Yα} and {Zα} (indexed by a common directed sets A) in the category ModR are given and let F := {fα : Xα → Yα}, G := {gα : Yα → Zα} are maps of these direct systems. If for all α ∈ A (or for α is a set A0 coﬁnal in A), the sequence fα gα Xα −→ Yα −→ Zα is exact, then also the sequence f g X −→∞ Y −→∞ Z, where f := lim f , g : − lim g , X := lim X , Y := lim Y and Z := ∞ −→α∈A α ∞ −→α∈A α −→α∈A α −→α∈A α lim Z , is exact. −→α∈A α In particular, if for each α ∈ A, the sequence

fα gα 0 −→ Xα −→ Yα −→ Zα −→ 0 is exact then so is the sequence

f g 0 −→ X −→∞ Y −→∞ Z −→ 0.

Assume now that inverse systems {Xα, hαα0 }α∈A and {Yβ, gββ0 }β∈B in the category K are given. Let ψ : B → A be an increasing map. By a map of the above inverse systems we mean a 0 family F = {fβ : Xψ(β) → Yβ}β∈B such that if β 6 β , then the diagram

fβ0 Xψ(β0) / Yβ0

hψ(β)ψ(β0) gββ0 Xψ(β) / Yβ fβ is commutative. If {X, pα}α∈A and {Y, qβ}β∈B are inverse limits of these systems, then for each β ∈ B, the a unique morphism fβpψ(β) : X → Yβ is deﬁned. It is easy to see that the family {fβpψ(β)} is compatible with {Yβ, gββ0 }. Therefore, by the deﬁnition there is the unique morphism f ∞ : X → Y such that, for any β ∈ B, the diagram

f X / Y @ @@ @@ qβ f p @ β ψ(β) @ Yβ Wstęp 25 is commutative. The existing morphism f ∞ is called the inverse limit of the family F and denoted by

f ∞ = lim f : lim X → lim Y . ←− β ←− α ←− β β∈B α∈A β∈B

0.4.16 If A = B, then f ∞ may be deﬁned as follows: given (x ) ∈ lim X , we put Remark: α ←− α ∞ f (xα) = (fα(xα)). Clearly (f (x )) ∈ lim Y . α α ←− α 0 0.4.17 Theorem: Under the above assumptions assume that there are families {sα : X → 0 0 Xα}α∈A and {tβ : Y → Yβ}β∈B compatible with these systems that induce morphisms s : X → X and t : Y 0 → Y . If k : X0 → Y 0 is a morphism such that, for any β ∈ B, the diagram

sψ(β) X0 / Xψ(β)

k fβ Y 0 / Yβ tβ is commutative, then the diagram s X0 / X

k f ∞ 0 / Y t Y os commutative.

0.4.18 Theorem: Suppose that inverse systems {Xα}α∈A, {yα}α∈A and {Zα}α∈A in the category ModR are given. If F := {fα Xα → Yα}, G := {gα : Yα → Zα} are maps of these systems and such that for all α ∈ A (or, at least for α from a coﬁnal subset A0 ⊂ A) the sequence

fα gα 0 −→ Xα −→ Yα −→ Zα is exact, then so is the sequence

f ∞ g∞ (∗) 0 −→ X −→ Y −→ Z where X := lim X , Y := lim Y and Z := lim Z . ←−α∈A α ←−α∈A α ←−α∈A α This in particular implies that if fα is a monomorphism, then the exactness of the sequence

fα gα Xα −→ Yα −→ Zα implies the exactness of the sequence

f ∞ g∞ X −→ Y −→ Z.

In general (i.e., if fα is not a monomorphism) this does not hold true. Observe also the inverse limit of isomorphisms is an isomorphism.

0.4.19 Remark: In general if F = {fα} is a map of inverse systems {Xα} and {Yα}, then the following holds: ker f ∞ = lim ker f ; ←− α α∈A 26 Introduction

Im f ∞ ⊂ lim Im f . ←− α α∈A

This is correct since {ker fα} and {Im fα} form inverse systems. Therefore while proving the exactness of the sequence (∗) we have that ker g∞ ⊃ Im f ∞; it remains to show the reverse inclusion. To show it one needs the assumption that fα, α ∈ A, are monomorphisms (then the proof is easy if we recall Remark 0.4.16). In fact if fα is a monomorphisms for all α ∈ A, then

Im f ∞ ⊂ lim Im f ←− α α∈A and lim Im f = lim ker g = ker g∞. ←− α ←− α Now we shall study the behavior of some functors with respect to direct (inverse) systems.

0.4.20 Theorem: (1) Given direct systems {Mα, hαβ}α∈A and {Nγ, gγδ}γ∈B of modules over R, the family {Mα ⊗R Nγ}(α,γ)∈A×B, where A × B is directed by the relation (α, γ) 6 (β, δ) if and only if α 6 β and γ 6 δ, together with homomorphisms hαβ ⊗ gγδ : Mα ⊗ Nγ → Mβ ⊗ Nδ is a direct system. Moreover,

lim M ⊗ N ∼ lim M ⊗ lim N . −→ α γ = −→ α −→ γ (α,γ)∈A×B α∈A γ∈B

In other words the direct limit commutes with the tensor product. In particular, for any R-module M,

M ⊗ lim N ∼ lim(M ⊗ N ) −→ γ = −→ γ γ∈B γ∈B and, for any R-module N, (lim M ) ⊗ N ∼ lim(M ⊗ N). −→ α = −→ α α∈A α∈A The same holds for the functor Tor but, unfortunately, this does not hold in case of lim and ←− inverse systems. (2) Given a direct system {Mα, hαβ}α∈A and an inverse system {Nγ, gγδ}γ∈B of modules over R, the family {HomR(Mα,Nγ), Hom(hαβ, gγδ}(α,γ)∈A×B is an inverse system and

Hom (lim M , lim N ) ∼ lim Hom (M ,N ). R −→ α ←− γ = ←− R α γ α∈A γ∈B (α,γ)∈A×B

In particular, for any R-module N, we have an isomorphism

Hom (lim M ,A) ∼ lim Hom (M ,A) R −→ α = ←− R α α∈A α∈A and, for any module M, the following isomorphims holds

Hom (M, lim N ∼ lim Hom (M,N ). R ←− γ = ←− R γ γ∈B γ∈B

The same holds for Ext but, unfortunately does not hold in case either {Mα} is an inverse system or {Nγ} is a direct system (however in such a case {Hom(Mα,Nγ)} is a direct system). Wstęp 27

In what follows we shall study direct and inverse system of chain complexes. α Suppose that, for all α ∈ A, C∗(α) = {Cq(α), ∂q }q∈Z is a chain complex and α 6 β, α, β ∈ A, q q hαβ = {hαβ} : C∗(α) → C∗(β) is a chain map such that, for any q ∈ Z, {Cq(α), hαβ}α∈A is a direct system. Then we say that {C∗(α), hαβ} is a direct system of chain complexes. Therefore, for each q ∈ Z, we may deﬁne

C (∞) = lim C (α). q −→ q α∈A

α Since hαβ is a chain map, for any q ∈ Z, the family {∂q } is a map of the direct system {Cq(α)} to {Cq−1(α)} and we are in a position to deﬁne a homomorphism

∂∞ = lim ∂α : C (∞) → C (∞). q −→ q q q−1 α∈A

It is easy to check that, for any q ∈ Z,

∞ ∞ ∂q−1∂q = 0.

∞ Hence C∗(∞) := {Cq(∞), ∂q }q∈Z is a chain complex. It is not diﬃcult to see that also C∗(∞) is a direct limit of the direct system {C∗(α), hαβ} in the category of chain complexes and chain maps. We shall see that direct limits of chain complexes commute with the functor of homology. Precisely the following holds.

0.4.21 Theorem: Under the above assumptions

H (C (∞)) ∼ lim H (C (α)). q ∗ = −→ q ∗ α∈A

In order to study the similar results with respect to inverse systems of chain complexes. In order to do it we shall restrict our attention to countable inverse systems and begin with inverse systems of R-modules.

Let {Mn, hnm}n∈N be an inverse system indexed by N. It is easy to see that we may study the so-called tower of the form

ϕ1 ϕ2 ϕ3 (∗) M1 ← M2 ← M3 ← ... where ϕn := hn,n+1 and let ∞ 0 Y M := Mk. k=1 We deﬁne d : M 0 → M 0 by the formula

d(x1, x2, ...) = (x1 − ϕ1(x2), x2 − ϕ2(x3), ...).

Then (by deﬁnition) it is easy to see that

ker d = lim M . ←− k k∈N

Let us deﬁne 1 lim M := M 0/Im d. ←− k k∈N 28 Introduction

1 0.4.22 Remark: The lim Mk does not change if we take any inﬁnite subsequence (this is ←−k∈N quite diﬃcult).

If {χn}n∈N is a map of the tower (∗) to a tower

ψ1 ψ2 ψ3 M1 ← N2 ← N3 ← ..., i.e., χn ◦ ϕn = ψn ◦ χn+1 for all n ∈ N, the – clearly – a homomorphism

1 1 1 χ := lim χ : lim M → lim N ∞ ←− n ←− n ←− n n∈N n∈N n∈N is induced.

0 00 0.4.23 Theorem: If we have three inverse systems {Mk}, {Mk}, {Mk } such that for any k ∈ N there is a short exact sequence

0 00 (∗∗) 0 −→ Mk −→ Mk −→ Mk −→ 0 of maps of these towers, then there a homomorphism

1 ∆ : lim M 00 → lim M 0 ←− k ←− k such that the following sequence

1 1 1 0 −→ lim M 0 −→ lim M −→ lim M 00 −→∆ lim M 0 −→ lim M −→ lim M 00 −→ 0 ←− k ←− k ←− k ←− k ←− k ←− k is exact (the meaning of the other arrows is clear). Here are conditions implying that lim1 M = 0. ←− k

0.4.24 Fact: If in the tower (∗) all maps ϕk are epimorphisms, then

1 lim M = 0. ←− k

0.4.25 Definition: We say that an inverse system satisﬁes the Mittag-Loeﬄer condition if, for every n ∈ N, there is m > n such that for any i > m,

Im {Mi → Mn} = Im {Mm → Mn}.

0.4.26 Example: The Mittag-Loeffler condition is satisfied, e.g., if all Mk are finite-dimensional vector spaces.

0.4.27 If the Mittag-Loeﬄer condition is satisﬁed, then lim1 M = 0. Theorem: ←− k

0.4.28 Remark: (1) It is easy to see that if in the sequence (∗∗), tower {Mk, ϕk} satisﬁes the 00 Mittag-Loeﬄer condition, then so does the tower {Mk }: precisely this condition follows since all gk are epimorphisms. (2) More generally if we have a map F = {fk} of towers {Mk, ϕk} and {Nk, ψk}, then – according to Remark 0.4.19 –

ker f ∞ = lim ker f , Im f ∞ ⊂ lim Im f . ←− k ←− k Wstęp 29

In order to get the equality in the second inclusion above it is not suﬃcient to assume that {Nk} has the M.-L. property. To see this let Mk = Z for any k and let ϕk be a multiplication by 3, while Nk = Z2, k ∈ N, and ψk is the identity. Let fk(x) = x(mod 2). Then, for each k ∈ N, f is an epimorphism. Hence lim Im f = . But, since lim M = 0, f ∞ = 0. Hence there is k ←− k Z2 ←− k no equality but {Nk} has the Mittag-Loeﬄer property. Observe also that in this example {Mk} does not satisfy the M.-L. property. Therefore it seems that in order to get the equality one has to assume the M.-L. property for {ker fk}. We shall show it. Consider the exact sequence of towers

qn 0 −→ {ker fn} −→ {Mn} −→ {Mn/ ker fn} −→ 0, where qn : Mn → Mn/ ker fn is the quotient epimorphism. According to (1), the tower {Mn/ ker fn} satisﬁes the M.-L. property; therefore – by Theorem refcos2 – the sequence q∞ 0 −→ lim ker f → M −→ lim M / ker f −→ 0 ←− n ←− n n is exact. In particular q∞ is an epimorphism. On the other hand the homomorphism

f¯n : Mn/ ker fn → Im fn induced by fn is an isomorphism. Hence f¯∞ lim M / ker f ∼ lim Im f . ←− n n = ←− n

Observe that f¯nqn = fn : Mn → Im fn. Since f¯∞q∞ = (lim f¯ )(lim q ) = lim(f¯ q ) = lim f = f ∞ : M → lim Im f , ←− n ←− n ←− n n ←− n ←− n we see that f ∞ is onto lim Im f . ←− n 0.4.29 Corollary: If we have an exact sequence

0 F G 00 {Mn} −→ {Mn} −→ {Mn }, where F = {fn}, G = {gn}, of towers and the tower {ker fn} satisﬁes the M.-L. property, then the sequence f ∞ g∞ M 0 −→ M −→ M 00 is exact. This is an important statement. In particular given a long exact sequence of towers

... −→ {An} −→ {Bn} −→ {Cn} −→ ..., where all towers consisting of the kernels of the arrow satisfy the M.-L. satisfy M.-L. property, then the sequence consisting inverse limits ... −→ A −→ B −→ C −→ .. is exact. In particular if we have a long exact sequence of towers consisting of ﬁnite-dimensional vector spaces, then the corresponding sequence of inverse limits is exact.

0.4.30 Question: Suppose that F = {fk} : {Mk} → {Nk} is a map of towers. Suppose that {Nk} has the M.-L. property and, for each k ∈ N, fk is a monomorphism. Is it true that {Mk} has M.-L. property? 30 Introduction

0.4.G Limits and (co)homology

Now we shall consider a tower of chain complexes

(∗∗) C∗(1) ← C∗(2) ← C∗(3) ← ...

n q q−1 n+1 (this means that for any q ∈ Z and any n ∈ N, ∂q ◦ ϕn = ϕn ◦ ∂q ). The following result holds true

0.4.31 If, for each q ∈ , lim1 C (k) = 0, then for every q ∈ , there is a short Theorem: Z ←− q Z exact sequence

1 β 0 −→ lim H (C (k)) −→ H (C (∞)) −→α lim H (C (k)) −→ 0. ←− q+1 ∗ q ∗ ←− q ∗

The similar result holds for towers of cochain complexes. Precisely if a tower

C∗(1) ← C∗(2) ← C∗(3) ← ... of cochain complexes is given (no doubt the reader will explain what does it mean), then

0.4.32 Theorem: If, for each q ∈ , lim1 Cq(k) = 0, then there is a short exact sequence Z ←−k∈N

1 β 0 −→ lim Hq−1(C∗(k)) −→ Hq(C∗(∞)) −→α lim Hq(C∗(k)) −→ 0. ←− ←−

In particular consider a direct (countable) system

(∗ ∗ ∗) C∗(1) −→ C∗(2) −→ C∗(3) −→ ... of chain complexes, then we have the corresponding tower

HomR(C∗(1)) ← HomR(C∗(2)) ← Hom(C∗(3)) ← ... of cochain complexes. If, for each q ∈ , lim HomR(Cq(k)) = 0, then there is a short exact Z ←−k∈N sequence 1 β 0 −→ lim Hq−1(C (k)) −→ Hq(C (∞)) −→α lim Hq(C (k)) −→ 0. ←− ∗ ∗ ←− ∗

q If above, for all q ∈ Z and k ∈ N, Cq(k) (resp. C (k)) is a finite-dimensional vector space (i.e., R = R), then so HomR(Cq(k)). Therefore the Mittag-Loeffler condition is satisfied. Hence

1 lim C (k) = 0 ←− q and, in case of the direct system (∗ ∗ ∗),

1 lim Hom (C (k)) = 0 ←− R q and in case of the tower (∗∗) 1 lim Cq(k) = 0. ←− Wstęp 31

0.5 Functional and nonsmooth analysis

Let X,Y be metric spaces, A ⊂ X and x0 ∈ cl A. Moreover let ϕ : A ( Y be a set-valued map. The lower (resp. upper) set-limit is the subset of Y deﬁned as follows

y ∈ Lim inf ϕ(x) ⇐⇒ ∀ (xn), xn ∈ A, xn → x0 ∃ yn ∈ ϕ(xn) yn → y; x→x0

y ∈ Lim sup ϕ(x) ⇐⇒ ∃ (xn), xn ∈ A, xn → x0 ∃ yn ∈ ϕ(xn( yn → y. x→x0

Let E be a Banach space. Given a closed set K ⊂ E and x ∈ K, we deﬁne the Clarke tangent cone by dK (y + hw) CK (x) = {w ∈ E | lim = 0} y→x, y∈K, h→0+ h and the normal cone

NK (x) := {v ∈ E | hv, wi 6 0 for all w ∈ CK (x)}.

Equivalently we may put K − y CK (x) = Lim inf . y→x, y∈K, h→0+ h

Along with CK (x) it is useful to consider the Bouligand tangent cone K − x TK (x) := Lim sup . h→0+ h

It is easy to see that CK (x) ⊂ TK (x) and

Lim inf TK (x) ⊂ CK (x) y→x,y∈K with equality if dim E < ∞. Chapter 1 Categories, Functors and Algebraization of Topological Problems

1.1 Categories and Functors

A description of a mathematical object in terms of categories implies that this object (a group or a space) is considered as a member of a collection of similar objects rather than separately. Intu- itively, a category may be represented as a collection of sets (possibly with additional structure) together with mappings between them agreeing with this structure. Correspondences between elements of diﬀerent categories obeying special rules are called functors. Although the category theory (some authors say abstract nonsense) may be sometimes awful (see e.g. Example 1.3.40) and we are not going to use it frequently , it seems that some basic knowledge of its language is a must for a practicing mathematician.

1.1.1 Category: Category K consists of: (a) class (collection) of objects OK ; (b) for each objects X,Y ∈ OK , the set of morphisms MK (X,Y ); (c) the composition transformation ◦ such that, for each objects X,Y,Z ∈ OK , ◦ : MK (Y,Z)× MK (X,Y ) → MK (X,Z) (for any pair (α, β) ∈ MK (Y,Z) × MK (X,Y ), α ◦ β := ◦(α, β) ∈ MK (X,Z)) having the following properties: • if γ ∈ MK (Z,T ) (T ∈ OK , then γ ◦ (α ◦ β) = (γ ◦ β) ◦ α; • for each X ∈ OK , there is the identity morphism 1X ∈ MK (X,X) such that 1X ◦ α = α for α ∈ MK (Y,X) and α ◦ 1X = α for α ∈ MK (X,Y ) (Y ∈ OK ). 0 By a subcategory of a category K we understand a category K such that OK0 ⊂ OK and, for any X,Y ∈ OK0 , MK0 (X,Y ) ⊂ MK (X,Y ). 0 We say that a subcategory K is full if OK0 = OK (in other words only the set of morphisms is restricted).

It is easy to see that, for X ∈ OK , the identity morphism 1X is uniquely determined. For X,Y ∈ OK and α ∈ MK (X,Y ), we write also α : X → Y saying that X is the domain of α, while Y is its range (note that sometimes α may not be a function – see Example 1.1.2 (g)).

If, for two morphism α : X → Y and β : Y → X, α ◦ β = 1Y , then α is the left inverse of β 1.1. Categories and Functors 33 and β is the right inverse of α. A morphism being a left and right inverse of α is said to be a two-sided inverse of α. A morphism α : X → Y is an equivalence (or an isomorphism) if it has a two-sided inverse (equivalently: has a left and a right inverse). In this case those inverses coincide; consequently an existing two-sided inverse is uniquely determined and denoted by α−1 (clearly α−1 : Y → X).

1.1.2 Examples: There is a lot of examples. In principle any mathematical structure gives rise to a category. (a) Category of sets Sets: sets are its objects, and maps between sets are its morphisms; (b) category Gr of (abelian) groups and (group) homomorphisms; (c) category ModR of R-modules over a ring R and homomorphisms; (d) category of topological spaces T op and continuous maps; ∗ (e) pointed topological category T op : its objects consists of pairs (X, x0) where X is a topological space and x0 ∈ X is a distinguished point, while morphism f :(X, x0) → (Y, y0) is a continuous map f : X → Y such that f(x0) = y0; (f) more generally: category of pair of topological spaces T op2 its objects are pairs (X,A), where X is a space, A ⊂ X; f :(X,A) → (Y,B) is a morphism if f : X → Y is a continuous map and f(A) ⊂ B (1); (g) Assume that (X, 4) is a poset (partially ordered set). Let objects of a category K be points of X, i.e., OK = X; for any x, y ∈ X, put

(x, y) gdy x y; M (x, y) = 4 K ∅ otherwise.

1.1.3 Functors: Let A, B be categories. A functor (or precisely a covariant functor) T : A → B is a pair of correspondences OA 3 X 7→ T (X) ∈ OB,

MA(X,Y ) 3 α 7→ T (α) ∈ MB(T (X),T (Y )), where X,Y ∈ OA, such that, for any morphisms α ∈ MA(Y,Z), β ∈ MA(X,Y ),

T (α ◦ β) = T (α) ◦ T (β),

T (1X ) = 1T (X). A cofunctor ( or a contravariant functor) T : A → B is a pair of correspondences

OA 3 X 7→ T (X) ∈ OB,

MA(X,Y ) 3 α 7→ T (α) ∈ MB(T (Y ),T (X)), where X,Y ∈ OA, such that, for any morphisms α ∈ MA(Y,Z), β ∈ MA(X,Y ),

T (α ◦ β) = T (β) ◦ T (α),

T (1X ) = 1T (X).

Roughly speaking (co)functors are general rules to assign to objects of one category objects of another one in such a manner that morphisms of the ﬁrst category are transformed into morphisms of the second one preserving compositions.

1The space X is identiﬁed with the pair (X, ∅) 34 1. Categories, Functors and Algebraization of Topological Problems

For instance we may assign to any vector space V over a ﬁeld K its (algebraic) dual V ∗, i.e. the vector space of all linear forms V → K. It is easy to check, that this correspondence deﬁnes actually a cofunctor in the category of all vector spaces (and linear maps between them).

1.1.4 Fact: If α : X → Y is an isomorphism in a category A, T : A → B is a (co)functor, then T (α): T (X) → T (Y ) (resp. T (α): T (Y ) → T (X)) is an isomorphism in B and

−1 −1 T (α) = T (α ).

1.1.5 Natural Transformation: Suppose that T,S : A → B are (co)functors. By a natural transformation Φ of (co)functors T and S (we write Φ: T → S) we understand a family {ΦX ∈

MB(T (X),S(X))}X∈OA such that, for any α : X → Y in A (i.e., α ∈ MA(X,Y ), X,Y ∈ OA) one has ΦY ◦ T (α) = S(α) ◦ ΦX

(resp. ΦX ◦ T (α) = S(α) ◦ ΦY .) In other words a diagram ΦX T (X) / S(X)

T (α) S(α) T (Y ) / S(Y ) ΦY is commutative.

If, for each X ∈ OA, the morphism ΦX : S(X) → T (X) is an isomorphism, then we say that Φ is a natural equivalence of (co)functors T and S. The word ‘natural’ sometimes seems to be overused; on some occasions it is replaced be the word functorial. However the story is always the same: suppose that we are given two categories A, B and two functors T,S : A → B. This means that to each object A ∈ OA, object T (A),S(A) ∈ OB is assigned. Further on suppose, that we have a general rule to assign to each A ∈ OA a morphism ΦA ∈ MB(T (A),S(A)). The question is whether the correspondence A 7→ ΦA behaves well with respect to morphisms in A, i.e. what is the relation between α ∈ MA(A, B), B ∈ OA, T (α), S(α), ΦA and ΦB. Naturality (or functoriality) of Φ means that the diagram T (α) T (A) / T (B)

ΦA ΦB S(A) / S(B) S(α) is commutative for all morphisms α ∈ MA(A, B). If the reader ﬁnds somewhere a word ‘natural’ (in an appropriate context), then he should check what is going on: what categories, functors and what ‘natural transformation’ are involved.

1.2 Algebraization

For a topologist the topological category T op (and some derived categories) is the most important: given a space X, the questions concerning its (widely understood) structure are of interest and, given another space Y , the properties of the set C(X,Y ) of continuous maps X → Y are 1.2. Algebraization 35 studied. Similarly, for analyst, the most interesting are smooth manifolds and smooth maps between them, while for people working in (linear or nonlinear) functional analysis, questions concerning subsets of topological vector spaces and (linear or nonlinear) continuous maps are interesting. Generally speaking the method of algebraic topology relies on the study of these and related questions via the use of algebraic functors on T op (or on another suitable category), i.e., taking ‘values’ in some algebraic category, e.g. in Gr or MR. To an interesting, e.g. topological, object there corresponds an algebraic object in such a way that a topological question may be translated to a usually simpler algebraic problem. For instance: suppose that there is a method to assign to each (continuous) map f : X → (Y, y0) between spaces an integer number d(f) such that the nontriviality of d(f) implies the existence of x ∈ X such that f(x) = y0. Thus – the fundamental problem of the solvability of an equation f(x) = y0 – is replaced by the problem: d(f) 6= 0 ? If, for f, g : X → Y , d(f) = d(g) whenever f ' g (i.e., f and g are homotopic: this concept will be explained later), then in order to determine whether f, g are not homotopic it is suﬃcient to check whether d(f) 6= d(g). Suppose that X,Y,Z are (topological) spaces and f : X → Y , g : X → Z are given. It appears that many important topological problems concern the existence of a map h : Z → Y such that the following diagram f X / Y (1.2.1) @@ g ? @@ h @@ @ Z is commutative, i.e., hg = f (or homotopy commutative, i.e., hg ' f). This the so-called generalized extension problem. Conversely we may assume that f : X → Y and h : Z → Y are given and we dually ask of the existence of a map g : X → Z such that hg = f (or hg ' f), i.e., the (homotopy) commutativity of the diagram

f X / Y (1.2.2) g ~? h ~~ ~~ ~~ Z

This is the so-called generalized lifting problem.

1.2.1 Extension problem: Let X be a space, A ⊂ X and f : A → Y . We ask about the ∗ ∗ ∗ existence of f : X → Y such that f |A = f, i.e., f iA = f where iA : A,→ X is the inclusion map: f ∗ X o A /' Y iA f

1.2.2 Retraction problem: Under the above assumptions we ask about the existence of a (continuous) map r : X → A such that r(a) = a for all a ∈ A (in this case we say that r is a retraction and A is a retract of X). Observe that this is equivalent to the problem of the existence of r such that riA = 1A, i.e.the existence of an extension of the identity 1A:

r X o A /' A. iA 1A 36 1. Categories, Functors and Algebraization of Topological Problems

1.2.3 Lifting problem: Let E,B be spaces, p : E → B and f : X → B. By a lifting of f we mean a map f 0 : X → E such that pf 0 = f:

E f 0 > p f X / B We easily see that the existence of a lifting is a particular case of the generalized lifting problem. Let T : T op → Gr be a functor. If there is h such that diagram (1.2.1) is commutative (or homotopy commutative and T is such that T (k) = T (k0): T (X) → T (Y ) whenever k ' k0 : X → Y ), then we have the commutative diagram

T (f) T (X) / T (Y ) GG w; GG ww GG ww T (g) GG wwT (h) G# ww T (Z)

Thus if we study e.g. the generalized extension problem, we may ﬁrst ask for the existence of a (group) homomorphism ϕ : T (Z) → T (Y ) such that the diagram

T (f) T (X) / T (Y ) GG ; GG GG T (g) GG ϕ G# T (Z) is commutative. Similarly when studying the generalized lifting problem (1.2.2) we ask about the existence homomorphism ψ : T (X) → T (Z) such that the diagram

T (f) T (X) / T (Y ) w; ww ww ψ wwT (h) # ww T (Z) is commutative. If T is a cofunctor, then (e.g. in the case of the generalized extension problem) we look for a homomorphism ϕ : T (Y ) → T (Z) such that the diagram

T (X) o T (Y ) cGG T (f) GG GG T (g) GG ϕ G { T (Z) is commutative. In both cases, if ϕ (or ψ) does not exist, then the considered problem has no solution; if it does exist, then (at the moment) the algebraic attitude does not work.

1.2.4 Example: Let n > 1. Consider the functor of the reduced (n − 1)-dimensional singular n−1 homology Hen−1 : T op → Gr (it will be introduced later on). It is known that Hen−1(S ) = Z 1.3. Theory of homotopy 37

n n−1 n n−1 n−1 and Hen−1(D ) = 0. Therefore the sphere S is not a retract of D , i.e., id : S → S has no extension onto Dn. Indeed, if there is an extension f : Dn → Sn−1, then there is a n n−1 homomorphism ϕ : Hen−1(D ) → Hen−1(S ) (i.e., ϕ : {0} → Z) such that the diagram

n−1 id n−1 Hen−1(S ) / Hen−1(S ) NNN pp7 NNN ppp NNN pppϕ Hen−1(i) NN' ppp n Hen−1(D )

(where i : Sn−1 ,→ Dn is the inclusion) is commutative: this is impossible. Usually the ‘negative’ answers following the use of the algebraic method can be turned into positive results. For instance the non-existence of a retraction r : Dn → Sn−1 implies the existence of ﬁxed points of any map f : Dn → Dn (the Brouwer theorem).

1.2.5 Exercise: Prove the Brouwer theorem

1.2.6 Example: The sphere S := {x ∈ `2 | kxk = 1} is a retract of the ball D := {x ∈ `2 | ∞ kxk 6 1}. Indeed: let e0 := 0 and, for i ∈ N, ei := (δik)k=1. Consider h : R+ → D given by

h(t) = (1 − t + n)en + (t − n)en+1 for t ∈ [n, n + 1], n > 0. Observe that h is a homeomorphism R+ onto the set [ A := [en; en+1]. n>0

The set A is closed in D. By the Tietze theorem, there exists a map f : D → R+ such that −1 f|A = h . Hence f ◦ h = idR+ . Deﬁne ϕ : D → D by ϕ(x) = h(f(x) + 1), x ∈ D.

Note that ϕ(x) 6= x for any x ∈ D. If there is x0 ∈ D such that ϕ(x0) = x0, then x0 = −1 −1 −1 h(f(x0) + 1) ∈ A and h (x0) = f(x0) + 1. But also f(x0) = f ◦ h(h (x0) = h (x0). The existence of a ﬁxed-point-free map ϕ : D → D implies the existence of a retraction r : D → S.

1.2.7 Exercise: If (X,A) is a polyhedral pair, the X × 0 ∪ A × I is a retract of X × I. Hint: consider ﬁrst the case when X is a standard simplex σ and A = ∂σ and then proceed by induction.

1.3 Theory of homotopy

1.3.A The space of continuous maps

Consider the set C(X,Y ) of all (continuous) maps from a topological space X to a topological space Y (2). The properties of this set and many properties of spaces X,Y are strongly interre- lated, e.g. if X = {∗} is one-point space, then C(X,Y ) =∼ Y (meaning the bijection). Topologies on C(X,Y ) may be introduced in many ways. The most important are:

1.3.1 Topology of Pointwise Convergence: Consider the sets [xi,Ui] := {f ∈ C(X,Y ) | f(xi) ∈ Ui} where xi ∈ X and Ui ⊂ Y is open for i = 1, ..., n. The topology τp generated by the

2The set of all (not only continuous) functions X → Y will be denoted by M(X,Y ). 38 1. Categories, Functors and Algebraization of Topological Problems family of such sets considered as a subbase is called the topology of pointwise convergence.

1.3.2 Compact-Open Topology: Consider all possible sets of the form [K,U] := {f ∈ C(X,Y ) | f(K) ⊂ U} where K ⊂ X is compact and U ⊂ Y is open. The topology τc generated by the family of such sets considered as a subbase is called the compact-open topology.

It easy to see that τp ≺ τc (topology τp is weaker than τc).

1.3.3 Topology of Uniform Convergence: Assume that (Y, d) is a metric space (with a metric bounded by 1; this can always be achieved). Consider the function

C(X,Y )2 3 (f, g) 7→ d(f, g) := sup d(f(x), g(x)). x∈X

It is easy to see that d is a metric. The topology τu generated by this metric is called the topology of uniform convergence.

If Y is a metric space, then τc ≺ τu and if X is compact, then τu = τc. In case X is noncompact and Y is metric, then these topologies does not coincide.

1.3.4 Topology of the Almost Uniform Convergence: Let Y be a metric space. For any compact K ⊂ X, consider the semimetric

2 C(X,Y ) 3 (f, g) 7→ dK (f, g) := sup d(f(x), g(x)). x∈K

The topology τau introduced in C(X,Y ) by the family {dK } where K runs over the family of all compact subset in X (being metrizable whenever X is σ-compact) is called the topology of the almost uniform convergence (3).

In case Y is metrizable (or uniform), then τau = τc, i.e., in particular, the convergence of a sequence in τc is equivalent to the so-called almost uniform convergence, i.e., the uniform convergence on any compact subset of X.

1.3.5 Exponential law: If Z is a space, then there is a properly deﬁned and continuous (in fact, an embedding, i.e., a homeomorphism onto the image) map, the so-called exponential map,

C(X × Z,Y ) 3 f 7→ Ω(f) ∈ C(Z, C(X,Y )) where Ω(f)(z)(x) := f(x, z) (1.3.3) for z ∈ Z and x ∈ X (functional spaces are considered with the compact-open topologies) (4). If, additionally, X is locally compact, then Ω is a homeomorphism. The same holds if X and Z satisﬁes the ﬁrst axiom of countability (or, more generally if X × Z is compactly generated, i.e., a k-space (5). If X is locally compact, then the evaluation E : C(X,Y ) × X → Y given, for f ∈ C(X,Y )

3The same construction may be performed if Y is a uniform space – see e.g the book of Engelking. 4Given sets X,Y,Z, the map

M(X × Z,Y ) 3 f 7→ Ω(f) ∈ M(Z, M(X,Y )) given by formula (1.3.3) is a bijection. Therefore the question is whether we restrict Ω to f ∈ C(X × Z,Y ), Ω(f) is continuous; we may also ask whether, given any space Z and g ∈ C(Z, C(X,Y )), Ω−1(g) ∈ C(X × Z,Y ). 5Recall that a space X is a k-space provided B ⊂ X is closed if and only, for B ∩K is compact for any compact K ⊂ X. Note that a space satisfying the first axiom of countability is a k-space (the Hausdorff assumption is important). 1.3. Theory of homotopy 39 and x ∈ X, by E(f, x) := f(x) is continuous. Generally speaking the evaluation E is continuous (compact-open topology is considered) if and only if, for any space Z and any g ∈ C(Z, C(X,Y )), the function Ω−1(g): X × Z → Y given by Ω−1(g)(x, z) := g(z)(x) is continuous. If X is arbitrary and Y is metric, then the evaluation is continuous if on C(X,Y ) one considers the topology τu. If the topology of the almost uniform convergence is considered, then this holds true provided X is locally compact (this is trivial for then the almost uniform convergence and the compact-open topologies coincide). The Proof is not difficult but tedious. For the future reasons let as mention with regard the continuity if Ω(f) (i.e., the correctness of its definition), where f ∈ C(X × Z,Y ) and Y is metric, that it is easy to see that, for any compact set K ⊂ X and z0 ∈ Z,

lim sup d(f(x, z), f(x, z0)) = 0. (1.3.4) z→z0 x∈K

In other words if zn → z0 in Z, then Ω(f)(zn) converges almost uniformly to Ω(f)(z0). 1.3.6 Remark: It appears that if X is a completely regular space, such that the evaluation E : C(X, R) × X → R is continuous (with respect to the compact-open topology) , then X is locally compact. The proof is quite involved. Therefore in general when X is completely regular but not locally compact, then the evaluation is not continuous if on C(X,Y ) the compact-open topology is considered. Some authors write Y X instead of C(X,Y ). In this setting the exponential law has the form Y X×Z =∼ (Y X )Z (provided appropriate assumptions are satisﬁed).

1.3.B Extensions

The extension problem seems to be very important for all mathematics. In the present section we shall study some aspects of this problem carefully. The best know result is:

1.3.7 Tietze’s Theorem: If A is a closed subset of a normal space X, then any function ∗ f : A → I admits an extension f : X → I. The main tool to prove this theorem is the Urysohn lemma stating that: given closed disjoint subsets A, B of a normal space X, there is a continuous function λ : X → I such that λ|A ≡ 0 and λ|B ≡ 1.

1.3.8 Exercise: Show that the Tietze theorem and the Urysohn lemma are equivalent. More- over show that a (Hausdorﬀ) space X is normal if and only if, for any closed A ⊂ X, the assertion of the Tietze theorem holds true. The property of the interval I described in the Tietze theorem deserves generalization. Namely suppose that C is a topological class of spaces, i.e., such that if Y ∈ C and Y 0 =∼ Y , then Y 0 ∈ C, and that C is weakly hereditary, i.e., if Y ∈ C and A ⊂ Y is closed, then A ∈ C For instance, the class N of all normal spaces, the class P of all paracompact spaces, the class M of all metric spaces, the class Com of all compact spaces and, ﬁnally, the class MCom of all compact metric spaces are topological and weakly hereditary.

1.3.9 Absolute Extensors and Absolute Neighborhood Extensors: We say that a space Y is an absolute extensor for the class C if given X ∈ C and a closed A ⊂ X, any map 40 1. Categories, Functors and Algebraization of Topological Problems f : A → Y has an extension f ∗ : X → Y . We write Y ∈ AE(C). In view of the Tietze theorem, I ∈ AE(N ). It is easy to show that: (i) the class AE(C) is topological; (ii) this class is closed with respect 0 to Cartesian products; (iii) if Y is a retract of Y ∈ AE(C), then Y0 ∈ AE(C) and (iv) if D ⊂ C, then AE(C) ⊂ AE(D) (exercise !).

1.3.10 Fact: The open interval (a, b) ⊂ R and R belong to AE(N ). Hence the Tychonoﬀ cube 6 n 7 arbitrary weight ( ), R , Dn and a Hilbert space ( ) are absolute extensors for the class of normal spaces. Proof: It is suﬃcient to show that (−1, 1) ∈ AE(N ). Let X ∈ N and A ⊂ X is closed. A map f : A → (−1, 1) has an extension f 0 : X → [−1.1].The set B := f 0−1({−1, 1}) is closed and A ∩ B = ∅. The Urysohn Lemma provides a function λ : X → R such that λ|A ≡ 1 and λ|B ≡ 0. ∗ 0 Then f := λf is a desired extension of f with values in (−1.1). n n n n n 1.3.11 Exercise: Show that D ,B , ∆ ∈ AE(N ), n > 1, where D := {x ∈ R | |x| 6 1}, n n n n+1 Pn+1 B := {x ∈ R | |x| < 1} and ∆ := {x ∈ R | i=1 xi = 1, xi > 0}. Is it true that Sn ∈ AE(N ) ? The next theorem (with a quite elaborate but instructive proof) is of importance.

1.3.12 The Dugundji Theorem: If C is a convex subset of a locally convex space E, then C ∈ AE(M). In analytical practice the most important is the case when E is a normed space.

1.3.13 Corollary: Any closed convex subset C of a normed space E is a retract. More precisely, for any ε > 0, there is a retraction r : E → C such that, for any x ∈ E,

kr(x) − xk 6 (1 + ε)d(x, A).

The ﬁrst part trivially follows from the Dugundji theorem (a retraction r : E → C is an extension of 1C ). In order to get the second part one has to modify the Dugundji proof a bit.

1.3.14 The Michael Extension Theorem: If C is a closed subset of the Fréchet space E, i.e., a locally convex complete metric space, then C ⊂ AE(P). In order to provide a proof we need to make use of another remarkable result of Michael. Let X,Y be spaces. A relation Φ ⊂ X × Y such that, for each x ∈ X, the value

Φ(x) := {y ∈ Y | (x, y) ∈ Φ}= 6 ∅, is lower semicontinuous if, for any open U ⊂ Y ,

Φ−1(U) := {x ∈ X | Φ(x) ∩ U 6= ∅}

6 m S Q Recall that the Tychonoﬀ cube I of weight m > ℵ is I = s∈S Is where Is = I for all s ∈ S and #S = m. In particular, the Hilbert cube I∞ := IN. It is easy to see that I∞ =∼ Q, where

2 Q := {x ∈ ` | 0 6 xi 6 1/i, i ∈ N}.

7 S A (real) Hilbert space having an orthonormal basis of cardinality m > ℵ is homeomorphic to R where 2 ∞ #S = m. In particular, a separable Hilbert space is (linearly) homeomorphic to ` =∼ R . 1.3. Theory of homotopy 41 is open (in X).

The Michael Selection Theorem: If E is as above, X ∈ P, Φ ⊂ X ×E is lower semicontinuous and, for each x ∈ X, the set Φ(x) is (nonempty) closed convex, then there is a continuous selection of Φ, i.e., a continuous map f : X → E such that f(x) ∈ Φ(x) for all x ∈ X. Moreover 8 one may achieve that f(X) ∈ cl conv Φ(X) ( ). Proof of theorem 1.3.14: Let A be a closed subset of a paracompact space X and f : A → C. Deﬁne Φ ∈ X × E by {f(x)} for x ∈ A; Φ(x) := C for x ∈ X \ A. The relation Φ has closed convex values and is lower semicontinuous. Indeed, if U ⊂ E is open, then ∅ if U ∩ U = ∅; Φ−1(U) = f −1(U) ∪ (X \ A) if C ∩ U 6= ∅. ∗ ∗ The selection theorem provide a map f : X → E such that f (x) ∈ Φ(x) for x ∈ X, i.e., ∗ ∗ f (x) ∈ C on X and f (x) = f(x) if x ∈ A. From the viewpoint of many diﬀerent applications is the following result.

1.3.15 The Graves Theorem: Let T : E → F be a continuous linear surjection of a Fréchet spaces. There is a continuous map f : F → E such that f(0) = 0 and T ◦ f = 1F. As a =∼ consequence, there is a homeomorphism h : E −→ ker T × F deﬁned by

h(x) = (x − f ◦ T (x),T (x)), x ∈ E such that h(x) = (x, 0) if x ∈ ker T and p ◦ h = T where p : ker T × F → F is the projection. −1 Proof: Consider a relation Φ ⊂ F × E deﬁned by Φ(y) := T (y) for y ∈ F.Clearly values of −1 Φ are closed convex and if U ⊂ E is open, then Φ (U) = T (U) is open in view of the Banach Open Mapping theorem, i.e., Φ is lower semicontinuous. In view of the Michael selection, there is −1 f : F → E such that f1(y) ∈ T (y) on F. The map f := f1 − f1(0) satisfy the desired properties (check the properties of h !). 1.3.16 Exercise: How the Graves theorem is related to the lifting problem ? There are obviously some other relevant extension results; we shall dwell upon them in the future if necessary.

1.3.C Homotopy

1.3.17 Homotopy: Let X,Y are spaces and Z ⊂ X. We say that maps f, g : X → Y are homotopic relative Z if there is a homotopy joining f to g relative Z, i.e., a map h : X × I → Y (here and below I := [0, 1] ⊂ R) such that h(·, 0) = f, h(·, 1) = g and h(z, t) = f(z) = g(z) for all z ∈ Z and t ∈ I (9). We write h : f ' g (rel Z). In case Z = ∅, then we say that f and g are homotopic and write h : f ' g or h : f ' g : X → Y in order to emphasis the domain and the range of h(·, ), t ∈ I.

8It is interesting that the assertion of the Michael theorem is a suﬃcient condition for a space X to be paracompact. 9In what follows continuous any map h : X × I → Y will be called homotopy. 42 1. Categories, Functors and Algebraization of Topological Problems

Sometimes we say that f : X → Y is null-homotopic if it is homotopic to a constant map.

1.3.18 Fact: (i) If maps f, g : X → Y are homotopic, then there is a path σ : I → C(X,Y ) such that σ(0) = f and σ(1) = g. If X is locally compact, or a k-space or satisfies the first axiom of countability, then the existence of a path σ having the above properties, implies that f ' g. (ii) The relation ' (rel Z) of relative homotopy is an equivalence relation. (iii) Compositions of homotopic maps are homotopic. Proof: is very simple. The first part follows from the exponential law (having the homotopy h : f ' g, σ(t)(x) := h(x, t) for (x, t) ∈ C × I, i.e., σ = Ω(h)). The remaining parts are straightforward. Observe that, for maps f, g : X → Y , the problem of homotopy, i.e., the existence of h : f ' g, is a particular case of the extension problem. Indeed: we actually ask for the existence of an extension of the map: ϕ : X × {0, 1} → Y given by

f(x) for t = 0 ϕ(x, t) = g(x) for t = 1.

The similar definitions concern maps of pairs of spaces. Recall that by a topological pair we mean a pair (X,A) where X is a space and A ⊂ X. Any space X is identified with the pair (X, ∅). For any pairs (X,A) and (Y,B), their cartesian product is defined

(X,A) × (Y,B) := (X × Y,A × Y ∪ X × B).

In particular (X,A) × Y = (X,A) × (Y, ∅) = (X × Y,A × Y ). We say that a pair (X,A) is closed if A = cl A. We say that a pair (X,A) is compact if X is compact and A = cl A. We write f :(X,A) → (Y,B) if f : X → Y and f(A) ⊂ B; we say then that f is a map of pairs. Given maps f, g :(X,A) → (Y,B) and a set Z ⊂ X, we say that a map h :(X,A) × I → (Y,B) provides a homotopy relative to Z joining f to g if h : f ' g(rel Z); sometimes one writes h : f ' g(rel Z):(X,A) → (Y,B), too.

1.3.19 Homotopy Extension Property (HEP): It appears that in many cases the existence of an extension of a given map f : A → Y , where A ⊂ X depends only on the homotopy class [f] of f. We say that A ⊂ X has the homotopy extension property with respect to Y if, for any map g : X → Y and a homotopy h : A × I → Y such that h(·, 0) = g|A, there is a homotopy H : X × I → Y such that H(·, 0) = g. In case A ⊂ X satisﬁes HEP with respect to any space Y , then we say the inclusion iA : A,→ X is a coﬁbration.

1.3.20 Exercise: The inclusion A ⊂ X is a cofibration if and only if T := X × 0 ∪ A × I is a retract in X × I. In particular any polyhedral pair A ⊂ X is a cofibration. If A ⊂ X is a cofibration, then T is a closed subset of X ×I (retracts of a Hausdorff space are closed). The set T 0 = T ∩X ×[1/2, 1] = A×[1/2, 1] is closed. Since the projection p : X ×I → X is closed, A = p(T 0) is closed.

