Algebraic Topology Lecture Notes, Winter Term 2016/17 OTSDAM P EOMETRIE in G

Home , Alexander horned sphere, Comb space, Homotopy group

Christian Bär

Algebraic Topology Lecture Notes, Winter Term 2016/17 OTSDAM P EOMETRIE IN G

Picture on title page taken from www.sxc.hu. Contents

Preface 1

1 Set Theoretic Topology 3 1.1 Typical problems in topology ...... 3 1.2 Some basic deﬁnitions ...... 7 1.3 Compactness ...... 9 1.4 Hausdorﬀ spaces ...... 10 1.5 Quotient spaces ...... 11 1.6 Product spaces ...... 12 1.7 Exercises ...... 13

2 Homotopy Theory 17 2.1 Homotopic maps ...... 17 2.2 The fundamental group ...... 21 2.3 The fundamental group of the circle ...... 28 2.4 The Seifert-van Kampen theorem ...... 40 2.5 The fundamental group of surfaces ...... 52 2.6 Higher homotopy groups ...... 58 2.7 Exercises ...... 75

3 Homology Theory 79 3.1 Singular homology ...... 79 3.2 Relative homology ...... 84 3.3 The Eilenberg-Steenrod axioms and applications ...... 87 3.4 The degree of a continuous map ...... 96 3.5 Homological algebra ...... 105 3.6 Proof of the homotopy axiom ...... 112 3.7 Proof of the excision axiom ...... 116 3.8 The Mayer-Vietoris sequence ...... 123 3.9 Generalized Jordan curve theorem ...... 126 3.10 CW-complexes ...... 130 3.11 Homology of CW-complexes ...... 135 3.12 Betti numbers and the Euler number ...... 140 3.13 Incidence numbers ...... 143 3.14 Homotopy versus homology ...... 146 3.15 Exercises ...... 152

Sources list for pictures 157

i Contents

Bibliography 159

Index 161

ii Preface

These are the lecture notes of an introductory course on algebraic topology which I taught at Potsdam University during the winter term 2016/17. The aim was to introduce the basic tools from homotopy and homology theory. Choices concerning the material had to be made. Since time was too short for a reasonable discussion of cohomology theory after homology had been treated, I decided to skip cohomology altogether and instead included more material from homotopy theory than is often done. In particular, there is a detailed discussion of higher homotopy groups and the long exact sequence for Serre ﬁbrations.

The necessary prerequisites of the students were rather modest. The course contains a quick introduction to set theoretic topology but a certain acquaintance with these concepts was certainly helpful. Familiarity with basic algebraic notions like rings, modules, linear maps etc. was assumed.

These notes are based on the lecture notes of a course I taught in 2010. I am grateful to Volker Branding who wrote a ﬁrst draft of those lecture notes and produced most of the pstricks-ﬁgures and to Ramona Ziese who improved those notes considerably.

Potsdam, February 2017

Christian Bär

1 Set Theoretic Topology

1.1 Typical problems in topology

Topology is rough geometry. For example, a sphere is topologically the same as a cube, even though the sphere is smooth and curved while the cube is piecewise ﬂat and has corners. In more precise mathematical terms this means that they are homeomorphic. On the other hand, the sphere is diﬀerent from a torus even topologically.

≈

Here are four versions of a typical question that one asks in mathematics. Fix the integers 1 n < m. ≤ 1.) Does there exist a linear isomorphism ϕ : Rm Rn ? → No, since linear algebra tells us that isomorphic vector spaces have equal dimension. 2.) Does there exist a diffeomorphism ϕ : Rm Rn ? → No, otherwise the differential dϕ(0) : Rm Rn would be a linear isomorphism. → 3.) Does there exist a homeomorphism ϕ : Rm Rn ? → We cannot answer that question yet, but we will develop the necessary tools to find the answer. 4.) Does there exist a bijective map ϕ : Rm Rn ? → Yes. We will now explicitly construct an example for such a map in the case n = 1 and m = 2.

Example 1.1. Since the exponential function maps R bijectively onto R+ = (0, ) it suﬃces to ∞ construct a bijective map ϕ : R+ R+ R+. → ×

3 1 Set Theoretic Topology

Given x > 0 write x as inﬁnite decimal fraction.

x = ... a3a2a1, b1b2b3 ...

Here aj 0, 1,..., 9 and almost all aj are equal to 0. The bj are blocks of the form 0 ... 0cj ∈ { } with cj 1,..., 9 . This representation of x is unique. Now deﬁne ϕ(x) = (ϕ (x),ϕ (x)) ∈ { } 1 2 where ϕ1(x) = ... a5a3a1, b1b3b5 ... ϕ2(x) = ... a6a4a2, b2b4b6 ...

One easily checks that ϕ maps R+ bijectively onto R+ R+. × Let us evaluate the construction for an example. Let x = 1987, 30500735 .... Then we can read oﬀ the aj and the bj as

a = 7, a = 8, a = 9, a = 1, aj = 0 for j 5 1 2 3 4 ≥ b1 = 3, b2 = 05, b3 = 007, b4 = 3, b5 = 5,...

Hence ϕ1(x) = 97, 30075 ... and ϕ2(x) = 18, 053 ...

Remark 1.2. In the continuous world counter-intuitive things can happen which are not possible in the diﬀerentiable world. For example, there exist continuous maps

ϕ : [0, 1] [0, 1] [0, 1] → × which are surjective. Such a map is called a plane-ﬁlling curve.

Example 1.3. The ﬁrst example goes back to Peano [6]. The following example was shortly after given by Hilbert [3] and is known as the Hilbert curve. It is deﬁned by

ϕ(x) := lim ϕn (x) n →∞ where the ϕn are deﬁned recursively as indicated by Hilbert in the pictures:

4 1.1 Typical problems in topology

ϕ1 ϕ2 ϕ3

ϕ4 ϕ5 ϕ6

Remark 1.4. This cannot happen for smooth curves due to a Theorem by Sard which states that for smooth ϕ : [0, 1] R2 the image ϕ([0, 1]) R2 is a zero set. → ⊂

Many typical problems in topology are of the following form:

1.) Given two spaces, are they homeomorphic? To show that they are, construct a homeomorphism. To show that they are not, ﬁnd topological invariants, which are diﬀerent for given spaces.

2.) Classify all spaces in a certain class up to homeomorphisms.

3.) Fixed point theorems.

Example 1.5 (Classification theorem for surfaces). The classification theorem for surfaces states that each orientable compact connected surface is homeomorphic to exactly one in the following infinite list:

5 1 Set Theoretic Topology

S2 (sphere), genus 0

T2 (torus), genus 1

F2 (Pretzel surface), genus 2

F3 (true pretzel), genus 3

b b b

The genus is the number of “holes” in the surface. We will give a more precise deﬁnition later. So the classiﬁcation theorem says that to each g N there exists an orientable compact connected ∈ 0 surface Fg with genus g and each orientable compact connected surface is homeomorphic to exactly one Fg.

Example 1.6 (Brouwer ﬁxed point theorem). Any continuous map f : Dn Dn with → Dn = x Rn x 1 has a ﬁxed point, i.e., there exists an x Dn such that { ∈ |k k ≤ } ∈ f (x) = x.

We give a proof for n = 1. Consider the continuous function g : [ 1, 1] R deﬁned by − → g(x) := f (x) x. Now f 1 implies − | | ≤ g( 1) 1 ( 1) = 0, g(1) 1 1 = 0. − ≥− − − ≤ − By the intermediate value theorem we can ﬁnd an x with g(x) = 0 which is equivalent to f (x) = x.

6 1.2 Some basic deﬁnitions

1.2 Some basic deﬁnitions

First of all let us recall the following

Deﬁnition 1.7. A subset U Rn is called open iﬀ ⊂ x U r > 0 : B(x, r) U ∀ ∈ ∃ ⊂ n r where B(x, r) = y R d(x, y) = x y < r . b x { ∈ | k − k }

The set of open subsets of Rn is called the standard topology of Rn. We write

n n Rn := U R open (R ). T { ⊂ } ⊂ P One easily checks

Proposition 1.8. Rn has the following properties T n (i) , R Rn ∅ ∈ T

(ii) Uj Rn, j J j J Uj Rn ∈T ∈ ⇒ ∈ ∈ T (iii) U , U Rn U S U Rn 1 2 ∈T ⇒ 1 ∩ 2 ∈T

The importance of the concept of open subsets comes from the fact that one can characterize continuous maps entirely in terms of open subsets. Namely, a map f : Rn Rm is continuous 1 1 → iﬀ for each U Rm the preimage f (U) is open, f (U) Rn . This motivates taking open ∈ T − − ∈ T subsets axiomatically as the starting point in topology.

Deﬁnition 1.9. A topological space is a pair (X, X ) with X (X) such that T T ⊂ P (i) , X X ∅ ∈ T

(ii) Uj X, j J j J Uj X ∈T ∈ ⇒ ∈ ∈ T (iii) U , U X U S U X 1 2 ∈T ⇒ 1 ∩ 2 ∈ T

7 1 Set Theoretic Topology

Deﬁnition 1.10. Let (X, X ) be a topological space. Elements of X are called open subsets T T of X. Subsets of the form X U with U X are called closed. \ ∈T

Examples 1.11. 1.) X arbitrary set, X = (X) (discrete topology). All subsets of X are open T P and they are also all closed.

2.) X arbitrary set, X = , X (coarse topology). Only X and are open subsets of X. They T {∅ } ∅ are also the only closed subsets.

3.) Let (X, X ) be a topological space, let Y X be an arbitrary subset. The induced topology T ⊂ or subspace topology of Y is deﬁned by Y := U Y U X . T { ∩ | ∈ T } 4.) Let (X, d) be a metric space. Then we can imitate the construction of the standard topology on Rn and deﬁne the induced metric as the set of all U X such that for each x U there ⊂ ∈ exists an r > 0 so that B(x, r) U. Here B(x, r) = y X d(x, y) < r is the metric ball ⊂ { ∈ | } of radius r centered at x.

Remark 1.12. Let two metrics d1 and d2 be given on a set X. If d1 and d2 are equivalent, i.e., C 0 with ∃ ≥ 1 C− d (x, y) d (x, y) Cd (x, y) x, y X, 2 ≤ 1 ≤ 2 ∀ ∈ then d1 and d2 induce the same topology.

Remark 1.13. Openness is a relative concept. If Y X be an arbitrary subset of a topological ⊂ space X, then Y is in general not an open subset of X but it is always open as a subset of itself (w.r.t. the induced topology). The same remark applies to closed subsets.

It is now clear how to deﬁne continuous maps between topological spaces.

Deﬁnition 1.14. Let (X, X ) and (Y, Y ) be topological spaces. Then a map f : X Y is T 1 T → called continuous iﬀ U Y : f (U) X . ∀ ∈ T − ∈T

Example 1.15. If X carries the discrete topology then every map f : X Y is continuous. →

Example 1.16. If Y carries the coarse topology then every map f : X Y is continuous. →

Remark 1.17. In general, a continuous bijective map need not have a continuous inverse. For example, let #X > 1 and consider f = idX : (X, ) (X, ). Tdiscrete → Tcoarse

Remark 1.18. f : X Y is continuous iﬀ f 1 (A) X is closed for all closed subsets A Y. → − ⊂ ⊂

8 1.3 Compactness

Deﬁnition 1.19. Let X and Y be topological spaces. A bijective continuous map f : X Y is → called a homeomorphism iﬀ f 1 : Y X is again continuous. If there exists a homeomorphism − → f : X Y then X and Y are called homeomorphic. We then write X Y. → ≈

Remark 1.20. If f : X Y and g : Y Z are continuous then the composition g f : X Z → → ◦ → is also continuous.

1.3 Compactness

Deﬁnition 1.21. Let X be a topological space. A subset Y X is called compact iﬀ for ⊂ any collection Ui i I , Ui X open, with Y i I Ui there exist i1,..., in I such that { } ∈ ⊂ ⊂ ∪ ∈ ∈ Y Ui Ui . ⊂ 1 ∪···∪ n

Examples 1.22. 1.) Finite sets are always compact. Namely, et Y = y1,..., yn and let Ui i I { } { } ∈ be an open cover of Y. Then choose i j such that yi Ui . Then Ui ,..., Ui still covers Y. ∈ j { 1 n } 2.) If X carries the discrete topology then a subset Y X is compact if and only if it is finite. ⊂ We have already seen that finite sets are always compact. Conversely, let Y be a compact subset of X. We cover Y by y y Y . These one-point sets are open because X carries {{ } | ∈ } the discrete topology. Since Y is compact this cover must be finite and therefore #Y < . ∞ 3.) If X carries the coarse topology then every Y X is compact ⊂ 4.) A subset Y Rn is compact iff Y is closed and bounded (Heine-Borel theorem). ⊂

Example 1.23. In particular, R is not compact. We can see this directly by looking at the open cover (x 1, x + 1) x R of R. Since all intervals in this cover have length 2, any ﬁnite { − | ∈ } subcover can cover only a bounded subset of R.

Proposition 1.24. Let X be a compact topological space. Let Y X be a closed subset. Then ⊂ Y is compact.

Proof. Let Ui i I be an open cover of Y. Put U := X Y X . Then U, Ui i I is an open { } ∈ \ ∈ T { } ∈ cover of X. Since X is compact, there exist i ,..., in such that X U Ui Ui and we 1 ⊂ ∪ 1 ∪···∪ n conclude that Y Ui Ui . ⊂ 1 ∪···∪ n

9 1 Set Theoretic Topology

Proposition 1.25. Let f : X Y be continuous. Let K X be compact. Then f (K) Y is → ⊂ ⊂ compact.

Proof. Let Ui i I be an open cover of f (K), i.e. f (K) i I Ui . Then we have { } ∈ ⊂ ∪ ∈ 1 1 1 K f − ( f (K)) f − ( i I Ui ) = i I f − (Ui ) . ⊂ ⊂ ∪ ∈ ∪ ∈ open

Since K is compact there exist i1,..., in such that for | {z }

1 1 1 K f − (Ui ) f − (Ui ) = f − (Ui Ui ) ⊂ 1 ∪···∪ n 1 ∪···∪ n and we conclude that

1 f (K) f ( f − (Ui Ui )) Ui Ui . ⊂ 1 ∪···∪ n ⊂ 1 ∪···∪ n

1.4 Hausdorﬀ spaces

Definition 1.26. A topological space X is called Hausdorff iff x , x X with x , x ∀ 1 2 ∈ 1 2 Ui X open with xi Ui , such that U U = . ∃ ⊂ ∈ 1 ∩ 2 ∅

The Hausdorﬀ property says that any two distinct points b b can be separated by disjoint open neighborhoods. x1 x2

U1 U2

Examples 1.27. 1.) Spaces with discrete topology are Hausdorff spaces, simply put Ui = xi . { } 2.) Let #X 2. Then X with the coarse topology is not a Hausdorff space. ≥ 3.) If the topology of X is induced by a metric d, then X is Hausdorff. Namely, for x1 , x2 put r := d(x1, x2) > 0. Then the open balls Ui = B(xi, r/2) separate x1 and x2. 4.) Let X be Hausdorff and let Y X any subset. Then Y with its induced topology is again ⊂ Hausdorff.

10 1.5 Quotient spaces

Proposition 1.28. Let X be a Hausdorﬀ space. Let Y X be a compact subset. Then Y is a ⊂ closed subset.

Proof. Let p X Y. Then for every y Y the Hausdorﬀ property tells us that there exist open ∈ \ ∈ subsets Uy, p, Vy, p X such that y Uy, p, p Vy, p and Uy, p Vy, p = . Since Y is compact ⊂ ∈ ∈ ∩ ∅ n there exist y ,..., yn Y such that Y Uy , p Uy , p. Now put Vp := Vy , p. Since 1 ∈ ⊂ 1 ∪···∪ n ∩j=1 j Vp is a ﬁnite intersection of open subsets it is open itself. Moreover, p Vp. Now ∈ Y Vp (Uy , p Uy , p ) Vp (Uy , p Vy , p ) (Uy , p Vy , p ) = , ∩ ⊂ 1 ∪···∪ n ∩ ⊂ 1 ∩ 1 ∪···∪ n ∩ n ∅ hence Vp X Y. Therefore ⊂ \ X Y = p X Y Vp X is open. \ ∪ ∈ \ ⊂ Thus Y X is closed. ⊂

Corollary 1.29. Let X be compact, Y Hausdorﬀ. If f : X Y is continuous and bijective, → then f is a homeomorphism.

Proof. We have to show that A X closed f (A) Y is closed. Now let A X be closed. ∀ ⊂ ⊂ ⊂ Since X is compact, A is compact and also the image f (A) is compact. Since Y is a Hausdorﬀ space we conclude that f (A) Y is closed. ⊂

1.5 Quotient spaces

Let X be a topological space. Let be an equivalence relation on X. For any x X let [x] be ∼ ∈ the equivalence class of x. Denote the set of equivalence classes by

X/ := [x] x X ∼ | ∈ 1 Let π : X X/ , x [x]. Define U X/ to be open iff π− (U) X is open. One can → ∼ 7→ ⊂ ∼ ⊂ easily check that this defines a topology on X/ . Observe that π : X X/ is continuous by ∼ → ∼ definition. We now state the universal property of the quotient topology: For any topological space Y and for any maps f : X Y and f¯ : X/ Y such that the diagram → ∼ → f X / Y ⑤⑤= π ⑤⑤ ⑤⑤ ¯ ⑤⑤ f X/ ∼ commutes, f is continuous iff f¯ is continuous.