1.3.21 Lemma: Let A ⊂ X be closed. The following conditions are equivalent: (a) A ⊂ X is a coﬁbration; 1.3. Theory of homotopy 43

(b) There exists a deformation D : X × I → X(rel A), i.e., D(x, 0) = x and D(a, t) = a for x ∈ X, a ∈ A and t ∈ I, and a function ϕ : X → I such that A = ϕ−1(1) and D(ϕ−1(0, 1] × {1}) ⊂ A; (c) there is a neighborhood N of A and a homotopy H : N × I → X such that H(x, 0) = x, H(a, t) = a and H(x, 1) ∈ A for any x ∈ N, a ∈ A and t ∈ I. Moreover there is a function ϕ : X → I such that ϕ|A ≡ 1 and ϕ|X\N ≡ 0. Proof: (a) ⇒ (b) Let R : X × I → T := X × {0} ∪ A × I be the retraction. Let R(x, t) = (D(x, t), ϕe(x, t)) where D : X × I → X and ϕe : X × I → I. If x ∈ X, then R(x, 0) = (x, 0), i.e., D(x, 0) = x. For a ∈ A, t ∈ I, R(a, t) = (a, t), i.e., D(a, t) = a. Now let

ϕ(x) = 1 − max |t − ϕ(x, t)|, x ∈ X. t∈I e

Since I is compact, this definition makes sense. Observe that if ϕ(x) = 1, then ϕe(x, t) = t for all t ∈ I. Thus x ∈ A: if x 6∈ A, then there is a neighborhood V of x such that V ∩ A = ∅, V × {0} is open in T , so (x, 0) ∈ R−1(V × {0}) and the latter set is open in X × I, so there is t > 0 such that R(x, t) ∈ V × {0}, i.e., ϕe(x, t) = 0: contradiction. If x ∈ A, then R(x, t) = (x, t) for all t ∈ I; hence ϕe(x, t) = t for t ∈ I, i.e., ϕe(x, t) = t and ϕ(x) = 1. −1 Let x ∈ ϕ (0, 1]; then ϕ(x) > 0. This implies that λ := ϕe(x, 1) > 0. Hence R(x, 1) = (D(x, 1), ϕe(x, 1)) = (D(x, 1), λ) ∈ T . Thus D(x, 1) ∈ A. It remains to show that ϕ is continuous. For this reason it is enough to show that X 3 x 7→ ψ(x) := maxt∈I |t − ϕe(x, t)| is continuous. This is a general result: given a continuous function f : X × I → R, the map ψ(x) := maxt∈I f(x, t), x ∈ X, is continuous. Indeed, suppose that ψ(x0) ∈ (a, b). Obviously there is t0 ∈ I such that ψ(x0) = f(x0, t0). −1 Clearly, for all t ∈ I, f(x0, t) 6 f(x0, t0) < b. Thus {x0} × Isubsetf ((−∞, b)). The compact- 0 0 −1 ness of I implies that there is an open neighborhood V of x0 such that V × I ⊂ f ((−∞, b)). 0 00 Thus ψ(V ) ∈ (−∞, b). On the other hand there is a neighborhood V of x0 such that f(x, t0) > a 00 00 0 00 if x ∈ V . Therefore ψ(x) > a for x ∈ V . Taking V := V ∩ V as a neighborhood of x0, we see that ψ(x) ∈ (a, b) for x ∈ V . −1 (b) ⇒ (c) Let N := ϕ (0, 1]. Then N is an open neighborhood of A. Put H = D|N×I . Then H(x, 0) = x, H(a, t) = a and H(x, 1) ∈ A for x ∈ N, a ∈ A and t ∈ I. Moreover, for x ∈ X \ U, ϕ(x) = 0. (c) ⇒ (a) We can assume that ϕ ≡ 0 on a neighborhood of X \ N by replacing ϕ by 1 + max{2ϕ − 2, 1} if necessary. We shall show that there is a map h : N × I → T such that h(x, 0) = (x, 0) for x ∈ N and h(a, t) = (a, t) for a ∈ A and t ∈ I. Having such h we define the retraction R : X × I → T by the formula R(x, t) = h(x, tϕ(x)) if x ∈ N and h(x, t) = (x, 0) if x ∈ X \ N. To this aim define: for x ∈ N and t ∈ I,

H(x, t ) × {0} if 1 − ϕ(x) > t; h(x, t) = 1−ϕ(x) H(x, 1) × {t − 1 + ϕ(x)} if 1 − ϕ(x) 6 t. We need to show only that h is continuous at those points (x, 0) such that ϕ(x) = 1, i.e., at points (a, 0 for a ∈ A. Note that H(a, t) = a for all t ∈ I. Thus, given a neighborhood U of a, there is a neighborhood V of a such that V ⊂ U and H(V × I) ⊂ U. Therefore, for any ε > 0, if x ∈ V and t ∈ [0, ε), then h(x, t) ∈ U × [0, ε). This implies that h is continuous. 1.3.22 Homotopy Domination and Equivalence: We say that f :(X,A) → (Y,B) is a homotopy equivalence if there is a map g :(Y,B) → (X,A) such that f ◦ g ' 1(Y,B) :(Y,B) → (Y,B) and g ◦ f ' 1(X,A) :(X,A) → (X,A). We say that pairs (X,A) and (Y,B) have the same 44 1. Categories, Functors and Algebraization of Topological Problems homotopy type if there is a homotopy equivalence f :(X,A) → (Y,B). It is easy to see that the relation of the homotopy equivalence is an equivalence relation. We say that a pair (X,A) homotopy dominates a pair (Y,B) if there is f :(X,A) → (Y,B) and g :(Y,B) → (X,A) such that fg ' 1(Y,B). Of course, if (pairs of) spaces are homotopy equivalent, then they are homotopy equal, i.e. one homotopy dominates the other (and vice versa). But not conversely.

1.3.23 Example: If A ⊂ X is a weak retract, i.e. there is r : X → A such that riA ' 1A, then X homotopy dominates A. In particular if A is a retract of X, then X homotopy dominates A.

1.3.24 Deformation Retract: We say that A ⊂ X is a deformation retract if there is a retraction r : X → A such that iAr ' 1X and that A is a strong deformation retract if, additionally, iAr ' 1X (rel A). It is clear that (strong) deformation retract A has the same homotopy type as X.

1.3.25 Exercise: Show that if (X,A) is a polyhedral pair, then X × 0 ∪ A × I is a strong deformation retract in X × I.

1.3.26 Contractibility: A space X is contractible if it has the homotopy type of a one-point space pt, i.e., there is a homotopy equivalence f : pt → X.

1.3.27 Fact: A space X is contractible if and only if the identity is 1X null-homotopic if and only if, for any space Y and maps f, g : Y → X, f ' g. Any map f : X → Y of contractible spaces X,Y is a homotopy equivalence. Proof: The necessity is obvious (by the deﬁnition). Conversely: null-homotopicity (i.e., the existence of h : 1X ' c where c(x) ≡ x0 ∈ X for all x ∈ X) shows that X has the homotopy type of {x0}. The suﬃciency of the second condition is trivial. To see the necessity let h : 1X ' c. Let H : Y × I → X be given by H(y, t) = h(f(y), t) for y ∈ Y and t ∈ I. Then H : f ' c. The transitivity of ' completes the argument. The last statement follows because there is a homotopy equivalence between X and Y and f is homotopic to it; clearly map homotopic to a homotopy equivalence is a homotopy equivalence itself.

1.3.28 Exercise: Show that a contractible space X is path-connected (i.e., given x0, x1 ∈ X, there is a continuous path σ : I → X such that σ(i) = xi, i = 0, 1) and any point x ∈ X is a deformation retract. Is it true that a point in a contractible space is a strong deformation retract ?

1.3.29 Remark: Observe that if f, g : Y → X where X is contractible and f|A = g|A for some A ⊂ X, then, unfortunately, not necessarily f ' g(rel A).

1.3.30 Example: Consider the ‘comb‘, i.e., the set

2 G := {(x, y) ∈ R | 0 6 y 6 1, x = 0, 1/n or y = 0, 0 6 x 6 1}.

Let F : G × I → G be given by F ((x, y), t) = (x, (1 − t)y) for (x, y) ∈ G. Then F (·, 0) = 1G, F (·, 1) = π where π : G → G is deﬁned by π(x, y) = (x, 0) for (x, y) ∈ G. Certainly π ' c, where c : G → G is constant: c(x, y) = (0, 0) on G. Hence G is contractible. Let p : G → G be given by p(x, y) = (0, 1) on G. Then 1G ' p, but not 1G ' p(rel {(0, 1)}). The space G is a weak retract in I ×I. Indeed the spaces I ×I and G are contractible. Thus, by the above fact the inclusion iG : G → I × I is a homotopy equivalence and, in particular, has 1.3. Theory of homotopy 45 a right homotopy inverse.

1.3.31 Exercise: Show that G is not a retract in I2.

1.3.D Special Homotopies

In this section we shall discuss special kinds of homotopies together with some applications.

1.3.32 Smooth Homotopies: Let U be an open subset of a Banach space E and let h : U ×I → r F, where F is another Banach space, be a homotopy. We say that h is a C -homotopy, 1 6 r 6 ∞, if h is Cr-smooth. Let us mention an important approximation result

n m 1.3.33 The Whitney Approximation Theorem: Let U ⊂ R be open and let f : U → R be Cr-smooth. Given a continuous function η : U → (0, +∞), there is a real analytic (in particular ∞ m n C ) function g : U → R such that, for any α ∈ Z , |α| < k + 1, and x ∈ U, α α |D f(x) − D g(x)| < η(x).

The theorem of Whitney is a generalization of the Weierstrass approximation theorem and may be very useful on some occasions.

1.3.34 Remark: It is not know whether the analogue of the Whitney theorem holds for maps between inﬁnite dimensional spaces. However there is a version of this result due to J. Eells and J. MacAlpin which states that: If U is and open subset of a separable Hilbert space H, f : U → E, where E is a Banach ∞ space, is continuous, ε : U → (0, +∞) is continuous, then there is a C -map g : U → E such that, for each x ∈ U, kf(x) − g(x)k < ε(x). It is easy to derive a ‘parameterized’ version of this result, i.e., if h : U × I → E is continuous, ∞ then the is a C function H : U × I → E such that kh(x, t) − H(x, t)k < ε(x) for all x ∈ U and t ∈ I. Some other important improvements of this result are due do N. Moulis. In particular we have

1.3.35 Theorem: If U is open in H, f, g : U → V , where V is open in a Banach space E, are Cr-smooth and f ' g, then f and g are Cr-homotopic (as maps U → V , i.e., there is Cr-homotopy U × I → V . 0 1 Proof: Let h : f ' g and let, for x ∈ U, η (x) = 2 inft∈I d(h(x, t), E \ V ). It is not diﬃcult to 0 check that η : U → R is lower semicontinuous, η > 0 and that there is a continuous function 0 η : U → R such that 0 < η 6 η . m ∞ Let h2 : U × I → R be a C -homotopy provided by the Eells-MacAlpin theorem, i.e., such that, for all x ∈ U, |h(x, t) − h2(x, t)| < η(x). It is clear that h2(x, t) ∈ V for any x ∈ U and k t ∈ I. Let f1(x) := h2(x, 0) and g1(x) = h2(x, 1) for x ∈ U. Clearly f1, g1 are C -smooth. Now m deﬁne h1 : U × I → R by

h1(x, t) = (1 − t)f(x) + tf1(x), x ∈ U, t ∈ I, m and h3 : U × I → R by

h3(x, t) = tg(x) + (1 − t)g1(x), x ∈ U, t ∈ I. 46 1. Categories, Functors and Algebraization of Topological Problems

k It is clear that both h1 and h3 are C -homotopies and h1(x, t), h3(x, t) ∈ V for all x ∈ U, t ∈ I. Finally let h : U × I → V be deﬁned by   h1(x, α1(t)) for t ∈ [0, 1/3]; h(x, t) = h2(x, α2(t)) for t ∈ [1/3, 2/3];  h3(x, α3(t)) for t ∈ [2/3, 1],

∞ where x ∈ U and αi : [(i − 1)/3, i/3] → [0, 1] is a C -smooth increasing function onto such (k) (k) that α ((i − 1)/3) = α (i/3) = 0 for all 1 6 k < ∞, i = 1, 2, 3. It is easy to check that k h(x, 0) = f(x), h(x, 1) = g(x) on U and h is a C -homotopy. 1.3.36 Exercise: (i) Show that η0 above is lower semicontinuous and prove the existence of a 0 continuous η : U → R such that 0 < η 6 η .

1.3.37 Gradient Homotopies: Let U be an open subset of a (real) Banach space E. A map ∗ 1 f : U → E is a gradient map if there is a C -function F : U → R such that, for each x ∈ U, F 0(x) = f(x). If E = R, then any f : U → R is a gradient map; this notion is interesting only if dim E > 2. If H is a Hilbert space, U ⊂ H is open, then f : U → H is a gradient map (or a potential 1 field) if there is a C -function F : U → R such that f(x) = ∇F (x) for x ∈ U. Recall that, by definition, if F : U → R is differentiable at x ∈ U, then the gradient ∇F (x) ∈ H is given as a 0 unique vector in H such that F (x)(h) = h∇F (x), hi, where h·, ·i denotes the inner product in H, for any h ∈ H; the existence of ∇F follows by the Riesz theorem. ∗ In a similar way we define gradient homotopies: a map h : U ×I → E (or to H) is a gradient 1 homotopy if there is a C -homotopy H : U × I → R such that, for each x ∈ U and t ∈ I, 0 0 h(x, t) = Hx(x, t) (clearly Hx(x, t) is the derivative if H at (x, t) with respect to x). ∗ A function F satisfying the above condition with respect to a gradient map f : U → E is called an antiderivative or a potential. It is easy to see that if U is connected, then a potential F is determined uniquely up to a constant.

n 1.3.38 The Parusiński Theorem: If f, g : U → R are gradient maps such that h : f ' g, then there is a gradient homotopy h : f ' g The author does not know the inﬁnite dimensional version of the Parusiński result.

1.3.E Homotopy Sets

Let X,Y be spaces and Z ⊂ X. By [X; Y ]Z we denote the set of equivalence classes of the relation ' (rel Z), i.e., the set of elements of the form [f]Z := {g : X → Y | f ' g(rel Z)} where f : X → Y . Let [X; Y ] := [X; Y ]∅ and [f] := {g : X → Y | f ' g} for f : X → Y . Fact 1.3.18 (iii) shows that one may consider the category of homotopy of topological spaces HT op whose objects are spaces and, for objects X,Y , the class of morphisms is the set [X; Y ]. The transformation which to a map f : X → Y assigns the homotopy class [f] is a functor from T op to HT op.

The set of homotopy classes of maps of pairs is denoted by [X,A; Y,B]Z and [f]Z := {g : (X,A) → (Y,B) | f ' g(rel Z)} for f :(X,A) → (Y,B). If Z = ∅, then we write [X,A; Y,B]. Again we may consider the category HT op2 with objects being topological pairs and morphisms are classes [f] ∈ [X,A; Y,B] of maps f :(X,A) → (Y,B). It is easy to see that HT op is a subcategory in HT op2. 1.3. Theory of homotopy 47

Using the language of category we see that f :(X,A) → (Y,B) is a homotopy equivalence if [f] is an isomorphism in the category HT op2, i.e., there is a map g :(Y,B) → (X,A) such that −1 2 f ◦ g ' 1(Y,B) and g ◦ f ' 1(X,A); equivalently [g] = [f] in the category HT op . In particular, pairs (X,A) and (Y,B) have the same homotopy type if objects (X,A), (Y,B) are isomorphic in the category HT op2.

1.3.39 Induced transformations: Assume that a pair (P,Q) is ﬁxed. Consider a functor 2 π(P,Q) : HT op → Sets assigning to a pair (X,A) the set π(P,Q)(X,A) := [P,Q; X,A] and to the class [f] of a map f :(X,A) → (Y,B) assigns the function π(P,Q)([f]) = f# :[P,Q; X,A] → [P,Q; Y,B] given by f#([g]) = [f ◦ g] for any g :(P,Q) → (X,A). In a similar way we have a cofunctor π(P,Q) : HT op2 → Sets, π(P,Q)(X,A) = [X,A; P,Q] and π(P,Q)([f]) = f # :[Y,B; P,Q] → [X,A; P,Q] is deﬁned by f #([g]) = [g ◦ f] for g :(Y,B) → (P,Q).

1.3.40 Exercise: Fix a space P and consider two functors on T op: T T (1) OT op 3 X 7→ C(P,X) ∈ OSets and MorT op(X,Y ) 3 f 7→ T (f) ∈ MorSets(C(P,X), C(P,Y )) given by T (f)(g) := f ◦ g for g ∈ C(P,X); S S (2) OT op 3 X 7→ πP (X) ∈ OSets and MorT op(X,Y ) 3 f 7→ S(f) ∈ MorSets(πP (X), πP (Y )) given by S(f)([g]) =: f#(g) = [f ◦ g] for [g] ∈ πP (X).

Let Φ = {ΦX }X∈OT op , where ΦX : T (X) → S(X) is given by ΦX (f) = [f] for f ∈ T (X) = C(P,X). Show that Φ is a natural transformation of T into S. Note how complicated the language of categories can be. The statement of this exercise say nothing more but that f#([g]) = [g ◦ f]. But the reader should not be discouraged: sometimes this language is very convenient indeed.

1.3.41 Fact: If spaces Y and Z have the same homotopy type, then for any space X,

[X; Y ] =∼ [X,Z].

Proof: It is easy to see that if f : Y → Z is a homotopy equivalence, then f# is a desired bijection. n−1 ∼ n 1.3.42 Example: For any space [X; S ] = [X, R \ 0], n > 1. 1.3.43 Exercise: Is Fact 1.3.41 true if Y and Z are homotopy equal ?

n n 1.3.44 Fact: Let n > 0 and let p0 ∈ S . For a map f : S → X, where X is a space, the following conditions are equivalent: (i) the map f is null-homotopic; (ii) the is an extension f ∗ : Dn+1 → X of f; n (iii) f ' c (rel {p0}) where c : S → X is a constant map. n n+1 Proof: Let F : f ' c where c(x) = x0 ∈ X for x ∈ S . For x ∈ D , deﬁne

x if 0 kxk 1/2; f ∗(x) = 0 6 6 F (x/kxk, 2 − 2kxk) if 1/2 6 kxk 6 1.

∗ It is easy to see that f |Sn = f. Assume that f ∗ : Dn+1 → X is an extension of f and deﬁne h : Sn × I → X by h(x, t) = ∗ n ∗ n−1 f ((1 − t)x + tp0) for x ∈ S and t ∈ I. Then h(·, 0) = f, h(x, 1) = f (p0) for x ∈ S ; thus h(·, 1) is constant. The last implication is obvious. . 48 1. Categories, Functors and Algebraization of Topological Problems

This simple result is important and has some deep consequences. To state them we need another general notion

1.3.45 Essentiality of maps: Consider a pair (X,A), a space Y and let K ⊂ Y . We say that ∗ a map f :(X,A) → (Y,Y \ K) is inessential over K if there is an extension f : X → Y of f|A such that f ∗(X) ∩ K = ∅. Dually, the map f is essential over K if it is not inessential, i.e., for ∗ ∗ any extension f : X → Y of f|A, f (X) ∩ K 6= ∅. Essentiality is a homotopy invariant of the restriction of a map to A.

1.3.46 Lemma: Suppose that: (i) the inclusion A,→ X is a cofibration, or (ii) X is binormal, the pair (X,A) is closed and Y ∈ ANE(N ). If g :(X,A) → (Y \ K) and h : g|A ' f|A : A → Y \ K, then f is essential if and only if so is g. Proof: Since the relation of homotopy is reflexive it is sufficient to assume that f is essential and to prove that so does g. Suppose to the contrary that g is inessential, i.e., there is g∗ : X → Y \K ∗ 0 such that g |A = g|A. Consider h : C := X × {0} ∪ A × I → Y \ K given by g∗(x) if x ∈ X, t = 0; h0(x, t) := h(x, t) if x ∈ A. t ∈ I. If A ⊂ X is a cofibration, then C is a retract of X ×I and, hence, there is H : X ×I → Y \K 0 ∗ ∗ 0 such that H|C = h . Let f := H(·, 1) : X → Y \ K. Then, for x ∈ A, f (x) = h (x, 1) = f(x), i.e., f is inessential. Suppose now that X is binormal, A is closed and that Y is an absolute neighborhood extensor for normal spaces. Since C is a closed subset of the normal space X × I, there are an 00 00 0 open neighborhood U of C in X × I and h : U → Y such that h |C = h . Let p : X × I → X be the projection. The compactness of I implies that B := p(X × I \ U) is closed (in view of the Kuratowski theorem). Clearly A ∩ B = ∅. Let λ : X → I be a continuous Uryshon function such 0 0 00 that λ|A ≡ 1 and λ|B ≡ 0. Let H : X × I → Y be given by H (x, t) = h (x, λ(x)t) for x ∈ X and t ∈ I. This map is correctly defined, since for any x ∈ X and t ∈ I, (x, λ(x)t) ∈ U (if not then x ∈ B and λ(x) = 0 and (x, 0) 6∈ U: contradiction). Observe that, for all x ∈ X, H0(x, 0) = h00(x, 0) = h0(x, 0) = g∗(x) and, if x ∈ A, then H0(x, t) = h00(x, t) = h0(x, t) = h(x, t) 6∈ A for all t ∈ I. Let D := {x ∈ X | H0(x, t) ∈ K for some t ∈ I}. Again D = p(H0−1(K)) is closed and D ∩ A = ∅. Take another Urysohn function η : X → I such that η|A ≡ 1 and η|D ≡ 0 and, finally define H : X × I → Y by H(x, t) := H0(x, η(x)t) for x ∈ X, t ∈ I. Then, for x ∈ A, H(x, 1) = H0(x, 1) = h00(x, 1) = h0(x, 1) = h(x, 1) = f(x), i.e. f ∗ := H(·, 1) extends f. If f ∗(x) = H(x, 1) ∈ K for some x ∈ X, then H0(x, η(x)) ∈ K, i.e. x ∈ D and η(x) = 0. It is 0 ∗ impossible since H (x, 0) = g (x) 6∈ K. n n−1 k k 1.3.47 Corollary: Let f :(D ,S ) → (R , R \ 0), n, k > 1. The map f is essential (over n−1 k 0) if and only the restriction g := f|Sn−1 : S → R \ 0 is not null-homotopic. This results shows that the solvability of the equation f(x) = 0 depends on the ‘position’ of n−1 k ∼ n−1 k−1 [g] in [S ; R \ 0] = [S ; S ]: it has a solution if and only if the element [g] is nontrivial k (observe that all constant maps in R \0, k > 2, are homotopic since this space is path-connected.

1.3.48 Remark: In the context of this observation there are three cases: (i) 1 6 n < k; (ii) n−1 k−1 1 6 n = k and (iii) n > k > 1. The first case is easy: [S ,S ] is a singleton and consists of the homotopy class of constant maps. The proof follows below. The two other cases are much 1.3. Theory of homotopy 49 more interesting: it appears that if n = 1 = k, then [S0; S0] has 4 elements (describe them) n n ∼ n 0 and if n = k > 1, then [S ; S ] = Z; if n > k = 1, then [S ,S ] has two elements, while in general if n > k > 1, then [Sn−1; Sk−1] is a finite set (we shall get back to the problem of the characterization of this set). n k Assume now that f : S → S where 0 6 n < k. Then, by the use of the Whitney approximation theorem and partitions of unity it is possible to construct a C∞-smooth map g : Sn → Sk such that f ' g. Therefore, without loss of generality we may suppose that f is n smooth. Take a point x0 ∈ S ; the continuity of the tangent field T f implies that rx := rank Txf is constant for x in a neighborhood V of x0, i.e. rx = r 6 n < k. By the Rank Theorem, for each x ∈ V there are local coordinates (U, ϕ) and (U 0, ϕ0) around x and f(x), respectively, such that 0 −1 k ϕ ◦ f ◦ ϕ |ϕ(U) coincides with the map (x1, ..., xn) 7→ (x1, ..., xr, 0, ..., 0) ∈ R . This shows that n n k (locally), meas f(S ) = 0. This implies, in particular, that f(S ) ( S . Therefore f ' const. (show it carefully).

1.3.49 Example: As an example of application we shall derive the fundamental theorem of algebra from Corollary 1.3.47: any polynomial w ∈ C[z] of degree n > 1 has a root. n n−1 Let w(z) = a0z + a1z + ... + an−1z + an where ai ∈ C, i = 0, ..., n and a0 6= 0; clearly we may treat w as a map w : C → C. (Dividing by a0) without loss of generality we may assume that a0 = 1. Then, for z 6= 0, n ! X ai w(z) = zn 1 + . zi i=1 Pn −i Let 0 < ε < 1. If |z| = r and r is large enough, then i=1 aiz < ε. For z ∈ Sr := {z ∈ C | |z| = r} (10) to and t ∈ I, let ! n n X −i h(z, t) := (1 − t)z + tw(z) = z 1 + t aiz . i=1

n Then h(z, 0) = z and h(z, 1) = 1(z) for |z| = r. Moreover h(z, t) 6= 0 on Sr × I. Indeed, if P −i h(z, t) = 0, then −1 = t i=1 aiz : contradiction for the module of the right-hand side is less n than 1. In other word we have proven that maps z 7→ z and w are homotopic as maps form Sr to C \ 0. If w has no roots, then – in particular – the equation w|Dr (z) = 0 has no solutions and, n in view of Corollary 1.3.47, w|Sr as well as z is null-homotopic. Below we shall show that this is not true. To continue the proof we need some new machinery.

1.3.F The Index of a Function relative to a Curve

2 2 Let f = (f1, f2): U → R , where U ⊂ R , be continuous and let γ : I → U be a piecewise 1 C -smooth, i.e., there are points 0 = t0 < t1 < ... < tn = 1 such that the restriction σ|[ti−1,ti], i = 1, ..., n, is C1-smooth, and closed, i.e., γ(0) = γ(1), curve such that w(t) := f(γ(t)) 6= 0 for all t ∈ I. Taking a smaller set we may assume that f does not vanish on U. 2 Now treat R and C. According to Fact 5.0.3, there is a branch L : I → C of the logarithm of w, i.e., eL(t) = w(t) for any t ∈ I; at the same time there is a branch A(t) = Im L(t), t ∈ I, of the argument of w. It is clear that the diﬀerences L(1) − L(0) and A(1) − A(0) do not depend of the choice of L and A. It is clear that L(1) − L(0) = i(A(1) − A(1)) because L(t) = ln |w(t)| + iA(t) and w(0) = w(1).

10 1 2 Note that Sr is homeomorphic to S and Dr := {z ∈ C | |z| 6 r} is homeomorphic D . 50 1. Categories, Functors and Algebraization of Topological Problems

It is easy to see that L(1) ≡ L(0)(mod 2πi) since eL(1) = w(1) = w(0) = eL(0); in other 1 1 words 2πi (L(1) − L(0)) ∈ Z and 2π (A(1) − A(0)) ∈ Z. The integer number 1 1 ind f := (l(1) − L0)) = (A(1) − A(0)) γ 2πi 2π is called the index of f with respect to γ or the winding number of f on γ. The geometric intuition behind indγf is evident: it counts the number of times the vector f(γ(t)) turns around 2 0 as t ∈ [0, 1] (taking into account the anti-clockwise orientation in C = R ). If f = 1R2 , then indγf is the index of γ with respect to 0. There is a question how can one compute the index eﬀectively. For this reason we need to recall some more techniques.

1.3.50 Curvilinear integral: Suppose U is an open subset of a Banach space E. Let a r ∗ continuous curve σ : I → U be piecewise C -smooth, r > 1. Given a continuous f : U → E , we deﬁne the (oriented) curvilinear integral of f along σ (11)

Z Z 1 f := hf(σ(t)), σ˙ (t)i dt (12), σ 0 where on the right-hand side the Riemann integral is considered. It is evident that this integral is well-defined for the integrand on the right-hand side is continuous at all but finite number of points. Remark: It is not difficult to generalize the notion of the curvilinear integral to rectifiable curves. However we shall consider only piecewise smooth or more regular curves. The defined integral has the standard properties: it is linear with respect to the integrand r and with respect to the curve, i.e. if γ : I → U is another piecewise C -smooth curve, r > 1, then Z Z Z f = f + f σ+γ σ γ where σ + γ is the curve defined as

σ(2t) for t ∈ [0, 1/2]; (σ + γ)(t) := γ(2t − 1) for t ∈ [1/2, 1].

r Given a piecewise C -smooth curve σ, r > 1, by the negative curve −σ we mean a curve −σ(t) := σ(1 − t) for t ∈ I. It is clear that −σ is a piecewise Cr-smooth curve and Z Z f = − f. −σ σ Observe also that Z Z 1

f 6 sup kf ◦ σ(t)k kσ˙ (t)k dt. σ t∈I 0

1.3.51 Remark: There is a slight diﬀerence between the just recalled notion of a curvilinear integral in the real sense and that known from complex analysis. Given a C1-smooth curve

11The mentioned orientation reﬂects the fact that we can distinguish the beginning and the end of the curve as σ(0) and σ(1), respectively. 12 ∗ Notation: given p ∈ E and x ∈ E, hp, xi := p(x). 1.3. Theory of homotopy 51

γ : I → C and a continuous function f : U → C defined on U ⊃ γ(I), we put Z Z 1 f dz := f(γ(t))γ0(t) dt. γ 0 This integral has properties similar to those collected above, but the difference consists on the 0 2 0 action of f(γ(t)) on γ (t): if we treat C as R , then the complex product f(γ(t))γ (t), where 0 0 0 γ (t) = (Reγ) (t) + i(Im γ) (t) and f(γ(t)) ∈ C differs from the (real) scalar product considered above.

In order to compute indγf assume that f is differentiable. We shall see now that L is piecewise smooth. Without loss of generality we assume that γ is smooth (otherwise we shall proceed by pieces) and let t0 ∈ I. For an increment ∆t 6= 0, let L(t0 + ∆t) = L(t0) + ∆z. We may assume that ∆z 6= 0 for, otherwise, L ≡ const. We proceed using the algebra in C. By definition, L(t0+∆t) ∆z L(t0) ∆z w(t0 + ∆t) = e = e e = e w(t0). Hence w(t + ∆t) − w(t ) (e∆z − 1)∆z 0 0 = w(t ) , ∆t 0 ∆z∆t i.e., L(t0 + ∆t) − L(t0) ∆z 1 w(t0 + ∆t) − w(t0) ∆z = = ∆z . ∆t ∆t w(t0) ∆t e − 1 The continuity of L implies that ∆z → 0 as ∆t → 0. Taking into account the (complex) differentiability of the exponent we see that

0 0 w (t0) L (t0) = . w(t0)

Now, for t ∈ I, w0(t) = f 0(γ(t))(γ0(t)) and, hence,

f 0(γ(t))(γ0(t)) L0(t) = . f(γ(t))

If f is holomorphic, then f 0(γ(t))γ0(t) L0(t) = f(γ(t)) (f 0◦γ)γ0 for t ∈ I. In other words L is the primitive of the function f◦γ . As a corollary we see that if f is holomorphic, then

Z f 0(z) Z 1 f 0(γ(t))γ0(t) dz := dt = L(1) − L(0) = 2πiindγf. (∗) γ f(z) 0 f(γ(t))

In order to get the real expression for indγf, i.e., without assumptions concerning holomor- phicity of f, let us observe that the diﬀerentiability of L implies the diﬀerentiability of A and (since A = Im L)

w0(t) 1 1 A0(t) = Im = Im w(t)w0(t) = h(−f , f )(γ(t)), f 0(γ(t))(γ0(t))i. w(t) |w(t)|2 |f(γ(t))|2 2 1 52 1. Categories, Functors and Algebraization of Topological Problems

2 2 ∗ Let G : U → R = (R ) be given by 1 G(x, y) = f 0(x, y)T (−f (x, y), f (x, y)), (x, y) ∈ U. (1.3.5) |f(x, y)|2 2 1

Then Z Z 1 Z 1 1 1 0 0 1 0 G = 2 h(−f2, f1)(γ(t)), f (γ(t))(γ (t))i dt = A (t) dt = indγf. 2π γ 0 |f(γ(t))| 2π 0

This expression does not look as elegant as (∗), but – as we shall see later – is much more convenient and geometrically as intuitive as (∗).

Now we shall show that: if continuous f0, f1 : C → C are such that fi(γ(t)) 6= 0 for i = 0, 1 and t ∈ I and f0 ' f1 (more precisely f0|Γ ' f1|Γ :Γ → C \ 0 where Γ := γ(I) of γ), then

indγf0 = indγf1.

2 Indeed, let f :Γ × I → C be a homotopy joining f0|Γ to f1|Γ; then w : I → C deﬁned by

w(t, s) := f(γ(t), s), t, s ∈ I, joins w0 := f0 ◦ γ, to w1 := f1 ◦ γ. Let L0 be a branch of the logarithm of w0. The Homotopy Lifting Property of the covering exp : C → C \ 0 implies that homotopy may be lifted to the 2 L(t,s) 2 homotopy L : I → C such that e = w(t, s) on I and Ls := L(·, s) lifts ws := w(·, s) = f(γ(·, s); i.e., Ls is the a branch of the logarithm of ws. For each s ∈ I, then number

L(1, s) − L(0, s) = ∆γf(·, s) ∈ Z.

But the function s 7→ L(1, s) − L(0, s) is continuous; thus, in view of the connectedness of I it must be constant. We may now get back to the proof of the fundamental theorem of algebra. It is clear that n n if z 7→ z is null-homotopic as the map from Sr to C \ 0, then indγz = 0, where γ is the curve 2πit with support Γ = Sr (i.e., γ(t) = re , t ∈ I), since any constant map (into C \ 0) has the trivial index with respect to γ. However

Z (zn)0 Z 1 n dz = n dz = 2nπi; γ z γ z

n hence indγz = n. This completes the proof. n 1.3.52 Remark: Above n ∈ N. The same computation show that of n ∈ Z, then indγz = n. 1 1 ∼ 1 ∼ 1.3.53 The Bijection Theorem: There is a bijection [S ; S ] = [S ; C \ 0] = Z provided by 1 [f] 7→ indS1 f for any continuous f : S → C \ 0. 1 Proof: To check the injectivity of indS1 suppose that f : S → C \ 0 and indS1 f = n choose θ ∈ log f(1). and let f(z) h(z, t) = , z ∈ S1, t ∈ I. etθ 1 Then h(·, 0) = f and, if g = h(·, 1); S → C \ 0, then g(1) = f(1)/ exp θ = 1. Let σ : I → C \ 0 be given by

σ(t) = g e2πit , t ∈ I. 1.3. Theory of homotopy 53

τ(t) The Path-Lifting Property implies that there is a path τ : I → C such that e = σ(t) for all τ(1) 2πi 0 t ∈ I. Since e = σ(1) = g e = 1 = e we gather that τ(1) = 2πim where m ∈ Z. Now let η(s, t) = (1 − t)τ(s) + 2πitsm for s, t ∈ I. Then η(s, 0) = τ(s), η(s, 1) = 2πims and η(1, t) = 2πim for any t ∈ I. 1 Finally let H : S × i → C \ 0 be given by eη(q(z),t) for z ∈ S1 \ 1; H(z, t) := 1 for z = 1 ∈ S1 where q : S1\ → (0, 1) is deﬁned as follows: the map (0, 1) 3 t 7→ e2πit is a homeomorphism; q is the inverse homeomorphism, i.e., e2πiq(z) = z for any z ∈ S1 \ 1. The map H is clearly correctly deﬁned and continuous. Observe that, for z ∈ S1 \ 1,

H(z, 0) = eη(q(z),0) = eτ(q(z)) = σ(q(z)) = g(exp(2πiq(z))) = g(z) and H(1, 0) = 1 = g(1). On the other hand, for z ∈ S1 \ 1

H(z, 1) = eη(q(z),1) = e2πimq(z) = zm

m m m and H(1, 1) = 1 = 1 . This shows that f ' g ' z . Thus n = inds1 f = indS1 z = m and f ' zm. 1 As a consequence if f, g : S → C \ 0 and indS1 f = indS1 g, then f ' g. n The surjectivity of Ind S1 is obvious: for any n ∈ Z, indS1 z = n. The Bijection Theorem may be slightly strengthened by observing that S1 is an abelian group (with complex multiplication). For any f, g : S1 → S1 let

(f ⊕ g)(z) := f(z)g(z), (f g)(z) := f(g(z)), z ∈ S1.

The ‘addition’ ⊕ and ‘multiplication’ deﬁned above generates the ring structure in C(S1,S1). Clearly it induces the ring structure in [S1; S1] as well: [f] ⊕ [g] := [f ⊕ g] and [f] [g] := [f g].

1 1 1.3.54 Corollary: The map indS1 :[S ; S ] → Z is a ring isomorphism.

Proof: We are to show that indS1 (f g) = indS1 findS1 g and indS1 (f ⊕g) = indS1 f +indS1 g. To n m n m n+m this end, let indS1 f = n and indS1 g = m. Then f ' z and g ' z ; but clearly z ⊕ z = z n m nm and z z = z . This completes the proof immediately. 1 1 √ 1 1.3.55 Exercise: Consider f : S → S given by f(z) = z, for z ∈ S and let indS1 f = k. 2 Then 1 = indS1 (f ) = 2k. What is wrong here ? The use of the index is obvious.

2 2 1 1.3.56 Theorem: Let F : D → R be a continuous map such that F (x) 6= 0 for x ∈ S (i.e., 1 f := F |S1 : S → C \ 0). The equation F (x) = 0 has a solution if indS1 f 6= 0. Precisely F is essential if and only if indS1 f 6= 0. n 1.3.57 Corollary: Any map of the Euclidean sphere S , n > 0, into a contractible space X is null-homotopic and has an extension onto the ball. k As we have seen this is not the case if the space X is not contractible (e.g. X := R \ 0, 0 6 k 6 n. Hence the idea is to classify spaces X according to the behavior of maps of spheres.

1.3.58 n-Connectedness: Let n > 0 be an integer. We say that a space X is n-connected if k ∗ k+1 any map f : S → X, 0 6 k 6 n, is null-homotopic, i.e., admits an extension f : D → X. 54 1. Categories, Functors and Algebraization of Topological Problems

A contractible space is therefore ∞-connected, i.e. n-connected for all n > 0. 1.3.59 Fact: (i) A space X is 0-connected if and only if it is path-connected. (ii) A space X is 1-connected if and only if, given paths σ, τ : I → X such that σ(i) = τ(i) for i = 0, 1, we have that σ ' τ (rel {0, 1}). (iii) A space X is n-connected, n > 0, if and only if, given a polyhedron T , dim T 6 n, any two maps f, g : T → X are homotopic. Proof: The ﬁrst two parts are obvious; the last part is a bit more diﬃcult: one proceeds inductively on simplices of the underlying triangulation. k We see that if X is n-connected, then the set πk(X) := [S ,X] is a singleton for all 0 6 k 6 n. Therefore the complexity of the set πk(X) measures the complexity of the space.

1.4 Absolute Homotopy Groups of a Space

Let X be a space and let us distinguish a point x0 ∈ X. The pair (X, x0) is called a pointed space. Pointed spaces are sometimes denoted by (X, ∗). n 0 0 Recall that, for n ∈ N, I := I × ... × I , I := {0} and ∂I = ∅. | {z } n For an integer n > 0, let n n π(X, x0) := [(I , ∂I ); (X, x0)],

n n 13 i.e., the set of all homotopy classes of maps (I , ∂I ) → (X, x0) ( ). Thus if n = 0, then π0(X, x0) may be identiﬁed with the set of all path-components of X; by 0 ∈ π0(X) we mean the path-component of X that contains x0. If n > 1, then by 0 ∈ πn(X, x0) we mean the homotopy n n n class of the constant map x0 : I → {x0}; in other words [f] = 0, where f :(I , ∂I ) → (X, x0), if and only if f ' x0.

n n 1.4.1 Remark: Observe that, for any n > 0, any map f :(I , ∂I ) → (X, x0) may be identified n n+1 with a map g :(S , p) → (X, x0) where p = (1, 0, ..., 0) ∈ R . It would be welcome if the reader keeps it in mind. n n n For n > 1 and f, g :(I , ∂I ) → (X, x0), we define a map f g : I → X given for n t = (t1, ..., tn) ∈ I , by f(2t , t ..., t ) if t 1/2; f g(t) = 1 2 n 1 6 g(2t1 − 1, t2, ..., tn) if t1 > 1/2. n n It is easy to see that this formula correctly defines a continuous map (I , ∂I ) → (X, x0). n n Moreover it is quite easy to check that if f0 ' f1 :(I , ∂I ) → (X, x0) and g0 ' g1 : n n (I , ∂I ) → (X, x0), then

n n f0 g0 ' f1 g1 :(I , ∂I ) → (X, x0).

As a consequence we see that

1.4.2 Fact: For n > 1, the operator deﬁnes a binary operation

: πn(X, x0) → πn(X, x0)

13 n n n n Recall that f0, f1 :(I , ∂I ) → (X, x0) if there is a map f :(I , ∂I ) × I → (X, x0) such that f(·, i) = fi, n i = 0, 1, i.e., f : f0 ' f1(rel ∂I ). 1.4. Absolute Homotopy Groups of a Space 55

deﬁned as follows: for [f], [g] ∈ πn(X, x0), [f] [g] := [f g]. It is a routine to show

1.4.3 Theorem: For n > 1, the pair (πn(X, x0), ) is a group, i.e. the operation is associative, −1 0 is the neutral element and, for any [f] ∈ πn(X, x0) the left (and right) inverse [f] (i.e. such −1 −1 −1 that [f] [f] = 0 = [f] [f] ) is given by [f] = [fb] where fb(t1, ..., tn) := f(1 − t1, ..., tn) for n t = (t1, ..., tn) ∈ I . For n > 1, this group is called the n-th homotopy group of the space X with a distinguished point x0.

1.4.4 Exercise: If X is contractible, then πn(X, x0) = 0 for all n > 0 (of course if n > 1, this means that the group πn(X, x0) is a trivial group, while for n = 0, this means that X is path-connected).

m 0 1.4.5 Homotopy groups πn(S , x0), m > 1: If m = 0, then π0(S , x0) consists of two 0 1 elements and, for n > 1, πn(S , x0) = 0. So assume that m = 1. It is clear that S , m > 1 is 1 1 path-connected; hence π0(S ) = 0. We already know that π1(S , x0) = Z. Let m > 1. Then m again π0(S , x0) = 0 and if 1 6 n < m, then in view of Remark 1.3.48, we see that any map n m n n m S → S is null-homotopic. Hence so is any map (I , ∂I ) → (S , x0) (check it carefully) and, m m m therefore πn(S , x0) = 0 if 0 6 n < m. The problem to compute πm(S , x0) and πn(S , x0) for n > m is more complicated. We shall do it later on. n n 1.4.6 Connection with homology groups: Let n > 0 and σ :(I , ∂I ) → (X, x0). It is clear that ∂σ = 0, hence σ ∈ Zn(X). Chapter 2 Diﬀerential Forms, de Rham cohomology and singular cubical homology

2.1 Gradient Maps Revisited

We shall now study necessary and suﬃcient conditions for a function to be a gradient map.

∗ m 2.1.1 Theorem: Suppose that f : U → E is C -smooth, m > 0. R 1 (i) If is a gradient map, then σ f = 0 for any closed piecewise C -smooth curve (i.e. such that σ(0) = σ(1)). R ∞ (r) (r) (ii) If σ f = 0 for any C -smooth closed σ : I → U, i.e., such that σ+ (0) = σ− (1) for all m+1 0 0 6 r < ∞, then there is a C -function F : U → R such that F = f; in particular f is a gradient map. 0 1 1 Proof: (i) If f = F where F ∈ C (U, R), then for a closed piecewise C -smooth curve σ, F ◦ σ is piecewise differentiable and at points of differentiability (F ◦ σ)0(t) = hF 0(σ(t)), σ˙ (t)i = hf(σ(t)), σ˙ (t)i for all t ∈ I. Hence the integrand hf ◦ σ(·), σ˙ (·)i has the primitive F ◦ σ. Thus Z 1 f = F ◦ σ(t)|0 = 0 σ (check it carefully). R (ii) The first step (difficult) is to show that in fact the assumption implies that σ f = 0 for any closed piecewise C1-curve. Assume that U is connected (for otherwise we should consider 1 each component separately). Fix a point x0 ∈ U and for x ∈ U let γx be an arbitrary C -curve ∞ joining x0 to x (the existence of an even C -smooth curve joining x0 to x is readily verifiable). Let Z F (x) = f, x ∈ U. γx ∞ This definition is correct since if σx is another C -curve joining x0 to x, then γx + (−σx) is a closed piecewise C∞-smooth curve and Z Z Z 0 = f = f − f. γx−σx γx σx 2.1. Gradient Maps Revisited 57

m+1 We shall show that F ∈ C (U, R). For any (suﬃciently small) h ∈ E, Z F (x + h) − F (x) = f, [x;x+h] since as a choice for γx+h we may take γx + [x; x + h] where [x; x + h] is a segment joining x to x + h (1), and

Z 1 |F (x + h) − F (x) − f(x)(h)| 1 = hf(x + th) − f(x), hi dt 6 sup kf(x + th) − f(x)k → 0 khk khk 0 t∈I

0 as h → 0. This show that F = f. 2.1.2 Remark: The assumption on σ in part (ii) of Theorem 2.1.1 may be stated as follows: σ : S1 → U is C∞-smooth. Such map may be identified with a closed C∞-smooth curve and vice-versa. We shall now study a different sufficient condition for f to be a gradient map. We start with the following observation

∗ 0 ∗ 2.1.3 Fact: If f : U → E is a differentiable gradient map, then f (x) ∈ L(E, E ) is symmetric 0 0 on U, i.e., for any h1, h2 ∈ E, hf (x)(h1), h2i = hf (x)(h2), h1i. ∗ 2 Proof: It is clear that L(E, E ) 'L (E, R); if F denotes a potential of f, then F is twice 00 0 differentiable and F (x)(h1, h2) = hf (x)(h1), h2i for all x ∈ U and h1, h2 ∈ E. However, by the 00 2 well-known Schwarz theorem the second derivative F (x) ∈ L (E, R), x ∈ U, is symmetric. The problem is whether the necessary condition provided above is also sufficient for f to be a gradient map. It is not so.

2 y 2.1.4 Example: Let U1 := R \{(x, 0) | x 6 0} and let g(x, y) := Arg(x + iy) = arc tg x for 1 (x, y) ∈ U1, then g ∈ C (U1,R) (actually g(x, y) ∈ (−π, π) on U1). It is easy to see that, for (x, y) ∈ U1, y x ∇g(x, y) = − , . x2 + y2 x2 + y2 2 2 Let f : U → R , where U := R \{(0, 0)}, be given by the above formula, i.e., y x f(x, y) := − , x2 + y2 x2 + y2 for (x, y) ∈ U. Then the derivative f 0(x, y) is symmetric since

∂f ∂f 1 = 2 ∂y ∂x

1 on U but f is not a gradient map. Indeed if not, then there is a potential F ∈ C (U, R) such that ∇F = f on U. Clearly ∇g = f on U1; therefore F − g ≡ const. on U1. Unfortunately g has no continuous extension onto U. 2 R By the way, if σ is the unit circle in R , then σ f = 2π. In view of Theorem 2.1.1 this shows that f is not a gradient map. As a matter of fact, suggested by the above example, it is easy to give another condition 0 ∗ suﬃcient for the symmetry of f : if f : U → E is a diﬀerentiable locally gradient map (i.e., any

1 For a, b ∈ E, [a; b] := {a + t(b − a) | t ∈ [0, 1]}. 58 2. Differential Forms, de Rham cohomology and singular cubical homology

0 point x ∈ U has a neighborhood Ux such that f|Ux is a gradient map), then f (x) is symmetric for all x ∈ U – see also Corollary 2.1.7. In the next section we shall see however that the positive solution to the problem whether the symmetry of f 0 implies that f is a gradient map depends on the topology of the set U (or on a special behavior of curvilinear integrals).

1 ∗ 0 2.1.5 Fact: If U is convex or, more generally, starshaped, f ∈ C (U, E ) and f is symmetric is symmetric, then f is a gradient map. Proof; Assume that U is starshaped around a ∈ U and let

Z Z 1 F (x) = f = hf(a + t(x − a)), x − ai dt (2.1.1) [a;x] 0

0 for x ∈ U. We shall see that F (x) = f(x) on U. Let ϕ : U × I → R be deﬁned by ϕ(x, t) := hf(a + t(x − a)), x − ai for t ∈ [0, 1], x ∈ U. Then, for each t ∈ I, ϕ(·, t) is C1-smooth and, for all x ∈ U, ϕ(x, ·) is continuous. Precisely, for all t ∈ I and x ∈ U,

0 0 hϕ (x, t), hi = hf (a + t(x − a))(th), x − ai + hf(a + t(x − a)), hi, h ∈ E.

0 0 The continuity of f implies that kϕ (x, t)k 6 C for some constant C > 0. In view of the theorem 2 on diﬀerentiation of a parameterized integral ( ), for any x ∈ U and h ∈ E, Z 1 hF 0(x), hi = [hf 0(a + t(x − a))(th), x − ai + hf(a + t(x − a)), hi] dt = 0 Z 1 [hf 0(a + t(x − a))(x − a), thi + hf(a + t(x − a)), hi] dt = 0 Z 1 d hf(a + t(x − a)), thi dt = hf(x), hi, 0 dt

0 i.e., F (x) = f(x) on U. Fact 2.1.5 is true if we assume that less about f. For the sake of completeness we state the following Goursat lemma (whose ingenious and beautiful proof is worth studying).

2 2 2.1.6 Goursat Lemma: Let g : G → R , where G ⊂ R is open, is diﬀerentiable (not neces- 1 0 0 sarily C -smooth) and, for each x ∈ G, let g (x) be symmetric in the sense that hg (x)(h1), h2i = 0 2 hg (x)(h2), h1i for all h1, h2 ∈ R . Given points a, b, c ∈ G, suppose that conv {a, b, c} ⊂ G. Then Z g = 0 γ

2 It states that: Let ϕ : U × I → R. If, for almost all t ∈ I, the function ϕ(·, t) is (Gateaux) diﬀerentiable, for 0 each x ∈ U, ϕ(x, ·) is integrable and there is an integrable g : I → R such that kϕ (x, t)k 6 g(t) for almost all t ∈ I 0 0 R 1 (where ϕ (x, t) is the Gateaux derivative), then for each x ∈ U, ϕ (x, ·) is integrable. If F (x) := 0 ϕ(x, t) dt, then F is Gateaux diﬀerentiable and, for all x ∈ U.

Z 1 F 0(x) = ϕ0(x, t) dt, (∗) 0 i.e., for any h ∈ E, Z 1 hF 0(x), hi = hϕ0(x, t), hi dt. (∗∗) 0 If, additionally, for almost all t ∈ I, ϕ(·, t) is continuously differentiable, then F is continuously differentiable (and formulas (∗), (∗∗) hold). 2.1. Gradient Maps Revisited 59 where γ = [a; b] ∪ [b; c] ∪ [c; a]. ∗ 0 2.1.7 Corollary: Given a differentiable map f : U → E , where U ⊂ E is open, f (x) is symmetric for all x ∈ U if and only if, for any a, b, c ∈ U such that conv {a, b, c} ⊂ U, Z f = 0. [a;b]∪[b;c]∪[c;a]

Proof: (⇒) Let F be two-dimensional subspace in E containing points a, b and c. Then ∗ ∗ ∼ 2 ∗ ∗ ∗ conv {a, b, c} ⊂ G := U ∩ F. Let g := j ◦ f|G : G → F = R (j us the adjoint F → E of the inclusion j : F ,→ E). In view of the Goursat Lemma 2.1.6, Z Z f = g = 0. [a;b]∪[b;c]∪[c;a] [a;b]∪[b;c]∪[c;a]

(⇐) Take an arbitrary x ∈ U and a ball Ux := B(x, ε) ⊂ U. For each a, b, c ∈ Ux, R conv {a, b, c} ⊂ U; hence [a;b]∪[b;c]∪[c;a] f = 0. Arguing as in the proof of Theorem 2.1.1 we see that f is a locally gradient map. 2.1.8 Corollary: Suppose that f and U satisfy the hypotheses of Corollary 2.1.7. If U is convex or starshaped, then f is a gradient map and, in particular, the curvilinear integral of f along any closed piecewise C1-smooth path in U vanishes. Proof: Choose a ∈ U and, for x ∈ U, let Z F (x) = f. [a;x]

Fix x ∈ U. For each (suﬃciently small) h ∈ E, Z Z Z F (x + h) − F (x) = f − f = f [a;x+h] [a;x] [x;x+h]

0 in view of Corollary 2.1.7. Now it is easy to show that F (x) = f(x). We shall see the hypothesis on U may be relaxed. We are now going to show:

∗ 2.1.9 Theorem: If U ∈ E is open and 1-connected, then any diﬀerentiable function f : U → E such that, for each x ∈ U, the derivative f 0(x) is symmetric, is a gradient map. ∗ Proof: Let f : U → E be satisfy the assumptions. In view of Theorem 2.1.1 (ii) it is suﬃcient R ∞ to show that the integral σ f along a closed C -smooth curve σ vanishes. Now we will need the following lemma:

1 2.1.10 Lemma: If U is arbitrary and piecewise C -smooth closed paths γ0, γ1 are homotopic (i.e., there is γ : [0, 1] × I → U such that γ(·, i) = γi for i = 0, 1 and γ(0, t) = γ(1, t) for all t ∈ I (3)), then Z Z f = f. γ0 γ1 Suppose the lemma is true and let σ be a closed C∞-smooth curve. Since U is 1-connected any map γ : S1 → U is null-homotopic; hence so is σ (recall Remark 2.1.2). According to Lemma R 2.1.10, σ f = 0. 3Any closed curve σ in U may be identiﬁed with a map S1 → U. Therefore the assumed relation of homotopy 1 between γ0 and γ1 means that these maps are homotopic in the usual sense as maps S → U. 60 2. Differential Forms, de Rham cohomology and singular cubical homology

1 Proof of Lemma 2.1.10: Suppose that γ : S × I → U is a homotopy joining γ0 to γ1. Due to Corollary ?? and Remark 1.3.34 with obvious modiﬁcations, we may assume without loss of 1 generality that, for each t ∈ I, γt := γ(·, t) is a piecewise C -smooth closed curve. Observe that the function Z Z 1 I 3 t 7→ Γ(t) := f = hf(γ(s, t)), γ˙t(s)i ds γt 0 is continuous. Hence, in order to prove that Γ(0) = Γ(1) it is suﬃcient to show that Γ(·) is locally constant, i.e., for any t0 ∈ I, there is δ such that Γ(t) = Γ(t0) for |t − t0| < δ and then and appeal to connectedness of I. Let K := γ(S1 × I); clearly K is compact and K ⊂ U. There is ε > 0 such that, for each x ∈ K, B(x, ε) ⊂ U. Since γ is uniformly continuous, there is δ such that if |s−s0|, |t−t0| < δ, then 0 0 1 kγ(s, t) − γ(s , t )k < ε. Take an arbitrary t0 ∈ I and choose points s1, s2, ..., sn = s1 ∈ S such that s1 < s2 < ... < sn < s1 with respect to the anti clock-wise orientation and |si − si+1| < δ. 1 For i = 1, ..., n − 1, and t ∈ I, |t − t0| < δ, consider a closed piecewise C -smooth curve σi being the sum of γ(·, t0)|[s ,s ], the segment [γ(si+1, t0); γ(si+1, t)], −γ(·, t)|[s ,s ] and the segment i i+1 i i+1 R [γ(si, t); γ(si, t0)]. Since this curve is located in the ball B(γ(si, t0), ε), we see that f = 0 in σi view of Corollary 2.1.8. On the other hand it is self-evident that

n−1 Z Z Z Z X Z f − f = f + = f = 0. γ(·,t0) γ(·,t) γ(·,t0) −γ(·,t) i=1 σi

Using a bit diﬀerent terminology we may restate Theorem 2.1.9 in the following way:

2.1.11 Corollary: If a differentiable differential form of order 1 defined on a 1-connected open set is closed, then it is exact. That is if U ⊂ E, where E is a Banach space, is 1-connected, ∗ ω : U → E is a differentiable 1-form and dω = 0, then there is a function f : U → R such that ω = df. In general, the so-called de Rham-Poincaré Lemma states that if U is k-connected, the any closed differential k-form is exact. The proof of this result will be given and discussed below. Now we shall describe the terminology used in Corollary 2.1.11.

2.2 Diﬀerential Forms

2.2.1 Differential Forms: Let U be a subset of the Banach space E and F be a Banach k space. A map ω : U → A (E,F ), k > 0, is called a diﬀerential form of degree k with values in F. If x ∈ U, then we write

ω(x; e1, ..., ek) := ω(x)(e1, ..., ek) for e1, ..., ek ∈ E.

0 Remember that A (E, F) = F, i.e. 0-form is a function U → F. k We shall consider differential forms ω : U → A (E, F) being regular, i.e., continuous or, in r case U is open, differentiable or C -smooth, 1 6 r 6 ∞ (note that ω is a map from (an open) set k r U ∈ E to the Banach space A (E, F) so one can speak of differentiability). The set of C -smooth k k k k-forms is denoted by Λ(r)(U, F). The symbol Λ (U, F) := Λ(∞) will also be used to denote the set of all C∞-smooth k-forms.

2.2.2 Convention: By default we assume that when discussing k-forms on U ⊂ E involving differentiability assumptions (e.g. Cr-smoothness), the set U is supposed to be open or, at least, 2.2. Differential Forms 61 such that it is possible to differentiate F-valued functions defined on it (e.g. U is contained in the closure of its interior). k It is easily seen that Λ(r)(U, F) (as well as the set of differentiable k-form) is a (real) vector space: (ω1 + ω2)(x) := ω1(x) + ω2(x), (λω)(x) := λω(x) k k for ω, ω1, ω2 ∈ Λ (U, F) (or Λ(r)(U, F)) and λ ∈ R (obviously in the right-hand sides we have the k addition and the scalar multiplication in the sense of L (E, F)). k The vector space Λ(r)(U, F) may be endowed with linear topology making it a topological linear spaces. We shall not study this problem.

The simplest example of a 1-form is a map f : U → L(E, F) or, in the case of R-valued ∗ r 0 r−1 forms, a map f : U → E . In particular, if f : U → F is of class C , then f is a C -smooth 0 1 1-form , i.e., f ∈ Λ(r−1)(U, F). The operation of exterior multiplication extends immediately to diﬀerential forms: if ω and η are k-forms, then (ω ∧ η)(x) := ω(x) ∧ η(x), x ∈ U.

Observe that if forms ω and η are of class Cr and Cs, respectively, then ω ∧ η is of class Cmin{r,s}. This follows since the exterior multiplication may be viewed as the superposition of the k m k m k+m map (ω, η): U → A (E) × A (E) with ∧ : A (E) × A (E) → A (E) which, being continuous bilinear, is of class C∞. 1 For instance if ω, η ∈ Λ(r)(U), then

(ω ∧ η)(x; e1, e2) = [ω(x) ∧ η(x)](e1, e2) = ω(x; e1)η(x; e2) − ω(x; e2)η(x; e1) for e1, e2 ∈ E. 2.2.3 Exterior Derivative: Let a k-form ω be differentiable. We define the exterior derivative dω as the (k + 1)-form defined by

k X i 0 dω(x; e0, e1, ..., ek) := (−1) [ω (x)(ei)](e0, ..., ebi, ..., ek) i=0 0 for e0, ..., ek ∈ E (where ..., ebi, ... means that ei is omitted and ω (x) denotes the derivative of k 0 k k ω : U → A (E, F), i.e. ω (x) ∈ L(E, A (E, F))). It is clear that if ω ∈ Λ(r), r > 1, then k+1 dω ∈ Λ(r−1)(U, F). There are at least two ways to explain this definition. Here is the first one. Assume that ω 0 k k is differentiable; hence, for any x ∈ U, ω (x) ∈ L(E, A (E, F)) exists. Clearly L(E, A (E, F)) ⊂ k ∼ k+1 L(E, L (E, F)) = L (E, F). The last (linear continuous) bijection or identification assigns to ev- k k+1 ery T ∈ L(E, L (E, F)) a map (denoted by the same symbol) E 3 (e0, ..., ek) 7→ T (e0, ..., ek) := [T (e0)](e1, ..., ek). This, in particular allows us to consider a map 0 0 ω (x)(e0, ...., ek) := [ω (x)(e0)](e1, ..., ek)

0 k+1 0 k+1 for e0, e1, ..., ek ∈ E. Obviously ω : U → L (E, F) but, in general, ω (x) ∈/ A (E, F) for 1 1 x ∈ U (if k = 0, then L(E, F) = L (E, F) = A (E, F) and everything is alright). For k > 1 and x ∈ U, let us therefore ‘alternate’ ω0(x), i.e. consider 1 X k+1 3 (e , ..., e ) 7→ (k + 1)Alt(ω0(x))(e , ..., e ) = sgn (σ)ω0(x)(e , e , ..., e ). E 0 k 0 k k! σ(0) σ(1) σ(k) σ∈Sk+1 62 2. Differential Forms, de Rham cohomology and singular cubical homology

The set Sk+1 of all permutations of the set {0, 1, ..., k +1} may be decomposed into k +1 disjoint subsets i S := {σ ∈ Sk+1 | σ(i) = 0}, i i i = 0, ..., k. Observe that, for all i = 0, ..., k, S may be identiﬁed with Sk; precisely to any σ ∈ S , there corresponds in the one-to-one way a unique permutation τ(σ) of the set {0, ...,bi, ....k}; moreover sgn (σ) = (−1)isgn (τ(σ)). Hence

0 (k + 1)Alt(ω (x))(e0, ..., ek) = k 1 X X (−1)isgn (τ)[ω0(x)(e )](e , ..., e , e , ..., e ) k! i τ(0) τ(i−1) τ(i+1) τ(k) i=0 τ∈Sk k   X 1 X = (−1)i sgn (τ)[ω0(x)(e )](e , ..., e , e , ..., e ) k! i τ(0) τ(i−1) τ(i+1) τ(k)  i=0 τ∈Sk k X k 0 = (−1) Alt(ω (x)(ei))(e1, ..., ebi, ...., ek) i=0 k X i 0 = (−1) [ω (x)(ei)](e1, ..., ebi, ..., ek) = dω(x)(e0, ..., ek) i=0 0 0 since Alt(ω (x)(ei)) = ω (x)(ei) for all i = 0, 1, .., k. We thus see that the exterior derivative dω(x) is, up to the factor (k + 1), nothing else but the alternation of ω0(x) understood as the k ∼ k+1 element of L(E, L (E, F)) = L (E, F), i.e., we may redeﬁne dω in the following way: dω(x) = (k + 1)Alt(ω0(x)), x ∈ U, for any diﬀerentiable k-form on U. The other interpretation of d will be discusses later.

k k+1 2.2.4 Exercise: Show that d defined on Λ(r)(U, F), r > 1, is a linear map into Λ(r−1)(U, F). Moreover: (i) if a k-form ω is twice differentiable, then d(dω) = 0 (be careful: this is very tedious); it is also true that if ω and dω are differentiable, then d(dω) = 0. (ii) If two R-valued k-forms ω, η are differentiable, then d(ω ∧ η) = dω ∧ η + (−1)kω ∧ dη.

2.2.5 Example: Let f : U → F be diﬀerentiable, i.e., f is a diﬀerentiable 0-form on U. Then 0 df(x; e) = f (x)e, x ∈ U, e ∈ E.

If a 1-form ω is diﬀerentiable, i.e. the function ω : U → L(E, F) is diﬀerentiable, then dω = 0 if and only if the bilinear map

0 E × E 3 (e1, e2) 7→ [ω (x)(e1)](e2)(∗)

0 (and if F = R, the bilinear form (e1, e2) 7→ hω (x)(e1), e2i) is symmetric. ∗ Hence a differentiable f : U → E (i.e., a 1-form) satisfies the assumptions from Corollary 2.1.7 if and only if df = 0. Having this we may restate Theorem 2.1.9: If a 1-form ω is differentiable, dω = 0, U is 1-connected, then there a twice differentiable function f : U → R such that ω = df. 2.2. Differential Forms 63

On the other hand Fact 2.1.3 is a particular case of the above Exercise: if f : U → R is twice diﬀerentiable (in particular the 1-form ω := df is diﬀerentiable), then dω = d(df) = 0.

2.2.6 Closed and exact forms: We say that a k-form, k > 1, ω on U is exact if ω has an exterior primitive η, i.e., there is a differentiable (k − 1)-form η on U such that ω = dη.A differentiable k-form, k > 0, ω is closed if dω = 0. Hence we have a result: any exact differentiable k-form is closed; and if U is 1-connected, then any closed 1-form is exact.

2.2.7 Canonical form of k-forms: Assume that dim E = n, let {v1, ..., vn} be a basis in E 1 n ∗ j and let {p , ..., p } be the dual basis in E . For any j = 1, ..., n, by x : E → R we denote the j P j j j j function E 3 x 7→ hp , xi, i.e., if x = j=1 a vj, then x (x) = a . Clearly x : U → R is a 0-form j j k and dx (x) = p for all x ∈ E. Hence, if ω : U → A (E), then

X j1 jk ω(x)(e1, ..., ek) = ωj1...jk (x)(p ∧ ... ∧ p )(e1, ..., ek), x ∈ U, 16j1

X j1 jk dω = dωj1...jk ∧ dx ∧ ... ∧ dx . 16j1

2.2.8 The pull-back of a form: Let E1, E2 be Banach spaces, Ui ⊂ Ei, i = 1, 2, be open and let ϕ : U1 → U2 be a diﬀerentiable map. If ω is a k-form in U2, k > 0, then we deﬁne the # pull-back of ω via ϕ to be the k-form ϕ ω on U1 given by

# 0 0 (ϕ ω)(x; e1, ..., ek) := ω(ϕ(x); ϕ (x)(e1), ..., ϕ (x)(ek)), x ∈ U1, ei ∈ E, i = 1, ..., k.

r+1 k # k Observe that if ϕ is C -smooth and ω ∈ Λ(r)(U2, F) then ϕ ω ∈ Λ(r)(U1, F) (this follows since ϕ#ω may be considered as the composition of Cr-maps). # k k Therefore it is clear that ϕ is a linear map form Λ(r)(U2, F) to Λ(r)(U, F) provided ϕ is Cr+1-smooth.

2.2.9 Exercise: (i) If ϕ is twice diﬀerentiable and a k-form ω, k > 0, is diﬀerentiable, then

d(ϕ#ω) = ϕ#(dω).

(ii) If ω is an R-valued k-form on U2 and η is an R-valued m-form on U2, k, m > 0, and ϕ : U1 → U2 is diﬀerentiable then

(ϕ#ω) ∧ (ϕ#η) = ϕ#(ω ∧ η)

(iii) If ψ : U2 → U3, where U3 is an open subset of a Banach space E3, is diﬀerentiable, then

(ψ ◦ ϕ)#ω = ϕ#(ψ#ω) for any k-form ω on U3. 64 2. Differential Forms, de Rham cohomology and singular cubical homology

At the end of this section we shall provide some information on complex diﬀerential forms

2.2.10 Complex differential forms: Let U ⊂ E, where E is a complex Banach space, and F be a complex Banach space. By a k-differential form we understand a function ω : U → Ak ( , ). R E F (p,q) A k-form is of type (p, q), where p, q > 0 and p + q = k, is a map ω : U → A (E, F). Of course as above we shall speak of regular complex k-forms. In particular we say a k-form ω is holomorphic if ω it is type (k, 0) and (as a function from the complex Banach space E to (k,0) the complex Banach space A R(E, F)) is holomorphic. Exterior multiplication and exterior derivative for complex differential forms are defined exactly as before and have the same properties.

n 2.2.11 Canonical form of a complex differential form: Let E = C . For i = 1, ..., n, i n n by z we denote the projection C → C onto the i-th coordinate, i.e., C 3 z = (z1, ..., zn) 7→ zi. i n n i On sees easily that a map dz : C → LR(C , C) is C-linear, while dz is C-linear. Hence arguing as above in Remarks 0.4.1 and 2.2.7, we see that, for p, q > 0,p + q = k, k-forms of the form

dzI ∧ dzJ := dzi1 ∧ ...dzip ∧ dzj1 ∧ dzjq , where I = {i1, ..., ip}, J = {j1, ..., jq} are multiindices such that 1 6 i1 < ..., ip 6 n, 1 6 j1 < ..., J n, form a basis in A(p,q)( n, ). Hence if ω : U → A(p,q)( n, ) is a k-form of type q 6 R C C R C C (p, q), then X I J ω = ωIJ dz ∧ dz (∗) I,J n where and ωIJ : C → C. Having the representation (∗) we see that the form (∗) is holomorphic if and only if q = 0 (i.e., p = k) and, for each I = {i1, i2, ..., ik}, ωI : U → C is holomorphic. If f : U → C, then f is a 0-form and

n X j j df = ∂zfdz + ∂zdz := (∂zj fdz + ∂zj dz ). j=1

Therefore for ω of the form (∗) we have

X I J dω = dωIJ ∧ dz ∧ dz . I,J

n n n Remembering that LR(C , C) = L(C , R) ⊕ iL(C , R), it is easy to see that, for any 1 6 j 6 n, dzj = dxj + idyj

j j n where x (resp y ) denotes the function C 3 z = (z1, ..., zn) 7→ Rezj (resp. z 7→ Im zj). Pn j n In particular given a holomorphic 1-form ω = j=1 ωjdz on U ⊂ C , where ωj = uj + ivj : U → C, we see that ω = ω1 + iω2

1 2 k 2n where ω , ω : U → A (R , R) are (real) 1-forms on U and

n n 1 X j j 2 X j j ω = (ujdx − vjdy ), ω = (vjdx + ujdy ). j=1 j=1 2.3. De Rham’s Cohomology 65

2.3 De Rham’s Cohomology

As above let U be a subset of a Banach space E and let F be another Banach space (remember about the convention 2.2.2). Recall that Λk(U, F ) (resp. Λk(U)) denotes the space of C∞- k smooth F-valued (resp. R-valued) k-forms on U, k > 0. For k < 0, we let Λ (U, F) := {0} and k k+1 d :Λ (U, F) → Λ (U, F) be the zero operator. For any k ∈ Z, let

k k Z (U, F ) := Ker d = {ω ∈ Λ (U, F) | dω = 0},

k k−1 k B (U, F ) = Im (d :Λ (U, F) → Λ (U, F)). k ∞ k In other word Z (U, F) consists of C -smooth closed k-forms, while B (U, F) consists of exact ∞ C -smooth k-forms. Since d ◦ d = 0, we see that, for all k ∈ Z,

k k B (U, F) ⊂ Z (U, F),

k k i.e., B (U, F) is a subspace of Z (U, F). Therefore we can form the quotient space

k k k H (U, F) := Z (U, F)/B (U, F)

k called the k-th de Rham cohomology space of U, k ∈ Z. Elements of H (U; F) are called k k-dimensional (de Rham) cohomology classes of U and are denoted by [ω] where ω ∈ Z (U, F).

2.3.1 Remark: Sometimes, for r > 1, we write

k k Z(r)(U, F) := {ω ∈ Λ(r)(U, F) | dω = 0} and, for r > 0,

k k B(r)(U, F) := {ω ∈ Λ(r)(U, F) | ω = dη for some diﬀerentiable (k + 1)-form η on U}

k k It is clear that B(r)(U, F) ⊂ Z(r)(U, F) for r > 1 and k > 0. Therefore formally one could consider k a ‘more general’ object than H (U, F); namely, for 1 6 r < ∞, we may put

k k k H(r)(U, F) := Z(r)(U, F)/B(r)(U, F). Later we shall see however that this does not lead to anything new.

k 0 2.3.2 Example: For any k < 0, H (U, F) = 0. Since Z (U, F) consists of locally constant 0 functions U → F, we see that f ∈ Z (U, F) if and only if f ≡ const. on any connected component 0 of U. Observing that B (U, F) = {0}, if C denotes the set of all these components, then

0 #C H (U, F) = F .

2.3.3 Induced homomorphism: Now let ϕ : U1 → U2, where Ui ⊂ Ei is an open set, be ∞ C -smooth. Then, for any k ∈ Z,

# k k # k k ϕ (Z (U2, F)) ⊂ Z (U1, F) and ϕ (B (U2, F) ⊂ B (U1, F).

Hence ϕ induces a linear map

∗ k k k ∗ # k ϕ : H (U2, F) → H (U1, F), H (U2, F) 3 [ω] 7→ ϕ [ω] := [ϕ ω] ∈ H (U1, F). 66 2. Differential Forms, de Rham cohomology and singular cubical homology

r+1 ∗ In an obvious manner C -smooth map ϕ : U1 → U2, r > 1, deﬁnes a linear operator ϕ : k k H(r)(U2, F) → H(r)(U1, F). ∗ Obviously 1U = 1Hk(U,F). More generally: it is quite easy to check the validity of the following fact.

2.3.4 Fact: The de Rham cohomology is a cofunctor form the category of open subsets of Banach spaces and C∞-smooth maps between them into the category of vector spaces and linear maps. ∞ ∗ In particular, if ϕ is a C -diﬀeomorphism of U1 onto U2, then ϕ is an isomorphism and ∗ −1 −1 ∗ (ϕ ) = (ϕ ) . ∞ ∞ ∗ ∗ 2.3.5 Theorem: If C -smooth maps ϕ0, ϕ1 : U1 → U2 are C -homotopic, then ϕ0 = ϕ1. Proof: Let k ∈ Z (obviously only the case k > 0 is interesting) and consider maps

ji : U1 → U1 × I, i = 0, 1 given by ji(x) = (x, i) for x ∈ U1. ∞ It is clear that, or i = 0, 1, ji is C -smooth and determines a linear map

# k k ji :Λ (U1 × I, F) → Λ (U1, F).

k For ω ∈ Λ (U1 × I, F), x ∈ U1, and e1, ..., ek ∈ E1, we have

# (ji ω)(x; e1, ..., ek) = ω((x, i); (e1, 0), ..., (ek, 0)).

Deﬁne a third linear map

k k−1 ` :Λ (U1 × I, F) → Λ (U1, F)

k given by the formula: for ω ∈ Λ (U1 × I, F), Z 1 `ω(x; e1, ...ek−1) := ω((x, t); (0, 1), (e1, 0), ..., (ek−1, 0)) dt, e1, ..., ek−1 ∈ E1. 0

This map is correctly defined since, for each x ∈ U1 and e1, ..., ek−1 ∈ E1, the integrand on the right-hand side is continuous; moreover, in view of the differentiability of parameterized integrals, ∞ the (k − 1)-form `ω is C -smooth. In particular, for x ∈ U1 and e ∈ E1, Z 1 0 0 [(`ω) (x)(e)](e1, ..., ek−1) = [ω (x, t)(e)]((0, 1), (e1, 0), ..., (ek−1, 0)) dt. 0 Hence the above definition is correct. We are going to show that # # `dω + d`ω = j1 ω − j0 ω (∗) 2.3. De Rham’s Cohomology 67

k for any ω ∈ Λ (U1 × I, F). Indeed, for x ∈ U1 and e1, ..., ek ∈ E1, Z 1 (`dω)(x; e1, ..., ek) = dω((x, t); (0, 1), (e1, 0), ..., (ek, 0)) dt = 0 Z 1 0 ω (x, t)((0, 1))((e1, 0), ..., (ek, 0)) dt + 0 k Z 1 X i 0 (−1) ω (x, t)((ei, 0))((0, 1), (e1, 0), ..., (\ei, 0), ..., (ek, 0)) dt = i=1 0 Z 1 ∂ω (x, t)((e1, 0), ..., (ek, 0)) dt + 0 ∂t k Z 1 X i 0 (−1) [ω (x, t)((ei, 0))]((0, 1), (e1, 0), ..., (\ei, 0), ..., (ek, 0)) dt = i=1 0 k Z 1 # # X i 0 (j1 ω − j0 ω)(x; e1, ..., ek) + (−1) ω (x, t)((ei, 0))((0, 1), (e1, 0), ..., (\ei, 0), ..., (ek, 0)) dt. i=1 0 On the other hand

k X i−1 0 (d`ω)(x; e1, ..., ek) = (−1) [(`ω) (x)(ei)](e1, ..., ebi, ..., ek) = i=1 k Z 1 X i 0 − (−1) [ω (x, t)((ei, 0))]((e1, 0), ..., (\ei, 0), ..., (ek, 0)) dt. i=1 0 This establishes formula (∗). k Let ϕ : U1 × I → U2 be a smooth homotopy joining ϕ0 to ϕ1 and let [ω] ∈ H (U2), then

# # # # # # # # # # # d`ϕ ω = `dϕ ω + d`ϕ ω = j1 ϕ ω − j1 ϕ ω = (ϕ ◦ j1) ω − (ϕ ◦ j0) ω = ϕ1 ω − ϕ0 ω.

# # k ∗ ∗ Therefore ϕ1 ω − ϕ0 ω ∈ B (U1, F), i.e., ϕ0[ω] = ϕ1[ω]. r 2.3.6 Smooth contractibility: A U ⊂ E is C -smoothly contractible, 0 6 r 6 ∞, if the r identity 1U is C -homotopic to a constant map.

2.3.7 Exercise: Show that any starshaped open set UC∞-smoothly contractible and note r r that if U is C -smoothly contractible, then any two C -smooth maps ϕ0, ϕ1 : E1 ⊃ V → U are Cr-homotopic.

∞ 2.3.8 Poincaré Lemma: (comp. paragraph 2.2.6) If U ⊂ E is C -smoothly contractible, then k k k H (U; F) = 0 for all k ∈ Z. In other words, for all k ∈ Z, Z (U, F) = B (U, F), i.e., any closed k k-form ω ∈ Λ (U, F) is exact. ∗ ∞ Proof: Evidently if c : U → F is a constant map, then c = 0. Thus if 1U is C -smoothly ∗ ∗ k homotopic to c, then 1Hk(U,F) = 1U = c = 0, i.e., H (U, F) = 0, by Theorem 2.3.5. 2.3.9 Remark: The proof of Theorem 2.3.5 tells us that if ϕ is a C∞-smooth homotopy joining # # 1U to c, then ω = −d`ϕ ω, i.e., η := −`ϕ ω satisﬁes the requirements of the Poincaré Lemma.

2.3.10 Example: Suppose that an open U is starshaped around a ∈ U. Then ϕ : U × I → U given by ϕ(x, t) = x + t(a − x), x ∈ U, t ∈ I, 68 2. Differential Forms, de Rham cohomology and singular cubical homology

∞ provide a C -smooth homotopy joining 1U to the constant map c : U 3 x 7→ a. Hence the exterior primitive of ω has the form

Z 1 0 0 0 η(x)(e1, ..., ek−1) = − ω(ϕ(x, t); (ϕ (x, t)(0, 1), ϕ (x, t)(e1, 0), ..., ϕ (x, t)(ek−1, 0)) dt = 0 Z 1 Z 1 k−1 ω(a + t(x − a); x − a, te1, ..., tek−1) dt = t ω(a + t(x − a); x − a, e1, ..., ek−1) dt. 0 0 Formally Z 1 η(x) = tk−1ω(a + t(x − a))(x − a, ·) dt. 0 In case k = 1, we have

Z 1 Z η(x) = ω(a + t(x − a); x − s) dt = ω 0 [a;x]

1 ∗ where 1-form ω ∈ Λ (U; F) is actually a function U → E ; compare this formula with (2.1.1).

2.3.11 Remark: Carefully studying the proof of Theorem 2.3.5, we see that if ϕ : U1 × I → U2 r+1 k r is a C -smooth homotopy, r > 1, then for any ω ∈ Z(r)(U2, F), k > 1, the C -smooth k-form # # # # k # # # ϕ1 ω − ϕ0 ω is exact, i.e., ϕ1 ω − ϕ0 ω ∈ B(r)(U, F) and ϕ1 ω − ϕ0 ω = dη where η = `ϕ ω ∈ k−1 r+1 1 Λ(r) (U, F). Hence if U is C -smoothly contractible, r > 1, then any closed C -smooth k-form, 1 2 k > 1, on U is exact. In particular any closed C -smooth k-form on a C -smoothly contractible (e.g. starshaped) open set is exact.

2.3.12 Question: It is not clear whether closeness of a diﬀerentiable k-form on a C2-smoothly contractible (or even C∞-contractible) open set is suﬃcient for its exactness. Results of this type discussed earlier had an intermediate step that has something to do with integration of 1-forms along closed curves. In the next section we shall pursuit this approach.

2.4 Integration of diﬀerential forms and smooth singular homology

r r By a C -smooth singular k-cube, 0 6 r 6 ∞, k > 1, in U ⊂ E, we understand a C -map σ : Ik → U (recall also the definition of differentiability of functions defined on non-open sets). By the carrier of σ we mean the image |σ| := σ(Ik). For k = 0, Ik = {0} and by a 0-cube we mean any map {0} → U, i.e., a point in U (of course, 0-cubes are C∞-smooth). r (r) The set of C -smooth singular k-cubes, k > 0, will be denoted Σk (U). It is convenient (r) (r) to put Σk (U) = {0} for any r and k < 0. Clearly Σ0(U) = Σ0 (U) = U for any r and (s) (r) Σk (U) ⊂ Σk (U) whenever k ∈ Z and 0 6 r 6 s 6 ∞.

k (1) 2.4.1 The integral: Let ω ∈ Λ(0)(U; F), k > 1, and let σ ∈ Σk (U). Then we put Z Z Z # 0 0 ω := (σ ω)(x; v1, ..., vk) dx = ω(σ(x); σ (x)(v1), ..., σ (x)(vk)) dx = σ Ik Ik Z ∂σ ∂σ ω σ(x); (x), ..., (x) dx Ik ∂x1 ∂xk 2.4. Integration of differential forms and smooth singular homology 69

k where {v1, ..., vk} is the canonical basis in R . We may also deﬁne the integral along 0-cubes. If ω : U → F is a 0-form and σ is a 0-cube, then Z ω := ω(σ(0)). σ The following simple exercise plays an important theoretical role.

k (∞) R 2.4.2 Exercise: Show that if ω ∈ Λ(0)(U, F) and, for each σ ∈ Σk (U), σ ω = 0, then (1) ω = 0. Is the converse statement true ? Actually we ask whether if σ ∈ Σk (U) and, for all k R ω ∈ Λ(0)(U, F), σ ω = 0, then σ = 0 (whatever it means) ? 2.4.3 Degenerate singular cubes: No doubt the reader will see that the answer to the question above is negative, at least when the equality σ = 0 is understood literally. For if a 1 C -smooth k-cube, k > 1, is degenerate, i.e., there is 1 6 i 6 k, such that σ(t1, ..., tn) does not R k depend on ti, then σ ω = 0 for any ω ∈ Λ(0)(U, F). From this view point we see that a correct (and positive) answer to the question pose in (1) R k Exercise 2.4.2 is as follows: For σ ∈ Σk (U), σ ω = 0, for all ω ∈ Λ(0)(U, F), if and only if σ is degenerate. Thus the correct and reasonable meaning of the equality ‘σ = 0’ is that ‘σ is degenerate’. Below we shall provide a correct setting for such a statement. Observe that 0-cubes are never degenerate; a 1-cube is degenerate if an only if it is a constant map. In case of curves (i.e., 1-cubes) σ : I → U, we spoke of different type of regularity: continuity, r r r C -smoothness or piecewise C -smoothness, r > 1. Especially piecewise C -smoothness was convenient. In order to incorporate this type of regularity to the case of higher dimensional r cubes k-cubes, k > 2, we introduce the notion of a C -smooth singular k-chain. ¯(r) (r) For any k > 0 and 0 6 r 6 ∞, let Qk (U) := R(Σk (U)) be the vector space of ‘chains’, ¯(r) Pn that is c ∈ Qk (U) whenever c := i=1 aiσi (formal linear combination) where ai ∈ R and k r σi : I → U is a C -smooth singular k-cube in U. ¯(r) For k < 0 and all r, we set Qk (U) = {0}. ¯(s) ¯(r) Note that, for all k ∈ Z, Qk (U) ⊂ Qk (U) if 0 6 r 6 s 6 ∞. ¯(r) 2.4.4 Remark: Now let us think whether the definition of Qk (U) is correct. Of course, from a formal viewpoint everything is all right. However let us try to understand what does equality ¯(r) ¯(r) c = 0 in Qk (U), k > 0, mean ? Namely it means that c, as an element of Qk (U), i.e., a (r) (r) function c :Σk (U) → R, vanishes. However, it particular, this implies that, for any σ ∈ Σk , σ ¯(r) as the element of Qk (U) is never equal to zero. This is an obvious contradiction with Remark 2.4.3. Hence in order to get a correct definition of a chain we have to ‘factor out’ all degenerate k-cubes. For k = 0 this has no effect, but for k > 1 it matters a lot. r (r) 2.4.5 C -smooth singular k-chains: For any k > 0 and 0 6 r 6 ∞, let Dk (U) denote ¯(r) r the subspace of Qk (U) generated by the degenerate C -smooth singular k-cubes. Moreover, for (r) k 6 0, we let Dk (U) = 0 (this is OK since, as mentioned above, there are no degenerate 0-cubes and no k-cubes, k < 0, at all. Let (r) ¯(r) (r Qk (U) := Qk (U)/Dk (U). (r) ¯(r) (r) ¯(r) If k < 0, then Qk (U) = Qk (U) = 0; if k = 0, then Qk (U) = Qk (U) and if k > 1, then 70 2. Differential Forms, de Rham cohomology and singular cubical homology

(r) (r) ¯(r) elements of Qk (U) may be identiﬁed with the cosets {c} := c + Dk (U) where c ∈ Qk (U). (r) r Elements of Qk (U) are called C -smooth singular k-chains. Another, and perhaps easier way, to think about singular k-chains is to consider them as Pn r formal linear combination c = i=1 aiσi where ai ∈ R and σi is a nondegenerate C -smooth singular k-cube for all i = 1, ..., n.

2.4.6 Boundary of a chain: We are now going to define the boundary of a Cr-smooth singular (r) (r) k-chain, i.e., an operator ∂ = ∂k : Qk (U) → Qk−1(U), k ∈ Z. For k 6 0, we put ∂ ≡ 0. In (r) (r) order to define ∂ on Qk (U), k > 1, we will first define the boundary ∂ of σ ∈ Σk (U). Let σ be a singular k-cube in U. For i = 1, ..., k, we define the singular (k − 1)-cubes

k−1 Aiσ, Biσ : I → U, the so-called the front i-face and back i-face, respectively, of σ. Namely, if k = 1, then A1σ := σ(0), B1σ = σ(1). Hence A1σ and B1σ are 0-cubes corresponding to the endpoints of σ. k−1 If k > 1 and (x1, ..., xk−1) ∈ I , then

Aiσ(x1, ..., xk−1) := σ(x1, ..., xi−1, 0, xi, ..., xk−1);

Biσ(x1, ..., xk−1) := σ(x1, ..., xi−1, 1, xi, ..., xk−1).

(r) (r) Clearly, for each 1 6 i 6 k, Aiσ, Biσ ∈ Σk−1(U) provided σ ∈ Σk (U). It is easy to check the following relations: for k > 2 and 1 6 i < j 6 k,

Ai(Ajσ) = Aj−1(Aiσ); B (B σ) = B (B σ); i j j−1 i (2.4.2) Ai(Bjσ) = Bj−1(Aiσ); Bi(Ajσ) = Aj−1(Biσ)

k Finally, for a singular k-cube σ : I → U, k > 1, k X i ∂(σ) := (−1) [Aiσ − Biσ]. i=1

Thus ∂σ = ∂(σ) is the (k − 1)-chain such that c(τ) = 1 if and only if τ = Aiσ with even i or τ = Bjσ with odd j and c(τ) = −1 if and only if τ = Aiσ with odd i or τ = Bjσ with even j. ¯(r) (r) Obviously ∂σ ∈ Qk−1, if σ ∈ Σk (U). 2.4.7 Example: For instance if k = 2, then

∂σ = −A1σ + B1σ + A2σ − B2σ.

The signs attributed to the faces of σ determine the ‘orientation’ of each of them. Geometrically one can imagine ∂σ as the ‘contour’ of the carrier |σ| ‘oriented’ anti-clockwise. Clearly we may identify ∂σ with a piecewise Cr-smooth closed curve. It would be very good if the reader had it in mind. ¯(r) Pn Finally, for arbitrary c ∈ Qk (U), k ∈ Z, c = i=1 aiσi, and we put n X ∂c := ai∂σi. i=1 ¯(r) ¯(r) Thus ∂ : Qk (U) → Qk−1(U) is linear map. 2.4. Integration of differential forms and smooth singular homology 71

It takes a minute to prove the following fundamental result: ¯(r) 2.4.8 Fact: (i) For any k ∈ Z and c ∈ Qk (U), ∂(∂c) = 0. (r) (r) (ii) For any k ∈ Z, ∂(Dk (U)) ⊂ Dk−1(U). Proof: (i) is an immediate consequence of (2.4.2), while (ii) is (almost) obvious. (r) (r) In view of Fact 2.4.8 (ii), ∂ determines a linear map ∂ = ∂k : Qk (U) → Qk−1(U), k ∈ Z, ¯(r) which we denote by the same symbol ∂. For c ∈ Qk (U), ∂{c} = {∂(c)}. Pn Observe that if c = i=1 aiσi, where σi is nondegenerate, then (n − 1)-cubes appearing in ∂c may be degenerate.

2.4.A Singular homology

For k ∈ Z and 0 6 r 6 ∞ we deﬁne the vector subspaces (r) (r) (r) (r) (r) (r) Zk (U) := Ker {∂ : Qk (U) → Qk−1(U)}; Bk (U) = Im {∂ : Qk+1(U) → Qk (U)} whose elements are called Cr-smooth singular k-cycles and k-boundaries, respectively. In view (r) (r) of the Fact 2.4.8 (i), Bk (U) ⊂ Zk (U). r 2.4.9 Example: Observe that a C -smooth singular k-cube σ, where k > 1, is a cycle if σ : (Ik, ∂Ik) → (U, p) where p ∈ U (and, for k = 1, only if). Any Cr-smooth map (Ik, ∂Ik) → (U, p) may be identiﬁed with a Cr-smooth map Sk → U. This perhaps explains the word ‘cycle’. Given a Cr-smooth cycle σ : Sk → U, σ is a boundary if σ admits a Cr-smooth extension onto Dk+1. This again explains the word ‘boundary’.

1 2 ∞ 2.4.10 Exercise: (i) Show that S treated as a 1-cube in U = R \ 0 is a C -smooth 1-cycle but not a 1-boundary. ¯(r) ¯(1) (ii) Show that if c ∈ Qk (U), r > 1 and c = ∂(c1) where c1 ∈ Qk+1(U), then c = ∂(c2) ¯(r) where c2 ∈ Qk+1(U). The quotient space (r) (r) (r) Hk (U) := Zk (U)/Bk (U), k ∈ Z, is called the k-dimensional Cr-smooth singular homology (vector) space of U; its elements are r (r) k-dimensional C -smooth singular homology classes of U. Given {c} ∈ Zk (U), [{c}] := {c} + (r) Bk (U) is the homology class of {c}. (r) (r) (r) (r) It is clear that Hk (U) = {0} for k < 0 and H0 (U) = Q0 (U)/B0 (U) since, by the very (r) (r) deﬁnition Z0 (U) = Q0 (U). ¯(s) ¯(r) 2.4.11 Remark: (i) Note that if 0 6 r 6 s 6 ∞ and k ∈ Z, then obviously Qk (U) ⊂ Qk (U) (s) (r) (s) (r) and Dk (U) ⊂ Dk (U). This inclusions induce a homomorphism Qk (U) → Qk (U) which (s) ¯(s) (r) maps the coset c + Dk (U), c ∈ Qk (U), onto c + Dk (U). This correspondence is injective. ¯(s) (r) (s) (s) In fact, if c1, c2 ∈ Qk and c1 − c2 = d ∈ Dk (U), then d ∈ Dk , i.e., {c1} = {c2} in Qk (U). (s) (r) Therefore we may write Qk (U) ⊂ Qk . (s) (r) (s) (r) In a similar way, Zk (U) ⊂ Zk (U) and Bk (U) ⊂ Bk (U) (to see the latter inclusion one (s) (r) has to use Exercise 2.4.10). Hence there is a homomorphism γ : Hk (U) → Hk (U) assigning 72 2. Differential Forms, de Rham cohomology and singular cubical homology

(s) (s) to the homology class of a coset c + Dk (U) from Zk (U), the homology class of the coset (r) ¯(s) c + Dk (U), c ∈ Qk (U). (r) (r) We claim that γ is injective. For this reason that c + Dk (U) ∈ Bk (U), i.e., there is ¯(r) (r) ¯(s) d ∈ Qk+1(U) such that c − ∂(d) = c1 ∈ Dk (U). This means that ∂(d) + c1 ∈ Qk (U). We may 4 (s) suppose that supports of d1 and ∂(d) are disjoint ( ). Hence c1 ∈ Dk and, by Exercise 2.4.10 ¯(s) (s) (ii), c − c1 = ∂(d1) where d1 ∈ Qk+1, i.e. the element {c} in Qk (U) is the boundary of {c1} in (s) Qk+1. (s) (r) In a consequence we may write Hk (U) ⊂ Hk (U) whenever k ∈ Z and 0 6 r 6 s 6 ∞. (r) (r) (r) (ii) Above we have deﬁned Zk (U), Bk (U) and Hk (U) for all k ∈ Z and 0 6 r 6 ∞. (0) However only the ‘largest’ space Hk(U) := Hk (U), k ∈ Z will be of interest. In fact we shall (r) ∼ (0) (r) prove that Hk (U) = Hk (U) for all k ∈ Z and 0 6 r 6 ∞; thus all Hk (U), 1 6 r 6 ∞ play only auxiliary roles. ¯ ¯(0) (0) For brevity in the sequel we write also Qk(U) := Qk (U), Dk(U) := Dk (U), Qk(U) := (r) (0) (0) Qk (U), Zk(U) := Zk (U) and Bk(U) := Bk (U). (iii) For each k ∈ Z we may deﬁne

Z¯k(U) := Ker {∂ : Q¯k(U) → Q¯k−1(U)};

B¯k(U) := Im {∂ : Q¯k+1(U) → Q¯k(U)} and H¯k(U) := Z¯k(U)/B¯k(U). Observe that, for all k ∈ Z, ∼ −1 Hk(U) = ∂ (Dk−1(U))/[Dk(U) + B¯k(U)]. −1 Indeed, for each c ∈ ∂ (Dk−1(U)), let

ξ(c) := [c] ∈ Hk(U).

This definition is correct since ∂(c) ∈ Dk−1(U), we see that ∂c = 0 in Qk−1(U), i.e. {c} ∈ Zk(U). Moreover ξ is an epimorphism. Note that ξ(c) = 0 if and only if {c} ∈ Bk(U), i.e., there is d ∈ Q¯k+1(U) such that {c} = ∂({d}) = {∂(d)}; hence c ∈ ∂(d) + Dk(U) ⊂ B¯k(U) + Dk(U). r (iv) In order to define C -smooth singular k-cubes (and other related objects) for 1 6 r 6 ∞ we needed that U is a subset of a Banach space (in order to speak of differentiability issues of a k map σ : I → U one needs that U ⊂ E). However this is not necessary when r = 0. Hence we see that Hk(U) (and all its ingredients) are defined for an arbitrary topological space U.

2.4.12 Example: Suppose that X is a one-point space. We shall compute Hk(X) and H¯k(X), k ∈ Z. For any k > 1, there is only one (continuous) singular k-cube in X and it is degenerate; ¯ hence Qk(X) = 0 for any k > 1, i.e., Hk(X) = 0. For k = 0, Z0(X) = Qk(X) = Q0 = R and B0(X) = 0. Thus 0 if k 6= 0; Hk(X) = R if k = 0. ¯ ¯ On the other hand, for any k > 0, Zk(X) = R and Bk(X) = 0; hence 0 if k < 0; H¯k(X) = R if k > 0. 4Given c ∈ R(A), where R is a p.i.d. and A 6= ∅, by the support of c we mean the set {a ∈ A | c(a) 6= 0}. 2.4. Integration of differential forms and smooth singular homology 73

This is another, and perhaps more important reason, to ‘factor out’ degenerate cubes out of Q¯k(X). r Let ϕ : U1 → U2 be a C -smooth map (Ui, i = 1, 2, are open subsets of Banach spaces if P ¯(r) 1 6 r 6 ∞, or arbitrary spaces if r = 0). For any k ∈ Z and c = i=1 aiσi ∈ Qk (U1), we put n X ¯(r) ϕ#(c) = ai(ϕ ◦ σ) ∈ Qk (U2). i=1

(r) It is clear that, for any σ ∈ Σk (U1), ∂(ϕ ◦ σ) = ϕ#(∂σ), i.e., in general

∂ ◦ ϕ# = ϕ# ◦ ∂.

(r) (r) Moreover, ϕ#(Dk (U1)) ⊂ Dk (U2). (r) (r) Therefore, for all k ∈ Z, ϕ# : Qk (U1) → Qk (U2) given by ϕ#({c}) = {ϕ#(c)} for (r) (r) (r) (r) (r) {c} ∈ Qk (U1). Clearly ϕ#(Zk (U1)) ⊂ Zk (U2) and ϕ#(Bk (U1)) ⊂ Bk (U2). Hence ϕ induces a linear map

(r) (r) ϕ∗ : Hk (U1) → Hk (U2) (r) ϕ∗([{c}]) = [{ϕ#(c)}], {c} ∈ Zk (U). r If U3 is a space (U3 is open in a Banach space if r > 1) and ψ : U2 → U3 is C -smooth, then (ψ ◦ ϕ)# = ψ# ◦ ϕ# and hence (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗.

Moreover 1U∗ = 1 (r) . Hk (U) (r) 2.4.13 Fact: Hk , k ∈ Z, 1 6 r 6 ∞, (resp. Hk) is the functor from the category of open subsets of Banach spaces and their Cr-smooth maps (resp. T op) into the category of vector r spaces and their linear maps. In particular, if ϕ : U1 → U2 is a C -diﬀeomorphism (resp. a −1 −1 homeomorphism), then ϕ∗ is an isomorphism and ϕ∗ = (ϕ∗) . For a while we will not study the further properties of the Cr-smooth singular homology spaces – we shall do it in the sequel in a greater generality. Here we shall only state the following result.The proof is quite similar to that of Theorem 2.3.5.

r 2.4.14 Fact: If ϕ0, ϕ1 : U1 → U2 are C -homotopic, then ϕ0∗ = ϕ1∗. As a consequence, if an r (r) open U is C -smoothly contractible, then Hk (U) = 0 for all k ∈ Z. (r) 2.4.15 Homotopy of singular cubes: Let σi ∈ Σk (U), i = 0, 1 and k > 0. These cubes r r k are C -homotopic if there is a C -smooth map σ : I × I → U such that σi = σ(·, i), i = 0, 1. r For instance if k = 0, then two 0-cubes, i.e., points x0, x1 are C -homotopic if and only r r if there is a C -smooth path σ : I → U such that σ(i) = xi, i = 0, 1. If U is C -smoothly contractible, then any two Cr-smooth singular k-cubes are Cr-homotopic.

(r) r (r) 2.4.16 Corollary: If σ0, σ1 ∈ Σk are C -homotopic and {σ0}, {σ1} ∈ Zk (U), then their homology classes are equal [{σ0}] = [{σ1}]. k r Proof: Let σ : I × I → U be a C -smooth homotopy joining σ0 to σ1. Without loss of k generality we may assume that σ is actually deﬁned on V × I where V ⊂ R is open; then also σi : V → U. In view of Fact 2.4.14, σ0# = σ1#. But, for i = 0, 1,

σi#(1Ik ) = σi. (2.4.3) 74 2. Differential Forms, de Rham cohomology and singular cubical homology

Thus [{σ0}] = [{σ0#(1Ik )}] = σ0∗[{1Ik }] = σ1∗([1Ik ]) = [{σ1}].

2.4.B Integration over chains

1 The operation of integration may be easily extended to C -smooth singular k-chains. Let U ⊂ E. Pn ¯(1) k Given c := i=1 aiσi ∈ Qk (U) and ω ∈ Λ(0)(U; F) we deﬁne

n Z X Z ω := ai ω. c i=1 σi

R (1) Since c ω = 0 for all c ∈ Dk (U), the formula Z Z hω, {c}i := ω := ω {c} c

(1) k (1) correctly deﬁnes the integral of ω along any {c} ∈ Qk (U) and h·, ·i :Λ(0)(U, F) × Qk (U, F) 3 R (ω, {c}) 7→ {c} ω ∈ F, is a bilinear map.