11 1 Set Theoretic Topology

Example 1.30 Let X = [0, 1] and let the equivalence relation be given by x x x X and 0 1. ∼ ∀ ∈ ∼

This equivalence relation identiﬁes the end points of the interval. We expect to obtain a topological space homeomorphic to the circle. 0 1

Indeed, we can construct such a homeomorphism. Consider f : X S1 given by → f (x) = (cos(2πx), sin(2πx)). Since f (0) = f (1) there exists a unique f¯ : X/ S1 such ∼ → that the diagram f [0, 1] / S1 ①①; ①① π ①① ① f¯ ①① [0, 1]/ ∼ commutes. From the universal property we know that f¯ is continuous because f is continuous. Moreover, f¯ : X/ S1 is bijective. Since X is compact (Heine Borel) π(X) = X/ is ∼ → ∼ compact as well. Since R2 is Hausdorﬀ, S1 is also Hausdorﬀ. Hence Corollary 1.29 applies and f¯ : X/ S1 is a homeomorphism. ∼ →

If X is a topological space and Y X a subset then X/Y := X/ where ⊂ ∼ x x x = x or x , x Y . 1 ∼ 2 ⇔ 1 2 1 2 ∈ This equivalence relation identiﬁes all points in Y to one point and performs no further identiﬁ- cations. Example 1.30 is of this form.

Example 1.31. R/[0,1] is homeomorphic to R.

Example 1.32. R/(0, ) is not a Hausdorﬀ space because the points [0] and [1] cannot be separated. ∞

1.6 Product spaces

Let X1,..., Xn be topological spaces. We set

X := X Xn 1 ×···×

12 1.7 Exercises

Deﬁnition 1.33. A subset U X is called open (for the product topology) iﬀ for all ⊂ p = (p ,..., pn ) U there exist open subsets Ui Xi with pi Ui and U Un U. 1 ∈ ⊂ ∈ 1 ×···× ⊂

It is easy to check that this deﬁnes a topology on X.

Examples 1.34. 1.) If all Xi carry the discrete topology then X carries the discrete topology. Namely, the sets (p ,..., pn ) are open, hence all subsets of the product space are open. { 1 } 2.) If all Xi carry the coarse topology then X carries the coarse topology. 3.) Rn Rm Rn+m. To see this it is convenient to use the maximum metric on Rn to × ≈ characterize the standard topology.

Now we list some important properties of the product topology:

1.) The projection maps πi : X Xi, πi (x) = xi , are continuous because for any open subset → Ui Xi ⊂ 1 = πi− (Ui ) X1 Xi 1 Ui Xi+1 Xn ×···× − × × ×···× is open in X.

2.) Fix xi Xi , i = 1,..., i 1, i + 1,..., n . Then the map ι : Xi X is continuous where ∈ { 0 − 0 } 0 → ι(ξ) = (x1,..., xi 1,ξ, xi +1,..., xn ). 0− 0 3.) Universal property: for all topological spaces Y and for all maps

f = ( f ,..., f n ) : Y X = X Xn 1 → 1 ×···× the map f is continuous if and only if all fi : Y Xi are continuous. → 4.) If all Xi are compact then X is also compact.

5.) If all Xi are Hausdorﬀ then X is also Hausdorﬀ.

1.7 Exercises

1.1. Determine all topologies on the set 1, 2, 3 and investigate which ones are homeomorphic. { }

1.2. Let W n = [ 1, 1] [ 1, 1] Rn and Dn = x Rn x 1 . Show that Dn and − ×···× − ⊂ { ∈ | k k ≤ } W n are homeomorphic.

1.3. Let be a topology on R which contains all half-open intervals (x, y] and [x, y), x < y. T Show that is the discrete topology. T

1.4. Show that (0, 1) and [0, 1]

13 1 Set Theoretic Topology a) are not homeomorphic when equipped with the standard topologies induced by R; b) are homeomorphic when equipped with the discrete topology.

1.5. Let Dn be as in Exercise 1.2. Let x, y D˚ n, i.e., x , y < 1. Construct a homeomor- ∈ k k k k phism ϕ : Dn Dn with ϕ(x) = y and ϕ(z) = z for all z ∂Dn, i.e., z = 1. → ∈ k k

1.6. Let X be a topological space, let “ ” be an equivalence relation on X. On page 11 we ∼ deﬁned that U X/ is called open if and only if π 1(U) X is open where π : X X/ is ⊂ ∼ − ⊂ → ∼ the standard projection. a) Show that X/ equipped with this system of open sets is a topological space. ∼ b) Show that the universal property on page 11 determines the topology on X/ uniquely, i.e., ∼ if is a topology on X/ such that π : (X, X ) (X/ , ) is continuous and if the T ∼ T → ∼ T universal property holds for (X/ , ) then is the topology deﬁned above. ∼ T T

1.7. Let Dn and W n be as in Exercise 1.2. Show: a) Dn/∂Dn Sn; ≈ b) W n/∂W n Sn. ≈

1.8. Let X = R/Q, i.e., the quotient space of R under the equivalence relation x y iﬀ x y Q. ∼ − ∈ Show that X has uncountably many elements but carries the coarse topology.

1.9. Let X = R2/Z2, i.e., the quotient space of R2 under the equivalence relation x y iﬀ ∼ x y Z2. Let Y = S1 S1 with the product topology. Show that X and Y are homeomorphic. − ∈ ×

1.10. Let X be a topological space. One obtains the cone CX over X by considering the cylinder X [0, 1] and then passing to the quotient CX = (X [0, 1])/ where the equivalence relation × × ∼ is given by (x, t) (x , t ) if and only if (x, t) = (x , t ) or t = t = 1. Show: ∼ ∼ ′ ′ ′ ′ ′ a) CSn Dn+1. ≈ b) If X is compact so is CX. c) If X is Hausdorﬀ so is CX.

1.11. Let X be a topological space. One obtains the suspension ΣX of X as the quotient ΣX = (X [0, 1])/ where the equivalence relation is given by (x, t) (x , t ) if and only if × ∼ ∼ ∼ ′ ′ (x, t) = (x′, t′) or t = t′ = 1 or t = t′ = 0. If furthermore f : X Y is a map then f id : X [0, 1] Y [0, 1] induces a map → × × → × Σ f : ΣX ΣY. Show: → a) If f is continuous so is Σ f .

14 1.7 Exercises b) ΣDn Dn+1. ≈ c) ΣSn Sn+1. ≈

2 Homotopy Theory

2.1 Homotopic maps

Deﬁnition 2.1. Let X and Y be topological spaces. Let A X be a subset. Two continuous ⊂ maps f , f : X Y are called homotopic relative to A iﬀ there exists a continuous map 0 1 → F : X [0, 1] Y such that × → (i) F(x, 0) = f (x) x X 0 ∀ ∈ (ii) F(x, 1) = f (x) x X 1 ∀ ∈ (iii) F(a, t) = f (a) a A, t [0, 1] 0 ∀ ∈ ∀ ∈ In symbols, f A f . The map F is then called a homotopy relative to A. 0 ≃ 1

Remark 2.2. f A f f = f 0 ≃ 1 ⇒ 0 A 1 A

Remark 2.3. If A = , then we say “ f and f are homotopic” instead of “ f and f are ∅ 0 1 0 1 homotopic relative to ”. Similarly, we write “ f0 f1” instead of “ f0 f1”. ∅ ≃ ≃∅

Examples 2.4. 1.) Let f , f : X Rn be continuous with f = f . Put 0 1 → 0 A 1 A F : X [0, 1] Rn with F(x, t) := t f (x) + (1 t) f (x). Then F is a homotopy from × → 1 − 0 f to f relative to A, hence f A f . 0 1 0 ≃ 1 2.) Let f : Rn Y be continuous. Put F(x, t) := f ((1 t)x), then f const map. 0 → 0 − 0 ≃ 3.) Lef f = Exp : R S1 C, Exp(x) = e2πix . From the previous example we know → ∈ f const map, but we will shortly see that f ;Z const map. ≃ Given two topological spaces X,Y we set

C(X,Y ) := f : X Y f is continuous { → | }

Lemma 2.5. Let X,Y be topological spaces, let A X and let ϕ C(A,Y ). Then “ A” is ⊂ ∈ ≃ an equivalence relation on f C(X,Y ) f A = ϕ . { ∈ | | }

17 2 Homotopy Theory

Proof. a) “ A” is reﬂexive: ≃ f A f , because we can put F(x, t) := f (x). ≃ b) “ A” is symmetric: ≃ Let f A g. We have to show g A f . Let F : X [0, 1] X be a homotopy relative to ≃ ≃ × → A from f to g. Put G(x, t) := F(x, 1 t). This is a homotopy relative to A from g to f , − therefore g A f . ≃ c) “ A” is transitive: ≃ Let f A g and g A h. We have to show f A h. Let F : X [0, 1] Y be homotopy ≃ ≃ ≃ × → relative to A from f to g and let G : Y [0, 1] X be homotopy relative to A from g to h. × → Then put H : X [0, 1] Y, × → F(x, 2t), 0 t 1/2 H(x, t) := ≤ ≤ G(x, 2t 1), 1/2 t 1  − ≤ ≤  This is a homotopy relative to A from f to h, thus f A h.  ≃

Lemma 2.6. Let X,Y, Z be topological spaces. Let A X, B Y be subsets. Let f , f ⊂ ⊂ 0 1 ∈ C(X,Y ) be such that f A f , fi (A) B and let g , g C(Y, Z) be such that g B g . 0 ≃ 1 ⊂ 0 1 ∈ 0 ≃ 1 Then g f A g f . 0 ◦ 0 ≃ 1 ◦ 1

Proof. Let F : X [0, 1] Y be homotopy relative to A from f to f and G : Y [0, 1] Z × → 0 1 × → be homotopy relative to B from g to g Then H : X [0, 1] Z is a homotopy relative to A 0 1 × → from g f to g f where H(x, t) = G(F(x, t), t). 0 ◦ 0 1 ◦ 1

Deﬁnition 2.7. A map f C(X,Y ) is called a homotopy equivalence iﬀ there exists a g ∈ ∈ C(Y, X) such that g f idX and f g idY If there exists a homotopy equivalence f : X Y ◦ ≃ ◦ ≃ → then X and Y are homotopy equivalent. In symbols, X Y. ≃

Remark 2.8. This deﬁnes an equivalence relation on the class of topological spaces.

Remark 2.9. Obviously, homeomorphic implies homotopy equivalent, in short,

X Y X Y . ≈ ⇒ ≃

Example 2.10. Euclidean space is homotopy equivalent to a point, Rn 0 . Namely, put n n ≃ { } f : 0 R , f (0) = 0, and g : R 0 , g(x) = 0. Then g f = id 0 and f g idRn by { } → → { } ◦ { } ◦ ≃ Example 2.4.1. Remark 2.8 implies Rn Rm for all n, m N. ≃ ∈

18 2.1 Homotopic maps

Deﬁnition 2.11. A topological space is called contractible iﬀ it is homotopy equivalent to point . { }

Deﬁnition 2.12. Let A X and let ι : A X be the inclusion map. Then A is called ⊂ → 1.) a retract of X iﬀ there exists r C(X, A) such that r = idA, i.e. r ι = idA. Then r is ∈ A ◦ called a retraction from X to A.

2.) a deformation retract of X iﬀ there exists a retraction r : X A such that ι r idX . → ◦ ≃ 3.) a strong deformation retract of X iﬀ there exists a retraction r : X A such that → ι r A idX . ◦ ≃

Examples 2.13. 1.) Let X any topological space, let A = x X consist of just one point. { 0} ⊂ Then r : X A, r(x) = x , is a retraction, hence A is a retract of X. The one-pointed set A → 0 is a deformation retract of X iff X is contractible. 2.) Let X = Rn+1 0 and A = Sn. Consider the map r : X A with r(x) = x . \ { } → x The composition ι r : Rn+1 0 Rn+1 0 then satisfies ι r = x . The mapk k ◦ \ { } → \ { } ◦ x F C(Rn+1 0 [0, 1], Rn+1 0 ) given by k k ∈ \{ }× \{ } x F(x, t) = tx + (1 t) − x k k is continuous and satisfies F(x, 0) = (ι r)(x), F(x, 1) = id(x) ◦ for all x Rn+1 0 and ∈ \{ } F(a, t) = a, a Sn ∈ n We thus conclude that ι r S n idRn+1 0 , hence S is a strong deformation retract of + ◦ ≃ \{ } Rn 1 0 . \{ }

The diﬀerence between a deformation retract and a strong deformation retract is rather subtle.

Example 2.14 We consider the comb space given by 1 X = (x, y) R2 0 y 1 and (x = 0 or x = for some n N) n ( ∈ | ≤ ≤ ∈ ) (x, y) R2 y = 0 and 0 x 1 ∪ ( ∈ | ≤ ≤ ) The space X is a bounded and closed subset of R2, hence compact. Let the set A be given by A = (0, 1) . { }

19 2 Homotopy Theory

A 1

......

Z 1 1 1 04 3 2 1 First we show that A is a deformation retract of X. The map F : X [0, 1] X given by × → F(x, y, t) := (x, (1 t)y) is continuous and satisﬁes − F(x, y, 0) = (x, y), F(x, y, 1) = (x, 0) .

Therefore idX InclusionZ X π where π : X [0, 1] 0 =: Z is the projection π(x, y) = ≃ → ◦ → ×{ } (x, 0). Moreover, we have π InclusionZ X = idZ . This shows that π is a homotopy equivalence ◦ → between X and Z. Hence X Z [0, 1] pt , which means that X is contractible. Therefore ≃ ≈ ≃ { } A is a deformation retract of X. Now we show that A is not a strong deformation retract of X. Suppose it were, then there would exist a continuous map G : X [0, 1] X, such that for all t [0, 1] and all (x, y) X × → ∈ ∈ G(x, y, 0) = (x, y), G(0, 1, t) = (0, 1).

Since X [0, 1] is compact the map G would be uniformly continuous. Therefore for ε = 1/2 × we can ﬁnd δ > 0 such that 1 G(x, y, t) G(x′, y′, t′) < − 2 whenever x x < δ, y y < δ and t t < δ. | − ′| | − ′| | − ′|

1 A Now choose n so large that n < δ and consider 1 1 (x, y) = , 1 , x , y = (0, 1), t = t . n ′ ′ ′ ! Then 1 1 G , 1, t G(0, 1, t) < , t [0, 1] . n − 2 ∀ ∈ !

Hence G( 1 , 1, t) B((0, 1), 1 ) for all t [0, 1]. n ∈ 2 ∈ Z 1 0n 1

20 2.2 The fundamental group

On the other hand the mapping t G( 1 , 1, t) is a continuous path in X from ( 1 , 1) to (0, 1) and 7→ n n must for some t take values in Z.

Remark 2.15. We have the following scheme of implications:

A is a strong deformation retract of X

⇐

A is a deformation retract of X ⇐

⇐ : A is a retract of XA X ; ≃

That both possible implications in the bottom row do not hold in general can be seen by counterexamples. Let A = x be a point in X and let X be not contractible. Then A is a retract { 0} of X but A and X are not homotopically equivalent. This is a counterexample for the implication “ ”. ⇒ A counterexample for the other direction “ ” is given by X = [0, 1] [0, 1] R2 and ⇐ × ⊂ A = comb space. Then X and A are contractible, hence X A, but one can show that there is ≃ no retraction X A. →

2.2 The fundamental group

Deﬁnition 2.16. Let X be a topological space and let x , x , x X. Put 0 1 2 ∈ Ω (X; x , x ) := ω C([0, 1], X) ω(0) = x ,ω(1) = x and 0 1 { ∈ | 0 1} Ω (X; x0) := Ω (X; x0, x0) .

Elements of Ω(X; x0, x1) are called paths and elements of Ω(X; x0) loops. For ω Ω(X; x , x ) and η Ω(X; x , x ) deﬁne ω ⋆ η Ω(X; x , x ) by ∈ 0 1 ∈ 1 2 ∈ 0 2 ω(2t), 0 t 1/2, (ω ⋆ η)(t) = ≤ ≤ η(2t 1), 1/2 t 1.  − ≤ ≤  1 Ω  1 = Ω Moreover, we consider ω− (X; x1, x0) with ω− (t) ω(1 t) and εx0 (X; x0) where = ∈ − ∈ εx0 (t) x0.

21 2 Homotopy Theory

x 1 X

x x η 0 2 ω

0 1 0 1 t 2t t 2t 1 → → −

b 0 1

Deﬁnition 2.17. For ω Ω(X; x ) denote by [ω] the homotopy class of ω relative to 0, 1 . ∈ 0 { } Then π (X; x ) := [ω] ω Ω(X; x ) 1 0 { | ∈ 0 } is called the fundamental group of (X, x0).

Lemma 2.18. For ω, ω , η, η , ζ Ω(X, x ) we have ′ ′ ∈ 0 (i) If ω 0,1 ω′ and η 0,1 η′, then ω ⋆ η 0,1 ω′ ⋆ η′; ≃{ } ≃{ } ≃{ } (ii) εx ⋆ω 0,1 ω 0,1 ω⋆εx ; 0 ≃{ } ≃{ } 0 1 1 (iii) ω⋆ω− 0,1 εx 0,1 ω− ⋆ω; ≃{ } 0 ≃{ } (iv) (ω ⋆ η) ⋆ ζ 0,1 ω ⋆ (η ⋆ ζ). ≃{ }

Proof. The proof will be given graphically. In the following diagrams we draw the domain of the required homotopies. The horizontal axis denotes the loop parameter whereas the vertical axis represents the deformation parameter. The interpolations are piecewise linear. The red area

22 2.2 The fundamental group

gets mapped to x0.