1 (1) Let ϕ : U → V , where V is a subset of a Banach space, be a C -smooth map. If σ ∈ Σk (U), (1) k R then ϕ ◦ σ ∈ Σk (V ). If ω ∈ Λ(0)(V, F), then the integral ϕ◦σ ω is deﬁned and, from the very deﬁnition, we have Z Z Z ω = = ϕ#ω. (2.4.4) ϕ◦σ ϕ#(σ) σ Therefore, for all c ∈ Q¯(1)(U) k Z Z ω = ϕ#(ω) (2.4.5) ϕ#(c) c and, for any {c} ∈ Q(1)(U), k Z Z ω = ϕ#(ω). (2.4.6) ϕ#({c}) {c}

In view of the above and Exercise 2.4.2 we have the following:

2.4.17 Fact: The map k (r) h·, ·i :Λ(0)(U, F) × Qk (U) → F, r r > 1, is a duality operator such that, for any C -smooth map ϕ : U → V ,

# hϕ ω, {c}iU = hω, ϕ#({c})i

k (r) for all ω ∈ Λ(0)(V, F) and {c} ∈ Qk (U). 2.4.18 Remark: The word ‘duality’ was not used above entirely correctly. Recall that given two real (or complex) (resp. normed or, more generally topological linear spaces) vector spaces X, Y , we say that {X,Y } is a dual pair if there is a (resp. continuous) duality operator h·, ·i : X×Y → R (or C), i.e., a bilinear map such that: (i) if hx, yi = 0 for all x ∈ X, then y = 0; (ii) if hx, yi = 0 for all y ∈ X, then x = 0. k (1) We see that h·, ·i introduced above is a duality between Λ(0)(U) and Qk (U). 2.4. Integration of differential forms and smooth singular homology 75

ξ Observe that if {X,Y } is a dual pair, then there is a (continuous) embedding X ,→ Y ∗ (where Y ∗ stands for the dual of Y in case of ‘pure’ vector spaces and for the topological dual in case of topological vector spaces): for x ∈ X, let ξ(x) ∈ Y ∗ be deﬁned by ξ(x)(y) = hx, yi; then clearly ξ(x) is a (continuous) linear form on Y and ξ : X → Y ∗ is a monomorphism (injective). For if ξ(x) = ξ(x0), then hx − x0, yi = 0 for all y ∈ Y , i.e., x = x0. Note that if {X,Y } is a dual pair and dim X, dim Y < ∞, then ξ is an isomorphism.

(r) k ∗ 2.4.19 Corollary: For any 1 6 r 6 ∞, there is a monomorphisms Qk (U) → [Λ(0)(U)] .

k−1 ¯(1) 2.4.20 The Stokes Formula on Chains: Let ω ∈ Λ(1) (U, F), k > 1, and let c ∈ Qk (U). Then Z Z ω = dω. ∂c c

(1) Proof: It is suﬃcient to show that, for any σ ∈ Σk (U), Z Z ω = dω. ∂σ σ

k−1 k To this end, in view of formulae (2.4.4), it is suﬃcient to show that if ω ∈ Λ(1) (I ,F ), then Z Z dω = ω.

1Ik ∂1Ik Now Z Z k Z X i−1 0 dω = dω(x; v1, ..., vk) dx = (−1) [ω (x)(vi)](v1, ..., vi, ..., vk) dx = k b 1Ik I i=1 Ik k Z k Z X i−1 ∂ω X i−1 (−1) (x)(v1, ..., vi, ..., vk) dx = (−1) [ω(x1, ..., xi−1, 1, xi+1, ..., xk) k ∂x b k−1 i=1 I i i=1 I −ω(x1, ..., xi−1, 1, xi+1, ..., xk)](v1, ..., vbi, ..., vk) dx1dx2...dxci...dxk. On the other hand

k ! Z X Z Z ω = (−1) ω − ω = ∂1Ik i=1 Ai1Ik Bi1Ik k Z X i (−1) ω(x1, ..., xi−1, 0, xi, ..., xk−1)(v1, ..., vi, ..., vk) dx1...dxk−1− k−1 b i=1 I Z ω(x1, ..., xi−1, 1, xi, ...xk−1)(v1, ..., vbi, ..., vk) dx1...dxk−1 Ik−1

Hence the assertion. 2.4.21 Remark: We may now give another interpretation of the operator of the exterior deriva- 1 tive. Suppose that ω is a C -smooth k-form on U = E (for simplicity), k > 0. Let x ∈ E, take a k collection of e1, ..., ek+1E and let (k + 1)-cube σ(e1, e1, ..., ek+1): I → U be deﬁned by

k+1 X k+1 σ(e1, ..., ek+1)(t) = x + tiei, t = (t1, ..., tk+1) ∈ I . i=1 76 2. Differential Forms, de Rham cohomology and singular cubical homology

∞ It is clear that σ(e1, ..., ek+1) is a C -smooth (k + 1)-cube and, thus, we may deﬁne a map F : k+1 → by E F Z F (e1, ..., ek+1) := ω. ∂σ(e1,...,ek+1) We claim that

k+1 k+1 F (te1, , ...tek+1) = t dω(x; e1, e1, ..., ek+1) + o(t ) as ε → 0.

In other words Z F (te1, ..., tek+1) 1 dω(x; e1, ..., ek+1) = lim = lim ω. t→0 k+1 t→0 k+1 t t ∂σ(te1,...,tek+1)

Indeed, by the above deﬁnition

k+1 Z Z X i F (te1, ..., tek+1) = (−1) ω − ω i=1 Aiσ Biσ

k where we have set σ := σ(te1, ..., tek+1). For any i = 1, , ..., k + 1 and s = (s1, ..., sk) ∈ I ,

i−1 k X X Aiσ(s1, ...sk) = σ(s1, ..., si−1, 0, si, ..., sk) = x + tsjej + tsjej+1 j=1 j=i and i−1 k X X Biσ(s1, ..., sk) = x + tsjej + tei + tsjej+1. j=1 j=i

Observe that, for any i = 1, ...k + 1 and j = 1, ..., k, ∂Aiσ ∂Biσ tej if 1 6 j 6 i − 1; (s1, ..., sk) = (s1, ..., sk) = ∂sj ∂tj tej+1 if i 6 j 6 k. Hence Z Z Z ω = ω(Aiσ(s); te1, ...tei−, tei+1, ..., tek+1) ds = ω(Aiσ(s); te1, ..., teci, ..., tek+1) ds k k Aiσ I I and Z Z ω = ω(Biσ(s); te1, ..., teci, ..., tek+1) ds. k Biσ I We have then k+1 Z X i+1 k F (te1, ..., tek+1) = (−1) t [ω(Biσ(s)) − ω(Aiσ(s))](e1, ..., ebi, ..., ek+1) ds i=1 Ik and k+1 Z F (te1, ..., tek+1) X 1 = (−1)i+1 [ω(B σ(s)) − ω(A σ(s))](e , ..., e , ..., e ) ds . tk+1 t i i 1 bi k+1 i=1 Ik To complete the proof we need to compute Z 1 lim [ω(Biσ(s)) − ω(Aiσ(s))](e1, ..., ei, ..., ek+1) ds, t→0 b Ik t 2.4. Integration of differential forms and smooth singular homology 77 so, in fact – in view of the Lebesque theorem – the pointwise limit 1 lim [ω(Biσ(s)) − ω(Aiσ(s))]. t→0 t But now it is easy to see that in this limit

0 ω(Biσ(s)) − ω(Aiσ(s)) = ω (x)(ei) + o(t) as t → 0.

Therefore

k+1 F (te1, ..., tek+1) X i+1 0 lim = (−1) [ω (x)(ei)](e1, ..., ei, ..., ek+1) = dω(x)(e1, ..., ek+1). t→0 tk+1 b i=1

On the other hand, the Stokes formula allows to write Z Z F (te1, ..., tek+1) = ω = dω ∂σ(te1,...,tek+1) σ(te1,...,tek+1) Z k+1 k+1 X = t dω(x + ttiei; e1, ..., ek+1) dt k+1 I i=1 Now F (te1, ..., tek+1) lim = dω(x; e1, ..., ek) t→0 tk+1 again. However this argument is not legitimated from the viewpoint of our aim to characterize dω. Let us now collect several obvious corollaries of the Stokes theorem.

k R (1) 2.4.22 Corollary: (i) if ω ∈ Λ(1)(U, F) and dω = 0, then {c} ω = 0 for any {c} ∈ Bk (U). 1 1 (1) In particular, if C -smooth singular k-cubes σ0, σ1 are C -homotopic and {σ0}, {σ1} ∈ Zk (U) and , then Z Z ω = ω. σ0 σ1 k k−1 (ii) If ω ∈ Λ(0)(U, F) and ω = dη for some η ∈ Λ(1) (U, F), then Z ω = 0 {c}

(1) for any {c} ∈ Zk (U). Observe that the second part of (i) above is a direct generalization of Lemma 2.1.10. (1) ¯(1) Proof: (i) If dω = 0, {c} ∈ Bk (U), then {c} = ∂({c1}) for some c1 ∈ Qk+1(U) and Z Z Z Z = ω = ω = dω = 0. {c} ∂({c1}) ∂(c1) c1

(1) As concerns the second part: in view of Corollary 2.4.16, {σ0} = {σ1}+{c} where {c} ∈ Bk (U); hence Z Z Z Z Z ω = = ω = ω = ω. σ0 {σ0} {σ1}+{c} {σ0} σ1 78 2. Differential Forms, de Rham cohomology and singular cubical homology

(ii) If {c} ∈ Z(1)(U), then k Z Z Z ω = dη = η = 0. {c} c ∂c

R k (1) 2.4.23 Corollary: The integral c ω, where ω ∈ Z (U, F) and {c} ∈ Zk (U) depends only k (1) on [ω] ∈ H (U, F) and [{c}] ∈ Hk (U). More precisely, if continuous k-forms ω1, ω2 are such R R (1) 1 that ω1 − ω2 is exact, then c ω1 = c ω2 if {c} ∈ Zk (U). If ω is C -smooth and closed, then R ω = R for all c , c ∈ Q¯(1)(U) such that {c − c } ∈ B(1)(U). c1 c2 1 2 k 1 2 k The next formulation of the above corollary corresponds to Fact 2.4.17 and Remark 2.4.18:

k (r) 2.4.24 Corollary: There is a correctly deﬁned bilinear map h·, ·i : H (U) × Hk (U) → R, 1 r ∞, given by 6 6 Z h[ω], [{c}]i = ω c k (r) k (r) for [ω] ∈ H (U) and [{c}] ∈ Hk (U). This map is a duality between H (U) and Hk (U). Proof: The ﬁrst part is evident. The second follows from Theorem 2.4.25 (ii) and (iii) below. The following result, being a converse to Corollary 2.4.22 is of the greatest importance.

k ∞ 2.4.25 De Rham Theorem: (i) If ω ∈ Λ(1)(U, F) and, for any C -smooth singular (k+1)-cube R σ, ∂σ ω = 0, then dω = 0. k R (∞) (ii) If ω ∈ Λ(1)(U) and c ω = 0 whenever {c} ∈ Zk (U), then ω is exact. Moreover if ω is r r C -smooth, r > 1, then it is possible to ﬁnds a C -smooth exterior primitive η. (r) k R (iii) If {c} ∈ Zk (U), 1 6 r 6 ∞, is such that, for any ω ∈ Z (U, F), c ω = 0, then (1) {c} ∈ Bk (U). Proof: (i) For any singular C∞-smooth (k + 1)-cube σ, Z Z dω = ω = 0. σ ∂σ Hence, in view of Exercise 2.4.2, dω = 0. (ii) The proof is very diﬃcult: we shall try to provide a proof later on. The de Rham Theorem 2.4.25 (ii), (iii) is one of the deepest results of algebraic topology and a source of almost all modern developments of this theory. The original proof of this remarkable result was based on the notion of a current which generalizes that of the Schwartz distribution ∞ (i.e. a weakly continuous functional on the space C0 (U) of the so-called test functions). Until then there are other, simpler proofs. We shall only make several important comments which will explain the nature of the de Rham theorem

2.4.26 Remark: (i) It is easy to see that the statement (i) of de Rham theorem corresponds to Corollary 2.1.7 while (ii) corresponds to Theorem 2.1.1 (ii). (ii) Observe that assumptions of Theorem 2.4.25 (ii) and (iii) are quite weak and imply Corollary 2.4.24. In (ii) we assume that ω is a C1-smooth k-form whose integral vanishes along ∞ k (r) any C -cycle. Hence if [ω] ∈ H (U) and h[ω], [{c}]i = 0 for all [{c}] ∈ Hk (r is ﬁxed), then, in (∞) particular, hω, ci = hω, {c}i = 0 for all {c} ∈ Zk (U), so dω = 0, i.e. [ω] = 0. (r) ¯(r) Similarly, part (iii) of Theorem 2.4.25 implies that if [{c}] ∈ Hk (U) (i.e., c ∈ Qk (U) and 2.4. Integration of differential forms and smooth singular homology 79

(r) k ¯(1) {c} ∈ Zk (U)) and h[ω], [{c}]i = 0 for all [ω] ∈ H (U), then {c} = ∂({c1}) where c1 ∈ Qk+1(U). 0 0 ¯(r) Reasoning as in Remark 2.4.11 (i), we may assume that {c} = ∂({c1}) where c1 ∈ Qk+1(U). (r) Thus [{c}] = 0 in Hk (U). This completes the proof of Corollary 2.4.24. k (iii) Observe that under assumptions from Theorem 2.4.25 (ii), ω ∈ Z(1)(U, F). Indeed, for ∞ R (∞) (∞) any C -smooth singular (k+1)-cube ∂σ ω = 0 because {∂(σ)} = ∂({σ}) ∈ Bk (U) ⊂ Zk (U); hence, in view of 2.4.25 (i), dω = 0. (iv) If U is C2-smoothly contractible, then the proof of Theorem 2.4.25 (ii) is very simple. By (iii) above, we know that dω = 0. In view of the Poincaré Lemma (see Remark 2.3.11), ω is exact. The existence of Cr exterior primitive also follows from the Poincaré lemma (see again Remark 2.3.11). (v) On the other hand de Rham theorem implies the following generalization of the Poincaré Lemma: if U is C1-smoothly contractible and dω = 0 (ω is a C1-smooth k-form), then ω is R (1) exact. Indeed, in view of Corollary 2.4.22, c ω = 0 provided {c} ∈ Bk (U). By Fact 2.4.14, (1) (∞) (1) (1) Hk (U) = 0, i.e., Zk (U) ⊂ Zk (U) = Bk (U). By de Rham theorem (ii), ω is exact. (1) (vi) We see that above the decisive argument was that Hk (U) = 0 (not contractibility). For instance if U is k-connected, then, in view of the Hurewicz theorem (which will be discussed (1) 5 later on), Hk (U) = 0. Hence we obtain a direct (partial) generalization of Theorem 2.1.9 ( ): (1) 1 If Hk (U) = 0 (it holds e.g. if U is k-connected) and a C -smooth k-form is closed, then it is exact.

2.4.27 Question: (comp. Question 2.3.12) Is it true that if ω is diﬀerentiable, dω = 0 and U is k-connected, then ω is exact ?

k (r) 2.4.28 Remark: (i) In view of Corollary 2.4.24, for all k ∈ Z and for each r > 1, {H (U),Hk (U)} (r) ∼ (1) is a dual pair. This suggests the following: for all r > 1 and k ∈ Z, Hk (U) = Hk (U). This is (r) ∼ (0) true and even more: Hk (U) = Hk(U) := Hk (U) for all r > 1. Moreover, we are going to in- k (0) troduce a cofunctor H (·, F) (in a sense dual to Hk := Hk ) of the so-called singular cohomology k ∼ k with coefficients in F and show actually that H (U, F) = H (U, F). This implies that an analytical approach via C∞-smooth forms on U is actually topological showing intimate connection between analysis and topology. (ii) All our considerations concerned open subsets of Banach spaces. It is not difficult to replace them by Banach manifolds (i.e., in place of U ⊂ E take a manifold M modeled over E). This generalization might be a good source of nice exercises for a reader. Let us now make some remarks concerning integration of complex differential forms. r Let U ⊂ E where E = X ⊕ iX is a complex Banach space. C -smooth, 0 6 r 6 ∞, singular k-cubes in U exactly as are defined exactly as before: σ : Ik → U is a Cr-smooth singular k-cube in U if the real and the imaginary part Reσ, Im σ : Ik → X are Cr-smooth singular k-cubes r (r) in Y . Having this we define C -smooth singular chains exactly as before: c ∈ Q (U, C) is, by definition, an element of the quotient space of the space of formal linear combinations with complex coefficients of Cr-smooth singular k-cubes modulo degenerate cubes. (p,q) Given a continuous k-form ω : U → A (E, C),

ω 5Observe that we have assumed C1-smoothness instead diﬀerentiability. 80 2. Differential Forms, de Rham cohomology and singular cubical homology

1 Pn k and a C -smooth singular chain c = i=1 aiσi, where σi : I → U, we deﬁne Z Z ω = c Ik

2.5 Orientation of chains

¯(r) (r) So far when speaking of singular k-chains from Qk (U) (or Qk (U)), 0 6 r 6 ∞ and k > 0, we have neglected their orientation, except for Example 2.4.7 where the sign attributed to k-dimensional faces of a singular (k + 1)-cube were interpreted by means of ‘orientation’ (in quotation marks). Moreover we did not pay too much attention to their carriers (i.e., the unions of the carriers of cubes participating in the chain). From a geometric point of view this is a serious deﬁciency of this attitude. For instance consider the following example:

2.5.1 Example: Let a 2-chain c = σ + τ be given where σ(t1, t2) = (t2, 1 − t1) and τ(t1, t2) = 2 2 (t1 +1, 1−t2) for (t1, t2) ∈ I . Let D = |c| = |σ|∪|τ| = {(x1, x2) ∈ R | 0 6 x1 6 2, 0 6 x2 6 1}. It is clear that we are inclined to think about the topological boundary ∂D as a cycle. However if we compute, then

∂c = −A1σ + B1σ + A2σ − B2σ − A1τ + B1τ + A2τ − B2τ.

Observe however that A1τ = B2σ and thus the union of the carriers of 1-cubes participating in ∂c is larger than the (topological boundary) ∂D. So: is the boundary a cycle or not. It might be worse: a singular chain may not be a cycle even though its ’carrier’ is a ‘cycle’ from the geometric point of view (i.e., a compact manifold without boundary or a boundary of a 2 compact manifold with boundary). For instance, let 1-cubes (curves) σ, τ : [0, 1] → R be given by σ(t) = (cos πt, sin πt), τ(t) = (cos πt, − sin πt) for t ∈ I. Then |σ + τ| = |σ| ∪ |τ| = S1 but ∂c 6= 0. In order to circumvent these and other shortcomings we shall introduce several new concepts.

Let, as above, U be a subset of a Banach space E. (r) r 2.5.2 Equivalent cubes: In the collection Σk (U), 1 6 r 6 ∞ and k > 0, of all C -smooth singular k-cubes in U we introduce a relation: (1) if k = 0, then two 0-cubes are equivalent if they are equal; (2) if k > 1, then two k-cubes σ1, σ2 are equivalent σ1 ∼ σ2 if there exists a r k k k 0 C -smooth diﬀeomorphism ϕ : I → I such that, for each t ∈ I , det ϕ (t) > 0 and σ1 = ϕ ◦ σ2.

2.5.3 Remark: There is also a way to deﬁne the equivalence between continuous k-cubes; however it is not easy.

2 2.5.4 Example: Let σ(t1, t2) = (t2, 1 − t1), τ(t1, t2) = (t1, 1 − t2) for (t1, t2) ∈ I . Then σ is equivalent with 1I2 but σ and τ are not equivalent.

Observe that if σ1 ∼ σ2, then |σ1| = |σ2| and if σ1 is degenerate, then so is σ2. It is clear that ∼ is an equivalence relation; by [σ] we denote the oriented cube, i.e. the equivalence class of σ; (r) [σ] := {τ ∈ Σk | τ ∼ σ}.

(r) For the rest of this section ﬁx k > 0 and 1 6 r 6 ∞. Given σ ∈ Σk (U), by −σ we denote k the cube deﬁned by (−σ)(t) = σ(1 − t1, t2, ..., tk) for t = (t1, ..., tk) ∈ I . It is clear that cubes σ and −σ are not equivalent; so it makes sense to write −[σ] := [−σ]. 2.5. Orientation of chains 81

(r) (r) ¯ (r) For 1 6 r 6 ∞, let us denote [Σ]k (U) := Σk (U)/ ∼ and let [Q]k (U) be the R-vector (r) space generated by the set [Σ]k (U) with relations [σ1] + [σ2] = 0 whenever [σ2] = −[σ1] (in other words ¯ (r) (r) [Q]k (U) = R([Σ]k (U))/G (r) where G is the subspace of R([Σ]k (U)) generated by elements of the form [σ] + (−[σ]) where σ (r) runs through Σk (U). Similarly as before we have to factor out degenerate cubes, i.e., deﬁne

(r) ¯ (r) (r) [Q]k (U) := [Q]k (U)/[D]k (U) (r) ¯ (r) where [D]k (U) denotes the subspace of [Q]k (U) generated by oriented degenerate cubes. (r) (r) The boundary operator ∂ :[Q]k (U) → [Q]k−1(U) may be deﬁned as before. Namely given r an oriented C -smooth k-cube [σ], each face Aiσ, Biσ, i = 1, ..., n, inherits the orientation from σ: [Aiσ] and [Biσ], respectively. Thus we may consider [Aiσ] and [Biσ] and put

k X i ∂[σ] := (−1) ([Aiσ] − [Biσ]). i=1

2.5.5 Exercise: (not easy) Show that this deﬁnition is correct, for it does not depend on the choice of the representative, i.e., if τ ∈ [σ], then

n k X i X i (−1) ([Aiτ] − [Biτ]) = (−1) ([Aiσ] − [Biσ]). i=1 i=1

Hint: show that if ϕ : Ik → Ik is a Cr-diﬀeomorphism, then the image of a (k − 1)-dimensional face is an (k − 1)-dimensional face; face

2.5.6 Example: Appropriately using oriented cubes phenomena described in Example 2.5.1 cannot occur. Note that orientations of σ and τ in this example are different (in the sense that the 2 determinant of the matrix transferring the base {∂1σ(t), ∂2σ} of R to the base {∂1τ(t), ∂2τ(t)} is negative). However if instead of τ we take −τ, then everything would be alright: the carrier of the boundary of [σ] + (−[τ]) is equal to the topological boundary of D and, in the context of the another example provided in Example 2.5.1, ∂([σ] + (−[τ]) = 0 . From the viewpoint of homology this new construction brings nothing new: namely if we (r) define the oriented singular homology [H]k (U) exactly as before (with obvious modifications), then (r) ∼ (r) Hk (U) = [H]k (U).

As concerns integration of (continuous) k-form on oriented C1-smooth singular k-chains (i.e., ¯ (1) (1) (1) elements of [Q]k (U) or [Q]k (U)) nothing changes, too. Indeed if σ1, σ2 ∈ Σk (U) and σ1 ∼ σ2, k then for each ω ∈ Λ(0)(U, F), Z Z ω = ω. σ1 σ2 Therefore, from an algebraic point of view oriented chains bring nothing new. However they do in the geometric context as was seen above and will additionally be explained in the next section. 82 2. Differential Forms, de Rham cohomology and singular cubical homology

2.5.A The Stokes Theorem revisited

The Stokes theorem on continuously differentiable singular chains is not satisfying from many points of view. On many occasions one would like to integrate a differential forms over a compact set not being the carrier of any singular cube (if a given compact is the carrier of some cube, then it may be difficult even if possible to find this cube). There are three solutions of this problem. The first, but also not always very convenient, is to appropriately treat manifolds (with or without) boundary; the second is to introduce a notion of a block and, finally, the third one, and perhaps the most effective, is to study the so-called regular domains. r Let us first observe that a C -smooth, 0 6 r 6 ∞, singular k-cube serves as a generalization of a k-dimensional surface. However quite often, given such a cube σ, even when r > 1, its carrier |σ| may not resemble any reasonable surface at all. If there are no additional assumptions on σ, then the set |σ| may be very ‘singular’. Secondly, when studying a surface (or, more generally, a manifold) its parameterization (or an atlas of charts) plays an auxiliary role: usually one deals with a certain set which may be parameterized in an appropriate way. To the contrary: a singular cube is a parameterization itself. Cubes are different even though their carriers are the same. The notions of an oriented cube or an oriented chain allow to identify equivalent cubes and, therefore, to treat rather pairs (|σ|, [σ]). Now we shall go a step farther.

As usual let U be a subset of a Banach space E.

2.5.7 Regular ‘singular’ cubes: By default all 0-cubes are regular. Let k > 1 and consider r k r a C -smooth singular k-cube σ : I → U, 1 6 r 6 ∞. If σ is a C -diﬀeomorphism, i.e. σ is k invertible and, for any t ∈ I , the vectors ∂iσ(t), i = 1, ..., k, are linearly independent, then such a r 6 r Pm cube will be called a regular C -smooth k-cube ( ). A C -smooth singular k-chain c = i=1 aiσi r is regular if σi is a regular C -smooth k-cube for any i = 1, ..., m. Observe that regular k-cubes are never degenerate for if σ does not depend on the i-th k variable, then ∂iσ(t) = 0 for all t ∈ I .

r 2.5.8 Exercise: Show that if a C -smooth k-cube σ is regular, k > 1, then so is any of its (k − 1)-dimensional faces (to this aim observe that, for instance, for any i = 1, ..., n − 1, k−1 t = (t1, ..., tk−1) ∈ I and 1 6 j 6 k − 1, ∂tj (Aiσ)(t) = ∂tj σ(t1, ..., ti−1, 0, ti, ..., tn−1) if

1 6 j 6 i − 1 and ∂tj (Aiσ)(t) = ∂tj+1 σ(t1, ..., ti−1, 0, ti, ..., tn−1) if i 6 j 6 k − 1.

2.5.9 Blocks: A compact set D ⊂ U is called a geometric k-chain or a k-block, k > 0, if 1 ¯(1) Pm it admits a parameterization, i.e. a C -smooth singular k-chain c ∈ Qk (U), c = i=1 1 · σi, 1 where σi, i = 1, ..., m, is a regular C -smooth k-cube (in what follows we say that c is a regular C1-smooth k-chain such that: Sm (i) the carrier |c| := i=1 |σi| = D; i j (ii) for i 6= j, the set |σ | ∩ |σ | is either empty or is the common carrier of l-faces of both σi and σj, 0 6 l 6 k − 1. We say that a k-block is Cr-smooth if its parameterization is Cr-smooth. Pm Of course one may consider a block parameterization of the form i=1 aiσi where ai = ±1. It is understood that if σi appears with −1, then we consider a cube (−σ).

r 2.5.10 Example: If σ is a regular C -smooth (k + 1)-cube, k > 0, then |σ| is a (k + 1)-block. Sk r The set D = |∂σ| := i=1(|Aiσ| ∪ |Biσ| is a C -smooth k-block: as its parameterization we may 6It would be awkward to say: a regular C1-smooth singular k-cube. 2.5. Orientation of chains 83

Pn i i+1 take the k-chain i=1((−1) Aiσ + (−1) Biσ.

It is clear that a k-block may admit diﬀerent parameterizations. In particular, if τi ∈ [σi] is Pm , i = 1, ..., m, then the k-chain i=1 1 · τi is another parameterization. We say that this is an equivalent parameterization. A block may admit non-equivalent parameterizations. Recall that any Cr-smooth singular k-cube σ determines its orientation, i.e., the oriented k-cube [σ]. This orientation determines the orientation of any of its (k − 1)-faces. Precisely, the face Aiσ

2.5.11 Orientable k-blocks: The carrier K of a regular Cr-smooth k-cube σ admits two orientations: [σ] or −[σ]. We say that this carrier is oriented if one of its orientations is chosen. In what follows we always understand that |σ| has [σ] as its orientation (if we want to consider K with the opposite orientation, then we shall speak of | − σ|). The orientation of |σ| determines the canonical orientation of the carrier of any of its (k −1)- i i+1 faces. Precisely, for i = 1, ..., k, the face Aiσ (resp. Biσ) has (−1) [Aiσ] (resp. (−1) [Biσ]) as its orientation. r We say that two regular C -smooth k-cubes σ1, σ2 are contiguous if the intersection |σ1|∩|σ2| is the common carrier of two (k − 1)-faces of both σ1 and σ2. Suppose that σ1 and σ2 are contiguous and oriented. We say that the orientations on |σ1| and |σ2| (i.e., [σ1] and [σ2]) are compatible when the following condition is satisﬁed: if τ1 and τ2 are (k − 1)-faces of σ1 and σ2, respectively, having the common carrier, then τ1 ∼ −τ2. Pm We say that a k-block D is orientable if it has a parametrization c = i=1 σi such that if |σi| ∩ |σj| (i 6= j) is the carrier of a common (k − 1)-face, then Given a regular Cr-smooth k-cube σ, the block |σ| is orientable; by the orientation of σ we mean the class [σ] mean any choice of the class [σ] or −[σ].

It is clear that the relation of equivalence of parameterizations is an equivalence relation. We say that a k-block is oriented if the class of equivalent parameterizations have been chosen. The role of orientation of a k-block is important from the geometric view point.

2.5.12 Example: Let us return to Examples 2.5.1 and 2.5.6. Evidently the set D is a 2-block. If we take σ + τ as its parameterization, then the union of the carriers participating in ∂(σ + τ) is larger that the boundary ∂D. If, however, as in Example 2.5.6, we take the chain σ + (−τ) as its (non-equivalent) parameterization, then ∂D is equal to the union of the carriers of 1-cubes participating in ∂(σ + (−τ)). We see that in order to get rid of a bad phenomenon described in Example 2.5.1 one needs that k-cubes participating in a parameterization of a k-block are ‘oriented’ in a similar way. Let us discuss this issue in a greater detail. r First observe that given a regular C -smooth k-cube, 1 6 r 6 ∞, the carrier |σ| is a k- dimensional Cr-surface with corners. It is also clear that this surface may be oriented. Namely, 0 k for each x ∈ |σ|, the linear (and closed) subspace, the so-called tangent space Tx|σ| := σ (R ) has the orientation determined by the sequence ∂1σ(t), ..., ∂kσ(t) where x = σ(t). It is clear that the space Tx|σ| and its orientation does depend only on [σ]. Pm Given a k-block with its parameterization c = i=1 1 · σi, in order to arrange cubes Let us discuss two cases more precisely. 1 n Pm (1) Assume that a C -smooth regular (n − 1)-chain c in U ⊂ R is given, i.e., c = i=1 aiσi 84 2. Differential Forms, de Rham cohomology and singular cubical homology

1 where ai ∈ R and σi is a regular C -smooth (n − 1)-cube. As remarked above each surface |σi|, n i = 1, ..., m, is orientable. Thus, for all i = 1, ..., m, there exists a continuous map ni : |σi| → R such that |n(x)| = 1 and ni(x)⊥ Tx|σi| for all x ∈ |σi|, i.e., hni(x), vi = 0 for any v ∈ Tx|σ| (for instance we may take ni(x) := ∂1σi(t) × ... × ∂n−1σi(t) (the vector product) for x = σi(t) ∈ |σi|, t ∈ In−1). Pm Suppose that along with c we consider an oriented (n − 1)-chain [c] = i=1 ai[σi]. We say that the oriented chain [c] is compatible with the family {n1, ..., nm} of vector ﬁelds, if for any n−1 t ∈ I and i = 1, ..., m, the sequence {n(σ(t)), ∂1σi(t), ..., ∂n−1σi(t)} is a base equivalent with n the canonical base in R . 1 Pm It is obvious that given a regular oriented C -smooth (n − 1)-chain [c] = i=1 ai[σi], its initial orientation may be not compatible with a family {n1, ..., nm} of vector ﬁelds. But changing accordingly orientations of the cubes σi, without changing the carriers (and, therefore, the very carrier of c) one may always obtain a new chain that is compatible.

1 Pm 2.5.13 Exercise: An oriented regular C -smooth (n − 1)-chain [c] = i=1 ai[σi] is compatible n−1 with the family {n1, ..., nm} if and only if, for each i = 1, ..., m, there is ti ∈ int I such that n the base {ni(ti), ∂1σi(ti), ..., ∂n−1σi} is equivalent with the canonical base of R . 1 Pm n (2) Let a regular oriented C -smooth n-chain [c] = i=1 ai[σi] in U ⊂ R be given. We say that [c] is compatible with the canonical orientation if, for any t ∈ In and i = 1, ..., m, the n n sequence (a base in R ) {∂1σi(t), ..., ∂nσi(t)} is equivalent with the canonical base of R . As before given a regular oriented chain [c], its orientation may not be compatible with the n canonical orientation of R ; however if we change orientations of cubes σi accordingly, then we get a new chain c0 being compatible and |c0| = |c|. Let us now compare situations (1) and (2) described above: suppose that [σ] is an oriented 1 n regular C -smooth n-cube compatible with the canonical orientation of R . Of course n X i i+1 [∂σ] = ((−1) [Aiσ] + (−1) [Biσ]) i=1 1 is a regular C -smooth (n − 1)-chain in which, for even i, the face Aiσ keeps the original orientation and Biσ has the reverse orientation, while for an odd i, Aiσ has the reverse orientation and Biσ keeps its original orientation. A B n Consider the family of 2n vector ﬁelds {ni , ni }i=1 given by

A n A i −1 −1 −1 ni : |Aiσ| → R , ni (x) = (−1) ∂1σ(σ (x)) × ... × ∂iσ\(σ (x)) × ... × ∂nσ(σ (x)) for x ∈ |Aiσ| and

B n B i+1 −1 −1 −1 ni : |Biσ| → R , ni (x) = (−1) ∂1σ(σ (x)) × ... × ∂iσ\(σ (x)) × ... × ∂nσ(σ (x)) for x ∈ |Biσ|. It is not diﬃcult to see that the orientation of ∂σ is compatible with this family.

1 n 2.5.14 Fact: Let [σ], [τ] be two regular C -smooth n-cubes in U ⊂ R compatible with the canonical orientation. Assume that the intersection |σ| ∩ |τ| is the carrier of a common (n − 1)- dimensional face of σ and τ, i.e., there are i, j = 1, ..., n such that (i) |Aiσ| = |Ajτ| or (ii) |Aiσ| = |Bjτ| or (iii) |Biσ| = |Ajτ| or (iv) |Biσ| = |Bjτ|. Then the signs of these faces in ∂σ and ∂τ, respectively, are different. Proof: First observe that the situation described above can not occur both faces are front or back faces; otherwise the intersection of |σ| and τ| would be essentially larger than the carrier 2.5. Orientation of chains 85 of this common face. Therefore situations (i) and (iv) are excluded. Hence we may assume that there are i, j = 1, ..., n such that |Aiσ| = |Bjτ|. Next observe that these faces have the same sign in ∂σ and ∂τ, respectively, if and only if i + j is an odd number. Since if the parity of i and j is the same, then these signs must be different. So, for instance, reasoning to the contrary, suppose that i is an odd number and j is an even number and, without loss of generality, let i = 1 and j = 2 (other case can be treated analogously) and let A1σ = B2τ. We claim that this implies that orientations of σ and τ are different (but it was assumed that both these orientations agree with the canonical orientation n of R ); so the achieved contradiction will complete the proof. 2 To simplify the argument assume that n = 3. We know that for any t = (t1, t2) ∈ I , A1σ(t1, t2) = B2(t1, t2), i.e., σ(0, t1, t2) = τ(t1, 1, t2), thus – in view of Exercise 2.5.8,

∂1(A1σ)(t) = ∂2σ(0, t) = ∂1τ(t1, 1, t2) = ∂1(B2τ)(t);

∂2(A1σ)(t) = ∂3σ(0, t) = ∂3τ(t1, 1, t2) = ∂2(B2τ)(t).

This implies that, for all t ∈ I2, the sequences

{∂2σ(0, t), ∂2τ(t1, 1, t2), ∂3σ(0, t)} and {∂1σ(0, t), ∂1τ(t1, 1, t2), ∂3τ(t1, 1, t2)}

3 are both equivalent to the canonical base {e1, e2, e3} in R . Let us study the ﬁrst sequence. It follows that, for any t ∈ I2, the sequence

{−∂2τ(t1, 1, t2), ∂2σ(0, t), ∂3σ(0, t)} is equivalent to {e1, e2, e3} and, therefore, also to the base {∂1σ(0, t), ∂2σ(0, t), ∂3σ(0, t)}. Fix 2 t = (t1, t2) ∈ int I and let T be the subspace spanned by ∂2σ(0, t), ∂3σ(0, t) and let T+ be 3 the half-space in R in which lies ∂1σ(0, t). We thus see that the vector −∂2τ(t1, 1, t2) lies also in T+ for otherwise the sequence {−∂2τ(t1, 1, t2), ∂2σ(0, t), ∂3σ(0, t)} would not be equivalent to {∂1σ(0, t), ∂2σ(0, t), ∂3σ(0, t)}. In other words the vector ∂2τ(t1, 1, t2) lies in T−, where T− denotes the half-space opposite to T+. Observe that τ(t1, 1 − h, t2) − τ(t1, 1, t2) ∂2τ(t1, 1, t2) = lim . h→0+ −h Hence + τ(t1, 1 − h, t2) = τ(t1, 1, t2) − h∂2τ(t1, 1, t2) + o(|h|) as h → 0 .

This is evidently a contradiction.

2.5.15 Corollary: Let σ and τ be as above. Then |∂(σ + τ)| = ∂(|σ| ∪ |τ|). The situation described in Fact 2.5.14 and the above Corollary may be generalized.

2.5.16 Blocks We say that a compact D ⊂ E is an (n − 1)-dimensional (resp. n-dimensional) 1 ¯(1) ¯(1) block if there is a regular C -smooth (n−1)-chain (resp. n-chain) c ∈ Qn−1(U) (resp. c ∈ Qn (U)) such that: Pm i (i) c = i=1 1 · σ ; Sm i (ii) D = i=1 |σ |; (iii) for each i = 1, ..., m, σi is compatible with the canonical orientation; i j (iv) for i 6= j, the set |σ |∩|σ | is either empty or the carrier of a common k-face 0 6 k 6 n−2 (resp. 0 6 k 6 n − 1); moreover if this intersection is a common (n − 1)-face, then we assume that this common face is a front face of one of these cubes and the back face of the other one. We say that c is a parameterization of D. 86 2. Differential Forms, de Rham cohomology and singular cubical homology

P We may write ∂c = j ±τj where τj is an (n − 1)-face of some σi, i.e., for any j, τj = Akσi or τj = Bkσi for some i = 1, ..., m and k = 1, ..., n. If τj = τr for some j 6= r, then (geometrically) they would be faces of the same cube, so in view of (iv) they would cancel themselves. Moreover, the topological boundary [ ∂D = |τj|. j In view of the preceding argument, the compatible orientation of the boundary is given. n Now let a continuous ω : U → A (U, F) and a block D be given. Let us deﬁne

m Z Z X Z ω = ω = ω D c i=1 σi where c is a parameterization of D. This deﬁnition is correct, i.e., it does not depend on the parameterization of D. Indeed, let

ω(x) = a(x)dx1 ∧ ... ∧ dxn where a : U → F is continuous. For any i = 1, ..., m Z Z Z Z ω = a(σ(t)) det σ0(t) dt = a(σ(t))| det σ0(t)| dt = a(x) dx n n σi I I |σi| in view of the (improved version of the) theorem on the change of variables in the Lebesque integral. Thus m m X Z X Z Z ω = a(x) dx = a(x) dx. i=1 σi i=1 |σi| D

n−1 Now suppose that ω ∈ Λ(1) (U, F). In view of the Stokes theorem Z Z dω = ω. c ∂c R R If we define ∂D ω := ∂c ω, then again we see that this is a correct definition and finally Z Z ω = ω. (∗) D ∂D This is the desired Stokes formula for blocks. The above argument was only a step towards the ‘correct’ Stokes formula. We shall provide n only the discussion of a relatively simple case: we shall introduce a class of sets D in R for which the formula (∗) holds true. Namely we say that a compact set D is a regular domain if: (i) D = cl int D; Pm (ii) there is a parameterization of ∂D, i.e., an (n − 1)-chain c = i=1 1 · σi, where σi, i = 1, ..., m, is a diffeomorphic C1-smooth singular (n−1)-cube (7), such that ∂D = cl D\int D = Sm i=1 |σi| and, for all i 6= j, the intersection |σi| ∩ |σj| is empty or is the carrier of a common r-dimensional face (0 6 r 6 n − 2) of both cubes;

7 n−1 1 Evidently we mean that σi : I → U is a diﬀeomorphism, i.e., a regular C -smooth homeomorphism onto its image. 2.5. Orientation of chains 87

n−1 (iii) for any i = 1, ..., m and any t0 ∈ int I , if v ∈ ND(σi(t0)), then the sequence n {v, ∂t1 σi(t0), ..., ∂tn−1 σi(t0)} belongs to the orientation of E = R . 2 For instance a set D ⊂ R such that D = cl int D and ∂D is the carrier of a closed piece-wise C1-smooth curve is a regular domain. The following theorem holds:

n 2.5.17 Stokes’ Formula: If D is a regular domain in an open U ⊂ E, ω ∈ Λ(1)(U, F), then Z Z dω = ω. D ∂D

1 n 1 R Of course, if dω(x) = a(x)dx ∧ ... ∧ dx , where a : U → F is C -smooth, then ω = R R D D a(x) dx.. The meaning of ∂D ω is, of course understood as Z Z ω = ω ∂D c where c is a parameterization of ∂D. The justification of this definition is a part of the proof. The proof of this fundamental result is quite involved. It consists of several parts. First we fix a parameterization c of ∂D and show that R dω = R ω. This, obviously shows that the R D c above definition of ∂D ω is correct. Then in order to get the above formula one has to proceed in several steps – see the book of Birkholc for details. Roughly speaking it is necessary to show that D is the so-called singular manifold (or a pseudo-manifold with boundary: precisely for n−1 any i = 1, ..., m and t0 ∈ int I , there is a neighborhood V of σ(t0) ∈ ∂D, V ⊂ U and a n diffeomorphism ϕ : V → W ⊂ E such that ϕ(D ∩ V ) = W ∩ {x ∈ R | x1 6 0}. The using a series of lemmata (in particular one have to remove singularities, i.e. points of the form σi(t) where t ∈ ∂In−1) we derive the Stokes formula.

2.5.B Index of a cycle, topological degree of a continuous map

(1) n In this section we are going to compute Hn−1(U) where U = R \ 0), n > 2. To this end suppose (1) n n (1) n that c ∈ Zn (U). Since Hn−1(R ) = 0 (R is smoothly contractible), c ∈ Bn−1(R ), i.e. there (1) n Pm is d ∈ Qn (R ) such that c = ∂d. We may suppose that d = i=1 aiσi where ai ∈ R and n 1 n Pm σi : In → R is a nondegenerate C -smooth singular n-cube in R , i.e. c = i=1 ai∂σi. Without loss of generality we may suppose that, for each i = 1, ..., m, ∂σi is nondegenerate. The idea is to consider each σi separately and to show that each cycle ∂σi is homologous to (1) a some biτ where τ is a certain fixed cycle τ in Zn−1(U). If this is true, then c is homologous to Pm ( i=1 aibi)τ, i.e., (1) 2.5.18 Theorem: The vector space Hn−1(U) is one dimensional. n n 1 Proof: Let us assume that σ : I → R is a nondegenerate C -smooth singular n-cube in U. (1) −1 n If |σ| ⊂ U, then ∂D ∈ Bn (U). Hence we may assume that 0 ∈ |σ|. Of course σ (0) ∩ ∂I = ∅ since ∂σ is a cycle in U. Now assume additionally that: (∗) 0 is a regular value of σ, i.e., for any x ∈ σ−1(0), det σ0(x) 6= 0. −1 −1 Let {x1, ..., xk} = σ (0) (in view of the above assumption the set σ (0) is compact and consists of isolated points, thus this set must be finite. There is ε > 0 such that, for each i = 1, ..., k, there σ restricted to the ball B(xi, ε) is a diffeomorphism and Wi := σ(B(xi, ε)) is an 88 2. Differential Forms, de Rham cohomology and singular cubical homology

n 0 open neighborhood of 0 in R . Moreover we may assume that ηi = sgn det σ (x) = ± is constant for x ∈ B(x)i, ε), n √ Let σi be an n-cube defined as follows: for t ∈ P := [−1, 1] , let ξi(t) = nt + xi and ∞ n σi(t) = σ ◦ ξi. It is clear that ξi(P ) ⊂ B(xi, ε) and since P is C -diffeomorphic to I we may think that σi is a diffeomorphic n-cube. Sk Clearly In \ i=1 ξ(P ) is a block (one easily sees its parameterization; hence cycles ∂σ and Pk i=1 1 · ∂σi are homologous. Now consider the carrier |σi|. Since σi is a diffeomorphism, 0 ∈ int |σi for all i = 1, ..., k. Take a small ball D(0, δ), δ > 0, such that D(0, δ) ⊂ int |σi| for all i = 1, ..., k. For each i = 1, ..., k, we consider the set Di := |σi| \ B(0, δ) it is clear that ∂D = n−1 S (0, δ) ∪ |∂σi|. Thus D is a regular domain provided we orient the boundary in an appropriate way. First observe that Sn−1(0, δ) is the carrier of an (n − 1)-cycle (it is sufficient to parameterize it as the trigonometric sphere via polar coordinates), i.e., there is a C1-smooth (n − 1)-cycle τ : In−1 → U such that Sn−1(0, δ) = |τ|. It is easy to see that τ has the orientation n compatible with the canonical orientation of R , so we must consider −τ. Similarly with ∂σi. If n ηi = +, then ∂σi is oriented in accordance to the canonical orientation in R , while if ηi = −, then its orientation is reverse. In effect we have to consider (ηi∂σ)+(−τ) as the parameterization of ∂Di. 1 n−1 Given any closed C -smooth (n−1)-form ω ∈ Λ(1) (U), in view of the Stokes formula 2.5.17, we see that Z Z 0 = dω = ω = 0. Di ηi∂σi−τ

By the de Rham theorem we see that ηi∂σi is homologous to τ. Consequently ∂σ is homologous Pk to bτ where b = ηi and i=1 Z Z ω = ηi ω. ∂σi τ In order to remove assumption (∗) it is suﬃcient to recall the Sard theorem and instead of n 0 take an arbitrarily close point y ∈ R such that y is a regular value of σ. n n 1 2.5.19 The index of a cycle with respect to a point: Let σ : I → R be a C -smooth singular n-cube such that 0 6∈ |∂σ|. There is an integer number bσ such that [∂σ] = bσ[τ] where 1 τ is the trigonometric sphere (or any equivalent C -smooth singular (n − 1)-cube).