(i) We run the loop parameter with twice the speed.

ω′ η′

ω η (ii) The ﬁrst diagram shows εx 0,1 ω, the second proves that ω 0,1 εx . 0 ≃{ } ≃{ } 0 ω ω

εx0 ω ω εx0

1 (iii) The statement ω⋆ω− 0,1 εx is proven by the following diagram ≃{ } 0

εx0

ω(1 s) s −

1 ω ω−

23 2 Homotopy Theory

For any s [0, 1] the corresponding ”blue“ line segment gets mapped to ω(1 s). To prove the ∈ 1 1 − statement εx 0,1 ω− ⋆ω interchange the roles of ω and ω− . 0 ≃{ } (iv) The last assertion follows from ω η ζ

ω η ζ

Lemma 2.18 has the following implications:

(i) [ω] [η] := [ω ⋆ η] is well deﬁned; ⇒ · (ii) [εx ] [ω] = [ω] = [ω] [εx ]; ⇒ 0 · · 0 1 1 (iii) [ω] [ω ] = [εx ] = [ω ] [ω]; ⇒ · − 0 − · (iv) ([ω] [η]) [ζ] = [ω] ([η] [ζ]). ⇒ · · · · = Hence π1(X; x0) together with “ ” is a group with neutral element 1 : [εx0 ] and inverse element 1 1 · [ω]− = [ω− ]. To any topological space with a preferred point we have associated a group, the fundamental group of the space with that point. Now we consider continuous maps. Let f C(X,Y ) and ∈ x0 X. We put f (x0) =: y0 Y. If ω 0,1 ω′ and H is a homotopy between them relative to ∈ ∈ ≃{ } 0, 1 then f H is a homotopy between f ω and f ω′ relative to 0, 1 . Hence f ω 0,1 f ω′. { } ◦ ◦ ◦ { } ◦ ≃{ } ◦ Therefore we have a well-deﬁned map f : π (X; x ) π (Y; y ), f ([ω]) = [ f ω]. # 1 0 → 1 0 # ◦

Lemma 2.19. (i) The map f : π (X; x ) π (Y; y ) is a group homomorphism. # 1 0 → 1 0 (ii) It has the functorial properties a) ( f g) = f g ; ◦ # # ◦ # = b) (idX )# idπ1 (X;x0 ).

(iii) If f x f ′ then f# = f ′. ≃{ 0 } #

24 2.2 The fundamental group

Proof. (i) From the deﬁnitions we have f (ω ⋆ η) = ( f ω) ⋆ ( f η) and hence ◦ ◦ ◦ f ([ω] [η]) = f ([ω ⋆ η]) = [ f (ω ⋆ η)] = [( f ω) ⋆ ( f η)] = f ([ω]) f ([η]). # · # ◦ ◦ ◦ # · # (ii) Assertion b) being obvious we compute a): ( f g) ([ω]) = [( f g) ω] = [ f (g ω)] = f ([g ω]) = f (g ([ω])) = ( f g )([ω]) . ◦ # ◦ ◦ ◦ ◦ # ◦ # # # ◦ #

(iii) By Lemma 2.6, f 0,1 f ′ implies f ω 0,1 f ′ ω. We conclude ≃{ } ◦ ≃{ } ◦ f ([ω]) = [ f ω] = [ f ′ ω] = f ′([ω]) # ◦ ◦ # which proves the statement.

Corollary 2.20. If f : X Y is a homeomorphism with f (x ) = y then → 0 0 f : π (X, x ) π (Y, y ) # 1 0 → 1 0 is a group isomorphism.

Proof. We only use the functorial properties: 1 1 ( f − ) ( f ) = ( f − f ) = (idX ) = idπ (X x ) # ◦ # ◦ # # 1 ; 0 1 = 1 = and similarly one sees that ( f#) ( f − )# idπ1 (Y ;y0). Thus f♯ is an isomorphism with ( f#)− 1 ◦ ( f − )#.

Now we want to deal with the question to what extent π1(X; x0) depends on the choice of x0.

X To study this question let x , x X and assume 0 1 ∈ ω there exists a path γ Ω(X; x0, x1). If such a ∈ x1 path does not exist π1(X; x0) and π1(X; x1) are not related. γ x0

1 Look at the map Φγ : Ω(X; x ) Ω(X; x ) where ω (γ⋆ω) ⋆ γ . Applying 1 → 0 7→ − Lemma 2.18 twice we know that if ω 0,1 ω′ then γ⋆ω 0,1 γ⋆ω′ and hence 1 1 ≃{ } ≃{ } (γ⋆ω) ⋆ γ− 0,1 (γ⋆ω′) ⋆ γ− . Thus the map ≃{ } Φˆ γ : π (X; x ) π (X; x ), 1 1 → 1 0 1 [ω] [Φγ (ω)] = [(γ⋆ω) ⋆ γ− ], 7→ is well deﬁned.

25 2 Homotopy Theory

Proposition 2.21. Let X be a topological space, assume x , x X and γ, γ Ω(X; x , x ). 0 1 ∈ ′ ∈ 0 1 Then

1. The map Φˆ γ : π (X; x ) π (X; x ) is a group isomorphism. 1 1 → 1 0 2. If γ 0,1 γ′ then Φˆ γ = Φˆ γ . ≃{ } ′ 3. For β Ω(X; x , x ) we have Φˆ γ⋆β = Φˆ γ Φˆ β . ∈ 1 2 ◦ 4. For the constant path we have Φˆ = id . εx0 π1 (X;x0 )

1 1 5. For any [ω] π (X; x ) we have Φˆ γ ([ω]) = κ Φˆ γ ([ω]) κ where κ = [γ ⋆ γ ] ∈ 1 1 ′ · · − ′ − ∈ π1(X; x0).

Proof. a) Assertions 2., 3., and 4. follow directly from Lemma 2.18 and the deﬁnitions. b) The map Φˆ γ is a group homomorphism because

Φˆ γ [ω] [η] = Φˆ γ ([ω ⋆ η]) · = 1 (γ ⋆ (ω ⋆ η)) ⋆ γ− f 1 g 1 = γ ⋆ ((ω ⋆ (γ− ⋆ γ)) ⋆ η) ⋆ γ− = f 1 1 g (γ⋆ω) ⋆ γ− ⋆ (γ ⋆ η) ⋆ γ− f g = Φˆ γ ([ω]) Φˆ γ ([η]). ·

The map Φˆ γ is bijective, because

3. 2. 4. Φˆ γ Φˆ 1 = Φˆ 1 = Φˆ ε = idπ (X;x ) . ◦ γ− γ⋆γ− x0 1 0

and similarly Φˆ 1 Φˆ γ = idπ (X;x ). This proves 1. γ− ◦ 1 1 c) We compute

1 1 1 1 κ Φˆ γ ([ω]) κ− = γ′ ⋆ γ− γ⋆ω⋆γ− γ ⋆ (γ′)− · · · · f 1g f 1 g f 1 g = γ′ ⋆ γ− ⋆γ⋆ω⋆γ− ⋆ γ ⋆ (γ′)− f 1 g = γ′ ⋆ ω ⋆ (γ′)− = Φfˆ ( ) g γ′ [ω] .

26 2.2 The fundamental group

Proposition 2.22. Let X,Y be topological spaces and x X. Let f, g C(X,Y ) and let 0 ∈ ∈ H : X [0, 1] Y be a homotopy from f to g. Deﬁne η Ω(Y; f (x ), g(x )) by × → ∈ 0 0

η(s) := H(x0, s).

Then the following diagram commutes:

π1(X; x0) PPP PP f# g# PPP PPP P( π1(Y; g(x0)) / π1(Y; f (x0)) Φˆ η

Proof. Let [ω] π (X; x ) and put F : [0, 1] [0, 1] Y with F(t, s) := H(ω(t), s). ∈ 1 0 × → Y g ω ◦

η s F f ω ◦ t

The deformation indicated in the ﬁgure yields a homotopy relative to 0, 1 from f ω to η ⋆ (g ω) ⋆ η 1. We conclude { } ◦ ◦ − that

1 f [ω] = [ f ω] = [η ⋆ (g ω) ⋆ η− ] = Φˆ η (g ([ω])) . # ◦ ◦ #

1|2 | 3 3

Now we can improve Corollary 2.20 and show that homotopy equivalent spaces have isomorphic fundamental groups.

27 2 Homotopy Theory

Theorem 2.23. Let f : X Y be a homotopy equivalence. Then → f : π (X; x ) π (Y; f (x )) # 1 0 → 1 0 is an isomorphism for all x X. 0 ∈

Proof. Let g : Y X be a homotopy inverse of f , i.e., g f idX and f g idY . We know → ◦ ≃ ◦ ≃ by Proposition 2.22 that for a suitable η Ω(X; x , f (g(x ))) ∈ 0 0

g f = (g f ) = Φˆ η (idX ) = Φˆ η idπ (X x ) = Φˆ η . # ◦ # ◦ # ◦ # ◦ 1 ; 0 Hence g f is an isomorphism. In particular, f is injective. Similarly, we can show that # ◦ # # f g is an isomorphism, hence f is surjective. Therefore f is an isomorphism. # ◦ # # #

Corollary 2.24. If A X is a deformation retract then the inclusion ι : A X induces an ⊂ → isomorphism ι : π (A, x ) π (X; x ) for any x A. # 1 0 → 1 0 0 ∈

Proof. If A X is a deformation retract then the map ι : A X is a homotopy equivalence. ⊂ → Theorem 2.23 yields the claim.

Example 2.25. If X is contractible then the one-point set A = x X is a deformation retract. { 0} ⊂ Hence π (X; x ) π (A; x ) = [εx ] = 1 . Thus contactible spaces have trivial fundamental 1 0 ≃ 1 0 { 0 } { } group.

2.3 The fundamental group of the circle

Recall the map Exp : R S1 C where Exp(t) = e2πit . → ⊂

Lemma 2.26. Let t R and z = Exp(t ) S1. Then for all f C(S1, S1) with f (1) = z 0 ∈ 0 0 ∈ ∈ 0 there exists a unique fˆ C(R, R) with fˆ(0) = t such that the following diagram commutes: ∈ 0 fˆ R / R

Exp Exp S1 / S1 f

Proof. a) First we show uniqueness of the map fˆ.

28 2.3 The fundamental group of the circle

Assume that besides fˆ there is a second map f˜ C(R, R) with the same proper- ∈ ties as fˆ. The equality of Exp(x) = Exp(x ) is equivalent to x x Z. Since ′ − ′ ∈ Exp( fˆ(t)) = f (Exp(t)) = Exp( f˜(t)) we deduce that fˆ(t) f˜(t) Z for all t R. Since both − ∈ ∈ f˜, fˆ are continuous it follows that fˆ f˜ is constant. Finally, we know that fˆ(0) = t = f˜(0), − 0 hence f˜ = fˆ. b) Now we show existence of fˆ. Since S1 is compact, f is uniformly continuous. Since Exp is also uniformly continuous, the composition f Exp : R S1 is uniformly continuous. Hence there exists ε > 0 such that ◦ → f (Exp(I)) S1 is contained in a semi-circle for every interval I R with length ε. ⊂ ⊂ ≤ 1 1 For any closed semi-circle C S the pre-image Exp− (C) R is the disjoint union of 1 ⊂ ⊂ compact intervals of length 2 . More precisely,

1 1 Exp− (C) = t + k, t + k + . " 1 1 2 # k Z [∈ Moreover, the restriction of Exp to each of these intervals is a homeomorphism onto its image,

1 + + + ≈ Exp [t +k,t +k+ 1 ] : t1 k, t1 k C. 1 1 2 " 2 # −→

Its inverse can written down explicitly in terms of logarithms but we will not need this. Now we construct fˆ step by step.

Since f (Exp([0,ε])) is contained in a closed semi-circle C0 we can deﬁne fˆ on [0,ε] by

1 fˆ := (Exp I )− f Exp, | 0 ◦ ◦ where I R is the compact interval of length 1 with Exp(I ) = C and t I . This insures 0 ⊂ 2 0 0 0 ∈ 0 that

1 1 1 fˆ(0) = (Exp I )− f Exp(0) = (Exp I )− f (1) = (Exp I )− (z ) = t . | 0 ◦ ◦ | 0 ◦ | 0 0 0

Put t1 := fˆ(ε).

Next, f (Exp([ε, 2ε])) is contained in a closed semi-circle C1 and we deﬁne fˆ on [ε, 2ε] by

1 fˆ := (Exp I )− f Exp | 1 ◦ ◦ where I R is the compact interval of length 1 with Exp(I ) = C and t I . This insures 1 ⊂ 2 1 1 1 ∈ 1 that the two deﬁnitions of fˆ at ε agree so that we obtain a continuous function fˆ : [0, 2ε] R. → Repeating this procedure inﬁnitely many times we can extend fˆ continuously to [0, ) R. ∞ → The extension to the left is done similarly so that we obtain a continuous function fˆ : R R. → Commutativity of the diagram holds by construction.

29 2 Homotopy Theory

Deﬁnition 2.27. For a map f C(S1, S1) a map fˆ C(R, R) for which the diagram in ∈ ∈ Lemma 2.26 commutes is called a lift of f .

1 1 n Example 2.28. Consider the map f n : S S with f n = z , n Z. Then we have → ∈ n f n Exp(t) = Exp(t) = Exp(nt) .

Hence the map fˆn given by fˆn (t) = nt is a lift of f n.

Deﬁnition 2.29. For f C(S1, S1) we call ∈ deg( f ) := fˆ(1) fˆ(0) − the degree of f , where fˆ is a lift of the map f .

Remark 2.30 1.) We have seen that different lifts of f differ by an additive constant in Z. Hence the degree deg( f ) is well defined, independently of the choice of lift fˆ.

2.) The degree deg( f ) is an integer because

Exp fˆ(1) = f Exp(1) = f (1) = f Exp(0) = Exp fˆ(0) Hence we have deg( f ) = fˆ(1) fˆ(0) Z. − ∈ 3.) The map t fˆ(t + 1) fˆ(t) is continuous and takes values in Z, by the same argument as 7→ − above. We conclude that fˆ(t + 1) fˆ(t) = deg( f ) for all t R. − ∈ 4.) For k Z we compute ∈ fˆ(t + k) fˆ(t ) = fˆ(t + k) fˆ(t + (k 1)) 0 − 0 0 − 0 − + fˆ(t + (k 1)) fˆ(t + (k 2)) 0 − − 0 − + + ··· + fˆ(t + 1) fˆ(t ) 0 − 0 (3) = k deg( f ) .

5.) For f, g C(S1, S1) and lifts fˆ, gˆ we compute ∈ Exp fˆ + gˆ (t) = Exp fˆ(t) Exp (gˆ(t)) = f Exp(t) g Exp(t) . Hence fˆ + gˆ is a lift of fg and we get the following formula for the degree

deg( f g) = fˆ(1) + gˆ(1) fˆ(0) + gˆ(0) = deg( f ) + deg(g) . −

30 2.3 The fundamental group of the circle

6.) For f, g C(S1, S1) and lifts fˆ, gˆ we compute ∈ Exp fˆ gˆ(t) = f Exp gˆ(t) = f g Exp(t)

and therefore fˆ gˆ is a lift of f g. For the degree of f g this means ◦ ◦ ◦ (4) deg( f g) = fˆ gˆ(1) fˆ gˆ(0) = fˆ gˆ(0) + deg(g) fˆ gˆ(0) = deg(g) deg( f ) , ◦ − − hence deg( f g) =deg(g) deg( f ). ◦ 7.) Let f C(S1, S1) with deg( f ) , 0. We show that f must then be surjective. ∈ Namely, let fˆ be a lift of f . Then deg( f ) = fˆ(1) fˆ(0) is an integer, not equal to 0. Then, − if fˆ(1) > fˆ(0), the whole interval I := [ fˆ(0), fˆ(1)] must be contained in the image of fˆ by the intermediate value theorem. If fˆ(1) < fˆ(0) this holds for I := [ fˆ(1), fˆ(0)]. In either case I is an interval of length at least 1, hence Exp(I) = S1. We conclude

S1 = Exp(I) Exp(im( fˆ)) = im( f ). ⊂ Thus f is onto.

1 1 n Example 2.31. For the map f n : S S with f n (z) = z a lift is given by fˆ(t) = nt so that its → degree is deg( f n) = fˆn (1) fˆn (0) = n. −

Lemma 2.32. Let f, g C(S1, S1). If f g, then we have deg( f ) = deg(g). ∈ ≃

Proof. Let F : S1 [0, 1] S1 be a homotopy from f to g. Since S1 [0, 1] is compact the × → × map F is uniformly continuous. Hence, there exists a δ > 0 such that

F(z, s) F(z, s′) < 1 | − | whenever z S1 and s, s [0, 1] with s s < δ. For such s, s the map ∈ ′ ∈ | − ′| ′ F(z, s) z : S1 S1 7→ F(z, s′) → is continuous and not surjective because 1 is not contained in the image. Hence, by Remark 2.30, F( , s) = − deg F( ·, s ) 0. We now compute using 2.30.5: · ′ F( , s) F( , s) deg(F( , s)) = deg · F( , s′) = deg · + deg F( , s′) = deg F( , s′) . · F( , s ) · · F( , s ) · · · ′ ! · ′ ! We see inductively that

deg( f ) = deg F( , 0) = deg F( , s ) = = deg F( , 1) = deg(g) · · 1 ··· · where 0 = s < s < < sr = 1 is a partition of the unit interval [0, 1] satisfying 0 1 ··· si+ si < δ. | 1 − |

31 2 Homotopy Theory

1 1 2 1 Corollary 2.33. Let f C(S , S ) be such that f = g 1 where g C(D , S ) then ∈ S ∈ deg( f ) = 0.