This one of the most fundamental results of algebraic topology. The integer number bσ is called the index of σ with respect to 0 or, more precisely, the index of ∂σ. We shall discuss some consequences of this concept.

n n n 1 2.5.20 Computation of the index: Let U := R \ 0 and let σ : I → R be a C -smooth singular n-cube such that 0 6∈ |∂σ|; in other words ∂σ is a C1-smooth singular (n − 1)-cycle in U. We shall provide formulae for the index bσ. To this aim we shall consider some cases: n = 1, n = 2 and n > 2. 0 x (1) Let n = 1 and let ω1 ∈ Λ(0)(U), i.e., ω1 : U → R, be given by ω1(x) = sgn x = |x| , x ∈ U. Obviously ω is closed and smooth. Then 1 Z sgn σ(1) − sgn σ(0) bσ = ω1 = . 2 ∂σ 2 The second equality follows directly from the deﬁnition of the integral. To show the ﬁrst one observe that, in view of the argument used in the proof of Theorem 2.5.18 (we sustain the 2.5. Orientation of chains 89 notation employed there)

k k ! Z X Z X Z ω1 = ω1 = ηi ω = 2bσ. ∂σ i=1 ∂σi i=1 τ Moreover, if we assume that 0 is a regular value of σ, then

X 0 bσ = sgn det σ (x). x∈σ−1(0)

We see that a point was to choose an appropriate smooth closed form for which it is easy to R compute τ ω. (2) Let n = 2. Recall formula (1.3.5) and let us try to determine the underlying 1-form responsible for the vector ﬁeld G. It is easy to see (e,g. putting in (1.3.5) f := 1R2 ) that if we 1 take ω2 ∈ Λ (U) given by 1 ω (x, y) = (−ydx + xdy), (x, y) ∈ U, 2 x2 + y2 then 1 Z indγf = ω2. 2π f◦γ

It is easy to see that the form ω2 is closed and Z ω2 = 2π τ (as above τ is the trigonometric circle). Therefore, arguing as before we get that 1 Z bσ = ω2, 2π ∂σ i.e., bσ = ind∂σ1R2 . Again the proof of Theorem 2.5.18 shows that if 0 is a regular value of σ, the X 0 bσ = sgn det σ (x). x∈σ−1(0)

n−1 (3) Let us pass th general case n > 3. Namely we consider an (n − 1)-form ωn ∈ λ(1) (U) given by n X xj ω (x) = (−1)j−1 dx1 ∧ ... ∧ dxdj ∧ ... ∧ dxn, x ∈ U. n |x|n j=1 Note that n X ∂ xj dω (x) = dx1 ∧ ... ∧ dxn = n ∂x |x|n j=1 j n 2 ! 2 X 1 nxj n n|x| − dx1 ∧ ... ∧ dxn = − dx1 ∧ ... ∧ dxn = 0. |x|n |x|n+2 |x|n |x|n+2 j=1

Moreover, it is easy to see (direct computation remembering that τ is the trigonometric sphere) Z n−1 ωn = λ(S ) τ 90 2. Differential Forms, de Rham cohomology and singular cubical homology is the (n − 1)-dimensional measure of the sphere (e.g. λ(S2) = 4π). Hence, arguing as before we see that 1 Z bσ = n−1 ωn. λ(S ) σ This the so called Kronecker formula. As above if 0 is a regular value of σ, then

X 0 bσ = sgn det σ (x). x∈σ−1(0)

2.5.21 Geometric interpretation of the index: In paragraph ?? we have interpreted indγf geometrically as the winding number. Let us make a similar interpretation with respect to ωn. We restrict ourselves to the case 3 3 n = 3. Namely it is immediate to see that given x ∈ U = R \ 0 and v, w ∈ R , 1 ω (x)(v, w) = hv × w, xi 3 |x|3 where v × w stands for the vector product of v and w; recall that if z = (z1, z2, z3) = v × w, then

v2 v3 v1 v3 v1 v2 z1 = , z2 = − , z3 = w2 w3 w1 w3 w1 w2 and |v × w| = |v||w| sin α where α = ](v, w). Therefore |hv × w, xi| is the volume of the of the prism (pol. graniastosłup) constructed on x, v and w, |hv × w, xi||x|−1 = |v × w| cos β, where β = ](v × w, x), is the surface are of the normal section of this prism and, ﬁnally, −2 |ω3(x)(v, w)| = |x| cos β|v × w| is the surface are of the radial projection of this section onto R the unit sphere. This explains geometrically why τ ω3 = 4π. At the same time this shows 1 R that 4π ∂σ ω3 is the (algebraic) number of how many times the sphere is covered be the radial retraction of the surface ∂σ. Chapter 3 General Constructions of (Co)homology of complexes

3.1 Homology and cohomology of complexes

3.1.1 Homology of chain complexes: A chain complex of R-modules (i.e. groups if R = Z and vector spaces if R is a ﬁeld, e.g. R = R or C) is a family (the so-called graded R-module) C∗ = {Cq, ∂q}q∈Z of R-modules Cq, q ∈ Z, together with a family of homomorphisms ∂q : Cq → Cq−1, q ∈ Z, such that, for any q ∈ Z, ∂q−1∂q = 0. We say that a complex C∗ is non-negative if Cq = 0 for q < 0 and free if Cq is a free R-module for all q ∈ Z. For q ∈ Z, let Zq(C∗) := Ker ∂q,Bq(C∗) := Im ∂q+1.

By a graded R-module of the homology of a chain complex C∗ we mean the graded R-module

H∗(C∗) = {Hq(C∗)}q∈Z where, for q ∈ Z,

Hq(C∗) := Zq(C∗)/Bq(C∗).

If M is an R-module, then a graded module of the homology of a chain complex C∗ with coeﬃcients in M is called the homology module H∗(C∗; G) := H∗(C∗ ⊗ G) of the chain complex

C∗ ⊗ C := {Cq ⊗ G, ∂q ⊗ 1G}q∈Z. 3.1.2 Examples: In the previous section we introduced the following chain complexes of vec- ¯(r) ¯(r) (r) tor spaces: Q∗ (U) := {Qk (U), ∂}k∈Z (the subscript k in ∂k was omitted) and Q∗ (U) := (r) {Qk (U), ∂}k∈Z where 1 6 r 6 ∞ and U was a subset of a Banach space if r > 1 and U was an arbitrary space if k = 0. The corresponding homology

(r) (r) H¯∗ (U) := H∗(Q¯∗ (U));

(r) (r) H∗ (U) := H∗(Q∗ (U)) and (0) H∗(U) := H∗ (U).

3.1.3 Cohomology of cochain complexes: By a cochain complex of R-modules we call a ∗ q q q q q+1 graded R-module C = {C , δ }q∈Z together with a family of homomorphisms δ : C → C , 92 3. General Constructions of (Co)homology of complexes

q+1 q q q ∈ Z, such that δ δ = 0 for all q ∈ Z. If C = 0 for all q < 0, then we speak of a non-negative cochain complex. For q ∈ Z,

Zq(C∗) := Ker δq,Bq(C∗) := Im δq+1.

By the cohomology module of a cochain complex C∗ we understand the graded R-module H∗(C∗) = q ∗ {H (C )}q∈Z where, for q ∈ Z,

Hq(C∗) := Zq(C∗)/Bq(C∗).

Previously we have introduced the de Rham complexes of vector spaces

∗ k Λ (U, F) := {Λ (U, F), d}k∈Z, Λ∗(U) := {Λk(U), d}k∈Z and the corresponding cohomology

∗ ∗ ∗ ∗ ∗ ∗ H (U, F) := H (Λ (U, F)), H (U) := H (Λ (U)).

3.1.4 Cohomology of chain complexes: Let C∗ = {Cq, ∂q}q∈Z be a chain complex of R-modules and let M be an R-module. There exists a cochain complex

q Hom(C∗,M) := {Hom(Cq,M), δ }q∈Z

q where, for q ∈ Z and u ∈ Hom(Cq,M), [δ u](c) = u(∂q+1c) for c ∈ Cq+1. Then, by the cohomology module of the chain complex C∗ with coeﬃcients in M we mean the cohomology module of the cochain complex Hom(C∗,M), i.e.,

∗ ∗ H (C∗,M) := H (Hom(C∗,G)).

If M = R, then ∗ ∗ H (C∗) := H (C∗,R).

3.1.5 Example: Let U be a subset of a Banach space (resp. U is an arbitrary space) and let M r be a vector space. For 1 6 r 6 ∞, we deﬁne the R-module of C -smooth singular cohomology of U with coeﬃcients in Mby

∗ ∗ (r) ∗ ∗ H(r)(U, M) := H (Q∗ (U),M). (resp.H (U, M) := H(0)(U, M)).

The de Rham Theorem asserts actually that: If U is an open subset of a Banach space, then

∗ ∼ ∗ H (U, F) = H (U, F).

0 0 0 3.1.6 Chain homomorphisms: Let C∗ = {Cq, ∂q}q∈Z and C∗ = {Cq, ∂q}q∈Z be chain com- 0 plexes. By a chain homomorphism we call a family ϕ := {ϕq : Cq → Cq}q∈Z of homomorphisms 0 0 such that ∂q ◦ ϕq = ϕq−1 ◦ ∂q for all q ∈ Z. We write ϕ : C∗ → C∗. 0 It is a routine to check that, for all q ∈ Z, ϕq(Zq(C∗)) ⊂ Zq(Cq) and ϕq(Bq(C∗)) ⊂ Bq(C∗). 0 Hence ϕ determine the family ϕ∗ := {ϕq∗ : Hq(C∗) → Hq(C∗)}q∈Z. We write

0 ϕ∗ : H∗(C∗) → H∗(C∗). 3.1. Homology and cohomology of complexes 93

00 00 00 0 00 Clearly, if C∗ = {Cq , ∂q }q∈Z is a third chain complex and ψ : C∗ → C∗ is a chain homomorphism, then ψ ◦ ϕ := {ψq ◦ ϕq}q∈Z is a chain homomorphism and

00 (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗ : H∗(C∗) → H∗(C∗ ).

3.1.7 Example: Let U, U 0 be two subsets of Banach spaces (resp. spaces) and let ϕ : U → U 0 r (r) (r) 0 be a C -smooth map, 1 6 r 6 ∞ (resp. continuous). Then ϕ# : Q∗ (U) → Q∗ (U ) is a chain (r) (r) 0 map and it determines ϕ∗ : H∗ (U) → H∗ (U ).

0 00 3.1.8 Theorem on connecting homomorphism: Let C∗,C∗,C∗ be chain complexes and assume that a short sequence of chain maps

0 α β 00 0 −→ C∗ −→ C∗ −→ C∗ −→ 0 is exact, i.e., for each q ∈ Z, the sequence

0 αq βq 00 0 −→ Cq −→ Cq −→ Cq −→ 0

00 0 is exact. Then, for each q ∈ Z, there is a homomorphism ∂∗ : Hq(C∗ ) → Hq−1(C∗) such that the sequence 0 α∗ β∗ 00 ∂∗ 0 ... −→ Hq(C∗) −→ Hq(C∗) −→ Hq(C∗ ) −→ Hq−1(C∗) −→ ... (∗) is exact. Moreover if a diagram

0 α β 00 0 / C∗ / C∗ / C∗ / 0

ϕ0 ϕ ϕ00 0 / D0 / D / D00 / 0, ∗ a ∗ b ∗ of chain complexes and chain maps has exact rows, then the diagram

00 ∂∗ 0 Hq(C∗ ) / Hq−1(C∗) (∗∗)

00 0 ϕ∗ ϕ∗ 00 0 Hq(D∗ ) / Hq−1(D∗) ∂∗ is commutative for any q ∈ Z.

Proof: To deﬁne ∂∗, ﬁx q ∈ Z and consider the following diagram:

βq 00 Cq / Cq

∂q

0 C / Cq−1 q−1 αq−1

00 00 in which βq is an epimorphism. Take [z] ∈ Hq(C∗ ); there is c ∈ Cq such that z := βq(c) ∈ Zq(C∗ ). We deﬁne −1 0 ∂∗([z]) := [αq−1(∂q(c))] ∈ Hq−1(C∗). 94 3. General Constructions of (Co)homology of complexes

00 00 This deﬁnition is correct, since βq−1(∂(c)) = ∂ (βq(c)) = ∂ (z) = 0, i.e., ∂(c) ∈ Ker βq−1 = Im αq−1. It is standard to check that ∂∗([z]) does not depend on the choice of c and that diagrams (∗) and (∗∗) are commutative. 0 0 0 3.1.9 Chain homotopy: Let C∗ = {Cq, ∂q}q∈Z and C∗ = {Cq, ∂q}q∈Z be chain complexes and 0 let ϕ0, ϕ1 : C∗ → C∗ be chain homomorphisms. By a chain homotopy we mean a collection 0 Φ = {Φq : Cq → Cq+1}q∈Z of homomorphisms such that, for any q ∈ Z, 0 ∂q+1 ◦ Φq + Φq−1 ◦ ∂q = ϕ1q − ϕ0q.

In this case we write Φ: ϕ0 ' ϕ1 and say that ϕ0, ϕ1 are chain homotopic. It is easy to see that the relation ' of chain homotopy is an equivalence relation and, given 0 00 chain maps ψ0, ψ : C∗ → C∗ , if ψ0 ' ψ1, then

ψ0 ◦ ϕ0 ' ψ1 ◦ ϕ1.

It is immediate to get:

0 3.1.10 Homotopy invariance: If chain maps f, g : C∗ → C∗ are chain homotopic, then

0 f∗ = g∗ : H∗(C∗) → H∗(C∗).

Proof: Indeed, for any z ∈ Zq(C∗),

0 0 f∗[z] − g∗[z] = [fq(z) − gq(z)] = [∂q+1(Φq(z)) + Φq−1(∂q(z))] = [∂q+1(Φq(z))] = 0.

3.1.11 Example: Consider two subsets of Banach spaces (resp. spaces if r = 0) U, U 0 and two r 0 C -homotopic maps ϕ0, ϕ1 : U → U . We claim that ϕ0#, ϕ1# are chain homotopic. ¯(r) ¯(r) To this end we construct the so-called prism operator Pq : Qq (U) → Qq+1(U × I), q > 0, by the formula: n ! n X X q+1 Pq aiσi = (−1) ai[σ × 1I ], i=1 i=1 where [σ × 1I ] is a (q + 1)-cube in U × I given by

q+1 [σ × 1I ](t1, ..., tq, tq+1) = (σ(t1, ..., tq), tq+1), (t1, ..., tq+1) ∈ I .

(r) (r) This deﬁnition is correct for all 0 6 r 6 ∞ for [σ × 1I ] ∈ Σq+1(U × I) if σ ∈ Σq (U). Moreover (r) (r) (r) (r) it is clear that Pq(Dq (U)) ⊂ Dq+1(U × I). Therefore Pq : Qq (U) → Qq+1(U × I). It is routine to check that P := {Pq}q∈Z (where Pq = 0 for q < 0) is a chain homotopy joining j0# to j1#, i.e., ∂ ◦ Pq + Pq−1 ◦ ∂ = j1# − j0# r where ji : U → U × I,ji(x) = (x, i) for x ∈ U and i = 0, 1. Now if ϕ : ϕ0 ' ϕ1 is a C -homotopy (r) (r) 0 joining ϕ0 to ϕ1, then ϕ∗ ◦ P = {ϕ∗ ◦ Pq : Qq (U) → Qq+1(U )}q∈Z is a chain homotopy joining ϕ0# to ϕ1# because ϕi# = (ϕ ◦ ji)# = ϕ# ◦ ji#, i = 0, 1.

∗ q q ∗ q q 3.1.12 Cochain homomorphisms: If C = {C , δ },C1 = {c1, δ1} are cochain complexes, q q q then by a cochain homomorphism we mean a family ϕ = {ϕ : C → C1 }q∈Z of homomorphisms q+1 q q q such that ϕ ◦ δ = δ1 ◦ ϕ for all q ∈ Z. As before a cochain homomorphism induces the homomorphism ∗ ∗ ∗ ∗ ∗ ϕ : H (C ) → H (C1 ). 3.1. Homology and cohomology of complexes 95

For instance if ϕ : U 0 → U is a C∞-smooth map between open subsets of Banach spaces, # 0 ∗ q q 0 then ϕ :Λq(U, F) → Λq(U , F), q ∈ Z, is a cochain map and induces ϕ : H (U, F) → H (U , F). 0 0 If C∗, C∗ are chain complexes, M is an R-module and ϕ : C∗ → C∗ a chain map, then the conjugate 0 Hom(ϕ, M) : Hom(C∗,M) → Hom(C∗,M), 0 Hom(ϕ, M) = {Hom(ϕq,M) : Hom(Cq,M) → Hom(Cq, m)}q∈Z, Hom(ϕq,M)(u)(c) := u ◦ ϕq(c) 0 for u ∈ Hom(Cq,M) and c ∈ Cq, is a cochain homomorphism. ∗ ∗ 0 ∗ It is easy to see that the homomorphism ϕ : H (C∗,M) → H (C∗,M) induced by Hom(ϕ, M) (i.e. ϕ∗ := (Hom(ϕ, M))∗) satisﬁes the identity

∗ hϕ ([u]), [c]i = h[u], ϕ∗([c])i

0 for all [u] ∈ H(C∗,M) and [c] ∈ H∗(C∗). 0 In the context of the universal coeﬃcients, if C∗ and C∗ are free complexes, and e.g. R is a ﬁeld (then M is a vector space over R), then

∗ ϕ = Hom(ϕ∗,M).

The notion of cochain homotopy may be introduced similarly. By a cochain homotopy joining ∗ ∗ q q q−1 two cochain maps ϕo, ϕ1 : C → C1 we mean a family Φ = {Φ : C → C1 }q∈Z such that, for any q ∈ Z, q−1 q q+1 q q q δ1 ◦ Φ + Φ ◦ δ = ϕ1 − ϕ0. ∗ ∗ Again it is immediate to see that if vp0, ϕ1 are cochain homotopic, then ϕ0 = ϕ1. 0 Note that if chain homomorphisms ϕ0, ϕ1 : C∗ → C∗ are chain homotopic and Φ: ϕ0 ' 0 ϕ1is a chain homotopy between them, then Hom(Φ,M) := {Hom(Φq−1,M) : Hom(Cq,M) → Hom(Cq−1,M)} is a cochain homotopy joining Hom(ϕ0,M) to Hom(ϕ1,M). Hence again if 0 chain maps ϕ0, ϕ1 : C∗ → C∗ are chain homotopic, then

∗ ∗ 0 ∗ ϕ0 = ϕ1 : H(C∗,M) → H (C∗,M).

There is also an obvious analogue of the connecting homomorphism (see Theorem 3.1.8) in the case of cohomology:

3.1.13 Connecting homomorphism for cohomology of cochain complexes: Let C0∗,C∗,C00∗ be cochain complexes and assume that a short sequence of cochain homomorphisms

β 0 −→ C0∗ −→α C∗ −→ C00∗ −→ 0 is exact, i.e., for each q ∈ Z, the sequence

q βq 0 −→ C0q −→α Cq −→ C00q −→ 0

∗ q 00∗ q+1 0∗ is exact. Then, for each q ∈ Z, there is a homomorphism δ : H (C ) → H (C ) such that the sequence

∗ β∗ ∗ ... −→ Hq(C0∗) −→α Hq(C∗) −→ Hq(C00∗) −→δ Hq+1(C0∗) −→ ... 96 3. General Constructions of (Co)homology of complexes is exact. Moreover if a diagram

α β 0 / C0∗ / C∗ / C00∗ / 0

ϕ0 ϕ ϕ00 0 / 0∗ / D∗ / 00∗ / 0, D a b D of cochain complexes and cochain maps has exact rows, then the diagram

∂∗ Hq(C00∗) / Hq+1(C0∗) (∗∗)

ϕ00∗ ϕ0∗ Hq(D00∗) / Hq+1(D0∗) ∂∗ is commutative for any q ∈ Z. In case of cohomology of chain complexes one has to be more careful.

3.1.14 Connecting homomorphism for cohomology of chain complexes: Assume that a short sequence of chain complexes and chain homomorphisms

0 α β 00 0 −→ C∗ −→ C∗ −→ C∗ −→ 0 (∗) is exact and split, i.e., for any q ∈ Z, the sequence

0 αq βq 00 0 −→ Cq −→ Cq −→ Cq −→ 0

∗ is exact and split. Then, for all q ∈ Z and an R-module, there is a homomorphism δ : q 0 q+1 00 H (C∗,M) → H (C∗ ,M) such that the sequence

∗ q 00 β q α∗ q 0 δ∗ q+1 00 ... −→ H (C∗ ,M) −→ H (C∗,M) −→ H (C∗,M) −→ H (C∗ ) −→ ... is exact. Again if a diagram

0 α β 00 0 / C∗ / C∗ / C∗ / 0

ϕ0 ϕ ϕ00 0 / D0 / D / D00 / 0, ∗ a ∗ b ∗ of chain complexes and chain maps has exact rows, then the diagram

∗ q 0 δ q+1 00 H (C∗) / H (C∗ ) (∗∗) O O ϕ0∗ ϕ00∗

Hq(D0 ) / Hq+1(D00) ∗ δ∗ ∗ is commutative for any q ∈ Z. Proof: follows immediately from Theorem 3.1.13 if we observe that since the sequence (∗) is exact and split, then the sequence of cochain complexes and cochain homomorphisms

00 Hom(β,M) Hom(α,M) 00 0 −→ Hom(C∗ ,M) −→ Hom(C∗,M) −→ Hom(C∗ ,M) −→ 0 (∗) 3.1. Homology and cohomology of complexes 97 is exact.

3.1.15 Kronecker duality: Let C∗ be a chain complex of R-modules and M be an R-module. q For q ∈ Z, [u] ∈ H (C∗,M) and [c] ∈ Hq(C∗), let h[u], [c]i := u(c) ∈ M.

This deﬁnition is correct since it does not depend on the choice of c ∈ [c] and u ∈ [u] (for- q q mally everything is OK since c ∈ Zq(C∗) ⊂ Cq and u ∈ Z (C∗,M) := Z (HomR(C∗,M)) ⊂ HomR(Cq,M)). The map ∗ h·, ·i : H (C∗,M) × H∗(C∗) → M is bilinear; it is called the Kronecker ‘duality’ (the apostrophe because it is not the full duality in the sense of 2.4.18).

q 3.1.16 Theorem: If C∗ is a free complex ξ : H (C∗,M) → HomR(Hq(C∗),M), given by

q ξ([u])([c]) := h[u], [c]i, [u] ∈ H (C∗,M), [c] ∈ Hq(C∗), is an epimorphism. q Proof: To shorten the notation, let Hq := Hq(C∗), Zq := Zg(C∗), Bq := Bq(C∗), H := q q q q q ∗ H (C∗,M), Z := Z (C∗,M), B := B (C∗,M) and, for each R-module A, A := Hom(A, M). The sequence i p 0 −→ Bq −→ Zq −→ Hq −→ 0, where p is the quotient map, is exact. Hence the sequence

∗ ∗ p ∗ i∗ ∗ 0 −→ Hq −→ Zq −→ Bq

∗ ∗ ∗ is exact. Hence i p = 0. Observe that i : Bq → Zq is the inclusion; hence i (u) = u|Bq for any ∗ ∗ ∗ u ∈ Zq . In particular p (u)|Bq = 0 for any u ∈ Hq . We have another exact sequence

j ∂q 0 −→ Zq −→ Cq −→ Bq−1 → 0, where j is the inclusion. Now Bq−1,as a submodule of a free R-module (R is p.i.d.) is free. Thus this sequence is split. Therefore, the sequence

∗ ∗ ∗ ∂q ∗ j ∗ 0 −→ Bq−1 −→ Cq −→ Zq −→ 0

∗ ∗ is exact and split. Again j : Zq → Cq is the inclusion, so j (u) = u|Zq for any u ∈ Cq . Since this ∗ ∗ ∗ ∗ sequence is split, there is k : Zq → Cq such that j k = idZq . Since the third sequence is split, there is a homomorphism h : Bq−1 → Cq such that

Cq = Zq ⊕ h(Bq−1).

In other words, any c ∈ Cq has a unique representation c = (z, h(b)) where z ∈ Zq and b ∈ Bq−1. ∗ ∗ For each u ∈ Zq , k(u) ∈ Cq . This is deﬁned, by

∗ k(u)(c) = u(z), u ∈ Zq , c(z, h(b)) ∈ Cq.

∗ Therefore, for u ∈ Hq , kp∗(u)(c) = p∗(u)(z). 98 3. General Constructions of (Co)homology of complexes

∗ ∗ Hence if c ∈ Bq ⊂ Zq, then c = (c, 0) and kp (u)(c) = p (u)(c) = 0. q ∗ q It is easy to see that Z is equal to {u ∈ Cq | u|Bq = 0}, i.e., u ∈ Z if and only if u(b) = 0 q q ∗ for any b ∈ Bq. Indeed, u ∈ Z = ker δ = ker ∂q+1. If b ∈ Bq = Im ∂q+1, i.e., b = ∂q+1d where ∗ ∗ d ∈ Cq+1, then u(b) = u(∂q+1d) = ∂q+1u(d) = 0. Conversely, if u ∈ Cq and u(b) = 0 for any q q b ∈ Bq, then δ u(c) = u(∂q+1c) = 0 for any c ∈ Cq+1, i.e, u ∈ Z . ∗ ∗ q Thus we see that, for any u ∈ Hq , kp (u) ∈ Z . ∗ ∗ q Now, for u ∈ Hq and any [c] ∈ Hq, i.e., c ∈ Zq, [kp (u)] ∈ H and ξ([kp∗(u)]([c]) = h[kp∗(u)], [c]i = kp∗(u)(c) = p∗(u)(c) = u(p(c)) = u[c].

We thus see that ξ is an epimorphism. q However the kernel Ker ξ ⊂ H (C∗,R) may be non-trivial. In order to have the full duality, one needs Ker ξ = 0. Let us study this problem more carefully.

3.1.17 Universal coefficients for cohomology of chain complexes: Assume that the complex C∗ is free. For q ∈ Z, let us consider a short exact sequence

j ∂q 0 −→ Zq −→ Cq −→ Bq−1 −→ 0, where j is the inclusion. This sequence is split because Bq−1 is free. Letting Z∗ := {Zq}, B∗−1 := {Bq−1}, the sequence

∗ ∗ ∂q ∗ j ∗ 0 −→ Bq−1 −→ Cq −→ Zq −→ 0

∗ ∗ is exact and split. Observe additionally that j (u) = u|Zq for any u ∈ Cq . By the theorem on the existence of the connecting homomorphism for cohomology of chain complexes, there is a long exact sequence

0 q−1 δ∗ q q q ξ q δ∗ q+1 .. −→ H (Z∗; M) −→ H (B∗−1,M) −→ H (C∗; M) = H −→ H (Z∗; M) −→ H (B∗−1; M) −→ ...

Since Z∗ and B∗−1 have trivial boundary homomorphisms,

q ∗ q ∗ H (Z∗; M) = Zq and H (B∗−1; M) = Bq−1. Moreover, the connecting homomorphism

∗ q q+1 δ : H (Z∗; M) → H (B∗−1; M) is equal to the homomorphism ∗ ∗ ∗ γq : Zq → Bq , where γq : Bq → Zq is the inclusion. Indeed, consider a diagram

∗ ∗ j 00 Cq / Cq .

∗ ∂q+1 B∗ q ∗ / Cq−1 ∂q+1

∗ ∗ ∗ ∗ Then δ is deﬁned as follows: for any v ∈ Zq there is u ∈ Cq such that j (u) = v, i.e., v = u|Zq ; then ∗ ∗ −1 ∗ δ (v) = [(∂q+1) ∂q+1(u)] = [u]. 3.1. Homology and cohomology of complexes 99

∗ ∗ Thus δ (v) = v|Bq = γq (v). It is also important to see how 0 q ∗ ξ : H → Zq ∗ ∗ ∗ is deﬁned By deﬁnition it is induced by the inclusion j : Zq → Cq or rather by j : Cq → Zq . Precisely 0 ∗ ∗ ∗ q ∗ ξ [u] = [j (u)] = j (u) ∈ Zq for u ∈ Z ⊂ Cq . 0 ∗ 0 ∗ Hence ξ [u][c] = j (u) = u(c) for c ∈ Zq ([c] = c). So we see that ξ [u]|Zq = p ξ[u] where p : Zq → Hq is the quotient map. Thus we have an exact sequence

0 ∗ q ξ ∗ 0 −→ Coker γq−1 −→ H −→ ker γq −→ 0. (∗)

For the sequence γq p 0 −→ Bq −→ Zq −→ Hq −→ 0 we have the exact sequence ∗ ∗ ∗ p ∗ γq ∗ 0 −→ Hq −→ Zq −→ Bq . ∗ ∗ ∗ ∗ ∗ ∗ 0 This means that Im p = ker γq and p : Hq → p (Hq ) is an isomorphism. This means that ξ in (∗) is equal to ξ (up to the isomorphism). In the sequence γq 0 −→ Bq −→ Zq −→ Hq −→ 0 the middle term is free, hence we have an exact sequence

∗ ∗ ∗ γq ∗ 0 −→ Hq −→ Zq −→ Bq −→ Ext(Hq,M) −→ 0.

Therefore ∗ ∼ ∗ ∗ ∼ ker γq = Hq and Coker γq = Ext(Hq,M). Putting this into the sequence containing Hq we get

q ξ ∗ 0 −→ Ext(Hq−1; M) −→ H −→ Hq −→ 0.

We have to check only that there is ξ indeed.

If Hq−1 is free or R is a ﬁeld, then Ext(Hq−1; M) = 0 and then

q ∼ ∗ H = Hq .

3.1.18 Example: Taking unto account that Q∗(U) is a free chain complex (as a graded vector space over R = R), in view of de Rham Theorem ∗ ∼ H (U, F) = Hom(Hq(U), F) = L(Hq(U), F).

3.1.19 Universal coefficients for homology of chain complexes As usual we assume that R is a p.i.d. Let C∗ be a free chain complex over R and let B be an R-module. Then there is an exact split sequence

µ 0 −→ Hq(C∗) ⊗ B −→ Hq(C∗ ⊗ B) −→ Tor(Hq−1(C∗),B) −→ 0. 100 3. General Constructions of (Co)homology of complexes

This sequence is functorial.

To prove this result let Z∗ := {Zq}q∈Z and, as before B∗−1 = {Bq−1}. Both complexes Z∗ and B∗−1 (with trivial boundary homomorphisms) are free. Moreover there is an exact and split sequence αq βq 0 −→ Zq −→ Cq −→ Bq−1 −→ 0, where βq = ∂q and αq : Zq → Cq is the inclusion. The existence of a connecting homomorphism implies that there is an exact sequence

i∗ β∗ ∂∗ ... −→ Hq(Z∗ ⊗ B) −→ Hq(C∗ ⊗ B) −→ Hq(B∗−1 ⊗ B) −→ Hq−1(Z∗ ⊗ B) −→ ...,

−1 −1 where ∂∗[b] = [αq ∂q∂q b] for b ∈ Bq−1. Clearly complexes Z∗ ⊗ B and B∗−1 ⊗ B have trivial boundary homomorphisms, too. Hence

Hq(Z∗ ⊗ B) = Zq ⊗ B,Hq(B∗−1 ⊗ B) = Bq−1 ⊗ B.

Therefore the above sequence looks like that

γq⊗1B γq−1⊗1B ... −→ Bq ⊗ B −→ Zq ⊗ B −→ Hq(C∗ ⊗ B) −→ Bq−1 ⊗ B −→ Zq−1 ⊗ B −→ ..., where γq : Bq → Zq is the inclusion. The exactness of this sequence implies that the sequence

0 ≤ Coker (γq ⊗ 1B) → Hq(C∗ ⊗ B) −→ Ker (γq−1 ⊗ 1B) −→ 0 is exact. We need to explore this sequence more carefully. Since the complexes B∗, Z∗ are free, the sequence 0 −→ Bq −→ Zq −→ Hq(C∗) −→ 0 is exact, we have the exact sequence

γq⊗1B 0 −→ Tor(Hq(C∗),B) −→ Bq ⊗ B −→ Zq ⊗ B −→ Hq(C∗) ⊗ B −→ 0.

Hence ∼ ∼ Coker (γq ⊗ 1B) = Hq(C∗) ⊗ B, Ker (γq ⊗ 1B) = Tor(Hq(C∗),B). The homomorphism µ : Hq(C∗) ⊗ B → Hq(C∗ ⊗ B) has the following form µ([c] ⊗ b) := [c ⊗ b], c ∈ Zq, b ∈ B.

0 3.1.20 The Künneth formula Let us assume that C∗ and C∗ are graded R-modules, i.e., 0 0 0 C∗ = {Cq}q∈Z and C∗ = {Cq}q∈Z, where Cq,Cq are R-modules for any q ∈ Z. We deﬁne

0 0 0 M 0 C∗ ⊗ C∗ = {(C∗ ⊗ C∗)q}q∈Z where (C∗ ⊗ C∗)q := Ci ⊗ Cj. i+j=q

0 0 The torsion product Tor(C∗,C∗) = {Tor(C∗,C∗)q}q∈Z is deﬁned similarly:

0 M Tor(C∗,C∗)q := Tor(Ci,Cj). i+j=q 3.2. General homology and cohomology theories 101

0 0 0 0 If C∗ = {Cq, ∂q} and C∗ = {Cq, ∂q} are chain complexes, then so are both C∗ ⊗ C∗ and 0 Tor(C∗,C∗). The respective boundary homomorphisms are given by the following formulae:

00 0 0 j 0 0 ∂q (c ⊗ c ) := ∂ic ⊗ c + (−1) c ⊗ ∂jc

0 for c ∈ Ci and c ∈ Cj, i + j = q, and

j 0 ∂ | 0 = Tor(∂ , 1) + (−1) Tor(1, ∂ ). q Tor(C∗,C∗)q i j

0 It is easy to check that the above definitions correctly define boundary homomorphisms in C∗⊗C∗ 0 and Tor(C∗,C∗), respectively. 0 0 0 Moreover, if Cq = B 6= 0 only if q = 0, then C∗ ⊗C∗ = C∗ ⊗B and Tor(C∗,C∗) = Tor(C∗,B). Now we define a homomorphism

0 0 µ : Hq(C∗) ⊗ Hq(C∗) → Hq(C∗ ⊗ C∗)

0 0 0 0 as follows. Let c ∈ Zi(C∗), c ∈ Zj(C∗). If c or c is a boundary, then so is c ⊗ c . There exists a homomorphism µ such that µ([c] ⊗ [c0]) = [c ⊗ c0].

We have a following result: 0 Assume that C∗ is a free complex, R is a p.i.d.. There is a functorial exact sequence

0 µ 0 0 0 −→ [H∗(C∗) ⊗ H∗(C∗)]q −→ Hq(C∗ ⊗ C∗) → [Tor(C∗,C∗)]q−1 −→ 0.

Moreover, if C∗ is free then this sequence is split. 0 The assumption that C∗ and C∗ is free may be weakened. Namely if we assume that the 0 complex Tor(C∗,C∗) is acyclic. Recall that a complex C∗ is acyclic, if Hq(C∗) = 0 for any q ∈ Z.

3.2 General homology and cohomology theories

Let G be a fixed R-module. By a homology theory with coefficients in G consists of: • a covariant functor h from the category of pairs of topological spaces to the category of graded R-modules (H is called the homology functor), i.e., h(X,A) = {hq(X,A)}q∈Z for topological any pair (X,A), where hq(X,A) ∈ ModR, q ∈ Z; • a natural homomorphisms ∂∗ := {∂q}q∈Z, where ∂q : hq(X,A) → hq−1(A) := hq(A, ∅) such that the following axioms are satisfied: Dimension: If P is a one-point space, then

G if q = 0; h (P ) = q 0 if q 6= 0.

Homotopy: If f ' q :(X,A) → (Y,B), then

h(f) = h(g): h(X,A) → h(Y,B), i.e., for any q ∈ Z, hq(f) = hq(g): hq(X,A) → hq(Y,B). Excision: If U is open in X and cl U ⊂ int A, then the inclusion i :(X \ U, A \ U) → (X,A) induces an isomorphism h(i): h(X \ U, A \ U) → h(X,A). 102 3. General Constructions of (Co)homology of complexes

Exactness: For any pair (X,A) and maps

j (A, ∅) −→i (X, ∅) −→ (X,A)

(with obvious inclusions), the following homology sequence of the pair (X,A)

∂q+1 hq(i) hq(j) ∂q ... −→ hq+1(X,A) −→ hq(A) −→ hq(X) −→ hq(X,A) −→ hq−1(A) −→ ... is exact. In what follows we shall show important examples of homology theories.

3.2.A Singular homology revisited

Let X be a topological space (we consider only Hausdorﬀ spaces). By a singular q-cube, q > 1, we mean a continuous map σ : Iq → X (with regard to our previous considerations, let us remark, that now we are not interested in any Cr-smoothness of singular cubes). q For q = 0, I = {0} and a singular 0-cube is simply a point in X, i.e., Σ0(X) = X. For q < 0 there are no singular q-cubes. (0) The set of all singular q-cubes, q > 0, is denoted by Σq(X) (i.e., Σq (X) in our previous notation). q The singular q-cube σ : I → X is degenerate it there is 1 6 i 6 q such that σ(t1, ..., tq) does not depend on ti. For any q ∈ Z, let Q¯q(X) := R(Σq(X)), ¯ Pn i.e., Qq(X) consists of formal linear combinations of the form c = i=1 aiσi, where ai ∈ R, σi ∈ Σq(X). It is convenient to put Q¯q(X) = {0} for any q < 0. Let Dq(X) denotes the submodule in Q¯q(X) generated by all degenerate singular q-cubes; for q 6 0, we let Dq(X) = {0} (this is alright since, there are degenerate 0-cubes singular and no q-cubes for q < 0). We put ¯ Qq(X) := Qq(X)/Dq(X), q ∈ Z. ¯ For q 6 0, Qq(X) = Qq(X), i.e., in particular Qq(X) = {0} for q < 0.

Remark: The module Qq(X) may be identified with the submodule in Q¯q(X) generated by nondegenerate singular q-cubes. Then Q¯q(X) = Qq(X) ⊕ Dq(X). In particular Qq(X) is a free R-module. The next definition is exactly the same as before. Namely we define the boundary operator ¯ ¯ ¯ ∂q : Qq(X) → Qq−1(X), q > 1, as follows: for any σ ∈ Σq(X),

q X i ∂q(σ) := (−1) [Aiσ − Biσ], i=1

q−1 where, for 1 6 i 6 q, Aiσ, Biσ : I → X are given by

Aiσ(t1, ..., tq−1) := σ(t1, ..., ti−1, 0, ti, ..., tq−1);

Biσ(t1, ..., tq−1) := σ(t1, ..., ti−1, 1, ti, ..., tq−1). 3.2. General homology and cohomology theories 103

Further on we skip the index q an write simply ∂(σ) instead of ∂q(σ). For q 6 0, ∂q ≡ 0. Pn ¯ If c = i=1 aiσi ∈ Qq(X), then n X ∂¯(c) := ai∂(σi). i=1

¯ Since ∂(Dq(X) ⊂ Dq−1(X), q ∈ Z, we see that ∂ induces

∂ : Qq(X) → Qq−1(X).

It is easy to see that ¯ ¯ ¯ 3.2.1 Lemma: The families Q∗(X) = {Qq(X), ∂q}q∈Z and Q∗(X) := {Qq(X), ∂q}q∈Z are chain ¯ ¯ complexes, i.e. ∂q−1 ◦ ∂q = 0 and ∂q−1 ◦ ∂q = 0 for all q ∈ Z. Now assume that f : X → Y is a continuous map. For any q > 0, it induces a transformation Σq(f):Σq(X) → Σq(Y ) given by

Σ(f)(σ) = f ◦ σ, σ ∈ Σq(X).

It is easy to see that ∂q(Σq(f)(σ)) = Σq−1(f)(∂q(σ)). In other words Σq−1(f)◦∂q = ∂q◦Σq(f). The homomorphisms Q¯q(f): Q¯q(X) → Q¯q(Y ) Pn is deﬁned for c = i=1 aiσi by n n X X Q¯q(f)(c) = aiΣ(f)(σi) = ai(f ◦ σi). i=1 i=1

It is easy to see that ∂¯q ◦ Q¯q(f) = Q¯q−1(f) ◦ ∂¯q.

3.2.2 Lemma: We have that Q¯q(f)(Dq(X)) ⊂ Dq(Y ). Hence f induces a homomorphism

Qq(f): Qq(X) → Qq(Y ) such that ∂q ◦ Qq(f) = Qq−1(f) ◦ ∂q. This means that {Qq(f)}q∈Z is a chain map

For q < 0, Qq(f) is a trivial homomorphism. Now let (X,A) be a topological pair and let i : A → X be the inclusion. Clearly one may ¯ ¯ treat any singular q-cube in A as a cube in X. Then Qq(A) is a submodule in Qq(X), q > 0 and Dq(A) is a submodule in Dq(X). Thus one can treat Qq(A) as a submodule of Qq(X) and Qq(i): Qq(A) → Qq(X) is the inclusion. We deﬁne Qq(X,A) := Qq(X)/Qq(A), q > 0.

Obviously, for q < 0, Qq(X,A) = {0}.

3.2.3 Remark: For any q > 0, Qq(X,A) is a free R-module that may be identiﬁed with a q submodule in Q¯q(X) generated by nondegenerate singular q-cubes σ such that σ(I ) 6⊂ A.

Since ∂q ◦ Qq(i) = Qq−1(i) ◦ ∂q, the homomorphisms

∂q : Qq(X,A) → Qq−1(X,A) 104 3. General Constructions of (Co)homology of complexes is well-deﬁned.

3.2.4 Lemma: The family Q∗(X,A) = {Qq(X,A)}q∈Z is a chain complex, i.e., ∂q−1 ◦ ∂q = 0. 3.2.5 Definition: We define singular (cubic) homology of the pair (X,A) with coefficients in R by Hq(X,A) := Hq(Q∗(X,A)), q ∈ Z. Moreover, we define singular homology with coefficients in G (recall that G is a fixed R-module), by Hq(X,A; G) := Hq(Q∗(X,A) ⊗ G), q ∈ Z.

Remember that the boundary homomorphism in Q∗(X,A) ⊗ G is, by definition, equal to ∂q ⊗ 1G, q > 0. Using the general construction we may also define the cohomology of (X,A) with coefficients in G by q q ∗ H (X,A; G) := H (Q (X,A; G)), q ∈ Z where ∗ Q (X,A; G) := Hom(Q∗(X,A),G), q i.e., Q (X,A; G) := Hom(Qq(X,A),G) for any q ∈ Z. q ∗ The boundary homomorphism in Hom(Q∗(X,A),G) is given by δ = ∂q+1, q ∈ Z.

For any f :(X,A) → (Y,B), f induces a chain map Q∗(f) = {Qq(f): Qq(X,A) → Qq(Y,B)}. Thus f induces the homomorphism

H∗(f) := {Hq(f): Hq(X,A) → Hq(Y,B) | q ∈ Z},

Hq(f)([c]) = [Qq(f)(c)], c ∈ Zq(X,A) := Zq(Q∗(X,A)), q ∈ Z.

It is clear that

H∗(f; G) = {Hq(f; G): Hq(X,A; G) → Hq(Y,B; G)} is deﬁned similarly. In the case of cohomology we deﬁne

∗ q q q H (f; G) := {H (f; G): H (Y,B; G) → H (X,A; G)}q∈Z

q ∗ where H (f; G), q ∈ Z, is given as the homomorphism induced by Qq(f) , i.e.,

q ∗ q q ∗ H (f; G)([u]) = [Qq(f) (u)], u ∈ Z (X,A; G) := Z (Q (X,A; G)).

It is sometimes convenient to describe R-modules Zq(X,A) and Bq(X,A), q > 0, in a bit diﬀerent way. (X,A) Let c ∈ Qq(X), then [c] ∈ Qq(X)/Qq(A). Suppose that [c] ∈ Zq(X,A), i.e., ∂q [c] = X X [∂q (c)] = 0 which means that ∂q (c) ∈ Qq−1(A). Putting X Zq(Xmod A) := {c ∈ Qq(X) | ∂q (c) ∈ Qq−1(A)} and observing that Zq(Xmod A) has the structure of an R-module, we may write ∼ Zq(X,A) = Zq(Xmod A)/Qq(A). 3.2. General homology and cohomology theories 105

Elements of Zq(Xmod A) are called q-dimensional cycles modulo A. (X,A) Similarly let [c] ∈ Bq(X,A), i.e., [c] ∈ Im ∂q+1 . Therefore there is a chain d ∈ Qq+1(X) such that (X,A) X [c] = ∂q+1 [d] = [∂q+1(d)]. Hence X c − ∂q+1(d) ∈ Qq(A). Putting

X Bq(Xmod A) := {c ∈ Qq(X) | ∃ d ∈ Qq+1(X) c − ∂q+1(d) ∈ Qq(A)} = Bq(X) + Qq(A), where the las expression denotes the smallest submodule in Qq(X) that contains Bq(X) and Qn(A), we see that ∼ Bq(X,A) = Bq(Xmod A)/Qn(A) = Bq(X) + Qq(A)/Qn(A) = Bq(X)/Bq(X) ∩ Qn(A).

The last equality follows from the following Noether theorem on an isomorphisms: If M and N are submodules of some module, then the inclusion M → M + N induces an isomorphism M/M ∩ N =∼ M + N/N.

It is immediate to see that Bq(Xmod A) ⊂ Zq(Xmod A). This implies that ∼ Hq(X,A) = Zq(Xmod A)/Bq(Xmod A).

Now we shall verify all the axioms for homology:

3.2.6 Dimension axiom: By the very deﬁnition we see that for a one-point space Qq(P ; G) = {0} for q 6= 0 and Q0(P ; G) = R ⊗ G = G. Indeed, for q > 1, there is only one singular q-cube in P and it is degenerate. For q = 0, Zq(P ) = Q0(P ) = Q¯0(P ) = R and B0(P ) = 0. Thus Hq(P ; G) = 0 for q 6= 0 and H0(P ; G) = G. ¯ ¯ 3.2.7 Remark As mentioned above {Qq(X), ∂q}q∈Z is chain complex. Therefore this is possible to deﬁne ¯ ¯ Hq(X) := Hq(Q∗(X)), q ∈ Z. ¯ ¯ ¯ ¯ Let us compute Hq(P ) for q ∈ Z. It is clear that Zq(P ) := Zq(Q∗(P )) = R and Bq(P ) := ¯ ¯ ¯ ¯ Bq(Qq(P )) = 0; hence Hq(P ) = 0 for q < 0 and Hq(P ) = R for all q > 0. Therefore H∗ does not satisfy the dimension axiom.

3.2.8 Exactness axiom: We prove something more general. Let (X, A, B) be a topological triple, i.e., B ⊂ A ⊂ X. Consider a sequence

j (A, B) −→i (X,B) −→ (X,A), where i and j are the respective inclusions. One checks that the sequence of chain complexes and chain maps Q∗(i) Q∗(j) 0 −→ Q∗(A, B) −→ Q∗(X,B) −→ Q∗(X,A) −→ 0 is exact. Since Q∗(X,A) is a free complex, this sequence is split. Therefore also the sequences

0 −→ Q∗(A, B; G) −→ Q∗(X,B; G) −→ Q∗(X,A; G) −→ 0 106 3. General Constructions of (Co)homology of complexes and 0 −→ Q∗(X,A) −→ Q∗(X,B; G) −→ Q∗(A, B; G) −→ 0 are exact and split. By the general construction (involving the existence of connecting homomorphisms) we have long homology and long cohomology exact sequences

Hq(i;G) Hq(j;G) ∂∗ ... −→ Hq(A, B; G) −→ Hq(X,B; G) −→ Hq(X,A) −→ Hq−1(A, B; G) −→ ... and

Hq(j;G) Hq(i;G) ∗ ... −→ Hq(X,A; G) −→ Hq(X,B; G) −→ Hq(A, B) −→δ Hq+1(X,A; G) −→ ...

Putting B = ∅ we the usual long homology sequence of the pair (X,A)

∂∗ ... −→ Hq(A; G) −→ Hq(X; G) −→ Hq(X,A; G) −→ Hq−1(A; G) −→ ...

It is easy to see that ∂∗ : Hq(X,A) → Hq−1(A, B) is the composition

∂∗ Hq−1(k;G) Hq(X,A; G) −→ Hq−1(A; G) −→ Hq−1(A, B; G) where k : A → (A, B) is the inclusion. 0 00 Of course δ∗ is functorial, i.e., given f :(X,A) → (Y,B), let f : X → Y and f : A → B be induced (in an obvious manner) by f. Then we have a diagram of chain complexes and chain maps 0 Q∗(i) Q∗(f ) 0 / Q∗(A) / Q∗(X) / Q∗(X,A) / 0

00 Q∗(f ) Q∗(f) Q∗(f) 0 / Q∗(B) / Q∗(Y ) / Q∗(Y,B) / 0 Q∗(i) Q∗(j) having exact sequences. Hence by the theorem on the existence of the connecting homomorphism, the following diagram commutes

∂∗ Hq(X,A; G) / Hq−1(A; G)

00 Q∗(f) Q∗(f ) Hq(Y,B; G) / Hq−1(B; G). ∂∗

3.2.9 Homotopy axiom If f ' g :(X,A) → (Y,B), then

H∗(f; G) = H∗(g; G): H∗(X,A); G) → H∗(Y,B; G).

Let jk :(X,A) → (X × I,A × I) be given by jk(x) = (x, k), x ∈ X. We shall deﬁne an q+1 operator P : Qq(X,A) → Qq+1(X × I,A × I). For any σ ∈ Σq(X), let σ × 1I : I → X be given by q (σ × 1I )(t1, ..., tq, t) = (σ(t1, ..., tq), t), (t1, ..., tq) ∈ I , t ∈ I. Let n ! n n X q+1 X X Pq(c) = Pq aiσi := (−1) ai(σi × 1I ), c = aiσi ∈ Q¯q(X). i=1 i=1 i=1 3.2. General homology and cohomology theories 107

This deﬁnition is correct and Pq(Dq(X)) ⊂ Dq+1(X × I). Hence

Pq : Qq(X) → Qq+1(X × I).

Moreover clearly Pq(Qq(A)) ⊂ Qq+1(A), i.e.,

Pq : Qq(X,A) → Qq+1(X × I,A × I).

For any σ ∈ Σq(X),

q+1 X i ∂q+1Pq(σ) = (−1) [AiPq(σ) − BiPq(σ)] = i=1 q ! X i q+1 q+1 −Pq−1 (−1) (Aiσ − Biσ) + (−1) (−1) (σ × 0 − σ × 1) = i=1 −Pq−1(∂q(σ)) + j0 ◦ σ − j0 ◦ σ.

In this way we see that chain maps Q∗(j0) and Q∗(j1) are chain homotopic. Therefore

H∗(j0) = H∗(j1).