Proof. The map F C(S1 [0, 1], S1), F(z, s) := g(sz), deﬁnes a homotopy from a constant ∈ × map to f . By Lemma 2.32 we conclude that deg( f ) = deg(const) = 0.

Let us give several applications of the concept of the degree.

Theorem 2.34 (Fundamental theorem of algebra). Every non-constant complex polynomial has a root.

Proof. Suppose we are given a non-constant polynomial

n n 1 p(z) = an z + an 1 z − + + a1z + a0, aj C, an , 0, n 1. − ··· ∈ ≥

Since dividing by an does not change the roots, we assume without loss of generality that an = 1. 1 = p(z) Now assume that p has no roots. Then the map f : C S with f (z) p(z) is a well-deﬁned → | | continuous map. To compute deg( f S1 ) consider

n n n 1 n s p( z/s) z + san 1z − + + s a0 F(z, s) := = − ··· . n n + n 1 + + n s p(z/s) z san 1z − s a0 | | | − ··· | 1 1 n The map F C(S [0, 1], S ) is a homotopy from f n (z) = z to f 1 . Computing its degree ∈ × S we ﬁnd deg( f 1 ) = deg( f n ) = n 1. S ≥ On the other hand, f is a continuous map deﬁned on all of C, hence Corollary 2.33 implies = deg( f S1 ) 0. This is a contradiction, thus p must have a root.

Lemma 2.35. Suppose the map f C(S1, S1) satisﬁes f ( z) = f (z) for all z S1. Then ∈ − − ∈ the degree deg( f ) is odd.

Proof. Lef fˆ be a lift of the map f . We compute

f Exp(t) = f Exp 1 Exp(t) = f Exp t + 1 = Exp fˆ t + 1 . − 2 2 2 Moreover,

f Exp(t) = Exp fˆ(t) = Exp 1 Exp fˆ(t) = Exp fˆ(t) + 1 . − − 2 2

32 2.3 The fundamental group of the circle

From f ( Exp(t)) = f (Exp(t)) we conclude Exp fˆ t + 1 = Exp fˆ(t) + 1 and hence − − 2 2 fˆ t + 1 fˆ(t) + 1 =: k(t) 2 − 2 is an integer for every t. Due to the continuity of the map, k(t) it is constant, k(t) = k. Hence fˆ(t + 1 ) fˆ(t) = k + 1 for all t R. Now we can compute 2 − 2 ∈ deg( f ) = fˆ(1) fˆ(0) = fˆ(1) fˆ 1 + fˆ 1 fˆ(0) − − 2 2 − = + 1 + + 1 = + (k 2 ) (k 2 ) 2k 1 which proves the assertion.

Theorem 2.36 (Borsuk-Ulam). Let f C(S2, R2) satisfy f ( x) = f (x) for all x S2. ∈ − − ∈ Then f has a zero.

Proof. Assume that the map f has no zero. Then the map g : S2 S1 with g(x) := f (x) → f (x) is defined and continuous. Moreover, g satisfies g( x) = g(x) for all x S2. Now definek k a − − ∈ map G : D2 S1 by G(y) = g(y, 1 y 2). The map G C(D2, S1) has the property that = → − k k = ∈ G S1 g S1 . By Corollary 2.33 we know that deg(g S1 ) 0. On the other hand we know by | p Lemma 2.35 that deg(g S1 ) is odd, which gives a contradiction.

Corollary 2.37. Let f C(S2, R2). Then there exists a point x S2 with f ( x ) = f (x ). ∈ 0 ∈ − 0 0

Proof. Put g(x) := f (x) f ( x). Then the map g C(S2, R2) satisﬁes g( x) = g(x) − − ∈ − − for all x S2. Hence by the Borsuk-Ulam Theorem 2.36 there exists an x S2 with ∈ 0 ∈ 0 = g(x ) = f (x ) f ( x ), which proves the theorem. 0 0 − − 0

Remark 2.38. In particular, the map f C(S2, R2) cannot be injective. Thus the sphere S2 ∈ cannot be homeomorphic to a subset of R2. This also shows that R3 cannot be homeomorphic to R2.

Now suppose you have a sandwich consisting of bread, ham, and cheese. You want to share it evenly with your friend. Can you cut the sandwich into two pieces such that each piece contains the same amount of bread, the same amount of ham and the same amount of cheese? The following theorem tells us that it is possible.

33 2 Homotopy Theory

Theorem 2.39 (Ham-Sandwich-Theorem). Let A, B, C R3 be open and bounded. Then ⊂ there exists an aﬃne hyperplane H R3 such that each of the sets is divided into pieces of ⊂ equal volume.

Proof. The proof consists of four steps.

2 a) For each x S and t R we deﬁne the aﬃne Hx,t ∈ ∈ hyperplane

= 3 = tx + Hx,t : y R y, x t . • Hx,t { ∈ | h i } S2 x It is clear that H x, t = Hx,t . We deﬁne the • half space H+ by− − 0 x,t • H+ := y R3 y, x t . x,t { ∈ | h i ≥ }

Hx,t Hx,t ′ b) Now ﬁx x S2. Look at the function ∈ ax : R R deﬁned by → + ax (t) := vol(A H ). B(0, RA) ∩ x,t

It satisﬁes a x (t) + ax ( t) = vol(A) A − − and is monotonically decreasing. Since A is bounded there exists R > A B( , R ) A 0 such that 0 A . t t ⊂ | − ′|

For t < t R we have ′ ∈ = + + 2 ax (t′) ax (t) vol(A (Hx,t H )) πR t t′ . | − | ∩ \ x,t ′ ≤ A · | − | Thus ax is Lipschitz continuous.

vol(A) αx Moreover, ax (t) = vol(A) for t 0 ≪ and ax (t) = 0 for t 0. ≫

+ t−x tx

34 2.3 The fundamental group of the circle

c) By continuity and monotonicity the pre-image of any value under αx is a closed interval. In 1 = + particular, a−x (vol(A)/2) [t−x, tx ]. + + = t x− t x Now put α(x) : 2 . Hence Hx,α(x) divides A into two pieces of equal volume. Moreover, α( x) = α(x) for all x S2 and it is not hard to check that α is continuous. Similarly, − − ∈ deﬁne the functions β for B and γ for C. d) Consider the map f C(S2, R2) with f (x) = (α(x) β(x),α(x) γ(x)). We have ∈ − − f ( x) = (α( x) β( x),α( x) γ( x)) − − − − − − − = ( α(x) + β(x), α(x) + γ(x)) − − = f (x). − Thus the Borsuk-Ulam Theorem 2.36 applies and there exists a point x S2 with 0 ∈ (0, 0) = f (x ) = (α(x ) β(x ),α(x ) γ(x )). 0 0 − 0 0 − 0 = = Hence α(x0) β(x0) γ(x0). Thus the hyperplance Hx0,α(x0) does the job.

Remark 2.40. The ham-sandwich theorem is optimal in the sense that it fails for more than three sets in R3.

Example 2.41. A ball B(x, r) R3 with center x is cut into two halves of equal volume exactly ⊂ by those planes that contain x. If you choose four balls in R3 in such a way that their centers are not contained in one plane, then no plane will cut them all into halves of equal volume.

1 In the following we want to use the concept of degree to determine π1(S ; 1).

fω

1 Exp ω ! a) For ω Ω(S ; 1) and I = [0, 1] consider the fol- S1 o I / S1 ∈ a❈❈ ④= lowing diagram: ❈❈ ④④ ❈≈❈ ④④ ❈ ④ ω Exp ❈ ④④ I/∂I Here ω and Exp are the continuous maps induced on the quotient space. They exist because ω(0) = ω(1) and Exp(0) = Exp(1), compare Section 1.5. By Example 1.30 we know that Exp : I/∂I S1 is a homeomorphism. Now put → 1 1 1 fω := ω (Exp)− C S , S ◦ ∈

and deﬁne deg(ω) := deg( fω ). We have obtained a map

deg : Ω(S1; 1) Z. →

35 2 Homotopy Theory

1 b) Now suppose ω 0,1 ω′. We choose a homotopy F : I I S from ω to ω′ relativ to ≃{ } × → 0, 1 . Then the map G : S1 I S1 deﬁned by G(z, s) := F((Exp) 1(z), s) is a homotopy { } × → − from fω to fω , hence fω fω . Therefore ′ ≃ ′ ( ) = ( ) = ( ) = ( ) deg ω deg fω deg fω′ deg ω′ . Hence we get a well-deﬁned map deg : π (S1; 1) Z where [ω] deg(ω). 1 → 7→ c) This map is surjective because for n Z we can consider the map ω(t) = Exp(nt). The ∈ commutative diagram

n z z =fn (z) S1 7→ / S1 f◆◆ O ◆◆◆ ◆◆◆ t ω(t )=Exp(nt )=Exp(t )n t Exp(t ) ◆◆ 7→ 7→ ◆◆◆ I n shows fω (z) = z = f n (z) and we get deg(ω) = deg( f n) = n. d) Now let ω, ω Ω(S1; 1) and consider the map ′ ∈ ( ( )) = ( ) fω⋆ω′ Exp t ω⋆ω′ t ω(2t), 0 t 1/2 = ≤ ≤ ω′(2t 1), 1/2 t 1  − ≤ ≤  fω (Exp(2t)), 0 t 1/2 =  ≤ ≤ f (Exp(2t 1)), 1/2 t 1  ω′  − ≤ ≤  ˆ ˆ ˆ () = ˆ ( ) Let fω, fω′ be lifts of fω, fω′ with fω′ 0 fω 1 . Then we have

Exp( fˆω (2t)), 0 t 1/2 fω⋆ω (Exp(t)) = ≤ ≤ ′ Exp( fˆ (2t 1)), 1/2 t 1  ω′  − ≤ ≤  fˆω (2t), 0 t 1/2 = Exp ≤ ≤ fˆ (2t 1), 1/2 t 1  ω′ !  − ≤ ≤  =:g(t )  ( ) | ˆ ( ) = {zˆ ( ) }( ) Note that the map g t is continuous because of fω′ 0 fω 1 . Thus g t is a lift of fω⋆ω′ . Now we compute the degree of ω⋆ω′.

deg(ω⋆ω′) = g(1) g(0) − = fˆω (1) fˆω (0) ′ − = fˆω (1) fˆω (0) + fˆω (1) fˆω (0) ′ − ′ − = deg(ω′) + deg(ω) Hence the map deg : π (S1; 1) (Z, +) is a group homomorphism. 1 →

36 2.3 The fundamental group of the circle

1 e) Finally we compute its kernel. Let ω Ω(S ; 1) with deg(ω) = 0. Let fˆω be the lift of fω ∈ with fˆω (0) = 0. Since 0 = deg(ω) = fˆω (1) fˆω (0) we have fˆω (1) = fˆω (0) = 0. Next 1 − consider the continuous map F : I I S with F(t, s) := Exp(s fˆω (t)). It satisﬁes: × →

F(t, 0) = 1 = ε1(t) ,

F(t, 1) = fω (Exp(t)) = ω(t) , F(0, s) = Exp(s 0) = 1, · F(1, s) = Exp(s fˆω (1)) = Exp(s 0) = 1 . · 1 We conclude that ω 0,1 ε1, hence [ω] = [ε1] = 1 π1(S ; 1). Therefore the kernel is ≃{ } ∈ trivial and the map deg : π (S1; 1) Z is injective. 1 → We summarize the discussion in the following

Theorem 2.42. The map deg : π (S1; 1) Z is a group isomorphism. 1 →

Example 2.43. We already know that S1 is a strong deformation retract of C 0 . Hence the \{ } inclusion ι : S1 C 0 induces an isomorphism → \{ } ι : π (S1; 1) π (C 0 ; 1) # 1 → 1 \{ } and therefore π (C 0 ; 1) Z. 1 \{ }

S1 = ∂D2 Example 2.44 We show that S1 = ∂D2 is not a retract of D2. Suppose the map D2 r : D2 S1 is a retraction and denote the inclusion by ι : S1 D2. → →

Then we would have r ι = id 1 , hence r ι = (r ι) = (id 1 ) = id 1 . We then get a ◦ S # ◦ # ◦ # S # π1 (S ;1) contradiction because of the following diagram

ι r Z π(S1; 1) # / π(D2; 1) 1 # / π(S1; 1) Z { } :

id 1 π1(S ;1)

As a corollary we get a proof of Brouwer’s ﬁxed point theorem in dimension two. See page 92 for the theorem in general dimensions.

37 2 Homotopy Theory

Theorem 2.45 (Brouwer’s ﬁxed point theorem in 2 dimensions). Let f : D2 D2 be a → continuous map. Then f has a ﬁxed point, i.e., there exists an x D2 such that f (x) = x. ∈

Proof. Assume that f C(D2, D2) has no ﬁxed point. Then f (x) , x for all x D2 so that we ∈ ∈ can consider the half line emanating from f (x) through x. We let r(x) be its intersection point with ∂D2 = S1 as indicated in the picture.

x r(x) f (x) •

This yields a retraction r : D2 S1 and we get a contradiction. →

Example 2.46. We show that the system of equations 1 1 + sin(x)y 2x = 0, 2 − 1 x2 cos(x)y + 2y = 0, (2.1) 2 2 − has a solution. We rewrite this as a ﬁxed point equation and apply the Brouwer ﬁxed point theorem. To do this we put

1 sin(x)y cos(x)y x2 f (x, y) := + , + 2 4 4 4 ! Fixed points of f (x, y) are then the same as solutions to the above system of equations (2.1). To apply Brouwer’s ﬁxed point theorem we have to show that f (D2) D2. Let x, y D2. Then ⊂ ∈ 1 sin(x)y 2 cos(x)y x2 2 f (x, y) 2 = + + + k k 2 4 4 4 ! ! ( ) ( )2 2 ( )2 2 ( ) 2 4 = 1 + sin x y + sin x y + cos x y + cos x yx + x 4 4 16 16 8 16 2 2 4 = 1 + sin(x)y + y + cos(x)yx + x 4 4 16 8 16 1 y y2 y x2 x4 + | | + + | | + ≤ 4 4 16 8 16 1 + 1 + 1 + 1 + 1 ≤ 4 4 16 8 16

38 2.3 The fundamental group of the circle

= 3 4 1. ≤

1 1 n Example 2.47. Consider f n C(S , S ) with f n (z) = z . We check that the diagram ∈ 1 deg π1(S ; 1) / Z

(f ) n n # · 1 deg π1(S ; 1) / Z commutes. Let [ω] π (S1; 1). We compute ∈ 1

deg ( f n) ([ω]) = deg [ f n ω] = deg( f n ω) # ◦ ◦ = deg( f n fω ) = deg( f n ) deg( fω ) = n deg [ω] . ◦ Remark 2.48. From Exercise 2.5 we know that for X , X and x X , x X we have 1 2 1 ∈ 1 2 ∈ 2 π (X X ; (x , x )) π (X ; x ) π (X ; x ) . 1 1 × 2 1 2 1 1 1 × 1 2 2 For the two-dimensional torus T2 = S1 S1 we get π (T2; x) Z Z = Z2. More generally, we × 1 × get inductively for the n-torus T n = S1 S1 that π (T n; x) Zn. In particular, T n ; T m ×···× 1 n , for n m. | {z }

Deﬁnition 2.49. Let X be a topological space.

1.) We call X connected iff X and are the only subsets of X which are both open and closed. ∅ 2.) We call X path-connected iff for all x , x X there exists a path ω Ω(X; x , x ). 1 2 ∈ ∈ 1 2 3.) Let X be path-connected. Then X is called simply-connected (or 1-connected) iff π (X; x ) = 1 for some (and hence all) x X. 1 0 { } 0 ∈

Example 2.50. The interval [0, 1] is connected. To see this let I [0, 1] be open and closed ⊂ and assume that I is neither empty nor all of [0, 1]. Then there exists t I and t [0, 1] I. 0 ∈ 1 ∈ \ W.l.o.g. let t < t , the other case being analogous. We put T := sup(I [0, t )). Then 0 1 ∩ 1 0 t T t 1. Since I is closed T I. ≤ 0 ≤ ≤ 1 ≤ ∈ If T = 1 then T = t1 which contradicts t1 < I. If T < 1 then there exists ε > 0 such that [T, T + ε) I because I is open. This contradicts the maximality of T. ⊂

Remark 2.51. If X is 1-connected then it is path-connected by deﬁnition but the converse is not true. For example, S1 is path-connected but not 1-connected.

39 2 Homotopy Theory

Remark 2.52. If X is path-connected then X is connected.

Proof. Let X be path-connected and let U X be open and closed, U , . We show U = X. ⊂ ∅ Since U is non-empty we can ﬁnd x U. Let x X be any point. 1 ∈ 2 ∈ Since X is path-connected there is a path ω Ω(X; x , x ). Then I := ω 1 (U) is an open and ∈ 1 2 − closed subset of [0, 1]. Since ω(0) = x U we have 0 I and hence I is non-empty. Thus 1 ∈ ∈ I = [0, 1] because [0, 1] is connected. We conclude 1 I and hence x = ω(1) U. This shows ∈ 2 ∈ that U contains all points of X.

Again, the converse implication does not hold in general. Consider for example the space

X := (t, sin(1/t)) t > 0 (0, s) 1 s 1 { | }∪{ |− ≤ ≤ }

Then X is connected but not path-connected.