If h : X × I → Y is a homotopy joining f to g, then h ◦ j0 = fand h ◦ j1 = g. Hence

H∗(f) = H∗(h) ◦ H∗(j0) = H∗(h) ◦ H∗(j1) = H∗(g).

3.2.10 Excision axiom: Given a pair (X,A) and a set W such that cl W ⊂ int A, the inclusion i :(X \ W, A \ W ) → (X,A) induced an isomorphism

H∗(i): H∗(X \ W, A \ W ) → H∗(X,A).

The proof is quite complicated. S Let U = {Uλ}λ∈Λ be a family of sets such that λ∈Λ int Uλ = X. We say that σ ∈ Σq(X) is q U-small if there is λ ∈ Λ such that the carrier σ(I ) ⊂ Uλ. ¯ ¯ Let Qq(X, U), q > 0, denotes the submodule of Qq(X) generated by singular q-cubes that are U-small. Similarly Dq(X, U) := Q¯q(X, U) ∩ Dq(X) and

Qq(X, U) = Q¯q(X, U)/Dq(X, U).

Exactly the same if A ⊂ X, i.e., we deﬁne Q(A, U) and, ﬁnally

Qq(X, A, U) := Qq(X, U)/Qq(A, U), q > 0. It is easy to see that ∂q(Qq(X, A, U)) ⊂ Qq−1(X, A, U). Therefore we have deﬁned a chain complex

Q∗(X, A, U) := {Qq(X, A, U), ∂q}q∈Z. This, as usual, gives possibility to deﬁne

Hq(X, A, U) := Hq(Q∗(X, A, U), q ∈ Z. 108 3. General Constructions of (Co)homology of complexes

If q = 0, then Q0(X, A, U) = Q0(X,A) and Z0(Q∗(X, A, U)) = Z0(X,A) = Q0(X,A). Thus

H0(X, A, U) = Q0(X,A)/B0(X, A, U), where Bq(X, A, U) := Bq(Q∗(X, A, U)), q ∈ Z.

It is clear that the inclusion Q¯(X, U) ⊂ Q¯q(X) induced a homomorphism

ϕq : Qq(X, A, U) → Qq(X,A), q ∈ Z.

Moreover ϕq−1 ◦ ∂q = ∂q ◦ ϕq for all q ∈ Z; thus ϕ := {ϕq}q∈Z is a chain homomorphism of the complex Q∗(X, A, U) to the complex Q∗(X,A). This map induces

H∗(ϕ): H∗(X, A, U) → H∗(X,A).

3.2.11 Theorem: The induced homomorphism H∗(ϕ) is an isomorphism, i.e., Hq(ϕq): Hq(X, A, U) → Hq(X,A) is an isomorphism for any q ∈ Z. The proof of this theorem will postponed for a while Having the above theorem we may prove the excision axiom for the singular homology. We shall threat a bit more general situation. Suppose that we have two sets A, B ⊂ X such that X = int A ∪ int B (then of course X = A ∪ B). We are going to show that the inclusion (B,A ∩ B) ,→ (A ∪ B,A) induces an isomorphism in homology. Observe that if cl W ⊂ int A, then putting B := X \ U, we have

int A ∪ int B = X since int B = int (X \ W ) = X \ cl W ⊃ X \ int A. Moreover A ∪ B = X, A ∩ B = A \ W . Hence the theorem asserts that ∼ H∗(X \ W, A \ W ) = H∗(X,A).

In what follows we are going to say that a pair {A, B} is excisible if the inclusion (B,A∩B) → ∼ (A ∪ B,A) induces an isomorphism H∗(A, A ∩ B) = H∗(A ∪ B,B). The above theorem asserts actually that a pair {A, B} such that A ∪ B = int A ∪ int B is excisible. Let U := {A, B}. According to our assumption U satisﬁes hypotheses of the theorem; therefore ∼ H∗(ϕ): H∗(A ∪ B,B, U) = H∗(A ∪ B,B). It is suﬃcient to show that

∼ H∗(A ∪ B,B, U) = H∗(A, A ∩ B).

Consider the following commutative diagram

1 Qq(A, A ∩ B) / Qq(A ∪ B,B) RR O RRR RRR ϕq 2 RRR RR( Qq(A ∪ B,B, U) 3.2. General homology and cohomology theories 109 where unmarked arrow are induced by inclusions. After passing to homology we have the commutative diagram 1 Hq(A, A ∩ B) / Hq(A ∪ B,B) RRR O RR ∼ RRR = Hq(ϕ∗) 2 RRR RR) Hq(A ∪ B,B, U). Thus we are to show that the arrow marked 2 is an isomorphism. To this end we have to study the homomorphism marked 2 in the ﬁrst diagram. By deﬁnition

Qq(A, A ∩ B) = Qn(A)/Qn(A ∩ B) = Qn(A)/[Qq(A) ∩ Qq(B)].

By the Noether theorem

∼ Qq(A)/[Qq(A) ∩ Qq(B)] = [Qq(A) + Qq(B)]/Qq(A).

But observe that Qq(A ∪ B, U) = Qq(A) + Qq(B)

(this is not a direct sum) and Qq(A) = Qq(A, U). This means that ∼ Qq(A, A ∩ B) = Qq(A ∪ B, U)/Qq(A, U).

Since this chain map consists of isomorphisms, the induced map

∼ Hq(A, A ∩ B) = Hq(A ∩ B,B, U) as required. Sketch of the proof of theorem: 1. The ﬁrst step is to introduce the following subdivision operator sd q : Qq(X) → Qq(X), q > 1. q By Vq we denote the set of all vertices of I , i.e., point of the form v = (v1, ..., vq), where vi = 0 or 1 for each i = 1, ..., q. For σ ∈ Σq(X) we deﬁne

X 1 sd (σ)(t) := σ (t + v) , t ∈ Iq. 1 2 v∈V

Hence, to each singular q-cube in X, sd q assigns the chain. This map extends by linearity to the map sd q : Q¯q(X) → Q¯q(X). If q = 0, then sd 0(σ) = σ, σ ∈ Σ0(X).

Properties of sd q: (1) sd q(Dq(X)) ⊂ Dq(X); thus

sd q : Qq(X) → Qq(X) is properly deﬁned;

(2) sd ∗ = {sd q}q∈Z is a chain homomorphism, i.e., ∂q ◦ sd q = sd q−1 ◦ ∂q for q > 1. (3) sd q preserves augumentation. n (4) By sd q we denote the n-th iteration of \q. For any chain c ∈ Qq(X), there is n = n(c) such that n sd q (c) ∈ Qq(X, U). 110 3. General Constructions of (Co)homology of complexes

To see this assume that m X c = aiσi. i=1

ni If we prove that for each σi, i = 1, ..., m, there is ni > 1 such that sd q (σi) ∈ Qq(X, U), then we are done by taking n = max{ni | i = 1, ..., n}. q Therefore let σ ∈ Σq(X). The carrier σ(I ) is compact and covered by {int Uλ}λ∈Λ. Let L be the Lebesgue coeﬃcient of this cover. It is easily seen that there is a number ε > 0 such that, for any K ⊂ Iq with diameter less then ε is mapped by σ onto the set with diameter less that L. Now we see that if n is suﬃciently large, then cubes of the n-th subdivision of Iq have diameter less than ε.

2. The second step is to show that the identity on Qq(X) and sd q are chain homotopic, i.e., there is a chain homotopy Φ∗ = {Φq}q∈Z, where Φq : Qq(X) → Qq+1(X) such that

sd q(c) − c = ∂q+1Φq(c) + Φq−1∂q(c), c ∈ Qq(X).

2 To deﬁne Φq, we put Φ0 = 0. Let q > 0. We deﬁne two auxiliary maps η0, η1 : I → I by the formulae t1 η0(t1, t2) := ; 2 − t2

t1+1 2−t when t1 + t2 6 1; η1(t1, t2) = 2 1 when t1 + t1 > 1. 2 for each t = (t1, t2) ∈ I . Let, for σ ∈ Σq(X) q+1 X Φq(σ) := (−1) σv, v∈V where

q+1 σv(t1, ..., tq+1) := σ(ηv1 (x1, xq+1), ..., ηvq (tq, tq+1)), t = (t1, ..., tq+1) ∈ I .

Therefore, by linearity, we have deﬁned Φq : Q¯q(X) → Q¯q+1(X). Observe that if σ is degenerate, then so is σv for any v ∈ V ; thus Φq(Dq(X)) ⊂ Dq+1(X). Therefore Φq : Qq(X) → Qq+1(X).

Now we show that {Φq}q∈Z is a chain homotopy between the identity and sd q. It is suﬃcient to show that, for any σ ∈ Σq(X),

∂q+1Φq(σ) = sd q(σ) − σ − Φq−1∂q(σ) + degenerate cubes.

Moreover it is easy to see that if c ∈ Qn(X, U), then Φq(c) ∈ Qq+1(X, U). 3. In the next step we deﬁne

n Ψq : Qq(X) → Qq+1(X), q > 0, n ∈ N, by the following formula: n−1 X n Ψq(c) = sd q Φq(c), c ∈ Qq(X). i=0 3.2. General homology and cohomology theories 111

n n By step 2, for any n ∈ N, {Ψq }q∈Z is a chain homotopy between sd q and the identity on Qq(X), i.e., n n n sd q (c) − c = ∂q+1Ψq + Ψq−1∂q, c ∈ Qq(X).

We have the following property:

n • If c ∈ Qq(X, U), then Ψq (c) ∈ Qq+1(X, U).

4. Now we are ready to show that H∗(ϕ∗): H∗(X, U) → H∗(X) is an isomorphism. First we show that it is an epimorphism. Let [c] ∈ Hq(X), i.e., c ∈ Zq(X). We shall show that there is n d ∈ Zq(X, U) such that Hq(ϕ∗)([d]) = [ϕq(d)] = [c]. There is n ∈ N such that sd q (c) ∈ Qq(X, U). n n Since sd q is a chain map (i.e., it commutes with the boundary operator), sd q (c) ∈ Zq(X, U). n But d := sd q and c belong to the same homology class. Therefore [c] = [ϕq(d)] as desired. To show that Hq(ϕ∗) is a monomorphism, assume that [c] ∈ Hq(X, U), where c ∈ Zq(X, U), and Hq(ϕ∗)([c]) = [ϕq(c)] = 0. We shall show that [c] =, i.e., c ∈ Bq(X, U). Since [ϕq(c)] = 0, there is u ∈ Qq+1(X) such that ∂q+1(u) = ϕq(c). Moreover there is n ∈ N such that n sd q+1(u) ∈ Qq+1(X, U). By the homotopy property

n n n n n sd q+1(u) − u = ∂q+2Ψq+1(u) + Ψq ∂q+1(u) = ∂q+2Ψq+1(u) + Ψq ϕq(c).

Applying ∂q+1 to both sides of this equality we get

n n ∂q+1(sd q+1(u)) − ϕq(c) = ∂q+1Ψq (ϕq(c)) or n n ϕq(c) = ∂q+1(sd q+1(u) − Ψq (ϕq(c))). n n n Since ϕq(c) ∈ Qq(X), Ψq (ϕq(c)) ∈ Qq+1(X, U). Thus sd q+1(u) − Ψq (ϕq(c)) ∈ Cq+1(X, U) and ϕq(c) ∈ Bq+1(X, U). This completes the proof for the absolute case A = ∅. In general case we have the commutative diagram of chain complexes and chain maps

0 / Q∗(A, U) / Q∗(X, U) / Q∗(X, A, U) / 0

00 0 ϕ∗ ϕ∗ ϕ∗ 0 / Q∗(A) / Q∗(X) / Q∗(X,A) / 0

00 0 where ϕ∗ is deﬁned as before with regard to A and ϕ∗ is the original homomorphism under question. Applying homology we get the commutative „ladder” of the form

∂∗ ... / Hq(A, U) / Hq(X, U) / Hq(X, A, U) / Hq−1(A, U) / Hq−1(X, U) // ...

00 00 Hq(ϕ∗ ) Hq(ϕ∗) Hq(ϕ∗) Hq−1(ϕ∗ ) Hq−1(ϕ∗) ... / Hq(A) / Hq(X) / Hq(X,A) / Hq−1(A) / Hq−1(X) / ... ∂∗

By the ﬁve lemma we complete the proof. 112 3. General Constructions of (Co)homology of complexes

3.2.B Reduced singular homology theory

Suppose that C∗ = {Cq, ∂q} is a nonnegative chain complex of R-modules, i.e., Cq = 0 for q < 0. By an augumentation (over R) we mean an epimorphism ε : C0 → R such that Im ∂1 ⊂ Ker ε and by an augumented chain complex we mean the pair (C∗, ε), where C∗ is a nonnegative chain complex and ε : C0 → R is an augumentation. Augumented (nonnegative) chain complexes form a category. Morphisms of this category consists of chain maps that preserve augumentations.

If (C∗, ε) is an augumented chain complex, then

epi ∼ H0(C∗) = Ker ∂0/Im ∂1 = C0/Im ∂1 −→ C0/Ker ε = R.

Hence H0(C∗) 6= 0. By the reduced chain complex Ce∗ we understand the complex

∂q ∂q−1 ∂2 ∂1 0 Ce∗ : ... −→ Ceq −→ Ceq−1 −→ ... −→ Ce1 −→ Ce0 −→ 0 where Ceq := Cq for q > 1, Ce0 := Ker ε. We write He∗(C∗) = H∗(Ce∗).

3.2.12 Lemma: The following equalities hold: ( Heq(C∗) if q > 1; Hq(C∗) = He0(C∗) ⊕ R if q = 0.

ε Proof: Since the sequence 0 −→ Ker ε −→ C0 −→ R −→ 0 is exact and split (the last ∼ ∼ module is free), we have C0 = Ker ε ⊕ R. Therefore Z0(C∗) = Ker ∂0 = C0 = Ce0 ⊕ R and H0(C∗) = C0/Im ∂1 = He0(C∗) ⊕ R.

The singular complex Q∗(X), where X is a topological space, admits an augumentation: let ∼ P be any one-point space and let η : X → P . Then ε := Q0(η): Q0(X) → Q0(P ) = R is an augumentation. As usual He∗(X) := H∗(Qe∗(X)).

By the above lemma ( Heq(X) if q > 1; Hq(X) = He0(X) ⊕ R if q = 0.

Consider the pair (X,P ) and its homology sequence

∂∗ H0(γ) ... −→ H0(P ) −→ H0(X) −→ H0(X,P ) −→ 0, where γ : P → X is the inclusion. Obviously η ◦ γ = 1P , i.e. H0(γ) is the right inverse to H0(η). In particular H0(γ) is a monomorphism. This means that the sequence

H0(γ) 0 −→ H0(P ) −→ H0(X) −→ H0(X,P ) −→ 0 3.2. General homology and cohomology theories 113

∼ is exact and split. Thus H0(X) = H0(P ) ⊕ H0(X,P ). This implies that ∼ He0(X) = H0(X,P ).

As a corollary we see that for any topological pair, the reduced homology sequence

∂∗ ∂∗ ... −→ Heq(A) −→ Heq(X) −→ Hq(X,A) −→ Heq−1(A) −→ .. is exact.

3.2.C Eilenberg-MacLane theorem

0 0 We say that to chain complexes C∗, C∗ are chain equivalent if there are chain maps ϕ∗ : C∗ → C∗ 0 and ψ∗ : C∗ → C∗ such that the compositions ϕ∗ ◦ψ∗ and ψ∗ ◦ϕ∗ are chain homotopic to identities 0 idC∗ and idC∗ , respectively. The chain maps ϕ∗ and ψ∗ are called chain equivalences. Assume that K is a category and a family M of objects, the so-called models, in K is distinguished. Let F be a functor from K to the category of nonnegative augumented chain complexes; therefore, for each q ∈ Z, there is a functor Fq form the category K to the category of R-modules: if, for K ∈ K, F (K) = C∗, then Fq(K) = Cq, q ∈ Z. We say that F is a free functor if, for each q ∈ Z, has a base with respect o M, i.e., there exists a family {dj ∈ Fq(MJ )}j∈J , where {Mj} is a subfamily in M such that, for any K ∈ K, the module Fq(K) is generated by elements of the form Fq(uj)(dj), j ∈ J, where uj : Mj → K is a morphism in K. For example consider the category T op of topological spaces and the functor Q that assigns q to any space X, the singular chain complex Q∗(X). Let M consists of all cubes I , q > 0. Fix q q > 0. By 1q we denote the chain in Qq(Iq) determined by the identity on I . Observe that, q for any space X, Qq(X) is generated by elements of the form Qq(σ)(1q), where σ : I → X is a nondegenerate singular q-cube. Thus the functor Q is free.

We say that the functor is acyclic if, for any q > 0, Heq(F (M)) = 0 for any M ∈ M. We shall prove that in the above example the functor Q is acyclic.

n 3.2.13 Lemma: Suppose that X is a starshaped subset of R , then Heq(X) = 0 for any q > 0. Proof: Without loss of generality we may assume that X is starshaped around 0.

Let P = {Pq : Qq(X) → Qq+1(X)}q∈Z be deﬁned as follows, for σ ∈ Σq(X), let Pq(σ) ∈ Q¯q+1(X) be given by

q+1 q Pq(σ)(t1, .., tq+1) = (−1) (1 − tq+1)σ(t1, ..., tq), (t1, ..., tq) ∈ I .

Then P extends linearly onto Q¯∗(X) → Q¯∗+1(X), i.e.,

m ! m X X Pq aiσi = aiPq(σi). i=1 i=1

It is easy to see that Pq(Dq(X)) ⊂ Dq+1(X), i.e., Pq : Q∗(X) → Q∗+1(X). Let us compute 114 3. General Constructions of (Co)homology of complexes

∂q+1Pq(σ) where σ ∈ Σq(X). We have

q+1 X i ∂q+1Pq(σ) = (−1) [AiPq(σ) − BiPq(σ)] = i=1 q X q q+1 q+1 (−1) [AiPq(σ) − BiPq(σ)] + (−1) (−1) (σ − 0) = i=1 ! X i −Pq−1 (−1) [Aiσ − Biσ] + σ − 0, i=1 where 0 means the singular q-cube that maps Iq → 0. Hence

∂q+1Pq(σ) + Pq−1∂qσ = σ − 0.

The above equality holds if we replace σ ∈ Σq(X) by any chain c ∈ Qq(X), i.e., for q > 1

∂q+1Pq(c) + Pq−1∂q(c) = c

(observe that the last term vanishes since this is an element of Dq(X)), while for q = 0 and Pn 0 Pm c = i=1 aiσi, where σi : I → X, ∂1P0(c) = c − i=1 ai0. If q > 1 and z ∈ Zq(X), then z = ∂q+1Pq(z), i.e., z ∈ Bq(X). This means that Hq(X) = 0 for any q > 1. Let q = 0, then Z0(X) = Q0(X) and we see that H0(X) = R. 3.2.14 Theorem (Eilenberg-MacLane): Let (K, M) be a category with models. Assume that F,G are functors from K to the category of nonnegative augumented chain complexes. Moreover suppose that F is free and G is acyclic. Then there exists a unique (up to the chain homotopy) natural transformation Φ of F into G. This means that if Φ0 : F → G is another natural transformation, then for any object X 0 of K, the chain maps ΦX : F (X) → G(X) nad ΦX : F (X) → G(X) are chain homotopic. X Moreover this chain homotopy is natural. Recall that a chain homotopy between ΦX = {ϕq : 0 X Fq(X) → Gq(X)}q∈Z and ΦX = {ψq : Fq(X) → Gq(X)}q∈Z is a sequence of homomorphisms X {Dq : Fq(X) → Gq+1(X)}q∈Z such that

X X X X ∂q+1 ◦ Dq + Dq−1 ◦ ∂q = ϕq − ψq for any q ∈ Z. The naturality of this chain homotopy means that given another object Y in K and a morphism α : X → Y , a diagram

X Dq Fq(X) / Gq+1(X)

Fq(α) Gq+1(α) F (Y ) G (Y ) q Y / q1 Dq is commutative.

3.2.15 Corollary: Under the assumptions of the theorem, suppose additionally that both functors are free and acyclic. Then functors F and G are chain equivalent in a natural way; in fact any natural transformation of F into G is a natural chain equivalence. Indeed suppose that Φ: F → G is a natural transformation (it exists in view of the theorem). There also exists a natural transformation Φ0 : G → F . Since Φ ◦ Φ0 is a natural transformation 3.2. General homology and cohomology theories 115 of G → G and Φ0 ◦ Φ is a natural transformation of F to F , for each object X form K, there 0 0 must exist chain homotopies that join ΦX ◦ ΦX to the identity on G(X) and ΦX ◦ ΦX to the identity on F (X).

X X 3.2.16 Theorem: Let X be a space and jk : X → X × I be given by jk (x) = (x, k), k = 0, 1. X X Then H∗(j0 ) = H∗(j1 ). Proof: The functors F , G from the category T op to the category of nonnegative augumented chain complexes given by F (X) = Q∗(X), G(X) = Q∗(X × I) are free and acyclic with respect q 0 to models {I }q>0. Let Φ and Φ be the natural transformations of F to G deﬁned as follows: X 0 X for any space X, ΦX = {Qq(h0 )}q∈Z and ΦX = {Qq(j1 )}. By the above corollary, there is a chain homotopy joining ΦX to ΦX . Thus H∗(j0) = H∗(j1). Thus we see that the Eilenberg-MacLane theorem provides another proof of the homotopy axiom.

3.2.D Various results and consequences of axioms

λ Given a family {M }λ∈Λ of R-modules the direct product ( ) Y λ [ λ λ M := x :Λ → M | xλ = x(λ) ∈ M . λ∈Λ λ∈λ

Q λ Elements of λ∈λ M are denoted by (xλ)λ∈Λ. The direct sum ( ) M λ Y λ M := x = (xλ) ∈ M | xλ = 0 for all but ﬁnite λ ∈ Λ . λ∈Λ λ∈Λ These are R-modules with naturally deﬁned arithmetic operations. λ λ If {ϕλ : M → N }λ∈Λ is a family of homomorphisms, then

M M λ M λ ϕλ : M → N λ∈Λ λ∈Λ λ∈Λ and Y Y λ Y λ ϕλ : M → N λ∈λ λ∈Λ λ∈λ are deﬁned by (xλ) 7→ (ϕλ(xλ) are correctly deﬁned. L λ λ λ λ If Cq = λ∈Λ Cq , q ∈ Z, where C∗ = {Cq , ∂q }q∈Z is a chain complex, then C∗ = {Cq, ∂∗}q∈Z λ λ is a chain complex provided, for q ∈ Z, ∂q ((cλ)λ∈Λ) = ∂q (cλ) where cλ ∈ Cq , λ ∈ Λ. Moreover M λ Hq(C∗) = Hq(C∗ ). λ∈Λ λ L λ Indeed if z = (zλ) ∈ Zq(C∗), then zλ ∈ Zq(C∗ ), i.e., Zq(C∗) = λ∈Λ Zq(C∗ ). In a similar way L λ we show that Bq(C∗) = λ∈Λ Bq(C∗ ). Thus , M λ M λ M λ Hq(C∗) = Zq(C∗ ) Bq(C∗ ) = Hq(C∗ ). λ∈Λ λ∈Λ λ∈Λ 116 3. General Constructions of (Co)homology of complexes

Exactly the same result holds for the direct product of chain complexes. Observe also that ! M Y Hom M λ,B = Hom(M λ,B) λ∈Λ λ∈Λ for any R-module B. This implies that, for all q ∈ Z, ! q M λ Y q λ H C∗ = H (C∗ ). λ∈Λ λ∈Λ

3.2.17 Theorem: If X is an arcwise connected space and P is a one-point space, then γ : P → X induces an isomorphism H0(P ) → H0(X); in particular H0(X) = R and He0(X) = 0 . γ Proof: Let η : X → P be the constant map; then η◦γ = 1P . Consider maps P −→ X → (X,P ) and the exact homology sequence for q = 0 we have

∂∗ H0(γ) H1(X,P ) −→ H0(P ) −→ H0(X) −→ H0(X,P ) −→ 0.

Since H0(η)H0(γ) = id, H0(γ) is a monomorphism. Thus ∂∗ is trivial and we have

H0(γ) 0 → R −→ H0(X) −→ H0(X,P ) −→ 0.

We shall show that H0(γ) is an epimorphism. Let σx be a path joining x to P , i.e., σx : I → X is a singular 1-cube, σx(0) = P , σx(1) = x. Clearly ∂1σx = x − P . Take any cycle Pn z ∈ Z0(X) = Q0(X), i.e. z = i=1 aixi, ai ∈ R, xi ∈ X, i = 1, ..., n. Then

" n # " n # X X H0(γ) aiP = aiP ∈ H0(X). i=1 i=1 But n n n n ! X X X X aixi − aiP = ai(xi − P ) = ∂1 aiσxi . i=1 i=1 i=1 i=1 Pn P This means that [z] = [ i=1 aiP ] in H0(X), i.e., H0(γ)[ i=1 aiP ] = [z]. W 3.2.18 Theorem: The singular homology is additive, i.e., if X = λ∈Λ Xλ, then

M q Y q Hq(X) = Hq(Xλ),H (X) = H (X), q ∈ Z. λ∈Λ λ∈Λ

S Proof: It is easy to see that Σq(X) = λ∈Λ Σq(Xλ); therefore M Qq(X) = Qq(Xλ), λ L i.e., Q∗(X) = λ∈Λ Q∗(Xλ). Now the assertion follows from the above arguments.

In a similar way one proves that if A ⊂ X and Aλ := Xλ ∩ A, then M Hq(X,A) = Hq(Xλ,Aλ), q ∈ Z. λ∈Λ 3.2. General homology and cohomology theories 117

3.2.19 Corollary: If Λ is the set of arcwise components of a space X, then

Λ H0(X) = RΛ,H0(X) = R .

Now we are going to establish some direct consequences of the axioms of homology. We assume that a homology theory (h∗, ∂∗) with coeﬃcients in am R-module is given. • (Homology sequence of a triple) If (X, A, B) is a topological triple (i.e., B ⊂ A ⊂ X), then the following sequence

∆∗ ∆∗ ... −→ hq+1(X,A) −→ hq(A, B) −→ hq(X,B) −→ hq(X,A) −→ hq−1(A, B) −→ ..., where the unmarked arrows are induced by the respective inclusions and ∆∗ = h∗(i) ◦ ∂∗ and i : A → (A, B) is the inclusion, is exact. This follows immediately from the exactness homology of a pair (X,A) and the commutativity of a diagram

∂∗ hq+1(X,A) / hq(A) / hq(X) / hq(X,A) qq8 h (i) qq ∗ qqq qqq hq(A, B) / hq(X,B) in which the upper row is exact.

• h∗(∅) = 0; h∗(X,X) = 0 for any space. This follows form the exactness of the homology sequence of the pair (X, ∅), where X is an arbitrary space. The second part follows from the exactness of the homology sequence of the pair (X,X). • Assume that r : X → A is a retraction, i : A → X is the inclusion. Then

(h∗(r), h∗(i)) : h∗(X) → h∗(A) ⊕ h∗(X,A) is an isomorphism. Consider the homology sequence of the pair (X,A)

∂∗ h∗(i) ∂∗ ... −→ hq+1(X,A) −→ hq(A) −→ hq(X) −→ hq(X,A) −→ hq−1(A) −→ ...

Observe that r ◦i = 1A, i.e., h∗(r)h∗(i) = 1h∗(A). Thus h∗(i) is a monomorphisms and, therefore, Im ∂∗ = 0. This implies that we have the following exact sequence

h∗(i) 0 −→ hq(A) −→ hq(X) −→ hq(X,A) −→ 0.

This sequence is split because h∗(r) is the left inverse of h∗(i). This completes the proof. ∼ • If pairs (X,A) and (Y,B) have the same homotopy type, then h∗(X,A) = h∗(Y,B).

• If r : X → A is a deformation retraction (i.e. r ◦ i = 1A, where i : A → X is the inclusion, and i ◦ r ' 1X ), then h∗(i): h∗(A) → h∗(X) is an isomorphism.

It is easy to see that h∗(r)◦h∗(i) = 1h∗(A) and, by the homotopy axiom, h∗(i)h∗(r) = 1h∗(X). This means that h∗(i) is an isomorphism. ∼ If B ⊂ A is a deformation retract, A ⊂ X, then h∗(X,B) = h∗(X,A) is an isomorphism 118 3. General Constructions of (Co)homology of complexes

(induced by the inclusion). This follows form the exactness of the homology sequences and the ﬁve lemma:

∂∗ h∗(B) / hq(X) / hq(X,B) / hq−1(B) / hq−1(X)

h∗(i) =∼ = h∗(i) =∼ = h∗(A) / hq(X) / hq(X,A) / hq−1(A) / hq−1(X) ∂∗

• If A ⊂ X and there is B ⊂ A such that B is a deformation retract of A and X, then h∗(X,A) = 0. Indeed, the homology sequences of pairs (A, B) and (X,B) shows that h∗(A, B) = 0 = h∗(X,B). The assertion follows now from the exactness of the homology sequence of the triple (X, A, B). Now we shall discuss some aspects of the excision axiom.

3.2.20 Theorem: If A ⊂ X is closed and there exists an open neighborhood N of A such that the boundary ∂A is a strong deformation retract of N \ int A (1), then for any U ⊂ int A (in particular this holds if U ⊂ A and U is open), the inclusion i :(X \ U, A \ U) → (X,A) induces an isomorphism h∗(X \ U, A \ U) → h∗(X,A).

Proof: In the following diagram

=∼ h∗(X \ U, N \ U) o h∗(X \ U, A \ U) i4 =∼ iii iiii iiii iiii h∗(X \ int A, N \ int A) UUUU UUUU ∼UUU = UUUU U* =∼ h∗(X,N) o h∗(X,A) all arrow are induced by inclusions. The oblique arrows are isomorphisms in view of the excision axiom. The horizontal arrows are also isomorphisms because A \ U is a strong deformation retract of N \ A and A is a strong deformation retract of N. To see this retraction consider the strong deformation retraction r : N \ int A → ∂A and deﬁne R : N → A by

r(x) if x ∈ N \ int A, R(x) = x if x ∈ A,

It is clear that R is a (strong) deformation retraction of N onto A and R|N\U is a (strong) deformation retraction of N \ A onto A \ U. Hence all arrows are isomorphisms. 3.2.21 Remark: In practice we have A is a deformation retract of some neighborhood N of A. This is suﬃcient for the proof above.

n n+1 3.2.22 Corollary Let S , n > 0, denotes the unit sphere in R . The we have: (a) R ⊕ R when q = 0; h (S0) = q 0 otherwise.

1Recall that A ⊂ X is a strong deformation retract (of X) if there is a retraction r : X → A such that i ◦ r ' 1X (rel A). 3.2. General homology and cohomology theories 119

(b) If n > 1, then R when q = 0, n; h (Sn) = q 0 otherwise.

Proof: Part (a) is obvious and follows from Corollary 3.2.19. n+1 n n n (c) Let n > 1. We consider the triple (D ,S ,S+), where S± is the upper (resp. lower) (closed) hemisphere. We have the exact sequence

n+1 n n+1 n ∂∗ n n n+1 n ... −→ hq+1(D ,S+) −→ hq+1(D ,S ) −→ hq(S ,S+) −→ hq(D ,S+) −→ ...

n n+1 n+1 n n+1 n ∼ Since S+ is a deformation retract of D , we see that h∗(D ,S+) = 0 and hq+1(D ,S ) = n n hq(S ,S+). On the other hand, by the above excision theorem n n−1 n n n n ∼ n n hq(S−,S ) = hq(S \ int S+,S+ \ int S+) = hq(S ,S+).

homeo n ∼ n It is evident that S− = D . Thus, for any q ∈ Z, n+1 n ∼ n n−1 hq+1(D ,S ) = hq(D ,S ).

Therefore n n−1 ∼ 1 0 hq(D ,S ) = hq−n+1(D ,S ). For a triple (D1,S0, ∗), where ∗ is a point in S0, we have

0 1 1 0 0 1 ... −→ hs(S , ∗) −→ hs(D , ∗) −→ hs(D ,S ) −→ hs−1(S , ∗) −→ hs−1(D , ∗) −→ ... which gives 1 0 ... −→ hs(∗) −→ 0 −→ hs(D ,S ) −→ hs−1(∗) −→ 0 −→ ... Therefore R if s = 1 h (D1,S0) = s 0 otherwise by the dimension axiom. n (b) Let ∗ be an arbitrary point of S , n > 1. The exactness of the homology sequence of the triple (Dn+1,S,∗) shows that

n+1 n+1 n ∂∗ n n+1 0 = hq+1(D , ∗) −→ hq+1(D ,S ) −→ hq(S , ∗) −→ hq(D , ∗) = 0.

Therefore n ∼ n+1 n hq(S , ∗) = hq+1(D ,S ). n n n But ∗ is a retract of S , i.e., hq(S ) = hq(∗) ⊕ hq(S , ∗). This completes the proof. Much more diﬃcult are the following results

3.2.23 Corollary: If A, B ⊂ X are closed and A, B, A ∩ B ∈ ANR, then the dyad {A, B} is excisible, i.e., the inclusion i :(A, A ∩ B) → (A ∪ B,B) induces an isomorphism

h∗(i): h∗(A, A ∩ B) → h∗(A ∪ B,B). 120 3. General Constructions of (Co)homology of complexes

Proof: Without loss of generality we may assume that X = A ∪ B. Its is known that X is an ANR. Since B is a strong deformation of some of its neighborhoods, according to Theorem 3.2.20, for any open U, U ⊂ B, there is an excision isomorphism h∗(X \ U, B \ U) → h∗(X,B). Take U := X \ A; then U is open and U ⊂ B. Moreover A = X \ U and B \ U = A ∩ B.. 3.2.24 Corollary: If A, B are simplicial subcomplexes of a simplicial complex X, then the dyad {|A|, |B|} is excisible. Proof: for the proof it is enough to observe that |A|, |B| are ANR and |A| ∩ |B| – as the body of a subcomplex A ∩ B – is an ANR, too. . 3.2.25 Theorem: If a dyad {A, B} is excisible, then so is {B,A}. Proof: Let (without loss of generality) X = A ∪ B. Consider new sets: C = A ∩ B;

X0 = (A × {0}) ∪ (C × I) ∪ B × {1} ⊂ X × I; A0 = (A × {0}) ∪ C × I; B0 = (C × i) ∪ (B × {1}); C0 = C × I.

0 0 0 0 The projection p : X × I → X determines maps p :(X ,C ) → (X,C), p1 :(X ,A ) → (X,A), 0 0 0 0 0 0 p2 :(X ,B ) → (X,B), p3 :(A ,C ) → (A, C) and p4 :(B ,C ) → (B,C). Observe that p3 is 0 0 0 0 a homotopy inverse to (A, C) → (A ,C ) and p4 is a homotopy inverse to (B,C) → (B ,C ). Moreover int A0 ∪ int B0 = X0 (interiors taken with respect to X0); therefore i0 :(A0,C0) → (X0,B0) induces an isomorphism

0 0 0 0 0 h∗(i ): h∗(A ,C ) → h∗(X ,B ).

Consider a diagram 0 0 0 h∗(i ) 0 0 h∗(A ,C ) / h∗(X ,B ) =∼

=∼ h∗(p3) h∗(p2) h∗(A, C)h∗(i) / h∗(X,B).

It is commutative; hence h∗(p2) is an isomorphism if and only so is h∗(i). Now let us consider a diagram

0 0 ∂∗ 0 0 0 0 0 0 ∂∗ 0 0 hq+1(X ,B ) / hq(B ,C ) / hq(X ,C ) / hq(X ,B ) / hq−1(B ,C )

hq+1(p2) =∼ hq(p4) =∼ hq(p) =∼ hq(p2) =∼ h∗(p4) hq+1(X,B) / hq(B,C) / hq(X,C) / hq(X,B) / hq−1(B,C) ∂∗ ∂∗

0 0 The ﬁve lemma implies that h∗(p): h∗(X ,C ) → h∗(X,C) is an isomorphism if and only if so is 0 0 h∗(i). Arguing by changing places of A, B (and correspondingly A and B ) we see that h∗(p) is an isomorphism if and only if so is h∗(j) where j :(B,C) → (X,A). 3.2.26 Theorem: Assume that A ⊂ X is a coﬁbration. Then the projection p :(X,A) → (X/A, ∗) induces an isomorphisms

h∗(p): h∗(X,A) → h∗(X/A, ∗). 3.2. General homology and cohomology theories 121

Proof: Let CA = A × I/A × 1. Let i : A × 0 → X (inclusion). Then we have X ∪i CA. Lemma: (a) CA ⊂ X ∪ CA is a coﬁbration; (b) If A ⊂ X is a coﬁbration and A is contractible, then q : X → X/A is a homotopy equivalence; (c) If A ⊂ X, then there exists a homeomorphism ϕ : X ∪ CA/CA → X/A. Consider a diagram j (X,A) / (X ∪ CA, CA)

q q0 (X/A, ∗) ϕ / (X ∪ CA/CA, ∗) This diagram is commutative. The map q0 is a homotopy equivalence because CA is contractible. To show that h∗(q) is an isomorphism it is suﬃcient to show that h∗(j) is an isomorphism. To this end we shall show that the dyad (X,CA) is excisive. This is enough to show that X is a deformation retract of some of its neighborhoods (in X ∪ CA). Let Y = X ∪ {[t, a] | a ∈ A, t ∈ [0, 1/2)}. Then Y is a neighborhood of X in X ∪ CA. It is clear how can one deform Y to X (inside Y of course). Therefore X ∩ CA = A

h∗(j): h∗(X,A) → h∗(X ∪ CA, CA) is an isomorphism. We say that f :(X,A) → (Y,B) is a relative homeomorphism if f maps X \ A homeomorphically onto Y \ B.

3.2.27 Theorem: Suppose that A ⊂ X is a coﬁbration and f :(X.A) → (Y,B) a relative homeomorphism, where B is closed. Then

h∗(f): h∗(X,A) → h∗(Y,B) is an isomorphism. Proof: By lemma 1.3.21 There is a neighborhood N of A and a homotopy H : N × I → X such that H(x, 0) = x, H(a, t) = a and H(x, 1) ∈ A for x ∈ N, a ∈ A and t ∈ I. Moreover there −1 is a function ϕ : X → I such that A = ϕ (1) and ϕ|X\N ≡ 0. Let N 0 := f(N \ A); then N 0 ∪ B is a neighborhood of B. Deﬁne

y if y ∈ B, t ∈ I, H0(y, t) := f(H(f −1(y), t) if y ∈ y ∈ N 0 \ B, t ∈ I.

Then H0(y, 0) = y for y ∈ N 0, H0(y, t) = y if y ∈ B, t ∈ I and H0(y, 1) ∈ B for y ∈ N 0. Moreover let us deﬁne ϕ0 : Y → I by ϕ0(y) = ϕ(f −1(y)) if y ∈ Y \ B and ϕ0(y) = 1 if y ∈ B. Again by lemma 1.3.21, B ⊂ Y is a coﬁbration. Therefore ∼ ∼ ∼ h∗(X,A) = h∗(X/A, ∗) = h∗(Y,B) = h∗(Y,B), where the middle isomorphism is induced by the homeomorphism determined by f.

3.2.E Reduced homology

To any homology theory (h∗, ∂∗) there corresponds the reduced theory. Namely for a nonempty space X, let c : X → P be the map onto the one-point space P . For any q ∈ Z, let ehq(X) denote 122 3. General Constructions of (Co)homology of complexes

the kernel of hq(c). Since c has a right inverse, so does hq(c). Therefore

hq(X) = ehq(X) ⊕ hq(P ), i.e., eh0(X) ⊕ G if q = 0, hq(X) = hq(X) if q 6= 0.

Let x0 be an arbitrary point in X. By the exactness axiom, the following sequence

hq+1(i) hq+1(j) ∂∗ hq(i) hq(j) ... −→ hq+1(x0) −→ hq+1(X) −→ hq+1(X, x0) −→ hq(x0) −→ hq(X) −→ hq(X, x0) −→ ... is exact. The existence of the right inverse to h∗(i) implies that h∗(i) is a monomorphism. Thus we have a short split exact sequence

h∗(i) h∗(j) 0 −→ h∗(x0) −→ h∗(X) −→ h∗(X, x0).

This implies that h∗(X) = hq(x0) ⊕ h∗(X, x0). Thus ∼ eh∗(X) = h∗(X, x0).

3.2.F Mayer-Vietoris sequence

3.2.28 Theorem: Suppose that a dyad {A, B} in X is excisible and let C ⊂ A ∩ B. Then there exists an exact sequence of the form

α β ∆ ... −→ hq(A ∩ B,C) −→ hq(A, C) ⊕ hq(B,C) −→ hq(A ∪ B,C) −→ hq−1(A ∩ B,C) −→ ..., where α(c) = (h∗(i1)(c), −h∗(i2)(c)) for c ∈ h∗(A ∩ B,C), β(c, d) = h∗(j1(c) + h∗(j2)(d) for c ∈ hq(A, C), d ∈ hq(B,C) and ∆ is given as the composition

∼ hq(I) h∗(J)= ∂∗ hq(A ∪ B,C) −→ hq(A ∪ B,B) ←− hq(A, A ∩ B) −→ hq−1(A ∩ B,C), where ∂∗ is the boundary homomorphism of the homology exact sequence of the triple (A, A∩B,C), i1 :(A ∩ B,C) → (A, C), i2(A ∩ B,C) → (B,C), j1(A.C) → (A ∪ B,C), j2 :(B,C) → (A ∪ B,C),I :(A ∪ B,C) → (A ∪ B,B) and J :(A, A ∩ B) → (A ∪ B,B) are the inclusions. This sequence is functorial, i.e., all its homomorphism are the natural transformations of the respective functors. Proof: The proof is the diagram „chase”: one uses the homology exact sequences of the triples (A, A ∩ B,C) and (A ∪ B,B,C). It is important o observe the particular cases: If C = ∅, then the above sequence reduces to the following sequence

α β ∆ ... −→ h∗(A ∩ B) −→ hq(A) ⊕ hq(B) −→ h∗(A ∪ B) −→ h∗−1(A ∩ B) −→ ...

If C = ∗ is a one point space, then we have the reduced Mayer-Vietoris sequence

α β ∆ ... −→ eh∗(A ∩ B) −→ ehq(A) ⊕ ehq(B) −→ eh∗(A ∪ B) −→ eh∗−1(A ∩ B) −→ ... 3.2. General homology and cohomology theories 123

3.2.29 Remark: Exactly in a similar way we obtain the exact sequence

α β ∆ ... −→ hq(X,A ∩ B) −→ hq(X,A) ⊕ hq(X,B) −→ hq(X,A ∪ B) −→ hq−1(X,A ∩ B) −→ ...

The Mayer-Vietoris sequence is a good devise for computations.

3.2.30 Example: We shall again compute the homology of spheres. For instance consider the k k excisible dyad {S+,S−}, k > 1. We have

α k k β k ∆ k−1 α k k ... −→ ehq(S+) ⊕ ehq(S−) −→ ehq(S ) −→ ehq−1(S ) −→ ehq−1(S+) ⊕ ehq−1(S−) −→ ... k k Since S+,S− are contractible, ∆ is an isomorphism, i.e., for each q ∈ Z, k ∼ k−1 hq(S ) = hq−1(S ). Therefore 0 if q 6= k, h (Sk) = q G if q = k. The assertion follows by the induction with respect to k. Indeed, for k = 1 and any q ∈ Z we have 0 if q 6= 1, h (S1) = h (S0) = h (S0, −1) = h (+1) = eq eq−1 q−1 q−1 G if q = 1 (recall that S0 = {−1, +1} and the dyad {−1, +1} is excisible. The induction step is obvious. 0 Observe also that ehq(S ) = G if q = 0 and 0 otherwise.

3.2.31 Remark: (1) If h∗ does not satisfy the dimension axiom, then k ∼ ehq(S ) = hq−k(P ) for any q ∈ Z and k > 1. (2) For any homology theory (h∗, ∂∗) (with or without the dimension axiom) and k > 1 we have k k−1 k hq(D ,S ) = ehq(S ). Thus, for any k > 1, k k ∼ k h∗(R , R \ 0) = eh∗(S ). k−1 k k ∼ k Indeed the sphere S is a strong deformation retract of D \ 0, i.e. eh∗(D \ 0) = eh∗(S ). Using the reduced homology sequences of the pairs (Dk,Dk \ 0) and Dk,Sk−1) and the ﬁve lemma we k k get the assertion. On the other hand excising R \ D , we have k k ∼ k k h∗(R , R \ 0) = h∗(D ,D \ 0).

At last we shall study the availability of the Mayer-Vietoris sequence for the singular homology. It appears that dealing with chains we are in a position to get a better version of this fact.

3.2.32 Theorem: Suppose that dyads {X1,X2} and {A1,A2} are excisible (for the singular homology theory), where Ai ⊂ Xi, i = 1, 2. Then there exists a short exact sequence of chain complexes and chain maps

(i1,−i2) 0 −→ Q∗(X1 ∩ X2)/Q∗(A1 ∩ A2) −→ Q∗(X1,A1) ⊕ Q∗(X2,A2)

j1+j2 −→ [Q∗(X1) + Q∗(X2)]/[Q∗(A1) + Q∗(A2)] −→ 0, 124 3. General Constructions of (Co)homology of complexes where ik :(X1 ∩ X2,A1 ∩ A2) → (Xk,Ak), jk(Xk,Ak) → (X1 ∪ X2,A1 ∪ A2), k = 1, 2, are the inclusions. The long exact homology sequence corresponding to the above short sequence has the form

... −→ Hq(X1 ∩ X2,A1 ∩ A2) −→ Hq(X1,A1) ⊕ Hq(X2,A2) ∆ −→ Hq(X1 ∪ X2,A1 ∪ A2) −→ Hq−1(X1 ∩ X2,A1 ∩ A2) −→ ..., where ∆ is the connecting homomorphism.

3.2.G Compact supports and continuity

Let (h∗, ∂∗) be a homology theory. We say that this theory has compact supports if for any topological pair (X,A) and any z ∈ H∗(X,A) there is a compact pair (P,Q) such that z ∈ Im h∗(j) where j(P,Q) :(P,Q) → (X,A) is the inclusion.

It is clear that the singular homology H∗ has compact supports: given a pair (X,A) and z ∈ Hq(X,A), q > 0, the z is the homology class of a cycle c ∈ Zq(X,A). The chain c ∈ Qq(X,A) Pn is represented by a chain j=1 ajσj ∈ Qq(X), where aj ∈ R, σj ∈ Σq(X); some of its singular q-cubes, e.g. σ1, ...σk belong to Σq(A), i.e., has carriers in A, some – e.g. σk+1, ..., σn does not. Sn q Sk q Putting P = j=1 σj(I ) and Q := j=1 σj(I ) we show the assertion.

Let (h∗, ∂∗) be an arbitrary homology theory. For any pair (X,A), by C(X,B) we denote the family of all compact pairs (P,Q) such that (P,Q) ⊂ (X,A). Obviously C(X,A) is directed by the inclusion. It is clear that, for each q ∈ Z, {hq(P,Q), hq(j)}(P,Q∈C(X,A), where j :(P,Q) → (P 0,Q0), (P,Q) ⊂ (P 0,Q0), forms a direct system of R-modules. Moreover, the family {hq(j(P,Q)): hq(P,Q) → hq(X,A)}(P,Q)∈C(X,A) is compatible with this system. Therefore there is a uniquely determined homomorphism

i : lim h (P,Q) → h (X,A). −→ q q (P,Q)∈C(X,A)

The following results holds.

3.2.33 Theorem: The theory (h∗, ∂∗) has compact supports if and only if the homomorphism i is an isomorphism.

Proof: We shall appeal to theorem 0.4.11 (1) and the remark following it. Suppose that (h∗, ∂∗) has compact supports. Then [ hq(j(P,Q))(hq(P,Q)) = hq(X,A). (P,Q)∈C(X,A)

Suppose that z ∈ Hq(P,Q) and h∗(j(P,Q))(z) = 0, i.e. z ∈ Ker h∗(j(P,Q)). We shall show that 0 0 there is a compact pair (P ,Q ) ⊃ (P,Q) such that z is the kernel of the homomorphism h∗(j) induced by the inclusion j :(P,Q) → (P 0,Q0). Obviously z belongs to the kernel of the composition

h∗(i) h∗(P,Q) −→ h∗(X,Q) −→ h∗(X,A).

This means that h∗(i)(z) ∈ Ker {h∗(X,Q) → h∗(X,A)}. By the exactness of the homology sequence of the triple (X, A, Q) we see that h∗(i)(z) is in the image of the homomorphism 3.2. General homology and cohomology theories 125

0 h∗(A, Q) → h∗(X,Q). Since the theory has compact supports, there is a compact space Q such 0 that Q ⊂ Q ⊂ A and h∗(i)(z) is in the image of the composition

0 h∗(Q ,Q) → h∗(A, Q) → h∗(X,Q).