Remark 2.53. Let X be path-connected. Then the following are equivalent (see Exercise 2.2): (i) X is simply connected;

(ii) Every ω C(S1, X) is homotopic to a constant map; ∈ (iii) Every ω C(S1, X) has a continuous extension to a map D2 X. ∈ →

2.4 The Seifert-van Kampen theorem

The Seifert-van Kampen theorem will allow us to compute the fundamental group of spaces which are built out of simpler spaces whose fundamental groups we already know. We start with an excursion to group theory.

Deﬁnition 2.54. Let G be a group. A subgroup H G is called normal iﬀ ⊂ g H = H g for all g G . · · ∈

40 2.4 The Seifert-van Kampen theorem

The condition on a subgroup of being normal can be reformulated in various ways. It is equivalent to any of the following:

1.) g H g 1 = H for all g G; · · − ∈ 2.) g H g 1 H for all g G; · · − ⊂ ∈ 3.) g h g 1 H for all g G and h H. · · − ∈ ∈ ∈

Example 2.55. Consider the cartesian product of two groups G = G G = (g , g ) gj G j 1 × 2 { 1 2 | ∈ } with componentwise multiplication. Then (g , 1) g G G is a normal subgroup of G { 1 | 1 ∈ 1} 1 because 1 1 1 1 (g , g ) (g˜ , 1) (g , g )− = g g˜ g− , g 1g− = g g˜ g− , 1 . 1 2 · 1 · 1 2 1 1 1 2 2 1 1 1 Example 2.56. Let ϕ : G K be a group homomorphism. Then H = ker(ϕ) is a normal → subgroup because for h ker(ϕ) and g G we have ∈ ∈ 1 1 ϕ(g h g− ) = ϕ(g) ϕ(h) ϕ(g)− = 1, · · · · 1 hence g h g 1 ker(ϕ). |{z} · · − ∈

Remark 2.57. If H G is a normal subgroup then G/H is again a group via ⊂ (g H) (g˜ H) = (gg˜) H. · · · · Normality of H ensures that this multiplication is well deﬁned. The group H is then the kernel of G G/H with g g H. Thus the normal subgroups are exactly those which arise as kernels → 7→ · of group homomorphisms.

Now let S G be any subset. Then ⊂ (S) := H N H G normal subgroup, ⊂ H S \⊃ is the smallest normal subgroup containing S. We call (S) the normal subgroup generated by N S.

Example 2.58. ( ) = 1 . N ∅ { }

41 2 Homotopy Theory

Deﬁnition 2.59

Let G1 and G2 be groups. A group G is called free product G1 ⑥ ❆❆ of G1 and G2 iﬀ there exist homomorphisms i j : G j G such i1 ⑥ ❆ ϕ1 → ⑥⑥ ❆❆ that for all groups H and for all homomorphisms ϕj : G j H ⑥⑥ ❆❆ → ~⑥ ϕ1⋆ϕ2 there exists a unique homomorphism G ❆ / H ` ❆❆ ⑥⑥> ❆❆ ⑥⑥ ❆ ⑥ϕ ϕ ⋆ϕ : G H i2 ❆❆ ⑥⑥ 2 1 2 → ⑥ G2 such that the diagram commutes.

Remark 2.60. Here we have characterized free products by G their universal property. This universal property implies for 1 ❇ ⑦ ❇❇❇❇ example that the maps i : G G are injective. i1 ⑦⑦ ❇❇❇❇id j j ⑦⑦ ❇❇❇❇ = →= = ⑦⑦ ❇❇❇ Namely, choose H G1, ϕ1 idG1 and ϕ2(g2) 1 for all ~⑦ id ⋆ϕ2 G / G1 g2 G2. The diagram now tells us that the map i1 must be `❅❅ > ∈ ❅❅ ⑤⑤ injective because the identity is injective. Similarly, we see that ❅❅ ⑤⑤ ❅ ⑤⑤ϕ2 i2 ❅ ⑤⑤ i2 is injective. G2

Remark 2.61. The free product of G1 and G2 is unique up to isomorphism.

G1 ⑥ ❇❇ Namely, let G′ be another free product of G1 and G2 with i1 ⑥ ❇ i1′ ⑥⑥ ❇❇ i′ : G j G′ the corresponding homomorphisms. By the ⑥⑥ ❇❇ j ⑥ i′ ⋆i′ → = = ~ 1 2 universal property of G with H G′ and ϕj i′ we get the G / G′ j `❆❆ ⑤> following commutative diagram: ❆❆ ⑤⑤ ❆❆ ⑤⑤ i ❆ ⑤ i 2 ❆ ⑤⑤ 2′ G2

G1 i ⑤ ❆❆ 1′ ⑤⑤ ❆❆i1 ⑤⑤ ❆❆ ⑤⑤ ❆ Interchanging the roles of G and G we get another commutative ~ i1⋆i2 ′ G′ / G `❇❇ ⑥> diagram: ❇❇ ⑥⑥ ❇❇ ⑥⑥ i ❇ ⑥ i 2′ ❇ ⑥⑥ 2 G2

42 2.4 The Seifert-van Kampen theorem

G1 ⑥ ❆❆ i1 ⑥⑥ ❆❆i1 ⑥ i1′ ❆ ⑥⑥ ❆❆ Combining both diagrams we obtain ~⑥i1′ ⋆i2′ i1⋆i2 G ❆ / G′ / G ` ❆❆ O ⑥⑥> ❆❆ i ⑥⑥ ❆❆ 2′ ⑥⑥ i2 ❆ ⑥⑥ i2 G2

G1 ⑥ ❆❆ i1 ⑥ ❆ i1 On the other hand, this diagram commute as well. ⑥⑥ ❆❆ ⑥⑥ ❆❆ ~⑥ id G ❆ / G ` ❆❆ ⑥⑥> ❆❆ ⑥⑥ ❆❆ ⑥⑥ i2 ❆ ⑥⑥ i2 G1 By the uniqueness of the induced homomorphisms we have

(i ⋆ i ) i′ ⋆ i′ = idG 1 2 ◦ 1 2 and similarly i′ ⋆ i′ (i ⋆ i ) = idG . 1 2 ◦ 1 2 ′ Hence the map i ⋆ i : G G is a group isomorphism with inverse i ⋆ i . 1 2 → ′ 1′ 2′ Next we show the existence of the free product of two groups by a direct construction. For this purpose let G1 and G2 be groups. For formal reasons we assume without loss of generality that G G = .1 We deﬁne 1 ∩ 2 ∅

G ⋆ G := (x ,... xn ) n N , x j G 1G ) (G 1G such that 1 2 1 | ∈ 0 ∈ 1 \{ 1 } ∪ 2 \{ 2 } if xi G then xi+ G or conversely . ∈ 1 1 ∈ 2

The group multiplication in G1 ⋆ G2 is then inductively deﬁned as

(x ,..., xn ) (xn+ ,..., xn+m ) 1 · 1 (x1,..., xn 1, xn xn+1, xn+2,..., xn+m ) if (xn, xn+1 G1 or xn, xn+1 G2) − · ∈ ∈ and x x + , 1,  n n 1 =  · : (x1,..., xn 1) (xn+2,..., xn+m ) if (xn, xn+1 G1 or xn, xn+1 G2)  − · ∈ ∈  and xn xn+ = 1,  · 1  (x1,..., xn, xn+1,..., xn+m ) otherwise.   This turns G1 ⋆G2 into a group with neutral element the empty sequence (). The inverse element 1 1 for (x1,... xn ) is given by (x−n ,..., x1− ) because of 1 1 = 1 1 = = 1 = (x1,..., xn ) x−n ,..., x1− (x1,..., xn 1) x−n 1,..., x1− (x1)(x1− ) (). · − · − ··· 1 Otherwise replace G2 by an isomorphic group which is disjoint to G1.

43 2 Homotopy Theory

Now consider the map i j : G j G ⋆ G given by → 1 2 (x), x , 1 i (x) = j = (), x 1  For ϕj : G j H homomorphisms we put  →  (ϕ ⋆ϕ )(x ,..., xn ) := ϕi (x ) ϕi (x ) ... ϕi (xn ) 1 2 1 1 1 · 2 2 · · n where i j is chosen such that x j Gi . ∈ j Remark 2.62 1.) The subset i j (G j ) G ⋆ G is a subgroup isomorphic to G j . The intersection i (G ) ⊂ 1 2 1 1 ∩ i (G ) = 1 is trivial. The union i (G ) i (G ) generates i (G ) ⋆ i (G ) as a group. 2 2 { } 1 1 ∪ 2 2 1 1 2 2 2.) If G = 1 then G ⋆ G = i (G ) G . Similarly, if G = 1 then G ⋆ G G . 1 { } 1 2 2 2 2 2 { } 1 2 1 3.) If G , 1 and G , 1 then we may choose x G 1 and y G 1 . Then 1 { } 2 { } ∈ 1 \{ } ∈ 2 \{ } (x), (x, y), (x, y, x), (x, y, x, y),... yields inﬁnitely many pairwise diﬀerent elements in G ⋆ G , hence G ⋆ G = (even if 1 2 | 1 2 | ∞ G j < ). In addition, we have | | ∞ (x) (y) = (x, y) , (y, x) = (y) (x) , · · hence the group G1 ⋆ G2 is not abelian, even if G1 and G2 are.

Remark 2.63. Usually one identiﬁes G1 with i1 (G1) and G2 with i2 (G2) and writes

x x ... xn instead of (x , x ,..., xn ). 1 · 2 · · 1 2

Example 2.64. Consider G = Z/2Z = 1, 1 and G = Z/2Z = 1 , 1 . Elements of 1 { − } 2 { ′ − ′} Z/2Z ⋆ Z/2Z are for example

a = ( 1) ( 1′) ( 1) ( 1′) − · − · − · − b = ( 1′) ( 1) ( 1′) ( 1) ( 1′) . − · − · − · − · − Now we calculate

a b = ( 1) ( 1′) ( 1) ( 1′) ( 1′) ( 1) ( 1′) ( 1) ( 1′) · − · − · − · − · − · − · − · − · − 1′ = ( 1) ( 1′) ( 1) ( 1 ) ( 1 ′ ) ( 1) ( 1′) − · − · − ·|− {z· − }· − · − 1 = ( 1) ( 1′) ( 1 ′) ( 1 ) ( 1′) − · − ·|− {z· − } · − 1′ = ( 1) ( 1 ) ( 1 ′ ) − |· − {z· − } 1 = ( 1 ′) . |− {z }

44 2.4 The Seifert-van Kampen theorem

Now we are ready to return to topology.

Proposition 2.65. Let X be a topological space and let U, V X be open such that U V = X. ⊂ ∪ Let x U V. Furthermore, assume that U, V and U V are path-connected. Then X is 0 ∈ ∩ ∩ path-connected and the map

i ⋆ j : π (U; x ) ⋆π (V; x ) π (X; x ) # # 1 0 1 0 → 1 0 is onto where i : U X and j : V X are the corresponding inclusion maps. → →

Proof. First of all we note that the space X is path-connected because each point in X lies in U or in V and can therefore be connected to x0 by a path. The statement of the proposition is equivalent to saying that i π (U; x ) j π (V; x ) # 1 0 ∪ # 1 0 = generates π1(X; x0) as a group. Now let [ω] π1(X;x0) and subdivide the unit interval I [0, 1] ∈ by 0 = t t < tn = 1 such that ω([ti, ti+ ]) U or V. 0 ≤ 1 ··· 1 ⊂ ⊂

X ω0 ω2

U x0

U VUη3 V ∩ η1 η2 ∩ ω( ) ω( ) t1 t2 ω(t3)

ω1

By removing subdivision points if necessary we can assume that if ω([ti 1, ti ]) U then − ⊂ ω([ti, ti+ ]) V or conversely. Reparametrize ωi (t) = ω((1 t)ti + tti+ ). Then ωi 1 ⊂ − 1 ∈ Ω(U or V; ω(ti ),ω(ti+ )) and in particular ω(ti ) U V. Since by assumption U V is 1 ∈ ∩ ∩ path-connected there exist η j Ω(U V; x ,ω(t j )). Then ∈ ∩ 0 1 1 1 ω0 ⋆ η1− , η1 ⋆ω1 ⋆ η2− , η2 ⋆ω2 ⋆ η3− , . . ., ηn 1 ⋆ωn 1 − −

45 2 Homotopy Theory

are loops with base point x0 contained entirely in U or V. We then calculate

1 1 1 ω0 ⋆ η1− ⋆ η1 ⋆ω1 ⋆ η2− ⋆ η2 ⋆ω2 ⋆ η3− ⋆ ⋆ (ηn 1 ⋆ωn 1) ··· − − 0,1 ω0 ⋆ω1 ⋆ ⋆ωn 1 0,1 ω ≃{ } ··· − ≃{ } in X. Denoting the homotopy class of a loop ω in X, U, V, or U V by [ω]X , [ω]U , [ω]V , and ∩ [ω]U V respectively, we have ∩ = 1 1 [ω]X ω0 ⋆ η1− X η1 ⋆ω1 ⋆ η2− X ... ηn 1 ⋆ωn 1 X · · · − − and it follows that 1 1 1 [ω]X = i ω ⋆ η− j η ⋆ω ⋆ η− i η ⋆ω ⋆ η− ... # 0 1 U · # 1 1 2 V · # 2 2 3 U = 1 1 1 (i#⋆ j#) ω0⋆ η1− U η1 ⋆ω1 ⋆ η2− V η2⋆ω2 ⋆ η3− U... · · f π1 (U;x0 )⋆π1 (V ;x0) ∈ which proves the assertion. | {z }

Corollary 2.66. Let X is a topological space and let U, V X be open such that U V = X ⊂ ∪ and U V , . If U and V are 1-connected and U V is path-connected, then the space X is 1-connected.∩ ∅ ∩

Proof. Since 1 = 1 ⋆ 1 = π (U; x ) ⋆π (V; x ) π (X; x ) { } { } { } 1 0 1 0 → 1 0 is onto we get that π (X; x ) = 1 . 1 0 { }

Example 2.67. Consider X = Sn and put U = Sn e . The stereographic projection yields \{ 1} a homeomorphism U Rn. Hence U is 1-connected. Similarly, V = Sn e is also → \ {− 1} 1-connected. Now U V = Sn e , e Rn 0 . ∩ \{ 1 − 1}≈ \{ } For n 2 the space U V is path-connected. Corollary 2.66 shows that Sn is simply connected ≥ ∩ for n 2. ≥ 1 Recall that we know already that this is not true for n = 1 because π1(S ; 1) Z.

To determine π1(X; x0) more precisely we compute the kernel of the homomorphism i ⋆ j : π (U; x ) ⋆π (V; x ) π (X; x ). # # 1 0 1 0 → 1 0 i U / X O O Consider the inclusion maps i′ : U V U and j ′ : U V V. ∩ → ∩ → i j Clearly the diagram on the right commutes. This implies ′ U V / V ∩ j′ j j ′ = i i′ . # ◦ # # ◦ #

46 2.4 The Seifert-van Kampen theorem

For α π (U V; x ) we calculate ∈ 1 ∩ 0 1 1 1 1 = i i′ (α) j j ′ (α) − = i i′ (α) j j ′ (α)− = (i ⋆ j ) i′ (α) j ′ (α)− . # # · # # # # · # # # # # · # π (U;x )⋆π (V ;x ) ∈ 1 0 1 0

Hence i (α) j (α) 1 ker(i ⋆ j ) and it follows that | {z } #′ · #′ − ∈ # # 1 i′ (α) j ′ (α)− α π (U V; x ) ker(i ⋆ j ) . N # · # | ∈ 1 ∩ 0 ⊂ # # Theorem 2.68 (Seifert-van Kampen). Let X be a topological space and let U, V X be ⊂ open subsets such that U V = X and x U V. Let U, V and U V be path connected. ∪ 0 ∈ ∩ ∩ Then the map i# ⋆ j# induces an isomorphism π (U; x ) ⋆π (V; x ) 1 0 1 0 π (X; x ). i (α) j (α) 1 α π (U V; x ) 1 0 N #′ · #′ − | ∈ 1 ∩ 0

Proof. It remains to show that

1 ker(i ⋆ j ) i′ (α) j ′ (α)− α π (U V; x ) . # # ⊂ N # · # | ∈ 1 ∩ 0 Let ω ,ω , Ω(U; x ) and ω ,ω, Ω(V; x ) be such that 1 3 ···∈ 0 2 4 ···∈ 0

1 = i ⋆ j [ω ]U [ω ]V [ω ]U ... # # 1 · 2 · 3 · = i ([ω ]U ) j ([ω ]V ) i ([ω ]U ) ... # 1 · # 2 · # 3 · = [ω ]X [ω ]X [ω ]X ... 1 · 2 · 3 · = [ω1 ⋆ω2 ⋆ω3 ⋆... ]X s

1 − Then there exists a homotopy H : [0, 1] × [0, 1] X such that →

H(t, 0) = (ω1 ⋆ω2 ⋆ω3 ⋆... )(t),

H(t, 1) = x0,

H(0, s) = H(1, s) = x0.

In the picture the red area gets mapped to 0 − | | | x0. ω1 ω2 ωN

| | t 0 1

47 2 Homotopy Theory

Now subdivide [0, 1] [0, 1] further such × b that H maps each closed subsquare entirely v to U or to V.

| | | ω1 ω2 ωN

u >

Considering the edges of the subsquares we l r get homotopies ∧ ∧

> d and hence the relation d ⋆ r 0,1 l ⋆ u (2.2) ≃{ } resulting in (H d) ⋆ (H r) 0,1 (H l) ⋆ (H u) in U or in V . ◦ ◦ ≃{ } ◦ ◦ For each vertex v in this subdivision of [0, 1] [0, 1] choose ηv Ω(X; x , H(v)) in such a way × ∈ 0 that ηv Ω(W; x , H(v)) if H(v) W where W = U, V or U V. This is possible because ∈ 0 ∈ ∩ x W and W is path-connected by assumption. If H(v) = x choose ηv = εx . For each edge 0 ∈ 0 0 with endpoints v0 and v1 we obtain a loop in U or in V by

1 ηv ⋆ (H c) ⋆ η− Ω(U or V; x ) . 0 ◦ v1 ∈ 0 Now look at one row of the subdivision

48 2.4 The Seifert-van Kampen theorem

u1 u2 uN > > > = r = ε εx0 l1 N x0 ∧∧∧> > ∧∧> d1 d2 dN

We find that 1 Di := ηd (0) ⋆ (H di ) ⋆ η− Ω(U or V; x0). i ◦ di (1) ∈ Similarly we define Li, Ri, Ui Ω(U or V; x ) and we conclude that by (2.2) ∈ 0 [Di ]W [Ri ]W = [Li ]W [Ui ]W π (Wi; x ) (2.3) i · i i · i ∈ 1 0 1 where Wi = Uor V. We now compute in π (U; x ) ⋆π (V; x ) modulo ( i (α) j (α) α 1 0 1 0 N { #′ · #′ − | ∈ π (U V; x ) ): 1 ∩ 0 } [D ]W [D ]W ... [DN ]W = [D ]W [D ]W ... [DN ]W [RN ]W 1 1 · 2 2 · · N 1 1 · 2 2 · · N · N = [D1]W1 ... [DN 1]WN 1 [LN ]WN [UN ]WN · · − − · · = [D1]W1 ... [DN 1]WN 1 [RN 1]WN [UN ]WN . · · − − · − · If now WN 1 = WN then we can apply (2.3) once more and get − = [D1]W1 [D2]W2 ... [DN ]WN [D1]W1 ... [DN 2]WN 2 [LN 1]WN 1 [UN 1]WN 1 [UN ]WN · · · · · − − · − − · − − · (2.4) In case WN 1 , WN , say WN 1 = U and WN = V, then LN = RN 1 Ω(U V; x0). Hence − − − ∈ ∩ mod (... ) = =N = [RN 1]V j#′ ([RN 1]U V ) i#′ ([RN 1]U V ) [RN 1]U . − − ∩ − ∩ − In π (U; x ) ⋆π (V; x ) G := 1 0 1 0 ( i (α) j (α) 1 α π (U V; x ) ) N { #′ · #′ − | ∈ 1 ∩ 0 } the computation (2.4) is still possible. By induction on all squares in the row from right to the left we find that =1 = = [D1]W1 ... [DN ]WN [L1]W1 [D1]W1 ... [DN ]WN [U1]W1 ... [UN ]WN . · · · · · · · z }| {

49 2 Homotopy Theory

A second induction on all rows from bottom to top yields in G

mod (... ) [ω ]U [ω ]V [ω ]U ... =N [εx ]W [εx ]W [εx ]W ... = 1 . 1 · 2 · 3 · 0 1 · 0 2 · 0 3 ·

We have shown that in π1(U; x0) ⋆π1(V; x0)

1 [ω ]U [ω ]V [ω ]U ... ( i′ (α) j ′ (α)− α π (U V; x ) ) 1 · 2 · 3 · ∈N { # · # | ∈ 1 ∩ 0 } and hence 1 ker(i ⋆ j ) i′ (α) j ′ (α)− α π (U V; x ) . # # ⊂ N # · # | ∈ 1 ∩ 0

Corollary 2.69. Let X be a topological space and U, V X such that U V = X and ⊂ ∪ x U V. Assume that U, V are path connected and that U V is 1-connected. Then 0 ∈ ∩ ∩

π1(X; x0) π1(U; x0) ⋆π1(V; x0) where the isomorphism is induced by the inclusion maps.

Proof. By assumption π (U V, x ) = 1 and hence 1 ∩ 0 { } 1 ( i′ (α) j ′ (α)− α π (U V; x ) ) = 1 . N { # · # | ∈ 1 ∩ 0 } { } The assertion then follows from Theorem 2.68.

Example 2.70. Consider the ﬁgure 8 space and the two subsets U and V as indicated in the picture:

X V U

b b x0 x0

It is easy to see that U S1 and also V S1. Moreover, the intersection ≃ ≃ U V point ∩ ≃ is 1-connected. Hence we have

1 1 π1(X; x0) π1(S ; x0) ⋆π1(S ; x0) Z ⋆ Z .

50 2.4 The Seifert-van Kampen theorem

Example 2.71. Let X be a connected n-dimensional manifold with n 3. ≥

b V p

Let p X and U := X p . Now let V be an open neighborhood of p homeomorphic to Rn, ∈ \{ } which is contractible and hence 1-connected. Then the space

n n 1 U V R 0 S − ∩ ≈ \{ } ≃ is 1-connected. By Corollary 2.69 we then deduce

π (X; x ) π (U; x ) ⋆π (V; x ) = π (X p ; x ) . 1 0 1 0 1 0 1 \{ } 0 Removing a point from a manifold of dimension at least 3 does not change its fundamental group.

Example 2.72. Let M and N be two connected manifold of dimension n 3. ≥

M N

Let X = M#N.

U V ∩

= U = | {z } V

| {z }

51 2 Homotopy Theory

Now choose U and V as in the picture. Then we have that U M p and V N q . Since ≈ \{ } ≈ \{ } n 1 n 1 U V S − (0, 1) S − ∩ ≈ × ≃ is 1-connected we ﬁnd by Corollary 2.69 once again that

π1(M#N) π1(U) ⋆π1(V ) π1(M) ⋆π1(N) . 3 3 3 3 3 For example, if M = N = T then π1(T #T ) = (Z ) ⋆ (Z ).

Remark 2.73. We now see easily that the torus T n with n 3 cannot be homotopy equivalent ≥ to the connected sum T n M#N of two non-1-connected manifolds M and N.2 If it were n ≃ n n possible then π1(T ) π1(M) ⋆π1(N) would not be abelian but we know that π1(T ) Z , a contradiction.

2.5 The fundamental group of surfaces

Our aim is to prove that orientable compact connected surfaces of diﬀerent genus (as depicted by the pastries in Example 1.5) are not homotopy equivalent and therefore not homeomorphic.

2 2 Deﬁnition 2.74. We call Fg := T # #T a surface of genus g 1. ··· ≥ g times −

| {z } F = g ···

g times −

| {z } 2 Remark 2.75. We also put F0 := S .

We now want to compute π1(Fg ) and show that the fundamental groups for surfaces of diﬀerent genus are not isomorphic. This then shows in particular that they are not homotopy equivalent.

Proposition 2.76. For any g 1 ≥ 2 2 2 Fg = T # #T D / ··· ≈ ∼ g times −

| {z } 2If one allows simply connected summands then it is of course possible, T n T n#Sn . ≈

52 2.5 The fundamental group of surfaces

where x y iﬀ x = y or x, y ∂D2 and are identiﬁed according to the following scheme: ∼ ∈ 1 b−g a1 a 1 b1 g− 1 a1− 1 b1− D2 ··· ≈ a2

b2 1 a2−

This identiﬁcation is to be understood as follows: Each line segment labelled by ai (or bj 1 1 respectively) is identiﬁed with with ai− (or b−j ) with respect to the direction indicated by the arrow. Note that there are 2g labels ai , bi , i = 1, .., g.

Proof. We do an induction on g. Induction basis for g = 1: We see that the labelled disk D2/ is homeomorphic to the labelled 2 ∼ W / which is homeomorphic to the two-dimensional torus, i.e. to F1. ∼ a1

1 b1− a1 a1

b 1 b b 1 b ≈ 1− 1 ≈ 1− 1 ≈

1 b1 a1− 1 a1−

Inductive step, g 1 g: We perform a cut along the line c. − ⇒

53 2 Homotopy Theory

1 b−g a1 a1 1 b1 1 b1 ag− b− 1 g 1 a− 1 a− 1 ag− 1 c 1 1 b1− b1− ≈ c a2 a2 b2

1 b2 a2− 1 a2−

The endpoints of c in the two pieces are identiﬁed. This yields a homeomorphism where the interior of the now closed loop c is cut out of the two remaining disks.

a1 b1 1 b−g 1 a− a 1 1 1 −g b−g a2 a 1 b2 c 1 g− 1 b1− 1 b a2− 1− a1 c c 1 c b2− ≈ a2

1 a− b1 b2 1

1 a2− By induction hypothesis the left-hand disc is homeomorphic to a surface of genus g 1 with − a disc removed that is bounded by c. The right-hand disc becomes a torus, also with a disc removed that is bounded by c. Gluing these two spaces together along c, we obtain a space which is homeomorphic to a surface of genus g:

c c ≈

(g 1) times g times − − −

| {z } | {z }

54 2.5 The fundamental group of surfaces

Remark 2.77. We note that F D2/∂D2. 0 ≈

Remark 2.78. Let G be a group. Assume G is generated by g1,..., gn G as a group. We have k ∈ group homomorphisms ϕi : Z G, ϕi (k) = g . Repeated application of the universal property → i of free products of groups yields the group homomorphism

(...(ϕ ⋆ϕ ) ⋆ϕ ) ⋆ ⋆ϕn =: ϕ ⋆ ⋆ϕn : (...(Z ⋆ Z) ⋆ Z) ⋆ ⋆ Z =: Z ⋆ ⋆ Z G 1 2 3 ··· 1 ··· ··· ··· → with k1 kr (ϕ1 ⋆ ⋆ϕn )((k1, i1),..., (kr, ir )) = g g , ··· i1 ··· ir where i j 1, ..., n is an index used to make the Z-factors formally disjoint. ∈ { } The fact that g , ..., gn generate G is equivalent to the fact that ϕ ⋆ ⋆ϕn : Z ⋆ ⋆ Z G 1 1 ··· ··· → is onto. It follows that Z ⋆ ⋆ Z G ··· ker(ϕ ⋆ ⋆ϕn) 1 ··· If the normal subgroup ker(ϕ ⋆ ⋆ϕn) is also ﬁnitely generated as a group, with generators 1 ··· x1,..., xm, then we call G ﬁnitely presentable and

g ,..., gn x ,..., xm h 1 | 1 i a presentation of G.

Example 2.79. For the cartesian product Z2 of two copies of Z we have the presentation

2 1 1 Z a, b aba− b− . h | i The cyclic group Z/2Z of order 2 is presentable as

Z/2Z a a2 . h | i

Remark 2.80. A word of caution: isomorphic groups may have several, quite diﬀerent presentations. For example

1 1 1 1 1 1 1 1 1 x, y xyxy− x− y− x, y, w, z xyxy− x− y− , xyxw− , zy− x− h | i h | i 1 1 1 because the generators xyxw− and zy− x− on the right-hand side can be used to eleminate w and z. It is therefore often not obvious whether two presentations give rise to isomorphic groups.

Theorem 2.81. For any g N we have ∈ 1 1 1 1 π (Fg ) a , b , ..., ag, bg a b a− b− ag bg a− b− . 1 h 1 1 | 1 1 1 1 ··· g g i

55 2 Homotopy Theory

2 Proof. Recall that Fg D / as in Proposition 2.76. To apply the Seifert-van Kampen ≈ 2 ∼ 2 2 1 theorem 2.68 we put U := D˚ and V := (D / ) D ( ). We have Fg = U V. ∼ \ 2 ∪

1 b−g a1 1 b1 ag− 1 a1−

1 b1− V

1 a2−

The subset U is contractible and therefore π (U) = 1 . The subset V is homotopy equivalent 1 { } to the boundary ∂D subject to the identifications of the equivalence relation, V ∂D/ . The ≃ ∼ identifications induced by generate a bouquet of 2g circles, one for each relation ai and bi . ∼ The bouquet is denoted by S1 S1, where the wedge sum “ ” of two topological spaces X ∨···∨ ∨ and Y is defined to be the disjoint union of X and Y with identification of two base points x X, 0 ∈ y Y such that X Y := X Y/ x y . For the bouquet of circles all S1’s are joined at the 0 ∈ ∨ ∪ { 0 ∼ 0} same base point. Graphically we can depict the bouquet of circles as follows

S1 S1 = ∨···∨

So we have V ∂D2/ S1 S1 . ≃ ∼≈ ∨···∨ 2g times

| {z }

56 2.5 The fundamental group of surfaces

The fundamental group of V follows immediately from Example 2.70 by induction: π (V ) Z ⋆ Z ⋆ ⋆ Z 1 ··· 2g times with generators again denoted by a1, b1, a2, b|2, ..., {zag, bg.} For the intersection of U and V we have U V = D˚ 2 D¯ 2( 1 ) S1 and thus by Theorem 2.42 we have π (U V ) Z. A generator ∩ \ 2 ≃ 1 ∩ of π (U V ) is given by a loop c of degree 1. 1 ∩

1 b−g a1 1 b1 ag− 1 a1−

1 b1− c

1 a2−

The inclusion map i : U V U induces the trivial homomorphism i because π (U) is ′ ∩ → #′ 1 trivial. For the inclusion map j ′ : U V V we note that the induced isomorphism maps c 1 1 1 1 ∩ → onto a b a b ag bg a b . 1 1 1− −1 ··· g− −g By the Seifert-van Kampen theorem 2.68 we ﬁnd π (U) ⋆π (V ) π (F ) 1 1 1 g ( j (α) i (α) 1 α π (U V )) N #′ · #′ − | ∈ 1 ∩ ( ) = π1 V ( j (α) α π (U V )) N #′ | ∈ 1 ∩ Z ⋆ ⋆ Z = ··· ( j (c)) N #′ 1 1 1 1 = a , b , ..., ag, bg a b a− b− ag bg a− b− . h 1 1 | 1 1 1 1 ··· g g i

Example 2.82. For the two-dimensional torus T2 we ﬁnd with Example 2.79

2 1 1 2 π (T ) = π (F ) = a, b aba− b− Z , 1 1 1 h | i in agreement with Remark 2.48.

57 2 Homotopy Theory

Corollary 2.83. For g, g N, g , g we have Fg ; Fg and hence Fg 0 Fg . ′ ∈ ′ ′ ′

( ) ( ) Proof. The statement follows once we see that π1 Fg π1 Fg′ . Attention here: as noted in 2.80 diﬀerent presentations can yield isomorphic groups. 1 1 For any group G let [G, G] be the normal subgroup generated by all commutators aba− b− , a, b G. The abelian factor group G/[G, G] is called the abelianization of G. We now calculate ∈ the abelianization of π1(Fg ). π (F ) 1 g = 1 1 1 1 1 1 a1, b1,..., ag, bg a1b1a1− b1− ag bg ag− b−g , a1b1a1− b1− ,... [π1 (Fg ),π1(Fg )] h | ··· 1 1 ..., a1a2a1− a2− ,... 1 1 i 1 1 = a , b ,..., ag, bg a b a− b− , ..., a a a− a− ,... h 1 1 | 1 1 1 1 1 2 1 2 i Z2g, 1 1 = where the second equality follows because the simple commutators ai bi ai− b−i , i 1, ..., g 1 1 1 1 generate the relation a b a b ag bg a b . 1 1 1− −1 ··· g− −g Hence if Fg Fg then π (Fg ) π (Fg ) and thus ≃ ′ 1 1 ′ π1(Fg )/[π1 (Fg ),π1 (Fg )] π1(Fg′ )/[π1 (Fg′ ),π1(Fg′ )]. 2g 2g Thus Z Z ′ and therefore g = g′.

Remark 2.84. This proves the uniqueness part of the classiﬁcation for surfaces, see Example 1.5.

2.6 Higher homotopy groups

We now generalize the deﬁnition of π1(X; x0).

Deﬁnition 2.85. Let W n = [0, 1] ... [0, 1] be the n-cube. Let X be a topological space × × n times and x0 X. Then ∈ | {z } n n πn (X; x ) := [σ]∂W n σ C(W , X), σ(∂W ) = x 0 { | ∈ { 0}}

is called the n-th homotopy group of X with base point x0. Here [σ]∂W n denotes the homotopy class of σ relative to ∂W n.

n n The group structure on πn (X; x ) is obtained as follows: For σ, τ C(W , X) with σ(∂W ) = 0 ∈ τ(∂W n ) = x deﬁne σ ⋆ τ by { 0} σ(2t1, t2,..., tn ), 0 t1 1/2, (σ ⋆ τ)(t1,..., tn ) = ≤ ≤ τ(2t1 1, t2,..., tn ), 1/2 t1 1.  − ≤ ≤   58 2.6 Higher homotopy groups

Then σ ⋆ τ C(W n, X) with (σ ⋆ τ)(∂W n) = x . ∈ { 0} W n W n W n

σ σ ⋆ τ

X t2,..., tn b t1 x0

Now put [σ]∂W n [τ]∂W n := [σ ⋆ τ]∂W n . The proof that this yields a well-deﬁned group · multiplication on πn (X; x0) for n 2 is literally the same as in the case for n = 1. The neutral ≥ n n n = 1 element is represented by the constant map ε : W X, ε (t ,..., tn ) x , and [σ] n is x0 → x0 1 0 −∂W represented by σ(1 t , t ,..., tn ). − 1 2 Unlike for the case n = 1 the higher homotopy groups are abelian:

Proposition 2.86. Let X be a topological space and x X. Then for n 2 the group 0 ∈ ≥ πn (X; x0) is abelian.