Again by the exactness of the homology sequence of the triple (X, Q, Q0) the composition

0 0 h∗(Q ,Q) → h∗(X,Q) → h∗(X,Q )

0 is trivial. Therefore z belongs to the kernel of h∗(P,Q) → h∗(X,Q ). Since z is in the kernel of the composition

h∗(j) 0 0 0 h∗(P,Q) −→ h∗(P ∪ Q ,Q ) → h∗(X,Q )

0 0 so – by the exactness of the homology sequence of the triple (X,P ∪ Q ,Q ) – h∗(j)(z) belongs to the image of 0 0 0 ∂∗ : h∗+1(X,P ∪ Q ) → h∗(P ∪ Q ,Q ). 0 0 0 By the compact supports there is a compact space P such that P ⊃ P ∪ Q and h∗(j)(z) is the image of the composition

0 0 0 ∂∗ 0 0 h∗+1(P ,P ∪ Q ) → h∗+1(X,P ∪ Q ) −→ h∗(P ∪ Q ,Q ).

This composition is equal to

0 0 0 0 ∂∗ : h∗+1(P ,P ∪ Q ) → h∗(P ∪ Q ,Q ).

By the exactness of the homology sequence of the triple (P 0,P ∪ Q0,Q0) the composition

0 0 ∂∗ 0 0 0 0 h∗+1(P ,P ∪ Q ) −→ h∗(P ∪ Q ,Q ) → h∗(P ,Q )

0 0 is trivial. Thus z belongs to the kernel of the homomorphism h∗(P,Q) → h∗(P ,Q ). This implies that i is an isomorphism. Conversely, assuming that i is an isomorphism, [ hq(X,A) = hq(j(P,Q))(hq(P,Q)) (P,Q)∈C(X,A) in view of theorem 0.4.11 (1) part (i). This shows that (h∗, ∂∗) has compact supports.

3.2.34 Remark: Suppose that (h∗, ∂∗) is an arbitrary homology sequence. For any topological pair (X,A) let h¯ (X,A) := lim h (P,Q). ∗ −→ ∗ (P,Q)∈C(X,A)

It is not diﬃcult to show that (h¯∗, ∂¯∗), where ∂¯∗ : h¯∗(X,A) → h¯∗−1(A) is the direct limit of the family {∂∗ : h∗(P,Q) → h∗−1(Q)}(P,Q)∈C(X,A) (being the map of direct systems {h∗(P,Q)} and {h∗−1(Q)} is a homology theory. This new theory has compact supports.

Suppose that {(Xγ,Aγ), hγδ}γ∈Γ is an inverse system of compact topological pairs. In this category inverse limits exist. Let

(X,A) = lim(X ,A ) ←− γ γ γ∈Γ 126 3. General Constructions of (Co)homology of complexes

together with a compatible family {pγ :(X,A) → (Xγ,Aγ)}γ∈Γ. Let (h∗, ∂∗) be a homology theory. It is clear that {h∗(Xγ,Aγ), h∗(hγδ)}γ∈Γ forms an inverse system in the category ModR and the family {h∗(pγ): H∗(X,A) → h∗(Xγ,Aγ)}γ∈Γ is compatible with this system. Therefore, there is a uniquely determined homomorphism

` : h (X,A) → lim h (X ,A ). ∗ ←− ∗ γ γ γ∈Γ

We say that h∗ is continuous if, for each inverse system {(Xγ,Aγ)} of compact pairs the homomorphism ` is an isomorphism.

3.2.35 Theorem: Suppose that a homology theory (h∗, ∂∗) is continuous. Then given compact pairs (X,A), (Y,B) any relative homeomorphism f :(X,A) → (Y,B) induces an isomorphism

h∗(f): h∗(X,A) → h∗(Y,B). In particular any compact dyad {A, B} is excisive.

Proof: Let {Bγ}γ∈Γ be the collection of all closed neighborhoods of B in Y . It is clear that Γ T is directed by the relation of the inverse inclusion in the family {Bγ} and γ∈Γ Bγ = B. Thus {(Y,Bγ)} forms an inverse system; moreover h (Y,B) = lim(Y,B ). ∗ ←− γ γ∈Γ

−1 For each γ ∈ Γ, let Aγ := f (Bγ). Then {(X,Aγ)}γ∈Γ is an inverse system, (X,A) = T γ∈Γ(X,Aγ) (because f is a relative homeomorphism). Let fγ :(X,Aγ) → (Y,Bγ) be determined by f. We shall see that, for any γ ∈ Γ, h∗(fγ) is an isomorphism. To this end let V be an open neighborhood of B such that cl V ∈ int Bγ. In view of the excision axiom, the inclusion (Y \ V,Bγ \ V ) induces an isomorphism ∼ h∗(Y \ V,Bγ \ V ) = h∗(Y,Bγ).

−1 Let U := f (V ). Then U is open and A ⊂ U ⊂ cl U ⊂ int Aγ. The embedding induces again an isomorphism ∼ h∗(X \ U, Aγ \ U) = h∗(X,Aα). homeo∼ Since f determines a homeomorphism (X \ U, Aγ \ U) = (Y \ V,Bγ \ V ), we gather that h∗(fγ) is an isomorphism. Since h (f) = lim h (f ) we see that h (f) – as the direct limit of isomorphisms – is an ∗ ←−γ∈Γ ∗ γ ∗ isomorphism.

3.2.36 Corollary: Suppose that (h∗, ∂∗) is continuous and has compact supports. Then each closed dyad {A, B} is excisive. Proof: Take any topological space X and let A, B ⊂ X be closed. For simplicity assume that A ∪ B = X. We are to show that the inclusion (A, C) → (X,B), where C := A ∩ B, induces an ∼ isomorphism h∗(A, C) = h∗(X,B). Let {Xγ}γ∈Γ be the family of all compact subsets of X. Since A, B are closed, Aγ := A∩Xγ, Bγ := B ∩ Xγ, γ ∈ Γ, are compact. Families {Aα} and {Bα} are coﬁnal in the families of all compact subsets of A and B, respectively. Thus

h (A, C) ∼ lim h (A ,C ) ∼ lim h (X ,B ) ∼ h (X,B). ∗ = −→ ∗ γ γ = −→ ∗ γ γ = ∗ γ∈Γ γ∈Γ 3.2. General homology and cohomology theories 127

3.2.37 Corollary: If (h∗, ∂∗) is continuous, then the theory (h¯∗, ∂¯∗) satisﬁes the strong excision axiom, i.e., any closed dyad {A, B} is excisive.

3.2.H Eilenberg-Steenrod Theorem

0 0 We say that J is a natural transformation of a homology theory (h∗, ∂∗) into a theory (h∗, ∂∗) if 0 for any topological pair (X,A), there is a homomorphism J(X,A) : h + ∗(X,A) → h∗(X,A) such 0 0 that ∂∗ ◦ J(X,A) = jA ◦ ∂∗ and if f :(X,A) → (Y,B), then h∗(f) ◦ J(X,A) = J(Y,B) ◦ h∗(f).

3.2.38 Theorem: If J is a natural transformation between homology theories (h∗, ∂∗) and 0 0 (h∗, ∂∗) such that for any one point space JP is an isomorphism, then for any compact polyhedral pair (X,A), J(X,A) is an isomorphism. Proof: By the use of the five-lemma it is sufficient to show that, for any compact polyhedron X, JX is an isomorphism. Let K be a simplicial complex such that |K| = X. We use induction with respect to the number of simplices in K. If K consists of one simplex, then |K| is homotopy equivalent to a one-point space and the assertion holds true. Assume that K has m simplices, where m > 1 and that the assertion is true whenever a complex has less than m simplices. Assume that dim K = n and that σ is an n-dimensional simplex in K. Let L be a subcomplex in K consisting of all simplices different from σ. By 0 the exactness of the long homology sequences for h∗ and h∗ and the five-lemma it follows that J|K| is an isomorphism if and only if so is J(|K|,|L|). On the other hand – by the excision ∼ 0 h∗(|K|, |L|) = h∗(|σ|, |σ˙ |) and the same holds for h∗. Hence it is sufficient to how that J(|σ|,|σ˙ |) is an isomorphism. If n = 0, then σ is a vertex in K and |σ˙ | = ∅, so the assertion follows directly from the assumption. Suppose that n > 0; then |σ| is contractible and J(|σ|,|σ˙ |) is an isomorphism if and only if so is J|σ˙ |. But |σ˙ | is a polyhedron having less simplices that does K. 3.2.39 Corollary: Under the same assumption as in theorem 3.2.38, suppose that both theories are continuous. Then J(X,A) is an isomorphism for any compact pair.

Proof: Again it is enough to show that JX is an isomorphism for any compact space X. The Urysohn theorem implies that X may be embedded into a Tychonoff cube Im of an appropriate m A A a weight. Let #A = m. Then I = I . For any finite set a ⊂ A, let pa : I → I be the a projection and let U be a polyhedral neighborhood of pa(X) in I . It is easy to see that the −1 family {Xa,U := pa (U)} corresponding to all finite subsets a ⊂ A and polyhedral neighborhoods a −1 of pa(X) in I is directed by the inverse inclusion and has X as its intersection. Moreover pa (U) is a compact set in IA homeomorphic to U ×IA\a; therefore it is homotopy equivalent to U. Thus we have shown that h∗(X) is isomorphic to the inverse limit of h∗(Xa,U ). The Eilenberg-Steenrod theorem 3.2.38 completes the proof. 3.2.40 Corollary: Under the same assumption as before in 3.2.38, suppose additionally that both homology theories have compact supports. Then, for any polyhedral pair (X,A), J(X,A) is an isomorphism. If, additionally both theories are continuous, then they are isomorphic. Proof: It is enough to observe that compact polyhedra form the family cofinal in the family of all compact subsets of (X,A). The last assertion follows immediately from corollary 3.2.39. 128 3. General Constructions of (Co)homology of complexes

3.3 Diﬀerentiable singular chains and the De Rham theorem

m Let M be an n-dimensional C -manifold, 0 6 m 6 ∞ or more generally a manifold modeled (r) over a separable Banach space). Recall that by Σq (U), q > 0 and 0 6 r 6 m, we have denoted the collection of all singular q-cubes σ : Iq → M which are Cr-smooth. (r) r As before Q¯q (M) denotes the free group generated by singular C -smooth q-singular cubes, (r) i.e., the group Z(Σq (M)). Then (r) ¯(r) (r) Qq (M) := Qq (M)/Dq (M) where (r) ¯(r) Dq (M) := Dq(M) ∩ Qq (M). (r) ¯(r) ¯ (r) Moreover Q0 (M) = Q0 (M) = Q0(M) = Q0(M). For q < 0, Qq (M) = 0. (r) (r) It is clear that ∂q : Qq (M) → Qq−1(M). Therefore we have deﬁned the chain complex of Cr-smooth singular chains (r) (r) Q∗ (M) := {Qq (M), ∂q}q∈Z. As usual (r) (r) Hq (M) := Hq(Q∗ (M)).

3.3.1 Theorem: Let M be an n-dimensional C∞-manifold. Then the natural embedding

(r) Q∗(M) → Q∗ (M) induces an isomorphism (r) ∼ H∗ (M) = H∗(M).

Before we go into the proof of this theorem we shall systematically study some other prop- (r) erties of H∗ . m r (a) Let M1,M2 be to C -manifolds and let f : M1 → M2 be C -smooth map. Then f induces the chain map (r) (r) f# : Q∗ (M1) → Q∗ (M2) by (r) (r) Σq (M1) 3 σ 7→ f ◦ σ ∈ Σq (M2).

r r (b) Given C -smooth maps f0, f1 : M1 → M2, we say that they are C -homotopic if there is r a C -smooth map f : M1 × I → M2 such that f(·, i) = fi for i = 0, 1. It is immediate to show that the chain maps f0# and f1# are chain homotopic, i.e., f0∗ = f1∗. It has the same consequences. n ∞ (c) An open convex set U in R (or in a Banach space) is C -contractible to a point. Therefore if q = 0; H(r)(U) = Z q 0 if q 6= 0.

∞ (r) (d) If A is a C -submanifold of a manifold M, then Q∗ (A) may be treated as a subcomplex (r) of Q∗ (M). Hence we may consider

(r) (r) (r) Q∗ (M,A) := Q∗ (M)/Q∗ (A). 3.3. Differentiable singular chains and the De Rham theorem 129

q r (e) If σ : I → M is a singular C -smooth q-cube, then the subdivision sd q(σ) is clearly seen to be a Cr-smooth singular chain. Hence

(r) (r) sd : Q∗ (M) → Q∗ (M)

(r) is a chain map. It follows that the excision property holds for H∗ , too. (r) Altogether we have that the pair (H∗ , ∂∗), restricted to the category of manifolds and their submanifolds, satisﬁes all of the Eilenberg-Steenrod axioms. (f) If M is the union of two open subsetsM = U ∪ V , then we get a version of the Mayer- (r) Vietoris sequence for H∗ . (r) (g) The theory (H∗ , ∂∗) has compact supports. Proof: We divide the argument into several steps. Step 1: M is a single point: trivial. n Step 2: If M is an open convex subset of R (or a Banach space), then the assertion follows since then M is C∞-contractible to a point. Step 3: If M = U ∪ V , where U, V are open sets for which the assertion is true, then it is true for M. This follows from the Mayer-Vietoris sequence and the ﬁve lemma. Step 4: Suppose that M is the union of a nested family of open sets and the theorem is assumed to be true for each set of the family. This follows from the compact supports axiom and the use of direct limits. n Step 5: Let M be an open subset in R (or a separable Banach space). Then

∞ [ U = Ui, i=1 where Ui is convex (i ∈ N). By step 2, the theorem is true for any Ui; by step 3, the theorem Sm is true for i=1 Ui for any m (the induction with respect to m and by the use of elementary properties of convex sets). Hence, by step 4, the theorem holds for U. Step 6: Let us consider a general case. Any differentiable manifold can be covered by open domains of an atlas; each such a domain is diffeomorphic to an open subset of an Euclidean space. Using steps 4, 5 and the Kuratowski-Zorn lemma, we find that there exists an open subset U ⊂ M for which the assertion holds true and U is maximal among all open sets for which the theorem is true. If U ( M, then there is a domain V of a chart which is not contained in U. Hence, by step 3, the theorem is true for U ∪ V . This contradicts the maximality of U and shows that U = M. Now let M be an n-dimensional C∞-manifold. It is known that any connected paracompact manifold has a countable basis of open sets (see Spivak). Moreover if a manifold has a countable basis, as a locally compact space, is paracompact (see Auslander-MacKenzie). q ∞ q Recall that Λ (M) the vector space of C -smooth of degree q > 0: if ω ∈ Λ (M), then for each x ∈ M, q ω(x) ∈ A (TxM, R). As usual d :Λq(M) → Λq+1(M) denotes the exterior derivative. Since d ◦ d = 0,

Λ∗(M) := {Λq(M), d} 130 3. General Constructions of (Co)homology of complexes is a cochain complex (the so-called de Rham complex). ∞ If f : M1 → M2 is a C -map between (Banach) manifolds, then it induces a linear map

# q q f :Λ (M2) → Λ (M1) given by the formula

# 0 0 f ω(x)(v1, ..., vq) = ω(f(x))(f (x)v1, ..., f (x)vq), x ∈ M1, vi ∈ TxM1, i = 1, ..., q.

As we know already f # ◦ d = d ◦ f #. # ∗ ∗ Therefore f :Λ (M2) → Λ (M1) is a cochain map. Having these we know that one is in a position to deﬁne the de Rham cohomology

Hq(M) := Hq(Λ∗(M)).

∞ Any smooth (i.e., C ) map f : M1 → M2 induced a linear map

∗ ∗ ∗ H (f): H (M2) → H (M1).

For any σ ∈ Σ(1)(M) and ω ∈ Λq(M) the integral Z ω σ ¯(1) ¯(1) Pn is well-deﬁned. It extends by linearity to Qq (M), i.e., for any c ∈ Qq (M), c = i=1 aiσi, (1) where ai ∈ R, σi ∈ Σq (M), i = 1, ..., n, n Z X Z = ai ω. c i=1 σi (r) R Observe that hσ, ωi = 0 for any σ ∈ Σq (M) if and only if ω = 0. In particular, if c ω = 0 for (1) all c ∈ Q¯q (M), then ω = 0. Moreover, if σ ∈ Σ(1)(M), then hσ, ωi for all ω ∈ Λq(M) if and only if σ is degenerate. In R q particular if the singular q-chain is degenerate, then c ω = 0 for any ω ∈ Λ (M). Thus there is a bilinear map

(r) q h·, ·i : Qq (M) × Λ (M) → R, 1 r ∞, given by 6 6 Z (r) hc, ωi = ω, c ∈ Qq (M), ω ∈ Λq(M). c This deﬁnition is correct in view of the above remarks. Observe that if, for some ω ∈ Λq(M), (∞) hc, ωi = 0 for all c ∈ Qq (M), then ω = 0. (1) q I do not know whether it is true that if for some c ∈ Qq (M), hc, ωi = 0 for all ω ∈ Λ (M), then c = 0 ? Let us now collect some properties of the bilinear map h·, ·i.

(a) If f : M1 → M2 is smooth, then Z Z # q (r) f ω = ω, ω ∈ Λ (M2), c ∈ Qq (M1), c f#(c) 3.3. Differentiable singular chains and the De Rham theorem 131 i.e., # hc, f ωi = hf#c, ωi, (r) (r) where f# = Qq(f): Qq (M1) → Qq (M2).

(r) q−1 (b) For any c ∈ Qq (M) and ω ∈ Λ (M), q > 1, we have the Stokes formula Z Z ω = dω, ∂qc c i.e., h∂c, ωi = hc, dωi.

q (r) (c) If ω ∈ Λ (M) and dω = 0, then hc, ωi = 0 for all c ∈ Bq (M).

q q−1 (1) (d) If ω ∈ Λ (M) and ω = dη for some η ∈ Λ (M), then hc, ωi = 0 for all c ∈ Zq (M). Fact (b) implies that h·, ·i extends to the bilinear map

(1) q h·, ·i : Hq (M) × H (M) → R.

(1) q Moreover given a smooth f : M1 → M2, [c] ∈ Hq (M1) and [ω] ∈ H (M2),

q (1) h[c], H (f)(ω)i = hHq (f)(c), ωi.

The bilinear map h·, ·i deﬁnes a homomorphism

q ϕ :Λ (M) → Hom(Qq(M), R)

(∞) (from now on we treat Qq(M) as Qq (M)) given by

ϕ(ω)(c) = hc, ωi.

This is in fact a cochain map ∗ Λ (M) → Hom(Q∗(M), R).

3.3.2 Theorem: (de Rham) For any paracompact diﬀerentiable manifold M, the cochain map induces an isomorphism

∗ ∗ ∗ ∗ H (M) → H (Hom(Q∗(M), R)) = H (M) = H∗ (M).

The last isomorphism follows from the universal coeﬃcients for singular homology. ∗ The de Rham theorem implies that the bilinear map h·, ·i : H∗(M)×H (M) → R is a duality. Proof: Step 1 Let M be a convex open. Then

if q = 0; H (M) = R q 0 if q 6= 0.

By the Poincare lemma if q = 0; Hq(M) = R 0 if q 6= 0. This result for q = 0 follows by the very deﬁnition and the connectedness of U. 132 3. General Constructions of (Co)homology of complexes

Step 2 M is the union of two open sets M = U ∪ V ; the de Rham theorem is assumed to hold for U and V and U ∩ V . Then it holds for M. To prove it we use the Mayer-Vietoris sequences. It is suﬃcient to show it for the de Rham cohomology. Let i : U ∩ V → U, j : U ∩ V → V , k : U → M, l : V → M denote the respective inclusions. Deﬁne cochain maps

α :Λ∗(M) → Λ∗(U) ⊕ Λ∗(V ),

β :Λ∗(U) ⊕ Λ∗(M) → Λ∗(U ∩ V ) by the formulae α(ω) = (k#(ω), l#(ω)), ω ∈ Λ∗(M), # # β(ω1, ω2) = i (ω1) − j (ω2). We claim that the sequence

β 0 −→ Λ∗(M) −→α Λ∗(U) ⊕ Λ∗(V ) −→ Λ∗(U ∩ V ) −→ 0 is exact. The only difficult part is to show that β is epimorphic. Let {h, g} be a smooth partition of unity subordinated to the cover {U, V }. Now let ω ∈ Λ∗(U ∩ V ). Then gω may be extended to a differential form ωV on V by putting ωV (x) = 0 for all x ∈ V \ U. Similarly hω may be extended to a differential form ωU on U by putting ωU (x) = 0 for all x ∈ U \ V . Then β(ωU , −ωV ) = ω. After passing to cohomology we get the Mayer-Vietoris sequence. S∞ Step 3 Let M = i=1 Ui where Ui ⊂ Ui+2 for all i = 1, 2, ... and suppose that Ui is the finite union of convex open sets. Suppose that cl Ui is compact. We assume that the assertion holds true for each Ui. Then it holds for M. ∗ ∗ For each i there is a cochain map Λ (M) → Λ (Ui). They form a family compatible with the inverse system ∗ ∗ ∗ Λ (U1) ← Λ (U2) ← Λ (U3) ← ... Moreover Λ∗ = lim Λ∗(U ). ←− i

Observe that the above tower has the Mittag-Loeﬄer property because cl Ui is compact: the compactness of cl Un implies that there is m > n such that cl Un ⊂ Um. It is now easy to see ∗ ∗ that for any i > m, the image of the homomorphism Λ (Ui) → Λ (Un) induced by the inclusion ∗ ∗ Un → Ui is equal to that of the homomorphism Λ (Um) → Λ (Un). Therefore

1 lim Λq(U ) = 0 ←− i and there is a short exact sequence

1 0 −→ lim Hq−1(U ) −→ Hq(M) −→ lim Hq(U ) −→ 0. ←− i ←− i

We know that Q (U) = lim Q (U ). ∗ −→ ∗ i Thus Hom(Q (M), ) = lim Hom(Q (U ), ). ∗ R ←− ∗ i R 3.3. Differentiable singular chains and the De Rham theorem 133

Moreover, for each i, Hom(Q∗(Ui+1), R) → Hom(Q∗(Ui), R) is an epimorphism. Thus 1 lim C (U ) = 0. ←− q i This implies that there is an exact short sequence

1 0 −→ lim Hq−1(U ) −→ Hq(M) −→ lim Hq(U ) −→ 0. ←− i ←− i This proofs the assertion. n Step 4 M is an open subset of R . Then M is a union of a countable family of convex open sets Ui such that cl Ui is compact and cl Ui ⊂ M. Using step 2, the theorem holds for ﬁnite Sn unions i=1 Ui (this is done by induction and uses basic properties of convex sets). In view of step 3, theorem holds for M by passing with n → ∞. Step 5 Let M be a connected paracompact manifold. Any connected manifold has a countable basis of open sets. It follows that M is a countable union of coordinate neighborhoods Ui (each of which is diﬀeomorphic to an open convex set in E). Thus, by steps 2,4, the theorem is true for Vn := U1 ∪ U2 ∪ ... ∪ Un, n ∈ N. Observe that cl Vn is compact. Therefore, by step 3, the theorem is true for M. Step 6 Theorem is true for each component of M. This implies that it is true for M (by the additivity). Chapter 4 Various construction of homology and cohomology theories of Čech type

4.1 Čech-type homology

In order to construct an infinite-dimensional homology theory we shall need an appropriate ordi- s nary homology. The most commonly used singular homology theory H∗ (·, ·; G) (with coefficients in an abelian group G) is defined for arbitrary topological pairs and satisfies all the Eilenberg- Steenrod axioms. However, in singular theory there exist certain pathological examples which may not be desirable. For instance, there are connected spaces (like the so-called Warsaw circle, see Remark 4.1.7 below) which admit homotopically nontrivial maps into the circle but have 3 trivial 1-dimensional singular homology, and there are compact subsets of R having nontrivial homology groups in arbitrarily high dimensions (see e.g. [?]). Moreover, the singular homology does not satisfy the strong excision axiom but only the weaker one saying that given a topological pair (X,B) and a set A ⊂ X such that int A ∪ int B = X, the inclusion j :(A, A ∩ B) → (X,B) s s induces an isomorphism j∗ : H∗ (A, A ∩ B; G) → H∗ (X,B; G).

In [13, Chap. IX], the Čech homology Hˇ∗(X,A; G) with coefficients in an abelian group G for an arbitrary pair (X,A) of topological pairs has been defined. However, in this general situation the exactness axiom is not always satisfied. It holds if G is a field (or a compact group) and (X,A) a compact pair. In this latter case the setting presented in [13] is complete and satisfactory although not intuitive from the geometric viewpoint. Therefore we propose a different approach.

Let (P,Q), where Q ⊂ P , be a pair of compact metric spaces. It is well known that P can be embedded in the Hilbert cube and hence in any infinite-dimensional normed space E (in other words, we can consider P and Q as compact subsets of E). ˇ ˇ Let F be a fixed field and H∗(P,Q) := {Hq(P,Q)}q∈Z the graded vector space defined by

Hˇ (P,Q) := lim{Hs(U, V ):(U, V ) ∈ U}, q ∈ , (4.1.1) q ←− q Z where U stands for the family of all (pairs of) neighborhoods of (P,Q) in E directed by the s s inverse inclusion and H∗ (U, V ) = H∗ (U, V ; F) denotes the singular homology of (U, V ) ∈ U 4.1. Čech-type homology 135

with coeﬃcients in F. In other words, Hˇq(P,Q) is the inverse limit of the inverse system s {Hq (U, V )}(U,V )∈U of vector spaces together with homomorphisms induced by the inclusions (U, V ) ,→ (U 0,V 0). In what follows we shall make repeated use of standard properties of inverse and direct limits, see e.g. [9, Appendix].

4.1.1 Remark: Observe that if a family U 0 of neighborhoods of the pair (P,Q) is coﬁnal in U, then Hˇ (P,Q) = lim{Hs(U, V ):(U, V ) ∈ U 0}. ∗ ←− ∗ In particular, U 0 may consist of all pairs of open neighborhoods of (P,Q) in E.

4.1.2 Lemma: Deﬁnition (4.1.1) is correct, i.e., it does not depend on the choice of a normed space E.

Proof. To see this, suppose that E1, E2 are normed spaces and let Q1 ⊂ P1 ⊂ E1, Q2 ⊂ P2 ⊂ E2 be homeomorphic copies of the pair (P,Q) embedded into E1 and E2, respectively. 0 Thus there is a homeomorphism h :(P1,Q1) → (P2,Q2). Let E := E1 × E2 and let ji : Ei → E, i = 1, 2, be inclusions given by j1(x1) = (x1, 0) and j2(x2) = (0, x2) for x1 ∈ E1 and x2 ∈ E2. In 0 view of Lemma (2.4) from [?], there is a homeomorphism h : E → E such that hj1|P1 = j2h .

For i = 1, 2, let Ui be the family of all neighborhoods of the pair (Pi,Qi) in Ei and let Uei be the family of all neighborhoods of the pair (ji(Pi), ji(Qi)) in E. Since h is a homeomorphism, we easily see that Ue2 = {(h(U), h(V )) : (U, V ) ∈ Ue1}. This shows that

lim{Hs(U, V ):(U, V ) ∈ U } = lim{Hs(U, V ):(U, V ) ∈ U }. ←− ∗ e1 ←− ∗ e2

On the other hand, given a neighborhood (U, V ) of the pair (j1(P1), j1(Q1)) in E, there are (U1,V1) ∈ U1 and an integer n > 1 such that U1 × B1/n(0) ⊂ U and V1 × B1/n(0) ⊂ V , where B1/n(0) is a ball in E2. In other words, the family {(U1,V1) × B1/n(0) : (U1,V1) ∈ U1, n > 1} is s ∼ s coﬁnal in Ue1. Since H∗ ((U1,V1) × B1/n(0)) = H∗ (U1,V1), we see that

lim{Hs(U ,V ):(U ,V ) ∈ U } ∼ lim{Hs(U, V ):(U, V ) ∈ U }. ←− ∗ 1 1 1 1 1 = ←− ∗ e1 Similarly one shows that

lim{Hs(U ,V ):(U ,V ) ∈ U } ∼ lim{Hs(U, V ):(U, V ) ∈ U }. ←− ∗ 2 2 2 2 2 = ←− ∗ e2

Let f :(P,Q) → (P 0,Q0) be a continuous map of compact metric pairs and assume that P ⊂ E, P 0 ⊂ E0, where E and E0 are normed spaces. In order to deﬁne the induced homomorphism 0 0 0 0 f∗ : Hˇ∗(P,Q) → Hˇ∗(P ,Q ), let f : E → E be an arbitrary (continuous) extension of f which exists in view of the Dugundji theorem. For each (U 0,V 0) ∈ U 0, where U 0 stands for 0 0 0 0 the family of all neighborhoods of the pair (P ,Q ) in E , consider the homomorphism f∗ : s 0−1 0 0−1 0 s 0 0 0 H∗ (f (U ), f (V )) → H∗ (U ,V ) induced by f |f 0−1(U). It is easy to see that the family 0 0 0 0 s {f∗ :(U ,V ) ∈ U } forms a transformation of the inverse system {H∗ (U, V ):(U, V ) ∈ U} into s 0 0 0 0 0 the inverse system {H∗ (U ,V ):(U ,V ) ∈ U } and therefore determines a homomorphism

f := lim{f 0 :(U 0,V 0) ∈ U 0} : Hˇ (P,Q) → Hˇ (P 0,Q0). (4.1.2) ∗ ←− ∗ ∗ ∗

4.1.3 Lemma: Deﬁnition (4.1.2) is correct, i.e., it does not depend on the choice of an extension f 0 of f. 136 4. Various construction of homology and cohomology theories of Čech type

Proof. Suppose that f 00 : E → E0 is another extension of f. Deﬁne a map h :(E × {0, 1}) ∪ (P × [0, 1]) → E0 by  0  f (x) for x ∈ E, t = 0; h(x, t) := f(x) for x ∈ P, t ∈ [0, 1];  f 00(x) for x ∈ E, t = 1. As before, h admits an extension h0 : E×[0, 1] → E0. If (U 0,V 0) ∈ U 0, then there is a pair (U, V ) of neighborhoods of P and Q respectively, such that U ×[0, 1] ⊂ h0−1(U 0) and V ×[0, 1] ⊂ h0−1(V 0); 0−1 0 00−1 0 0−1 0 00−1 0 0 in particular, U ⊂ f (U )∩f (U ) and V ⊂ f (V )∩f (V ). Since the maps f |(U,V ) and 00 0 0 s s 0 0 f |(U,V ) are homotopic in (U ,V ), they induce the same homomorphism H∗ (U, V ) → H∗ (U ,V ). This completes the proof.

In order to deﬁne the boundary operator ∂ : Hˇ∗(P,Q) → Hˇ∗−1(Q) and show exactness of the homology sequence of (P,Q) we need some preparations. First observe that given a compact subset P of a normed space E and a neighborhood U of P in E, there is an integer n > 1 such that, for any x ∈ P , B1/n(x) ⊂ U. Since P is compact, there are points x1, . . . , xk ∈ P such that Sk P ⊂ i=1 B1/n(xi) ⊂ U.

In other words, we see that the family U0 of pairs of open sets (U, V ), where U (resp. V ) is the ﬁnite union of balls centered at points of P (resp. Q) and of radius 1/n, where n > 1, is coﬁnal in the family of all neighborhoods of (P,Q).

4.1.4 Lemma: If X ⊂ E is a ﬁnite union of open convex sets, then for any q ∈ Z, the vector s space Hq (X) is ﬁnite-dimensional. Proof. We shall proceed by induction on the number of open convex sets covering X. If s s X is open convex, then H0 (X) = F and Hq (X) = 0 for q 6= 0. Suppose that, for any q ∈ Z, s 0 0 Sk+1 dim Hq (X ) < ∞ whenever X is the union of k open convex sets. Let X = i=1 Ci where Ci is 0 Sk 0 open convex in E, i = 1, ..., k + 1, and put X := i=1 Ci; then X = X ∪ Ci+1. 0 Since the pair {X ,Ci+1} is excisive, the Mayer-Vietoris sequence

s 0 α1 s 0 s α2 s 0 ... −→ Hq (X ∩ Ci+1) −→ Hq (X ) ⊕ Hq (Ci+1) −→ Hq (X ∪ Ci+1) α3 s 0 α4 s 0 s −→ Hq−1(X ∩ Ci+1) −→ Hq−1(X ) ⊕ Hq−1(Ci+1) −→ ... s 0 s 0 s is exact. In view of the induction hypothesis, the spaces Hq (X ∩ Ci+1) and Hq (X ) ⊕ Hq (Ci+1) 0 are ﬁnite-dimensional for all q (note that X ∩ Ci+1 is the union of at most k convex open sets). Passing to subspaces and quotient spaces we see that the following sequence

0 α2 s 0 α3 0 −→ Coker (α1) −→ Hq (X ∪ Ci+1) −→ ker(α4) −→ 0 is exact. Since the above terms are vector spaces, this sequence is split and

s s 0 Hq (X) = Hq (X ∪ Ci+1) = Coker (α1) ⊕ ker(α4); s hence Hq (X) is ﬁnite-dimensional. s s The above lemma implies that if (U, V ) ∈ U0, then the vector spaces Hq (U), Hq (V ) are s ﬁnite-dimensional for all q, and therefore so are Hq (U, V ) in view of the homology sequence of (U, V ).

Let (P,Q) be a compact pair in a normed space E. Since the family U0 is coﬁnal in the family of all neighborhoods of (P,Q), we have Hˇ (P,Q) = lim{Hs(U, V ):(U, V ) ∈ U }. ∗ ←− ∗ 0 4.1. Čech-type homology 137

Let q ∈ Z and consider the inverse system of exact sequences

s s s ∂ s s ... −→ Hq (V ) −→ Hq (U) −→ Hq (U, V ) −→ Hq−1(V ) −→ Hq−1(U) −→ ...,

s s where (U, V ) ∈ U0 (∂ : Hq (U, V ) → Hq−1(V ) is the connecting homomorphism). Since all terms in these sequences belong to the category of ﬁnite-dimensional vector spaces, in view of Theorem VIII.5.7 from [13], we conclude that the limit sequence

∂ ... −→ Hˇq(Q) −→ Hˇq(P ) −→ Hˇq(P,Q) −→ Hˇq−1(Q) −→ Hˇq−1(U) −→ ... is exact.

4.1.5 Remark: It is important that we deal with ﬁnite-dimensional vector spaces since in general the inverse limit of an inverse system of exact sequences consisting of arbitrary groups is not exact; this is the reason why the Čech homology for arbitrary topological pairs with coeﬃcients in an arbitrary group is not exact. It is easy to see that the inverse limit homomorphism ∂ := lim{∂ : Hs(U, V ) → Hs (V ): ←− q q−1 (U, V ) ∈ U0}) is in fact the desired boundary operator for the pair (P,Q): if f :(P,Q) → 0 0 (P ,Q ) is a continuous map of compact pairs, then ∂f∗ = (f|Q)∗∂ because inverse limits preserve commutativity.

4.1.6 Theorem The above deﬁned Čech homology Hˇ∗ is a functor on the category of compact metric pairs which satisﬁes the following Eilenberg-Steenrod axioms: (i) (Functoriality) If id is the identity map on a compact metric pair (P,Q), then id∗ is the 0 0 0 0 00 00 identity on Hˇ∗(P,Q); if f :(P,Q) → (P ,Q ) and g :(P ,Q ) → (P ,Q ) are continuous maps of compact metric pairs, then (g ◦ f)∗ = g∗ ◦ f∗. (ii) (Naturality of ∂) If f :(P,Q) → (P 0,Q0) is a continuous map of compact metric pairs, then ∂ Hˇq(P,Q) / Hˇq−1(Q)

f∗ (f|Q)∗ Hˇ (P 0,Q0) / Hˇ (Q0) q ∂ q−1 (iii) (Exactness) For a compact metric pair (P,Q), let i : Q,→ P and j : P,→ (P,Q) be the inclusions. Then, for any q ∈ Z, the homology sequence

i∗ j∗ ∂ ... −→ Hˇq(Q) −→ Hˇq(P ) −→ Hˇq(P,Q) −→ Hˇq−1(Q) −→ ... is exact. (iv) (Strong excision) If P,Q are compact metric spaces, then the inclusion (P,P ∩ Q) ,→ (P ∪ Q, Q) induces the excision isomorphism

exc∼ Hˇ∗(P,P ∩ Q) = Hˇ∗(P ∪ Q, Q).

(v) (Homotopy invariance) If two continuous maps f, g :(P,Q) → (P 0,Q0) of compact metric pairs are homotopic, then f∗ = g∗. (vi) (Dimension) If ∗ is a one-point space, then

F for q = 0; Hˇ (∗) = q 0 otherwise. 138 4. Various construction of homology and cohomology theories of Čech type

Proof. The functoriality of Hˇ∗ is easy to see. Using arguments similar to those from the proof of Lemma 4.1.3, we show the homotopy invariance axiom (this time we extend the homotopy h : P × [0, 1] → P 0 to h0 : E × [0, 1] → E0). It is clear that the dimension axiom is satisﬁed. The naturality of ∂ and the exactness axiom have been discussed above. To show the excision property we may assume that P and Q are compact subsets of a normed space E. Given neighborhoods U, V of P and Q respectively, the excision axiom for the singular theory implies that the inclusion i :(U, U ∩ V ) → (U ∪ V,V ) induces an isomorphism s s i∗ : H∗ (U, U ∩ V ) → H∗ (U ∪ V,V ). Since the family U1 (resp. U2) of neighborhoods of the form (U, U ∩ V ) (resp. (U ∪ V,V )), where U is a neighborhood of P and V a neighborhood of Q, is coﬁnal in the family of all neighborhoods the pair (P,P ∩ Q) (resp. (P ∪ Q, Q)), we obtain that

∼ Hˇ (P,P ∩ Q) = lim{Hs(U, U ∩ V ): P ⊂ U, Q ⊂ V } −→= ∗ ←− ∗ lim{Hs(U ∪ V,V ): P ⊂ U, Q ⊂ V } = Hˇ (P ∪ Q, Q), ←− ∗ ∗ where the isomorphism is induced as above by the inclusion, i.e., it is given as the inverse limit map lim{i : Hs(U, U ∩ V ) → Hs(U ∪ V,V ): P ⊂ U, Q ⊂ V } ←− ∗ ∗ ∗ (remember the that the inverse limit of isomorphisms is an isomorphism).

4.1.7 Remark: Recall that the Warsaw circle X is the union of two sets A and B such that A 2 2 is the closure of {(x, y) ∈ R : y = sin(π/x), 0 < x < 1} and B is an arc in R which meets A only at (0, −1) and (1, 0). It is easy to see that each map S1 → X is homotopically trivial, s hence H1 (X) = 0. However, there exists a nested sequence of open annuli Un ⊃ X such that T∞ U = X. It follows that Hˇ (X) = lim Hs(U ) = F. n=1 n 1 ←− 1 n 4.1.8 Remark (i) Mardesič [11] (see also [10]) shows that our definition (4.1.1) coincides with the definition of the Čech homology introduced in [13]. Moreover, it can be shown (cf. [11], [10]) that if a compact space P is locally contractible (or – more generally – homologically locally ˇ ∼ s connected, see e.g. [9]), then H∗(P ) = H∗ (P ; F); therefore, if Q ⊂ P is also locally contractible, ˇ ∼ s then H∗(P,Q) = H∗ (P,Q; F). In particular, if (P,Q) is a pair of compact metric absolute ˇ ∼ s neighborhood retracts, then H∗(P,Q) = H∗ (P,Q; F). (ii) Another construction of a Čech type homology theory for compact pairs, with coefficients in an arbitrary group G, which satisfies the Eilenberg-Steenrod axioms including the exactness and the strong excision axioms, was provided in Massey [9]. The main advantage of his theory is that it is valid for arbitrary (not necessarily metric) compact pairs and coefficient groups. If G is a field, then the Massey and the Čech theories coincide. The disadvantage of the Massey approach is that it is less intuitive, especially in contrast to the formula (4.1.1) above.

The Čech homology Hˇ∗ introduced above is not sufficient for our purposes because it is defined on compact pairs only. Now we extend it to a more general situation. Let (X,A) be an arbitrary pair of metric spaces. It easy to see that the family C(X,A) of all ˇ compact pairs (P,Q) ⊂ (X,A) is directed by inclusion. The family {H∗(P,Q)}(P,Q)∈C(X,A) 0 0 together with the family Hˇ∗(P,Q) → Hˇ∗(P ,Q ) of homomorphisms induced by inclusions (P,Q) ⊂ (P 0,Q0) ∈ C(X,A) form a direct system of vector spaces. Following [13, Chap. IX, Exercise D] (comp. [9, Chap. 9]), we define the Čech homology with compact supports and coefficients in F by setting

Hˇ c(X,A) := lim Hˇ (P,Q). ∗ −→ ∗ (P,Q)∈C(X,A) 4.2. Alexander-Spanier cohomology and Massey homology 139

ˇ ˇ Given a continuous map f :(X,A) → (Y,B), the family {H∗(P,Q) → H∗(f(P ), f(Q))}(P,Q)∈C(X,A) ˇ of homomorphisms induced by f determines a map of the system {H∗(P,Q)}(P,Q)∈C(X,A) to the ˇ 0 0 direct system {H∗(P ,Q )}(P 0,Q0)∈C(Y,B). Therefore f determines a unique (graded) homomor- ˇ c ˇ c phism f∗ : H∗(X,A) → H∗(Y,B).

In a similar manner one checks that given a pair (X,A), the family {∂ : Hˇ∗(P,Q) → Hˇ∗−1(Q)} of boundary homomorphisms in the Čech theory determines a map between the di- ˇ ˇ rect systems {H∗(P,Q)}(P,Q)∈C(X,A) and {H∗−1(Q)}Q∈C(A) and therefore defines the boundary ˇ c ˇ c homomorphism ∂ : H∗(X,A) → H∗−1(A). ˇ c The Čech homology H∗ with compact supports and coefficients in F on the category of arbitrary metric pairs satisfies all the Eilenberg-Steenrod axioms. This follows from the properties of the direct limit and the respective properties of Hˇ∗. In particular, the exactness of the sequence

ˇ c i∗ ˇ c j∗ ˇ c ∂ ˇ c ... −→ H∗(A) −→ H∗(X) −→ H∗(X,A) −→ H∗−1(A) −→ ... is a consequence of the exactness of the functor of direct limit (see [9, Theorem A.7]). More precisely, the following holds true:

ˇ c 4.1.9 Theorem: The Čech homology with compact supports H∗ is a functor on the category of metric pairs which satisﬁes the Eilenberg-Steenrod axioms (i)-(iii) and (v),(vi) stated in Theorem 4.1.6. Axiom (iv) is satisﬁed for closed pairs, i.e., if X = A ∪ B, where A, B are closed in X, then the inclusion (A, A ∩ B) ,→ (X,B) induces the isomorphism

exc ˇ c ∼ ˇ c H∗(A, A ∩ B) = H∗(X,B).

Only the excision axiom requires some explanation. It is easy to see that the families of pairs (P,P ∩ Q) and (P ∪ Q, Q), where P ⊂ A, Q ⊂ B are compact, are coﬁnal in the families of all compact subsets of respectively (A, A∩B) and (A∪B,B). Indeed, if R ⊂ A∪B is compact, then so are R ∩A := P and R ∩B := Q (because A, B are closed), and if R ⊂ A∩B, then we can take ∼ P = Q = R. Hence passing to the direct limit in the isomorphisms Hˇ∗(P,P ∩ Q)=Hˇ∗(P ∪ Q, Q) gives the conclusion.

ˇ c ˇ 4.1.10 Remark: If a metric pair (X,A) is compact, then H∗(X,A) = H∗(X,A). In view of ˇ c ∼ the results of [11], if a metric space is locally compact and locally contractible, then H∗(X) = s ˇ c ∼ s H∗ (X; F). If A ⊂ X is closed and locally contractible, then H∗(X,A) = H∗ (X,A; F).

4.2 Alexander-Spanier cohomology and Massey homology

In this section we provide a rough description of the Massey approach to the Alexander-Spanier (co)homology theory (see [9]). It is closely related to the Čech theory. The reader should recall the construction of homology (resp. cohomology) groups for chain (resp. cochain) complexes.

4.2.A Alexander-Spanier cohomology

q Assume that P is a compact space and, as usual, R is a p.i.d. Given q > 0, by Φ (P ; R) we denote the set of all ﬁnite-valued functions ϕ : P q+1 → R, i.e., ϕ ∈ Φq(P ; R) if and only if #ϕ(P q+1) < ∞. 140 4. Various construction of homology and cohomology theories of Čech type

q q+1 At the same time Φ∞(P ; R) denotes the set of all functions ϕ : P → R. Clearly q q Φ (P ; R) ⊂ Φ∞(P ; R). q q It is also easy to see that both sets Φ (P ; R) and Φ∞(P ; R) carry the structure of an R-module; q q from this viewpoint Φ (P ; R) is a submodule of Φ∞(P ; R). Let us define δq :Φq(P ; R) → Φq+1(P ; R) by the formula q+1 q X i δ (ϕ)(x0, x1, ..., xq+1) = (−1) ϕ(x0, ..., xî, ..., xq+1) i=0 q where xi ∈ P , i = 0, ..., q + 1. We see immediately that this is correctly defined (i.e., δ (ϕ) ∈ Φq+1(P ; R) whenever ϕ ∈ Φq(P ; R)); moreover δq is a homomorphisms of modules and

δq+1δq = 0.

We may deﬁne a homomorphism

q q q+1 δ∞ :Φ∞(P ; R) → Φ∞ (P ; R)

q+1 q by the same formula; similarly as before δ∞ δ∞ = 0. For ϕ ∈ Φq(P ; R), let the support |ϕ| of ϕ consist of points x ∈ X such that, for any neighborhood V of x, there are x0, ..., xq ∈ V with ϕ(x0, ..., xq) 6= 0. In other words x 6∈ |ϕ| if and only if there is a neighborhood V of x such that ϕ(x0, ..., xq) = 0 for all xi ∈ V , i = 0, ..., q. It is easy to check the following fact.

4.2.1 Fact: The following properties are satisﬁed: (i) the support |ϕ| of ϕ ∈ Φq(P ; R) is closed (and hence compact); (ii) |0| = ∅; (iii) |ϕ ± ψ| ⊂ |ϕ| ∪ |ψ| for all ϕ, ψ ∈ Φq(P ; R); (iv) |δq(ϕ)| ⊂ |ϕ| for any ϕ ∈ Φq(P ; R).

4.2.2 Remark: If |ψ| = ∅, then |ϕ + ψ| = |ϕ|. q The same definition of a support and its properties are valid in case of ϕ ∈ Φ∞(P ; R). Let us define q q Φ0(P ) := {ϕ ∈ Φ (P ) | |ϕ| = ∅} This is a submodule of Φq(P ). q q In a similar way we may define the submodule Φ∞,0(P ; R) of Φ∞(P ; R) consisting of all functions with empty supports. Let ¯q q q C (P ; R) := Φ (P ; R)/Φ0(P ; R) and ¯q q q C∞(P ; R) := Φ∞(P ; R)/Φ∞,0(P ; R).

Observe that, in view of Remark 4.2.2, if ϕ, ψ ∈ Φq(P ; R) represent the same element u ∈ C¯q(P ; R), i.e., u = [ϕ] = [ψ], then |ϕ| = |ψ|; thus we may speak of the support |u| of u. 4.2. Alexander-Spanier cohomology and Massey homology 141

In view of the Nöbeling theorem (see page 16) we have

4.2.3 Theorem: The R-module C¯q(P ; R) is free. Proof: First observe that Φq(P ; R) is a set (R-module) of all finite-valued functions X := P q+1 → R. Thus the setting of the Nöbeling theorem is granted. Now it is sufficient to show q q that Φ0(P ; R) is a Specker module, i.e., given ϕ ∈ Φ0(P ; R) having representation n X ϕ = aiχXi , i=1 where Xi = {x ∈ X | ϕ(x) = ai} and ai 6= 0, all characteristic functions χXi , i = 1, 2, ..., n, have empty supports. If it is not so, then there is i = 1, ..., n, such that |χXi |= 6 ∅, i.e., there is x ∈ P such that any neighborhood V of x contains points x0, ...xq with χ(x0, ..., xq) = 1. It is clear that (x0, ..., xq) ∈ Xi. Hence ϕ(x0, ..., xq) = ai 6= 0. Thus x ∈ |ϕ|: contradiction. q q It is clear (in view of property (iv) from Fact 4.2.1) that δ (resp. δ∞) induces q ¯q ¯q+1 q ¯q ¯q+1 δ : C (P ; R) → C (P ; R), δ∞ : C∞(P ; R) → C∞ (P ; R) given by the formula δq(u) = [δq(ϕ)] q q where u = [ϕ], ϕ ∈ Φ (P ; R) (or ϕßΦ∞(P ; R)). Therefore we have define two cochain complexes

¯∗ ¯q q C (P ; R) := {C (P ; R), δ }q∈Z, ¯∗ ¯q q C∞(P ; R) := {C∞(P ; R), δ∞}q∈Z, where we have put ¯q ¯q q q C (P ; R) = C∞(P ; R) = 0 and δ = δ∞ = 0 for q < 0.