Proof. For σ, τ C(W n, X) with σ(∂W n ) = τ(∂W n) = x we need to show that σ ⋆ τ and ∈ { 0} τ⋆σ are homotopic relative to ∂W n. The homotopy is obtained by precomposing with the homotopy of the n-cube indicated in the following picture (which illustrates the case n = 2):

59 2 Homotopy Theory

σ τ σ σ σ deform deform deform −→ τ −→ τ −→ τ

τ σ deform −→

Remark 2.87. This proof also shows that replacing t1 by any other variable in the deﬁnition of σ ⋆ τ gives the same group multiplication on πn (X; x0).

For f C(X,Y ) with f (x ) = y we get a group homomorphism ∈ 0 0

f : πn (X; x ) πn (Y; y ) # 0 → 0 deﬁned by [σ]∂W n [ f σ]∂W n . 7→ ◦

Remark 2.88. Lemma 2.18, 2.19, Corollary 2.20, Propositions 2.21, 2.22, Theorem 2.23 and Corollary 2.24 also hold for πn (X; x ). In particular, if f : X Y is a homotopy equivalence 0 → then the map f : πn (X; x ) πn (Y; f (x )) is an isomorphism. For γ Ω(X; x , x ) there is # 0 → 0 ∈ 0 1 an isomorphism Φγ : πn (X; x ) πn (X; x ) 1 → 0 given by [σ]∂W n [σ ]∂W n where 7→ ′ 1 σ(2t), t max 2, σ′(t) = k k ≤ ( ) 1 γ 2 2 t max , 2 t max 1.  − k k ≤ k k ≤  

x0 x1 σ

60 2.6 Higher homotopy groups

Remark 2.89. By Exercise 1.7 we have the maps

n π n ϕ n W / W /∂W n / S ≈ n n n and we see that σ C(W , X) with σ(∂W ) = x0 corresponds uniquely to fσ C(S , X) ∈ n { } ∈ with fσ (s ) = x , where s = ϕ(∂W ) such that σ = fσ ϕ π. Thus there is a canonical 0 0 0 ◦ ◦ bijection 1:1 n πn (X; x0) [ f ] s f C(S , X), f (s0) = x0 . ←→ { { 0 } | ∈ }

Remark 2.90. The Seifert-van-Kampen theorem for πn works only under very restrictive assumptions. For this reason the computation of πn for explicit examples can be very diﬃcult. We m m will be able to compute πn (S ) for n m but many of the πn (S ) for n > m are actually still ≤ unknown.

0 Remark 2.91. The deﬁnition of πn (X; x ) also works for n = 0. A map σ C(W , X) 0 ∈ corresponds to the point σ(W 0) X. Two such maps are homotopic iﬀ the corresponding points ∈ can be joined by a path. Hence

π (X; x ) = path components of X . 0 0 { }

But there is no (natural) group structure on π0 (X; x0). More precisely, π0(X; x0) is a pointed set, i.e., a set with a distinguished point, namely the path component containing x0. This corresponds to the neutral element in πn (X; x ) for n 1. 0 ≥

Deﬁnition 2.92. Let W, E and B be topological spaces and let p C(E, B). We say that p has ∈ the homotopy lifting property, abbreviated by HLP, for W iﬀ for every f C(W, E) and every ∈ h C(W [0, 1], B) with h(w, 0) = p( f (w)) for all w W there exists H C(W [0, 1], E) ∈ × ∈ ∈ × such that H(w, 0) = f (w) w W and h = p H . ∀ ∈ ◦ In other words, there exists an H C(W [0, 1], E) such that the diagram ∈ × f w W / E ❴ ✉: ∈ ✉ H ✉ p ✉ ✉ h (w, 0) W [0, 1] / B ∈ × commutes.

61 2 Homotopy Theory

= = = e Example 2.93. Consider the spaces E point , B R, W b 0 { } W 0 and the map given by p(e) = 0. No h : W [0, 1] B except the constant path ε can be lifted × → 0 because it leaves the image of p. Here the problem is the lack of surjectivity of p. R | h 0

Example 2.94. Now consider E = [0, ) 0 ( , 0] 1 R2, B = R and let p be the ∞ ×{ } ∪ −∞ ×{ } ⊂ projection onto the ﬁrst factor, p(t, s) = t.

f b E

W = w { 0} B | h

The map p is surjective but still does not have the HLP for W = W 0. For example, choose h(w0, t) = t, f (w0) = (0, 1).

Definition 2.95. A map p C(E, B) is called a Serre fibration or weak fibration iff it has the ∈ HLP for all W n, n 0. The space E is called the total space and B is called the base space of ≥ the fibration. For b B we call p 1 (b ) the fiber over b . 0 ∈ − 0 0

Example 2.96. For topological spaces F and B put E := B F and p = pr , the projection on × 1 the B-factor. Then p has the HLP for all W. In particular, p is a Serre ﬁbration. Namely, let f C(W, B F) and h C(W [0, 1], B) be given such that p( f (w)) = h(w, 0) for all w W. ∈ × ∈ × ∈ Now write f (w) = ( β(w),ϕ(w)) with β C(W, B) and ϕ C(W, F). Hence ∈ ∈ h(w, 0) = p( f (w)) = β(w). Now put H(w, t) := (h(w, t),ϕ(w)). Then H C(W [0, 1], B F) and ∈ × × p(H(w, t)) = pr1 (h(w, t),ϕ(w)) = h(w, t) , H(w, 0) = (h(w, 0),ϕ(w)) = ( β(w),ϕ(w)) = f (w) , as required. Hence p has the HLP for any W.

62 2.6 Higher homotopy groups

Definition 2.97. A map p C(E, B) is called fiber bundle with fiber F iff for each b B there ∈ ∈ exists an open subset U B with b U and a homeomorphism Φ : p 1 (U) U F such ⊂ ∈ − → × that the diagram Φ 1 U F o p− (U) × ≈ tt pr tt 1 tt tt p p 1(U ) tt | − U z commutes.

Lemma 2.98. Every ﬁber bundle is a Serre ﬁbration.

Proof. Let f C(W n, E) and h C(W n [0, 1], B) such that h(w, 0) = p( f (w)) for all w W. ∈ ∈ × ∈ Now subdivide W n [0, 1] such that h maps each subcube entirely into an open subset U as in × the deﬁnition of the ﬁber bundle.

W n [0, 1] ×

[0, 1]

Q11 W n

By Example 2.96 products have the HLP, hence we can extend the map f to a continuous map H : (W n 0 ) Q E such that p H = h. 11 ×{ } ∪ 11 → ◦ 11

63 2 Homotopy Theory

W n [0, 1] ×

[0, 1]

Q11 Q12 W n

Next we want to extend the lift to Q12. Now there seems to be a problem because the required lift H need not only coincide with f along the edge Q (W n 0 ) but also with H along 12 12 ∩ ×{ } 11 Q Q . 11 ∩ 12 But there are homeomorphisms of a cube onto itself mapping two edges onto one as indicated in the picture.

≈

Thus we can apply the HLP and extend the lift to (W n 0 ) Q Q . Iteration of this ×{ } ∪ 11 ∪ 12 procedure proves the assertion.

Example 2.99. Let G be a Lie group, e.g. a closed subgroup of GL(n; K) with K = R or K = C. Let H G be a closed subgroup. We equip the space B = G/H = g H g G with the ⊂ { · | ∈ } quotient topology. Such a space is called a homogeneous space. Then G G/H with g g H → 7→ · is a fiber bundle with fiber H. The proof of this fact requires some technical work, see [8, p. 120 ff].

64 2.6 Higher homotopy groups

Example 2.100. Let G = SO(n + 1) and

B 0 H = B SO(n) . 0 1 ∈ ( ! )

Then H is a closed subgroup of G isomorphic to SO (n). We now show that G/H Sn. Consider n ≈ the map f : G S with A A en+ where en+ the (n + 1)-st unit vector of the canonical → 7→ · 1 1 basis. The map f is continuous and surjective. We observe

f (A) = f (A˜) A en+1 = A˜ en+1 ⇐⇒ ·1 · A˜− A en+ = en+ ⇐⇒ · · 1 1 1 ⋆ 0 A˜− A = ⇐⇒ · ⋆ 1 ! 1 A˜− A H ⇐⇒ · ∈ π(A) = π(A˜) ⇐⇒ where the map π : G G/H is the canonical projection. We conclude that the map f descends → to a bijective map f¯ : G/H Sn. By the universal property of the quotient topology the map → f¯ : G/H Sn is continuous. Since G is compact the space G/H is also compact. Moreover, → the sphere Sn is a Hausdorff space, hence the map f¯ is a homeomorphism. Thus we obtain a fiber bundle SO(n + 1) Sn with fiber SO(n). →

Let p : E B be any Serre ﬁbration and ﬁx e E. Put b := p(e ) B and let F := p 1 (b ). → 0 ∈ 0 0 ∈ − 0 Then e F. Let ι : F E be the inclusion map. We obtain the following two homomorphisms: 0 ∈ →

ι : πn (F; e ) πn (E; e ), # 0 → 0 p : πn (E; e ) πn (B; b ). # 0 → 0

Now we construct a map ∂ : πn (B; b0) πn 1(F; e0). We deﬁne → − k k Box := 1 and Boxk := (W 1 ) (∂W [0, 1]) for k 1. 0 { } ×{ } ∪ × ≥ Then we have

n n 1 ∂W = (W − 0 ) Boxn 1, ×{ } ∪ − n 1 n 1 (W − 0 ) Boxn 1 = ∂W − 0 . ×{ } ∩ − ×{ } k Consider the homeomorphism ηk : Boxk W obtained by the central projection from → ( 1,..., 1, 1). 2 2 −

65 2 Homotopy Theory

b x

b ηk (x)

b ( 1,..., 1, 1) 2 2 −

This homeomorphism maps the faces out of which Boxk is built onto the regions depicted n in the right ﬁgure. In order to deﬁne ∂ : πn (B; b0) πn 1(F; e0) let σ C(W , B) with → − ∈ σ(∂W n ) = b . { 0}

F E

tn b e0

σ p

b b0 t1,..., tn 1 −

Since (t1,..., tn 1) σ(t1,..., tn 1, 0) = b0 is constant we can lift it to the constant map − 7→ − n 1 n (t1,..., tn 1) e0. Now the HLP of p for W − yields a continuous map Σ : W E with − 7→ →

66 2.6 Higher homotopy groups

n n Σ(t1,..., tn 1, 0) = e0 and p Σ = σ. From σ(∂W ) = b0 we have Σ(∂W ) F. We put −1 n 1 ◦ { } ⊂ σ˜ := Σ η− : W − F. We want to deﬁne the map ∂([σ]∂W n ) := [σ ˜ ] n 1 . We have to ◦ n 1 → ∂W − check well-deﬁnedness− of this map:

a) We have to show that [σ ˜ ] n 1 does not depend on the particular choice of the lift Σ. ∂W − n 1 Let Σ′ C(W , E) be another lift of σ with Σ′(t1,..., tn 1, 0) = e0. Then Σ− Σ′ is a lift of 1 ∈ − 1• σ− σ. Here denotes the concatenation with respect to the variable tn, Σ− (t1,..., tn ) = • • 1 1 n Σ(t1,..., tn 1, 1 tn ) and similarly for σ− . Since σ− σ ∂W n ε we can ﬁnd a homotopy − − • ≃ b0 h : W n+1 B relative to ∂W n with → 1 h(t ,..., tn, 0) = (σ− σ)(t ,..., tn ) and h(t ,..., tn, 1) = b . 1 • 1 1 0 n+1 n+1 Then h(Boxn ) = b . We apply the HLP of p for W to get a lift H C(W , E) of h { 0} ∈ with

1 H(t ,..., tn, 0) = Σ− Σ′(t ,..., tn ). 1 • 1

From h(Boxn ) = b0 we have H(Boxn ) F. Then we get a homotopy in F relative to n 1 = Σ{ } 1 = Σ 1⊂ ∂W − fromσ ˜ η−n 1 toσ ˜ ′ ′ η−n 1 as shown in the picture. ◦ − ◦ −

Σ′

1 Σ−

tn+1 tn

t1,..., tn 1 −

b) We also have to show that σ ∂W n σ′ in B impliesσ ˜ n 1 σ˜ ′ in F. ≃ ≃∂W − Let h : W n [0, 1] B be a homotopy in B from σ to σ relative to ∂W n. × → ′

67 2 Homotopy Theory

σ′

B h b b0

tn+1 tn

t1,..., tn 1 − σ

n n+1 The HLP for W yields a lift H : W E of h with H(t1,..., tn 1, 0, tn+1) = e0. We thus → − obtain a homotopy ˆ = 1 H(t1,..., tn 1, s) H(η−n 1(t1,..., tn 1), s) − − − n 1 in F fromσ ˜ toσ ˜ ′ relative to ∂W − and hence

[σ ˜ ] n 1 = [σ ˜ ] n 1 . ∂W − ′ ∂W −

We have shown that the map ∂ : πn (B; b0) πn 1 (F, e0) is well deﬁned. → −

Lemma 2.101. For n 2 the map ∂ : πn (B; b0) πn 1 (F, e0) is a group homomorphism. ≥ → −

Proof. Let σ, τ C(W n, B) such that σ(∂W n ) = τ(∂W n) = b . Choose a lift H C(W n, E) ∈ { 0} ∈ of σ ⋆ τ with H(t1,..., tn 1, 0) = e0. Restriction yields lifts Σ of σ and T of τ up to stretching − in the t1-direction. Moreover, we have

Σ 1 1 1 ( η−n 1) ⋆ (T η−n 1) ∂W n 1 H η−n 1 ◦ − ◦ − ≃ − ◦ −

68 2.6 Higher homotopy groups

The homotopy is given by shrinking the marked region in the t1-direction.

b b b

We conclude that

= Σ 1 1 ∂([σ]∂W n ) ∂([τ]∂W n ) [ η−n 1]∂W n 1 [T η−n 1]∂W n 1 · ◦ − − · ◦ − − = Σ 1 1 [( η−n 1) ⋆ (T η−n 1)]∂W n 1 ◦ − ◦ − − = 1 [H η−n 1]∂W n 1 ◦ − − = ∂([σ ⋆ τ]∂W n )

= ∂([σ]∂W n [τ]∂W n ). ·

Hence ∂ : πn (B; b0) πn 1 (F, e0) is a homomorphism if n 2. For n = 1 this statement → − ≥ does not make sense because π0(F, e0) is not a group. But ∂ still maps the neutral element of π1(B, b0) to the distinguished element of π0(F, e0).

Theorem 2.102 (Long exact homotopy sequence of a Serre ﬁbration). Let p : E B be a → Serre ﬁbration, e E, b = p(e ) B and F = p 1 (b ). Let ι : E B be the inclusion 0 ∈ 0 0 ∈ − 0 → map. Then the following sequence is exact:

∂ ι# p# ∂ ι# ... / πn (F; eo ) / πn (E; e0) / πn (B; b0) / πn 1(F; e0) / ... −

∂ ι# p# ... / π0(F; e0) / π0(E; e0) / π0(B; b0)

69 2 Homotopy Theory

Remark 2.103. Exactness means that the image of the incoming map equals the kernel of the outgoing map. The question arises what this means on the π0-level where we do not have homomorphisms. The image is defined for an arbitrary map. For the kernel we recall that π0 is a set together with a distinguished element which corresponds to the neutral element of a group. It is therefore natural to define the kernel of a map to be the set of all elements in the domain of the map which are mapped to the distinguished element. Having clearified this, exactness of the above sequence also makes sense on the π0-level.

Proof. a) Exactness at πn (E; e ) for n 0: 0 ≥ i) im(ι ) ker(p ): # ⊂ # Since p ι is the constant map, we have for any [σ]∂W n πn (F; e ) ◦ ∈ 0 n p# (ι#([σ]∂W n )) = (p ι)#([σ]∂W n ) = [p ι σ]∂W n = [ε ]∂W n = 0 ◦ ◦ ◦ b0

ii) ker(p#) im(ι#): ⊂ n Let [τ]∂W n πn (E; e0) with p# ([τ]∂W n ) = [p τ]∂W n = 0. Hence p τ ∂W n ε . ∈ ◦ ◦ ≃ b0 Let h : W n [0, 1] B be a homotopy in B relative to ∂W n from p τ to εn . Lift × → ◦ b0 the map h to a homotopy H : W n [0, 1] E with initial conditions τ, i.e. H( , 0) = τ. × → · The red area in the diagram gets mapped to F by H because it gets mapped to b0 by h.

[0, 1]

W n

n 1 1 We obtain a homotopy in E relative to ∂W from τ to H η . Hence τ ∂W n H η ◦ −n 1 ≃ ◦ −n 1 and we conclude that − −

= 1 = 1 [τ]E,∂W n [H η−n 1]E,∂W n ι#([H η−n 1]F,∂W n ) im ι#. ◦ − ◦ − ∈ b) Exactness at πn (F; e ) for n 0: 0 ≥ i) im ∂ ker ι#: ⊂ = Σ 1 Σ Let [σ]∂W n+1 πn+1(B; b0). Then we have ∂([σ]∂W n+1 ) [ η−n ]F,∂W n where is ∈ n ◦ a lift of σ with initial conditions εe0 .