4.2.4 Remark: By Theorem 4.2.3, the complex C¯∗(P ; R) is free.

Now let f : P → Q, where Q is compact, be a continuous map. For any q > 0, f induces a homomorphisms

#q q q #q q q f :Φ (Q; R) → Φ (P ; R), f∞ :Φ∞(Q; R) → Φ∞(P ; R) deﬁned by the formula #q f (ψ)(x0, ..., xq) = ψ(f(x0, ..., f(xq)) q q for ψ ∈ Φ (Q; R) (resp. Φ∞(Q; R)) and x0, ...xq ∈ P . Clearly this deﬁnition is correct. q q Moreover it is standard to check that, for each ψ ∈ Φ (Q; R) (or ψ ∈ Φ∞(Q; R))

|f #q(ψ)| ⊂ f −1|ψ| and #q −1 |f∞ (ψ)| ⊂ f |ψ|. #q q #q q q q Therefore f maps Φ0(Q; R) (resp. f maps Φ∞,0) into Φ0(P ; R) (resp. Φ∞,0(P ; R)). Hence f induces homomorphisms

∗q ¯q ¯q ∗q ¯q ¯q f : C (Q; R) → C (P ; R), f : C∞(Q; R) → C∞(P ; R) 142 4. Various construction of homology and cohomology theories of Čech type given by f ∗q(v) = [f #q(ψ)] for v = [ψ] ∈ C¯q(Q; R), ψ ∈ Φq(Q; R). It is easy to see that

q #q #q+1 q q #q #q+1 q δ f = f δ , δ∞f∞ = f∞ δ∞.

∗q ∗q ¯∗ ¯∗ Thus {f }q∈Z (resp. {f∞ }q∈Z) is a cochain map between complexes C (Q; R) and C (P ; R) ¯∗ ¯∗ (resp. C∞(Q; R) and C∞(P ; R)), i.e., Let Q ⊂ P be closed. For any q > 0, the inclusion i : Q,→ P induces the homomorphism ∗q q q ∗q q q ∗q i : C¯ (P ; R) → C¯ (Q; R) (and i∞ : C¯∞(P ; R) → C¯∞(Q; R)). It is easy to see that i is an epimorphism. Let C¯q(P,Q; R) := ker i∗q. In a similar way we may deﬁne ¯q ∗q C∞(P,Q; R) := ker i∞.

4.2.5 Remark: There is another description of C¯q(P,Q). Namely let

Φq(P,Q; R) = {ϕ ∈ Φq(P ; R) | |i#q(ϕ)| = ∅}.

It is clear that ¯q q q C (P,Q; R) = Φ (P,Q; R)/Φ0(P ; R).

#q #q Since {i }q∈Z (resp. {i∞ }q∈Z) is a cochain map, we see that

δ(C¯q(P,Q; R)) ⊂ C¯q+1(P,Q; R) and q ¯q q+1 δ∞(C∞(P,Q; R)) ⊂ C∞ (P,Q; R); hence again we have two cochain complexes

¯∗ ¯q q ¯∗ ¯q C (P,Q; R) := {C (P,Q; R), δ }q∈Z , C∞(P,Q; R) := {C∞(P,Q; R), δ∞}q∈Z.

The cohomology H∗(C¯∗(P,Q; R)) of the (cochain) complex C¯∗(P,Q; R) is denoted by H¯ ∗(P,Q; R) and called the Alexander-Spanier cohomology of (P,Q) (with coeﬃcients in R) based on ﬁnite- valued cochains. We also put ¯ ∗ ∗ ¯∗ H∞(P,Q; R) := H (C∞(P,Q; R)). This is the Alexander-Spanier cohomology of (P,Q) (based on all cochains).

4.2.6 Theorem: (Keese-Gordon) Given compact pair (P,Q), the inclusion C¯∗(P,Q; R) → ¯∗ C∞(P,Q; R) indices an isomorphism ¯ ∗ ∼ ¯ ∗ H (P,Q; R) = H∞(P,Q; R). 4.2. Alexander-Spanier cohomology and Massey homology 143

¯ ∗ ¯ ∗ From now on we forget about H∞(P,Q; R) and will be concerned only with H (P,Q; R). ¯ ∗ 4.2.7 Remark: (1) The construction of H∞(P,Q; R) may be performed for arbitrary spaces P,Q (not necessarily compact). (2) Of course, given an arbitrary R-module G one may consider

∗ ∗ ∗ H¯ (P,Q; G) := H (C¯ (P,Q; R) ⊗R G)

¯ ∗ ∗ ¯∗ H∞(P,Q; G) := H (C∞(P,Q; R) ⊗R G) (the latter construction in case of arbitrary spaces yields the Alexander-Spanier cohomology of (P,Q) with coeﬃcients in G. The following result is of the utmost importance. For a compact pair (P,Q) and q > 0, let

T q(P,Q; R) := {u ∈ C¯q(P,Q) | |u| ∩ Q = ∅}.

Observe that T q(P,Q; R) ⊂ C¯q(P,Q; R). For if u ∈ C¯q(P ; R), u = [ϕ] where ϕ ∈ Φq(P ; R) (then |u| = |ϕ|), and |ϕ| ∩ Q = ∅, then |i#q(ϕ)| = ∅, i.e., i∗q(ϕ) = 0 in C¯q(Q). Indeed if there is x ∈ Q and x ∈ |i#q(ϕ)|, then for any neighborhood V of x in P , V ∩Q is a neighborhood of x in Q and there are points x0, ..., xq ∈ V ∩Q #q such that i (ϕ)(x0, ..., xq) 6= 0; thus ϕ(i(x0), ..., i(xq)) 6= 0; thus x ∈ |ϕ|: contradiction. In general modules T q(P,Q; R) and C¯q(P,Q; R) are diﬀerent. It is however clear that δq : q q+1 ∗ q q T (P,Q; R) → T (P,Q; R); thus T (P,Q; R) := {T (P,Q; R), δ }q∈Z is cochain complex. 4.2.8 Theorem: (Massey) The inclusion T ∗(P,Q; R) ⊂ C¯∗(P,Q; R) induces an isomorphisms

H∗(T ∗(P,Q; R)) =∼ H¯ ∗(P,Q; R).

We will provide no proof of this theorem (see [9, Thorem I.1.4]), but we shall use it quite often. Namely given a ∈ H¯ q(P,Q; R), a is treated as an element of Hq(T ∗(P,Q; R)), i.e., a is the cohomology class of some cocycle u ∈ T q(P,Q; R) (this means that u ∈ C¯q(P ), |u| ∩ Q = ∅} and δ(u) = 0 in C¯q+1(P ; R), i.e., δq(ϕ) = ∅ where u = [ϕ], ϕ ∈ Φq(P ; R)). The next observation explains the nature of H∗(T ∗(P,Q; R)). Let U := P \ Q. The set U is q not compact but it is locally compact. Let Φc(U; R) denotes the set of all ﬁnitely-valued functions q+1 q ϕ : U → R having compact supports. As before Φc0(U; R) is a submodule of functions with empty supports. We deﬁne

¯q q q Cc (U; R) := Φc(U; R)/Φc0(P ; R).

q q q+1 It is clear that δ : C¯c (U; R) → C¯c (U; R). Therefore we have another cochain complex

¯∗ ¯q q Cc (U; R) := {Cc (U; R), δ }q∈Z.

Its cohomology ∗ ¯∗ ¯ ∗ H (Cc (U; R)) =: Hc (U; R).

#q q q Let j : U → P . It is easy to see that j :Φc(P ) → Φ (U) and that

∗q q ¯q j : T (P,Q; R) → Cc (U; R). 144 4. Various construction of homology and cohomology theories of Čech type

Massey [9, Lemma I.1.3] shows that in fact j∗q is an isomorphism. Therefore

¯ ∗ ∼ ∗ ∗ ∼ ¯ ∗ H (P,Q; R) = H (T (P,Q; R)) = Hc (P \ Q; R).

0 0 0 0 If f :(P,Q) → (P ,Q ) is a continuous map of compact pairs, then f ◦ i = i ◦ f|Q where i : 0 0 ∗q ∗q ∗q 0∗q ∗q ¯q 0 0 ¯q Q ,→ P is the inclusion. Thus, for any q, i f = f|Q i . Thus f : C (P ,Q ) → C (P,Q). ∗ ∗q ¯∗ 0 0 ¯∗ This shows that f = {f }q>0 : C (P ,Q ) → C (P,Q) is a well-deﬁned cochain map and f induces the homomorphism

¯ q ¯ q 0 0 ¯ q H (f)H (P ,Q ; R) → H (P,Q; R), q > 0.

4.2.B Massey homology

Now we pass to the construction of the Massey homology. Suppose again that (P,Q) is a compact pair and let G be an arbitrary R-module. For any q ∈ Z, let q C¯q(P,Q; G) := HomR(C¯ (P,Q; R),G).

We also deﬁne for q > 0,

∂q : C¯q(P,Q; G) → C¯q−1(P,Q; G) by q−1 ∂q(u)(ϕ) = u(δ (ϕ)) q−1 for u ∈ C¯q(P,Q; G) and ϕ ∈ C¯ (P,Q), i.e.,

q−1 ∂q := HomR(δ , 1G).

It is clear that ∂q−1∂q = 0; hence ¯ ¯ C∗(P,Q; G) := {Cq(P,Q; G), ∂q}q>0 is a chain complex.

The corresponding homology H∗(C¯∗(P,Q; G)) of the (chain) complex C¯∗(P,Q; G) is denoted by H¯∗(P,Q; G) and called the Massey homology group with coeﬃcients in G. ∗q ¯ ¯ 0 0 Evidently f∗ := {Hom(f , 1G)}q∈Z : H∗(P,Q; G) → H∗(P ,Q ) is a well-deﬁned chain map and f induces a homomorphism

0 0 H¯q(f): H¯q(P,Q; G) → H¯q(P ,Q ; R).

4.2.C Eilenberg-Steenrod axioms

We are going to check the Eilenberg-Steenrod axioms for both H¯∗(·, ·; R) and H¯∗(·, ·; G). First observe that H¯ ∗(·, ·; R) is a cofunctor on the category of compact spaces with values in (graded) ModR. At the same time H¯∗(·, ·; G) is a functor. q (1) Let us observe that H¯ (P ; R) = H¯q(P ; G) = 0 if q < 0. (2) If P is a one point space, then for all q > 0 q ∼ q ¯q ∼ Φ (P ; R) = R, Φ0(P ; R) = 0, C (P ; R) = R. 4.2. Alexander-Spanier cohomology and Massey homology 145

This follows that R if q = 0; H¯ q(P ; R) = 0 if q > 0 and G if q = 0; H¯ (P ; G) = q 0 if q > 0. This establishes the dimension axiom for both (co)homology theories. (3) Let (P,Q) be an arbitrary compact pair and let i : Q → P be the inclusion. The very deﬁnition show that, for all q > 0, the sequence

∗q 0 −→ C¯q(P,Q; R) −→ C¯q(P ; R) −→i C¯q(Q; R) −→ 0 is exact. In other words, we have a short exact sequence of cochain complexes

∗q 0 −→ C¯∗(P,Q; R) −→ C¯∗(P ; R) −→i C¯∗(Q; R) −→ 0.

Therefore by the general theorem on the existence of connecting homomorphism for cohomology ¯ q ¯ q+1 of cochain complexes we get that, for each q > 0, there is δ : H (Q; R) → H (P,Q; R) such that the following (long) sequence

H¯ q(i) ... −→ H¯ q(P ; R) −→ H¯ q(Q; R) −→δ H¯ q+1(P,Q; R) −→ H¯ q+1(P ; R) −→ ... is exact. ∗ The last arrow is induced be the epimorphism C¯∗(P,Q; R) → C¯ (P ; R) induced by the inclusion Q ⊂ P . However it is worthwhile to observe that it is identical to the homomorphism on the homological level induced by the inclusion j : P = (P, ∅) → (P,Q). Moreover, again by the general theorem on the connecting homomorphism for cochain complexes we get that δ is natural in the following sense: given a continuous map f :(P,Q) → (P 0,Q0) the following diagram δ H¯ q(Q0; R) / H¯ q+1(P 0,Q0; R)

H¯ q(f¯) H¯ q+1(f) H¯ q(Q; R) / H¯ q+1(P,Q; R). δ This shows the exactness axiom for H¯ ∗(·, ·; R).

In order to show the corresponding axiom for H¯∗(·, ·; G) we need to recall some remarks. Let (P,Q) be a compact pair; in order to explain the existence of a natural transformation ∂ : H¯∗(P,Q; G) → H¯∗−1(Q; G) := H¯∗−1(Q, ∅; G) we shall proceed as follows. Even thought the cofunctor HomR(· ,G) is, in general, not exact, given a short exact sequence 0 → A → B → C → 0 of groups and homomorphisms, if C is a free group, then the sequence is split and, in particular, the corresponding sequence

0 → HomR(C,G) → HomR(B,G) → HomR(A, G) → 0

∗q ∗q ¯q j ¯q i ¯q is exact. For any q > 0, the sequence 0 → C (P,Q; R) −→ C (P ; R) −→ C (Q; R) −→ 0, where i∗, j∗ are induced by the inclusions i : Q → P , j : P → (P,Q), respectively, is exact. Since q i∗ j∗ the module C¯ (Q) is free, the sequence 0 −→ C¯q(Q; G) −→ C¯q(P ; G) −→ C¯q(P,Q; G) −→ 0 is exact, too. The usual construction of the connecting homomorphism for this sequence yields 146 4. Various construction of homology and cohomology theories of Čech type the existence, naturality of ∂(P,Q) and the exactness of the long homology sequence of the pair (P,Q). (4) Now we shall check the homotopy axiom. It is suﬃcient to show the given continuous q q map f, g : P → Q of compact spaces, if f ' g, then H¯ (f) = H¯ (g) and H¯q(f) = H¯q(g) for any q > 0.

As usual, for t ∈ I, let ht : P → P × I be given by ht(x) = (x, t) for x ∈ P . It is suﬃcient ∗ ∗ to show that H¯ (h0) = H¯ (h1). Let, for t ∈ I, jt : P × P × {t} be given by jt(x) = (x, t), x ∈ P , and let it : P × {t} → P × I be the inclusion. For each t ∈ I, the diagram

jt P / P × {t} GG GG GG it ht GG G# P × I is commutative; hence also the diagram

H¯ q(P ) o H¯ q(P × {t}) (∗) H¯ q(j ) fMMM t O M q MM H¯ (it) ¯ q MM H (ht) MMM H¯ q(P × I)

q is commutative. Clearly jt is a homeomorphism; thus H¯ (jt) is an isomorphism. This implies that q q ker H¯ (ht) = ker H¯ (it).

q q Let π : P × I → P be the projection. Then π ◦ ht = id and H¯ (ht) ◦ barH (π) is the identity on H¯ q(P ). This implies that

q ∼ q q q q 1 H¯ (P × I) = Im H¯ (π) ⊕ ker H¯ (ht) = Im H¯ (π) ⊕ ker H¯ (it)( ).

q q q q Observe also that, for any t, r ∈ I, H¯ (ht)H¯ (π) = H¯ (ir)H¯ (π), so

¯ q ¯ q H (ht)|Im H¯ q(π) = H (ir)|Im H¯ q(π). (∗∗)

q We claim that for any t ∈ I and a ∈ ker H¯ (ht), there is a neighborhood N of t such that q a ∈ ker H¯ (hr) for all r ∈ N. q q Choose t ∈ I, a ∈ ker H¯ (ht) = ker H¯ (it) and consider a part of the long cohomology sequence of the pair (P × I,P × {t}),

j H¯ q(i ) ... −→ H¯ q(P × I,P × {t}) −→ K¯ q(P × I) −→t H¯ q(P × {y}) −→ ..., where j is the homomorphism induced by the inclusion P × I → (P × I,P × {t}), The exactness implies that there is b ∈ H¯ q(P × I,P × {t}) such that j(b) = a.

1Observe that in general given modules A, B and homomorphisms α : A → B and β : B → A such that β ◦ α = idA, then B = Im α ⊕ ker β. This is implied by the following argument. Clearly α is a monomorphism and β is an epimorphism. Therefore the sequence 0 −→ A −→α B −→ B/Im α −→ 0 is exact and, according to the assumption, it is split. Thus B =∼ Im α ⊕ ker β since B/Im α =∼ ker β. This last isomorphism is deﬁned as follows [b]Im α 7→ b − α(β(b)). 4.2. Alexander-Spanier cohomology and Massey homology 147

Now we shall need the following lemma:

4.2.9 Lemma: Let (X,A) be a compact pair. For any a ∈ H¯ q(X,A) there is a compact neighborhood U of A such that a ∈ Im {H¯ q(X,U) → H¯ q(X,A)} induced by the inclusion (X,A) → (X,U). Proof: Let u ∈ T q(X,A) be a cocycle representing a, u = [ϕ], ϕ ∈ Φq(X) and |u| = |ϕ| ⊂ X \A. Then |ϕ| ∩ A = ∅. Therefore there is a compact neighborhood U such that U ∩ |ϕ| = ∅. Hence q q u ∈ T (X,U) and δ (u) = 0. Hence we have the assertion. Let us return to the proof. In view of the lemma, there is a compact neighborhood U of P × {t} such that b ∈ Im {H¯ q(P × I,U) → H¯ q(P × I,P × {t})}. Since P is compact, there is a compact neighborhood N of t such that P × N ⊂ U. Therefore b ∈ Im {H¯ q(P × I,P × N) → H¯ q(P × I,P × {t})}. Hence a belongs to the image of the homomorphism H¯ q(P × I,P × N) → H¯ q(P × I) induced by the inclusion P × I → (P × I,P × N). This implies that a belongs to q q q ker{H¯ (P × I) → H¯ (P × N). Therefore it also belongs to ker H¯ (ir) for any r ∈ N. This implies that, for each t ∈ I and any a ∈ H¯ q(P × I) there is a compact neighborhood N 0 of t such that q q H¯ (ir)(a) = H¯ (it) whenever r ∈ N 0. q q Indeed: observe that a = a1 + a2, where a1 ∈ Im H¯ (π) and a2 ∈ ker H¯ (it). In view of 0 q the above argument, there is a compact neighborhood N of t such that a2 ∈ ker H¯ (ir) for any r ∈ N 0. Hence – in view of (∗∗) – we get

q q q q q H¯ (ir)(a) = H¯ (ir)(a1) + H¯ (ir)(a2) = H¯ (ir)(a2) = H¯ (it)(a1) q q q = H¯ (it)(a1) + H¯ (it)(a2) = H¯ (it)(a).

The commutativity of the diagram (∗) and the connectedness of I completes the proof of the equality q q H¯ (h0) = H¯ (h1).

In order to prove the homotopy axiom for H¯∗(·, ·; G) we need some auxiliary lemmata.

0 0 4.2.10 Lemma: Suppose that C∗, c∗ are free chain complexes. If ϕ∗ : C∗ → C∗ is a chain map such that, for each q ∈ Z, Hq(ϕ∗) is an isomorphism, then ϕ∗ is a chain equivalence. The proof is purely algebraic and can be ﬁnd in [9].

4.2.11 Remark: It is important to note the same result holds for cochain complexes and cochain maps.

4.2.12 Lemma: Let ϕ∗, ψ∗ : C∗ → C0∗ be cochain homotopic cochain maps of cochain complexes. For any R-module G, chain maps

∗ ∗ 0∗ ∗ HomR(ϕ , 1G), HomR(ψ , 1G) : HomR(C ,G) → HomR(C ,G) are (chain) homotopic. ∗ q If ϕ is a cochain equivalence, then {HomR(ϕ , 1G)}q∈Z is a chain equivalence. q q+1 0q ∗ ∗ Proof: Let {D : C → C })q ∈ Z be a cochain homotopy joining ϕ to ψ . Then q 0q q+1 HomR(D , 1G) : Hom(C ,G) → HomR(C ,G)

∗ ∗ is a chain homotopy joining HomR(ϕ , 1G) to HomR(ψ , 1G). The last assertion is a simple corollary of the lemma. 148 4. Various construction of homology and cohomology theories of Čech type

Now we are ready to show the homotopy axiom for H¯ (·, ·; G). Suppose that P is compact and let, for t ∈ I, ht : P → P × I be given by ht(x) = (x, t), x ∈ P . Moreover let π : P × I → P be the projection. Let us consider the cochain maps

h# : C¯∗(P × I; R) → C¯∗(P ; R), π# : C¯∗(P ; R) → C¯∗(P × I; R).

Observe that π ◦ ht = idP and, for each (x, s) ∈ P × I, ht ◦ π(x, s) = (x, t). Thus ht ◦ π is homotopic to idP ×I . Therefore ¯ ∗ ¯ ∗ ¯ ∗ H (π ◦ ht) = H (ht)H (π) = idH¯ ∗(P ) and ¯ ∗ ¯ ∗ ¯ ∗ ¯ ∗ idH¯ ∗(P ×I) = H (idP ×I ) = H (ht ◦ π) = H (π)H (ht). ∗ ∗ In other words H¯ (π) and H¯ (ht), t ∈ I are isomorphisms. Taking into account the fact that ¯∗ ¯∗ # # complexes C (P ; R) and C (P × R) are free, we get that cochain maps π and ht are cochain equivalences. # # The second lemma implies that chain maps HomR(π , 1G) and HomR(ht , 1G) are chain equivalences, too. Therefore

∗ H¯ (π): H¯∗(P × I; G) → H¯∗(P ; R), H¯∗(ht): H¯∗(P ) → H¯∗(P × I; R) are isomorphisms. Moreover, for any t ∈ I,

−1 H¯∗(ht) = [H¯∗(π)] .

In particular H¯∗(h0) = H¯∗(h1). This concludes the proof of the homotopy axiom.

4.2.D Excision axiom

∗ We shall pay special attention to the excision axiom for H¯ (·, ·; R) and H¯∗(·, ·; G). Let X be a compact space and A ⊂ X be closed. Consider the system of all closed neighborhoods {Uα}. The set {α} of indices is directed by the relation α 6 β if and only if Uβ ⊂ Uα. ¯ ∗ ¯ ∗ Given α 6 β we have a homomorphism H (Uα) → H (Uβ) induced by the inclusion. It is ∗ easy to see that in this way we have obtained a direct system {H¯ (Uα)} of (graded) R-modules. ∗ ∗ For each α, A ⊂ Uα; thus the homomorphism H¯ (Uα) → H¯ (A) is well-deﬁned and the family ∗ ∗ {H¯ (Uα) → H¯ (A)} is compatible with this system. Hence there is a unique homomorphism

lim H¯ ∗(U ) → H¯ ∗(A). −→ α α We say that A is taut in X if this homomorphism is an isomorphism.

4.2.13 Theorem: Any closed set A ⊂ X is taut. Proof: For any α let us consider a diagram

... / H¯ q(X,A) / H¯ q(X) / H¯ q(A) / ... O O O

q q q ... / H¯ (X,Uα) / H¯ (X) / H¯ (Uα) / ... 4.2. Alexander-Spanier cohomology and Massey homology 149

in which rows are parts of long cohomology sequences of pairs (X,A) and (X,Uα), respectively, and vertical arrows are induced by the respective inclusions.This diagram is commutative, Similarly, if α 6 β, we have the diagram

q q q ... / H¯ (X,Uβ) / H¯ (X) / H¯ (Uβ) / ... O O O

q q q ... / H¯ (X,Uα) / H¯ (X) / H¯ (Uα) / ...

Suppose that lim H¯ q(X,U ) ∼ H¯ q(X,A). This together with the exactness of the functor of −→ α = direct limit and the ﬁve lemma will show the assertion. To show the above claim we have to show two things (comp. section concerning direct limits) : (a) For any cohomology class a ∈ H¯ q(X,A), there is a compact neighborhood U of A such that a ∈ Im {H¯ q(X,U) → H¯ q(X,A)}. This has been established already in Lemma 4.2.9. (b) Given a compact neighborhood V of A such that a ∈ ker{H¯ q(X,V ) → H¯ q(X,A)}, there is a compact neighborhood U of A, U ⊂ V , such that a ∈ ker{H¯ q(X,V ) → H¯ q(X,U)}. Let u ∈ T q(X,V ) be a cocycle representing a. Let j :(X,A) → (X,V ) be the inclusion. We know that H¯ q(j)(a) = 0. On the other hand H¯ q(j)(a) is the cohomology class of [j#(ϕ)]. Hence [j#(a)] is a coboundary, i.e., there is a cochain v ∈ T q−1(X,A) such that [j#(ϕ)] = δq−1(v). Clearly |v| ⊂ X \A. Therefore there is a compact neighborhood U of A such that |ψ|∩U = ∅. We may assume that U ⊂ V . It is clear that v ∈ T q−1(X,U). Now it is easy to see that δ(v) = i∗(u) ¯ q ¯ q where i :(X,U) → (X,V ) is the inclusion. Hence H (i)(a) = 0 in H (X,U). 4.2.14 Theorem: Suppose that (X,A) is a compact pair, U ⊂ A is open. Then the inclusion j :(X \ U, A \ U) → (X,A) induces an isomorphism

H¯ q(j): H¯ q(X,A) → H¯ q(X \ U, A \ U)

. Proof: First assume additionally that there is an open set W ⊃ U such that cl W ⊂ int A. We shall show that then j∗ : C¯∗(X,A) → C¯∗(X \ U, A \ U) is an isomorphism. Recall that for a compact pair (P,Q),

¯q ∗ q C (P,Q) = Φ (P,Q)/Φ0(P ). Therefore we have an exact sequence

q q ¯q 0 −→ Φ0(P ) −→ Φ (P, q)) −→ C (P,Q) −→ 0. Let consider the following diagram

q q ¯q 0 / Φ0(X) / Φ (X,A) / C (X,A) / 0

k j∗ q 0 / Φ (X \ U) / Φq(X \ U, A \ U) / C¯q(X \ U, A \ U) / 0 0 (1) λ 150 4. Various construction of homology and cohomology theories of Čech type where k is induced by j (on the level of Φ(·, ·)) and λ is the quotient homomorphism. Since the −1 q rows are exact, it is suﬃcient to prove that λ ◦ k is an epimorphism and that k (Φ0(X \ U)) = q Φ0(X). If ϕ ∈ Φq(X \ U, A \ U), then let us deﬁne ϕ¯ ∈ Φq(X) by the formula 0 if xi ∈ W for some 0 6 i 6 q; ϕ¯(x0, ..., xq) = ϕ(x0, ..., xq) if x0, ..., xq ∈ X \ W.

Since ϕ ∈ Φq(X \ U, A \ U), the restriction of ϕ to A \ U has empty support. This means that each point x ∈ A \ U has a neighborhood Vx (in A \ U) such that ϕ(x0, ..., xq) = 0 whenever x0, ..., xq ∈ Ux. Let Ux := W ∩ Vx, x ∈ A \ U.

It is clear that U := {Ux}x∈A\U is an open cover of A and ϕ¯(x0, ...xq) =) if x0, ..., xq ∈ Ux for some x ∈ A \ U. This means that the restriction to A of ϕ¯ has empty support. In other words ϕ¯ ∈ Φq(X,A). By the deﬁnition of ϕ¯ it follows that ψ := k(ϕ ¯) − ϕ has the following property: if x0, ..., xq ∈ Vx ∩ int A (for some x ∈ A \ U) or x0, ...xq ∈ X \ W , then ψ(x0, ..., xq) = 0. This shows that ψ belongs to the image of the arrow denoted by (1), i.e., it belongs to the kernel of λ. This implies that λ(k(ϕ ¯)) = λ(ϕ). Since λ is an epimorphism, we gather that so is λ ◦ k. Now we shall see the other part of the above assertion. Namely let ϕ ∈ Φq(X,A) and suppose q that k(ϕ) ∈ Φ0(X \ U). Since the restriction of ϕ to A has empty support, there is an open cover U1 = {U1} of A such that ϕ(x0, ..., xq) = 0 whenever x0, ..., xq ∈ U1 for some U1 ∈ U1. On the other hand there is there is an open cover U2 of X \ U such that k(ϕ)(x0, ...xq) = 0 if x0, ..., xq ∈ U2 for some U2 ∈ U2. Let V1 := {U1 ∩ int A | U1 ∈ U1}, V2 := {U2 ∩ (X \ cl U) | U2 ∈ U2}.

Then V := V1 ∪ V2 is an open cover of X and ϕ(x0, ..., xq) = 0 if x0, ..., xq ∈ V for some V ∈ V. q I.e., |ϕ| = ∅ and ϕ ∈ Φ0(X) Now let us pass to the general situation. Recall that (X,A) is a compact pair and U ⊂ A is open and j :(X \ U, A \ U) → (X,A) is the inclusion. Let {Uα} be the family of all compact neighborhoods of A (in X). Then {Vα := Uα \ U} consists of compact neighborhoods of A \ U (in X \ U) and this family is coﬁnal in the family of all compact neighborhoods of A \ U. It is clear that U ⊂ A ⊂ int Uα. Thus there is an open Wα such that

U ⊂ A ⊂ Wα ⊂ cl Wα ⊂ int Uα.

Hence – in view of the above part – we have

q ∼ q q H¯ (X,Uα) = H¯ (X \ U, Uα \ U) = H¯ (X \ U, Vα).

On the other, by tautness,

H¯ q(X,A) = lim H¯ q(X,U ) ∼ lim H¯ q(X \ U, V ) = H¯ q(X \ U, A \ U). −→ α = −→ α

4.2.15 Remark: The ﬁrst part of the proof implies the classical excision axiom for H¯ ∗. If (X,A) is a compact pair, U ⊂ A is open and cl A ⊂ int A, then there is an isomorphism

H¯ ∗(X,A) =∼ H¯ ∗(X \ U, A \ U). 4.2. Alexander-Spanier cohomology and Massey homology 151

Indeed, since cl U ⊂ int A, there is an open set W such that U ⊂ cl U ⊂ W ⊂ cl W ⊂ int A because X is normal.

4.2.16 Corollary: Suppose that a compact X = A ∪ B, where A, B are closed. Then the inclusion (B,A ∩ B) → (X,A) induces an isomorphism

H¯ q(X,A) =∼ H¯ q(B,A ∩ B).

Given closed sets A, B ⊂ X, we have an isomorphism

H¯ q(A ∪ B,A) =∼ H¯ q(B,A ∩ B).

In other words any closed dyad {A, B} in a compact space is excisive Proof: It is enough to put U := X \ B and to apply the above theorem. The second statement is just a reformulation of the ﬁrst one.

Now we shall discuss the excision property for H¯∗(·, ·; G). Using Lemmas 4.2.10 and 4.2.12 we get immediately that

4.2.17 Theorem: If (X,A) is a compact pair, U ⊂ A is open, then the inclusion (X\U, A\U) → (X,A) induces an isomorphism

∼ H¯∗(X \ U, A \ U) = H¯∗(X,A).

If A, B are closed subsets of a compact space X, then the inclusion (B,A ∩ B) → (A ∪ B,A) induces an isomorphism ∼ H¯∗(B,A ∩ B) = H¯∗(A ∪ B,A).

Eilenberg and Steenrod show that this sort of excision odes not hold for the singular homology. We will show something even stronger:

4.2.18 Theorem: Let X,Y be compact spaces, A ⊂ X, B ⊂ Y closed subsets. Suppose that f :(X,A) → (Y,B) is relative homeomorphism (i.e., continuous and such that f : X \A → Y \B is a homeomorphism (2)). Then

H¯∗(f): H¯∗(X,A) → H¯∗(Y,B) is an isomorphism.

Proof: Let {Nα} be the family of all compact neighborhoods of B. It is easy to see that −1 {Uα := f (Nα)} is a family of compact neighborhoods of A coﬁnal in the family of all compact neighborhoods of A.

2Observe that if f maps X \ A in 1-1 fashion onto Y \ B, then f is a relative homeomorphism. 152 4. Various construction of homology and cohomology theories of Čech type

4.2.E Continuity of the Massey homology

4.3 The Massey homology with compact supports

4.4 The Alexander-Spanier cohomology theory with compact supports

4.5 The de Rham-Čech Theorem

4.6 Final remarks

4.6.1 Remark: (i) The same approach as in section 1, allows to deﬁne the Massey homology ¯ c with compact supports H∗(X,A; G) for any topological pair and an arbitrary abelian group G. ¯ c Theory H∗ satisﬁes all the Eilenberg-Steenrod axioms. It is also worth to observe that, for any topological pair (X,A),

¯ c ¯c H∗(X,A; G) = H∗(C∗(X,A; G)) where the chain complex C¯c(X,A) := lim C¯ (P,Q) and the direct limit is taken with respect to ∗ −→ ∗ the directed set C(X,A). (ii) If (P,Q) is a compact pair, a group G is algebraically compact (e.g. G = F is a ﬁeld), ∼ then there is an isomorphism H¯∗(P,Q; F) = Hˇ∗(P,Q), where Hˇ∗(P,Q) is the Čech homology of (P,Q) introduced in Section ?? (see [12]) (3). Consequently, for any topological pair (X,A), ¯ c ∼ ˇ c ˇ c there is an isomorphism H∗(X,A; F) = H∗(X,A) where H∗ stands for the Čech homology with compact supports introduced in Section ??. ¯ c (iii) It may be shown, see [9, §9.6], that if a paracompact space X is HLC, then H∗(X; G) = s H∗ (X; G) (see also [16, Cor. 11.9]; the proof in [16] given in terms of the Borel-Moore homology is true for the Massey homology). In particular, if X is an open subset of an ANR, then ¯ c s H∗(X; G) = H∗ (X; G).

3 It is however not clear whether a formula analogous to (??) holds for H¯∗(P,Q; G). Chapter 5 Fiber Bundles and Coverings

5.0.2 Branches of the logarithm and the argument: The exponential function exp : C → C given by the formula z x exp z = e := e (cos y + i sin y), z = x + iy ∈ C z 0 z 0 z z is holomorphic, (e ) = e , e = 1 and e 1 = e 2 if and only if z1 ≡ z2(mod 2πi). It is clear that actually exp : C → C \ 0. We claim that exp is a covering. To this aim recall that if z ∈ C \ 0, then the set (the argument of z)

arg z := {ϕ ∈ R | z = |z|(cos ϕ + i sin ϕ} is nonempty. Note moreover that

arg z1z2 = arg z1 + arg z2

(in the sense of algebraic sum of sets) and if ϕ1, ϕ2 ∈ arg z, then ϕ1 ≡ ϕ2(mod 2π). Hence there is a unique ϕ ∈ (−π, π] ∩ arg z called the principal argument of z and denoted by Arg z. Let u ∈ C \ 0. The set (the logarithm of u) z −1 log u := {z ∈ C | e = u} = (exp) (u) is nonempty since e.g. z = ln |u| + iArg u ∈ log u and if z1, z2 ∈ log u, then z1 ≡ z2(mod 2πi); in other words log u = {z = ln |u| + iArg u + 2kπi | k ∈ Z}.

In order to show that exp is a covering, let u0 ∈ C \ 0 and let U := {u = vu0 | Arg v ∈ −1 (−π, π), v 6= 0}. Evidently U is an open neighborhood of u0. Let Z := (exp) (U); then W is open (as the preimage of an open set under a continuous map) and it is easy to see that [ Z = Zk k∈Z where, for k ∈ Z,

Zk := {z = ln |vu0| + i(Arg u0 + Arg v + 2kπ) | Arg v ∈ (−π, π), v 6= 0}.

For all k ∈ Z, Zk is open (since the function Arg : C \ R− → R is continuous) and Zk ∩ Zm = ∅ whenever k 6= m. Indeed if k 6= m and

z1 = ln |v1u0| + i(Arg u0 + Arg v1 + 2kπ) = z2 = ln |v2u0| + i(Arg u0 + Arg v2 + 2mπ) 154 5. Fiber Bundles and Coverings

where Arg vi ∈ (−π, π), vi 6= 0 for i = 1, 2, then

ln |v1| = ln |v2|, Arg v1 − Arg v2 = 2(m − k)π : contradiction for |Arg v1 − Arg v2| < 2π. Almost the same argument show that, for any k ∈ Z, exp |Zk : Zk → U is a homeomorphism. Let let X be a space and f : X → C \ 0. Any continuous function A : X → R such that f(x) eiA(x) = , x ∈ X, |f(x)| is called a branch of the argument of the function f on X. Similarly, a continuous function L : X → C such that eL(x) = f(x), x ∈ X, is called a branch of the logarithm of f on X. Let us remark that if A is a branch of the argument of f, then L : X → C given by

L(x) = ln |f(x)| + iA(x), x ∈ X, is a branch of the logarithm of f; conversely the function Im L is a branch of the argument provided L is a branch of the logarithm. As to the uniqueness we see immediately that if X is connected, then two branches of the argument (resp. logarithm) of f on X are congruent modulo 2π (resp. 2πi). The question of the existence of branches of the logarithm (equivalently of the argument) of a function f : X → C \ 0 is actually the lifting problem: f admits a branch L : X → C of the logarithm if and only if exp ◦L = f on X, i.e., the covering exp : C → C \ 0 admits a lifting of f. In the context of Path-Lifting property, we have that:

5.0.3 Fact: any path w : I → C \ 0 may be lifted, i.e., it has a branch of the logarithm. Chapter 6 Brouwer Theorem

6.1 Lemat Spernera

Skończony kompleks symplicjalny K jest n-wymiarową pseudo-rozmaitością jeżeli: (i) każdy jego sympleks jest ścianą pewnego n-wymiarowego sympleksu; (ii) każdy sympleks (n−1)-wymiarowy jest ścianą co najwyżej dwóch sympleksów n-wymiarowych; 0 0 (iii) Jeśli σ, σ ∈ Σ są n-wymiarowe, to istnieje taki ciąg σ = σ0, ..., σm = σ sympleksów n-wymiarowych, że σi−1 oraz σi mają wspólną (n − 1)-wymiarową ścianę dla i = 1, ..., m. Brzegiem takiej pseudo-rozmaitości K jest podkompleks symplicjalny K˙ kompleksu K gen- erowany przez zbiór sympleksów (n − 1)-wymiarowych, które są ścianami dokładnie jednego sympleksu n-wymiarowego. Niech K będzie n-wymiarową pseudo-rozmaitością, n > 0, o niepustym brzegu K˙ , K0 jest podpodziałem symplicjalnym kompleksu K (wiadomo, że K0 jest też pseudo-rozmaitością i wiadomo, że K˙ 0 = K0|K˙ ; oznacza to, że kompleks K˙ 0 jest podpodziałem kompleksu K˙ ), zaś ϕ : K0 → K przekształceniem symplicjalnym takim, że ϕ|K˙ 0 przeprowadza K˙ 0 w K˙ i jest aproksymacją symplicjalną identyczności |K˙ 0| = |K|. Przypomnijmy, że ϕ jest przekształceniem symplicjalnym, tzn. ϕ : V 0 → V i jeśli s0 jest sympleksem w K0, to ϕ(s0) jest sympleksem w K. Poza tym założyliśmy, że ϕ przeprowadza K˙ 0 w K˙ i ϕ|K˙ 0 jest aproksymacją symplicjalną tożsamości |K˙ 0| = |K˙ |, tzn. dla dowolnego x ∈ |K˙ 0| i sympleksu s w K˙ , jeśli x ∈ |s|, to |ϕ|(x) ∈ |s|. Oznacza to, że jeśli x ∈ |K˙ 0| jest punktem (n − 1)-wymiarowej ściany s pewnego n-wymiarowego sympleksu S w K (s jest wtedy ścianą (n − 1)-wymiarową tylko tego jednego sympleksu n-wymiarowego S w K), to |ϕ|(x) jest też punktem tej ściany s. W szczególności, jeśli v jest wierzchołkiem K0, który leży w |K˙ |, a zatem v ∈ |s|, gdzie s jest (n − 1)-wymiarowym sympleksem z K˙ , to ϕ(v) jest jednym z wierzchołków s. Zakładamy, że τ jest ustalonym sympleksem (n − 1)-wymiarowym w K˙ , zaś σ jest tym jedynym sympleksem n-wymiarowym w K, którego (n − 1)-wymiarową ścianą jest τ. (1) Niech σ0 będzie ustalonym sympleksem n-wymiarowym w K0. Oznaczmy a = a(σ0) liczbę jego (n − 1)-wymiarowych ścian, które ϕ przeprowadza na τ. Udowodnimy, że a = 1 wtedy i tylko wtedy, gdy ϕ przeprowadza σ0 na σ; w przeciwnym przypadku a = 0 lub a = 2. 0 Dostateczność: Jeśli ϕ przeprowadza σ na σ, to oczywiście a > 1 i gdyby a > 1, to ϕ nie 156 6. Brouwer Theorem mogłoby przeprowadzać σ0 na σ. Konieczność: jeśli a = 1, to dokładnie jedna ściana sympleksu σ0, powiedzmy τ 0, przechodzi na τ. Pokażemy, że ϕ przeprowadza σ0 ma σ. Oczywiście s = ϕ(σ0) jest sympleksem w K; jedną z jego ścian jest τ, poza tym s musi być ścianą pewnego n-wymiarowego sympleksu S ∈ Σ. Wtedy S = σ bo S ma ścianę s, którego ścianą jest τ, czyli ścianą S jest τ, a σ jest jedynym takim sympleksem. Zatem s jest ścianą σ. Jeśli jest właściwą ścianą, czyli co najwyżej (n − 1)-wymiarową, to musi być równa τ, a to oznacza, że przynajmniej jeszcze jedna ściana (n − 1)-wymiarowa sympleksu σ0 przechodzi na τ: sprzeczne. Zatem s jest n-wymiarowy, tzn. s = S = σ. To kończy dowód. Przypuśćmy, że ϕ(σ0) 6= σ i niech a > 0, tzn. nie mniej niż dwie różne (n − 1)-wymiarowe ściany τ 0 i τ 00 sympleksu σ0 przechodzą na τ. Ponieważ σ0 na n+1 wierzchołków, zaś τ 0 i τ 00 mają po n wierzchołków, to muszą mieć (n − 1) wspólnych wierzchołków. Stąd wynika, że każda inna (n − 1)-wymiarowa ściana sympleksu σ0 musi przejść na ścianę co najwyżej (n − 2)-wymiarową sympleksu σ, nie może więc to być τ. Zatem a = 2.

(2) Pokażemy, że liczba N sympleksów n-wymiarowych w Σ0, które przechodzą na σ jest tej samej parzystości co liczba M sympleksów (n − 1)-wymiarowych w K˙ 0, które przechodzą na τ. Udowodnimy, że obie liczby maja tę samą parzystość co liczba L = P a(σ0), gdzie sumowanie przebiega po zbiorze wszystkich n-wymiarowych sympleksów σ0 w K0. Rozważmy sympleks n-wymiarowy σ0 w K0. Są dwie możliwości: I. ϕ(σ0) = σ; czyli σ0 daje wkład 1 do liczby N. Wtedy też a(σ0) = 1; czyli σ0 daje wkład 1 do liczby L. II. ϕ(σ0) 6= σ. Wtedy wkład do liczby N jest 0; wkład do L jest 0 lub 2. Zatem po wyczerpaniu sympleksów o własności I, wyczerpiemy liczbę N, liczbę L zm- niejszymy o N, a reszta będzie podzielna przez 2. Tak więc N ≡ L(mod 2).

Podobnie analizujemy liczby M i L. Każdemu sympleksowi σ0 w K0 przyporządkowujemy jego wszystkie (n − 1)-wymiarowe ściany τ 0, które przechodzą na τ (o ile istnieją) (takich ścian jest a(σ0)). Wtedy każdy taki (n − 1)-wymiarowy sympleks τ 0 odpowiada jednemu lub dwóm n-wymiarowym sympleksom, w zależności od tego, czy jest on w K˙ 0, czy nie. Mamy więc L ≡ M(mod 2).

To kończy dowód, bo N ≡ L(mod 2) i L ≡ M(mod 2). (3) Pokażemy, że liczba N sympleksów n-wymiarowych σ0 w K0, które przechodzą na σ jest nieparzysta. Zastosujemy indukcję ze względu na n. Gdy n = 0, to natychmiast widać, że N = 1. Przypuśćmy, że twierdzenie jest prawdziwe dla n−1, gdzie n > 1. Rozważamy n-wymiarową pseudo-rozmaitość K i bierzemy ten sympleks τ 0; jest to (n − 1)-wymiarowa pseudo-rozmaitość, zaś K0|τ jest jej podpodziałem, a ϕ|τ jest przekształceniem symplicjalnym, którego obcięcie do brzegu jest aproksymacja symplicjalną identyczności. Zatem z założenia indukcyjnego liczba N 0 (n − 1)-wymiarowych sympleksów brzegu K0|τ, które przechodzą na τ jest nieparzysta. Z poprzedniego zadania liczba N jest tej samej parzystości co N 0. Stąd N jest nieparzysta. (4) Pokażemy, że bryła |K˙ | nie jest retraktem bryły |K|. Przypuśćmy przeciwnie, tzn. istnieje takie przekształcenie ciągłe r : |K| → |K˙ |, że r(x) = x dla x ∈ |K˙ |. Z twierdzenia o aproksymacji symplicjalnej istnieje podpodział K0 kompleksu 158 6. Brouwer Theorem

K oraz aproksymacja symplicjalna ϕ : K0 → K przekształcenia r (tym podpodziałem jest 0 m podpodział barycentryczny K = sd K, gdzie m jest dostatecznie duże, powiedzmy m > M). Oczywiście, skoro ϕ jest aproksymacją symplicjalną r, to ϕ|K˙ 0 przeprowadza K˙ 0 w K˙ i jest aproksymacją symplicjalną tożsamości |K˙ 0| = |K˙ |. Z powyższego wynika, że każdy n-wymiarowy sympleks σ z K jest obrazem jakiegoś n-wymiarowego sympleksu σ0 z K0 (bo liczba sympleksów n-wymiarowych σ0, które na niego przechodzą jest nieparzysta, czyli różna od zero). To jest oczywiście niemożliwe. n n (5) Pokażemy, że każde ciągłe przekształcenie f kuli D = {x ∈ R | kxk 6 1} w siebie ma punkt stały. Przypuśćmy, że tak nie jest, tzn. istnieje takie ciągłe przekształcenie f : Dn → Dn bez punktów stałych. Dla dowolnego x ∈ Dn, rozważmy półprostą o początku f(x) i przechodzącą n−1 n przez x. Niech R(x) oznacza punkt przecięcia tej półprostej ze sferą S = {x ∈ R | kxk = 1}. Wtedy R : Dn → Sn−1 jest odwzorowaniem ciągłym i R(x) = x dla x ∈ Sn−1. Jest to zatem retrakcja. Oczywiście Dn ma triangulację K, która jest pseudo-rozmaitością n-wymiarową (tzn. istnieje pseudo-rozmaitość K i homeomorﬁzm h : |K| → Dn) i której brzeg |K˙ | jest triangulacją sfery Sn−1 (dokładniej f||K˙ | : |K˙ | → Sn − 1 jest homeomorﬁzmem). Niech r := h−1◦R◦h : |K| → |K˙ | wtedy r jest retrakcją, której być nie może. Uzyskana sprzeczność dowodzi tezy. Bibliography

[1] R. Engelking, Topologia Ogólna; [2] W. Rudin, Analiza Funkcjonalna; [3] L. Górniewicz, Analiza matematyczna dla ﬁzyków; [4] W. Kryszewski, Wstęp do nieskończenie-wymiarowej analizy matematycznej i topologii; [5] W. Kryszewski, Metody topologii algebraicznej i geometrycznej w analizie nieliniowej. [6] A. Dold, Lectures on Algebraic Topology; [7] R. Sikorski, Rachunek rózniczkowy i całkowy. [8] S. V. Petkova, On the coincidence of homologies for homologically locally connected spaces (in Russian), Comp. Rend. Acad. Bulg. Sci, 35 (1982), 427–430. [9] W.S. Massey, Homology and cohomology theory, Dekker, New York, 1978. [10] S. Mardešić, J. Segal, Shape theory. The inverse system approach, North-Holland, Amster- dam, 1982. [11] S. Mardešić, Comparison of singular and Čech homology in locally connected spaces, Michi- gan Math. J. 6 (1959), 151–166. [12] V. I. Kuz’minov, I. A. Shvedov, Hyperhomology of the limit of a direct spectrum of complexes and homology groups of topological spaces, Sib. Math. J. 16 (1975), 49–58. [13] S. Eilenberg, N. Steenrod, Foundations of algebraic topology, Princeton University Press, Princeton, 1952. [14] A. Dold, Lectures on Algebraic Topology, Springer-Verlag, Berlin, 1972.

[15] T. Watanabe, The continuity axiom for the Čech homology. In: Geometric topology and shape theory, Lecture Notes in Mathematics 1283, Springer, Berlin, 1987, pp. 221–239. [16] G. E. Bredon, Sheaf Theory, Mc Graw-Hill, New York, 1967. [17] E. H. Spanier, Algebraic Topology, Mc Graw-Hill, New York, 1966.