70 2.6 Higher homotopy groups

n εe0

The map Σ yields a homotopy relative to ∂W n in E from εn to Σ η 1 . Hence e0 −n 1 ◦ − 1 1 n ι (∂( σ n+1 )) = ι ( Σ η n ) = Σ η n = ε n = . # [ ]∂W # [ −n 1]F,∂W [ −n 1]E,∂W [ e0 ]E,∂W 0 ◦ − ◦ − ii) ker ι# im ∂: ⊂ n n ( ) = ( n ) n Let [τ]∂W πn F; e0 with 0 ι# [τ]∂W . Hence τ ∂W εe0 in E. Let H be a ∈ n n ≃ = homotopy in E relative to ∂W from εe0 to τ. Then H is a lift of h : p H. Since n+1 n+1 ◦ H maps ∂W to F, the map h maps ∂W to b0. Thus h represents an element in 1 + n = n+ πn 1(B; b0). By deﬁnition of ∂ we have [H η−n 1]∂W ∂([h]∂W 1 ). ◦ −

[0, 1] εn e0

W n

1 In the image of Boxn under η−n we let the interior cube grow and thereby obtain a n 1 homotopy in F relative to ∂W from H η−n 1 to τ. ◦ −

71 2 Homotopy Theory

e0 e0

e0 τ e0 deform e0 τ e0 deform τ −→ −→

e0 e0

= 1 = + Therefore [τ]∂W n [H η−n 1]∂W n ∂[h]∂W n 1 im(∂). ◦ − ∈ c) Exactness at πn (B; b ) for n 1: 0 ≥ i) im p# ker ∂: ⊂ n 1 Let [τ]∂W n πn (E; e ). Then τ is a lift of p τ with initial conditions ε . Since τ ∈ 0 ◦ e0− maps the boundary of W n to e we have τ η 1 = εn 1. Thus 0 −n 1 e0− ◦ − 1 n 1 ( ( n ) = ( n )) = n 1 = n 1 = ∂ p# [τ]∂W ∂ [p τ]∂W [τ η−n 1]∂W [εe0− ]∂W 0. ◦ ◦ − − − ii) ker ∂ im p : ⊂ # Let [σ]∂W n πn (B; b0) with ∂([σ]∂W n ) = 0. Let Σ be a lift of σ with initial condition n 1 ∈ 1 ε . Then Σ η represents ∂([σ] n ) = 0. Hence we have in F e0− −n 1 ∂W ◦ − 1 n 1 Σ n 1 η−n 1 ∂W εe0− ◦ − ≃ − e0

maps to F

F F e0 Σ e0

Now choose a homotopy in F relative to ∂W n 1 from Σ η 1 to εn 1. Use this homotopy − −n 1 e0− ◦ − to continuously extend Σ to the larger cube with boundary values e0; call this extension τ. We then have

[τ]∂W n πn (E; eo ) and p ([τ]E,∂W n ) = [p τ]∂W n . ∈ # ◦ Now p τ is homotopic relative to ∂W n to σ as shown below. ◦

72 2.6 Higher homotopy groups

b0 deform deform −→ −→

b0 σ b0 σ σ

Thus [σ]∂W n = [p τ]∂W n im p . ◦ ∈ #

Definition 2.104. A fiber bundle with discrete fiber is called a covering.

Corollary 2.105. If p : E B is a covering with e F, b = p(e ) B, then the map → 0 ∈ 0 0 ∈

p : πn (E, e ) πn (B; b ) # 0 → 0 is an isomorphism for all n 2. ≥

Proof. The assertion follows from the long exact sequence:

ι# p# ∂ 0 = πn (F; eo ) / πn (E; e0) / πn (B; b0) / πn 1(F; e0) = 0 . { } − { }

Example 2.106. The map Exp : R S1 with t e2πit is a covering with ﬁber Z. Hence → 7→ 1 πk (S ; 1) πk (R, 0) = 0 { } for all k 2. More generally, Exp : Rn T n = S1 ... S1 with ≥ → × × 2πit1 2πit n (t ,..., tn ) (e ,..., e ) 1 7→ n n n is a covering with ﬁber Z . Hence πk (T ) πk (R ) = 0 for all k 2. { } ≥

Example 2.107. Consider the real projective space, deﬁned by RPn := Sn/ , where x y ∼ ∼ x = y or x = y. Then the map p : Sn RPn with x [x] is a covering with ﬁber ⇐⇒ n− n → 7→ ∼ Z/2Z. Hence πk (RP ) πk (S ) for all k 2. For n 2 we investigate the sequence: ≥ ≥ 0 = π (Sn ) / π (RPn ) / π (Z/2Z) / π (Sn ) = 0 { } 1 1 0 0 { }

73 2 Homotopy Theory

n n We deduce that π1(RP ) π0(Z/2Z) Z/2Z as sets. But then π1(RP ) Z/2Z also as groups because there is only one group of order 2 (up to isomorphism).

Example 2.108. We know that SO(n + 1)/SO(n) Sn from Example 2.99. In the case of n = 2 ≈ this means that SO(3)/SO(2) S2. We also know that SO(2) S1. For k 3 we consider the ≈ ≈ ≥ long exact sequence

2 / πk (SO(2)) / πk (SO(3)) / πk (S ) / πk 1 (SO(2)) ··· − 1 1 We know that πk (S ) = 0 and πk 1 (S ) = 0 . Hence we ﬁnd { } − { } 2 πk (SO(3)) πk (S ) for all k 3. ≥ The map f : S3 SO(3) given by → x2 + y2 u2 v2 2(yu xv) 2(yv xu) − − − − f (x, y, u, v) = 2(yu + xv) x2 y2 + u2 v2 2(uv xy) − − − * 2(yv xu) 2(uv + xy) x2 y2 u2 + v2+ . − − − / satisﬁes f ( x, y, z, v) = f (x, y, z, v) and therefore induces a continuous map − − − −, - f¯ : RP3 SO(3). This map is bijective and thus a homeomorphism. Hence we have → SO(3) RP3. It follows that ≈ 2 3 3 πk (S ) πk (SO(3)) πk (RP ) πk (S )

3 for all k 3. Later we will see (Example 3.121 on page 151) that π3(S ) Z and consequently 2 ≥ π3(S ) Z.

Example 2.109. By the same proof as for SO(n + 1)/SO(n) Sn we get that ≈ SU(n + 1)/SU(n) S2n+1. ≈ Therefore there is a ﬁber bundle SU(n + 1) S2n+1 with ﬁber SU(n). For n 1 we consider → ≥ the following part of the long exact homotopy sequence:

ι# p# + π (SU(n)) / π (SU(n + 1)) / π (S2n 1) = 0 . 0 0 0 { } Hence the map ι : π (SU(n)) π (SU(n+1)) is onto. Since SU(1) = 1 we have π (SU(1)) = # 0 → 0 { } 0 0 and thus π (SU(n)) = 0 by induction on n. Thus SU(n) is path-connected for all n 1. { } 0 { } ≥ Now let us analyze π1(SU(n)). Consider

ι# p# + π (SU(n)) / π (SU(n + 1)) / π (S2n 1) = 0 . 1 1 1 { } Again, we conclude that the map ι : π (SU(n)) π (SU(n+1)) is onto. By the same induction # 1 → 1 as before we ﬁnd π (SU(n)) = 0 for all n 1. Thus SU(n) is simply connected for all n 1. 1 { } ≥ ≥

74 2.7 Exercises

2.7 Exercises

2.1. Let X be a set and let x X. Determine π (X; x ) where 0 ∈ 1 0 a) X carries the discrete topology; b) X carries the coarse topology.

2.2. Let X be a topological space and let ω : S1 X be continuous. Show that the following → are equivalent:

(i) ω is homotopic to a constant map.

(ii) ω has a continuous extension D2 X. →

2.3. Let X be a topological space and let f, g : X Sn be continuous. Assume f (x) , g(x) → − for all x X. Show that f and g are homotopic. ∈

2.4. Let X and Y be topological spaces. Show that X Y is contractible if and only if X and Y × are contractible.

2.5. Let X , X be topological spaces, xi Xi. Put X := X X and x := (x , x ) X. Let 1 2 ∈ 1 × 2 1 2 ∈ pi : X Xi be the canonical projections. Show that → (p , p ) : π (X; x) π (X; x ) π (X ; x ) 1# 2# 1 → 1 1 × 1 2 2 is a group isomorphism.

2.6. Let X = [0, 1] [0, 1] R2 and let A X be the comb space. Show that there is no × ⊂ ⊂ retraction X A. →

1 2 2 1 1 1 2.7. For f C(S , R ) and p R f (S ) consider f p C(S , S ) given by ∈ ∈ \ ∈ f (z) p f (z) = − . p f (z) p | − | Then U( f, p) := deg( f p ) is called the winding number of f around p. a) Show that for p, q R2 f (S1) which can be joined by a continuous path in R2 f (S1) we ∈ \ \ have U( f, p) = U( f, q).

2 1 n b) Compute U( f n, p) for all p R f (S ) and all n Z where f n (z) = z . ∈ \ ∈

75 2 Homotopy Theory

2.8. Show that the system of equations

cos 1 + x2 y3 + sin(xy2) x2 = 0, − 1 y + = 0, cosh(x + y + 10) has a solution (x, y) R2. ∈

2.9. a) Compute π (D2 x ; x ) for all x , x D2. 1 \{ 0} 1 0 1 ∈ b) Show that each homeomorphism f : D2 D2 maps the boundary onto itself, f (∂D2) = → ∂D2.

2.10. On [0, 1] [ 1, 1] consider the equivalence relation given by (t, s) (t , s ) iﬀ (t, s) = × − ∼ ∼ ′ ′ (t , s ) or t t = 1 and s = s. The quotient space M := [0, 1] [ 1, 1]/ is called the ′ ′ | − ′| ′ − × − ∼ Möbius strip. The image S of [0, 1] 0 is called the chord of M, that of [0, 1] 1, 1 is the ×{ } × {− } boundary ∂M of M. a) Show that ∂M is homeomorphic to S1. b) Show that S is a strong deformation retract of M. c) Determine π (M; x ) for some x ∂M and the subgroup ι (π (∂M; x )) π (M; x ) 1 0 0 ∈ # 1 0 ⊂ 1 0 .where ι : ∂M ֒ M is the inclusion map → d) Show that ∂M is not a retract of M.

2.11. Let X = (t, t ) 0 t 1, n N (s, 0) 1 s 1 R2 equipped with the { n | ≤ ≤ ∈ }∪{ | 2 ≤ ≤ } ⊂ induced topology. Show that X is connected but not path-connected.

2.12. Let G and G be groups. Show that G G G G 1 2 1 ∗ 2 2 ∗ 1 a) using the construction of the free product; b) using the universal property.

2.13. Let G1 and G2 be groups. Show: a) If g G G has ﬁnite order then g is conjugate to an element in i (G ) or in i (G ). ∈ 1 ∗ 2 1 1 2 2 b) If G and G are nontrivial then G G contains elements of inﬁnite order. 1 2 1 ∗ 2

2.14. Let X be a path-connected topological space. Show that the suspension ΣX (see Exer- cise 1.11) is also path-connected.

2.15. Show that the map C C, z z2, has the homotopy lifting property for W 0 but not for → 7→ W 1.

76 2.7 Exercises

2.16. Let X = R3 (S1 0 ) = R3 (x, y, z) R3 x2 + y2 = 1, z = 0 . Show: \ ×{ } \{ ∈ | } π (X, 0 ) Z 1 { } and draw a generator of the fundamental group.

2.17. On Sn consider the equivalence relation given by x y x = y or x = y. The ∼ ∼ ⇔ − quotient RPn := Sn/ is called the n-dimensional real projective space. ∼ Show inductively using the Seifert-van Kampen theorem that for n 2 ≥ n π1(RP ) Z/2Z.

Hint: Exercise 2.10 may be helpful for the induction base n = 2.

2.18. Decide by proof or counter-example whether or not the following assertion holds true: The Seifert-van Kampen theorem also holds if one only assumes that U V is connected rather ∩ than path-connected.

2.19. Let E = B F and p = pr : E B be the product ﬁbration. Show that the boundary × 1 → map ∂ : πn+ (B; b ) πn (F; e ) is trivial in this case. 1 0 → 0

2.20. Let p : E B be a Serre ﬁbration. Show that for [σ]∂W n+1 πn+1(B; b0) the class 1 → ∈ [Σ η ]∂W n πn (F; e ) does not depend on the choice of lift Σ of σ with the “initial condition” ◦ −n ∈ 0 εn e0 .

Show that the inclusion SU(n) ֒ U(n) induces an isomorphism .2.21 →

πk (SU(n)) πk (U(n))

for all k 2. ≥

2.22. The complex projective space is defined as CPn := S2n+1/ where z w iff there exists + ∼ + ∼ u S1 C such that z = u w. Here we have regarded S2n 1 = z Cn 1 z = 1 as a subset ∈ + ⊂ + · { ∈ | | | } of Cn 1. The map p : S2n 1 CPn, z [z] , is called the Hopf fibration. → 7→ ∼ Compute n πk (CP ) for all k 2n. ≤ m Hint: You can use πk (S ) = 0 for all 1 k < m and m 2. { } ≤ ≥

2.23. Let p : X Y and q : Y Z be Serre ﬁbrations. Prove that q p : X Z is a Serre → → ◦ → ﬁbration as well.

2.24. Let p : E B be a Serre ﬁbration, e E, b = p(e ) and F = p 1 (b ). Show → 0 ∈ 0 0 − 0

77 2 Homotopy Theory

a) If F is simply connected then π1(E; e0 ) π1(B; b0). b) If E is contractible then πn+ (B; b ) πn (F; e ) for all n 1. 1 0 0 ≥

i p 2.25. Let 0 A A A 0 be an exact sequence of abelian groups. Show that the → −→ ′ −→ ′′ → following three conditions are equivalent:

(i) There exists an isomorphism Ψ : A A A such that the following diagram commutes: ′ → × ′′ i p 0 / A / A′ / A′′ / 0 = Ψ = 0 / A / A A / A / 0 × ′′ ′′ where the arrows A A A and A A A are given by the canonical maps a (a, 0) → × ′′ × ′′ → ′′ 7→ and (a, a ) a , respectively. ′′ 7→ ′′ (ii) There exists a homomorphism p : A A such that p p = idA . ′ ′′ → ′ ◦ ′ ′′ (iii) There exists a homomorphism r : A A such that r i = idA. ′ → ◦ If these conditions hold then we say that the exact sequence is split.

i p 2.26. a) Show that every exact sequence 0 A A A 0 of vector spaces (i.e. all → −→ ′ −→ ′′ → spaces are vector spaces over some ﬁxed ﬁeld and all homomorphisms are linear maps) is split.

2 b) Show that the exact sequence 0 Z Z Z/2Z 0 is not split. → −→ → →

78 3 Homology Theory

Homotopy groups are in general hard to compute. For instance, for spheres not all homotopy groups are known even now. In this chapter we introduce rougher invariants which are much easier to determine, the homology groups.

3.1 Singular homology

We will use the notation

e = (0,..., 0) Rn 0 ∈ e = (1,..., 0) Rn 1 ∈ . . n en = (0,..., 1) R ∈

Now we deﬁne

n ∆ := convex hull of e0,..., en n n = ti ei ti 0, ti 1 . ≥ ≤  i=0 i=0  X X    ∆n   Then is called n-dimensional standard simplex . 

Example 3.1. For n = 0,..., 3 the standard simplices can be visualized as:

n = 0 ∆0 = e = point { 0} n = 1 ∆1 = [0, 1] = line segment

e2 b

2 n = 2 ∆ = triangle b b e0 e1

79 3 Homology Theory

b ∆3

3 n = 3 ∆ = tetrahedron b b

Deﬁnition 3.2. Let X be a topological space. A singular n-simplex in X is a continuous map ∆n X. →

Now we ﬁx a commutative ring R with 1. The most important examples will be R = R, Q, C, Z, Z/nZ.

Deﬁnition 3.3. We deﬁne the set of singular n-chains by

Sn (X; R) := free R-module generated by all n-dimensional singular simplices in X m n = formal linear combinations αi σi αi R, σi C(∆ , X), n N0 . = ∈ ∈ ∈ Xi 1

Then Sn (X; R) carries an obvious structure as an R-module.

For n 0 we consider the aﬃne linear map Fi : ∆n ∆n+1 given by ≥ n+1 →

i = ej, j < i Fn+1 (ej ) (3.1) ej+1, j i  ≥ i ∆n ∆n+1  Note that Fn maps to the face of opposite to ei .

1 Example 3.4. Consider the special case F2 :

80 3.1 Singular homology

e2 σ b 1 F2

b b

e0 e1 e0 e1

σ(i)

Deﬁnition 3.5. If σ is an (n + 1)-dimensional singular simplex in X then σ(i) := σ Fi is ◦ n called the i-th face of σ.

The boundary of a singular n-simplex in X is given by:

n ∂σ := ( 1)i σ(i) . = − Xi 0 We see that the boundary of a singular n-simplex is a singular (n 1)-chain. We extend ∂ to − chains by linearity. The boundary of a singular n-chain in X is thus given by

m m ∂ α j σj = α j ∂σj . = = Xj 0 Xj 0

Hence we obtain a linear map ∂ : Sn (X; R) Sn 1 (X; R) and we set ∂(0-chain) := 0. → −

Lemma 3.6. ∂ ∂ = 0. ◦

Proof. It suﬃces to prove ∂∂σ = 0 for all n-simplices σ. For j < i we have

i j = j i 1 Fn Fn 1 Fn Fn−1. (3.2) ◦ − ◦ − We compute

n ∂∂σ = ∂ ( 1)i σ Fi − ◦ n i=0 n*X + = , ( 1)i ∂ σ Fi - − ◦ n i=0 X

81 3 Homology Theory

n n 1 = i − j i j ( 1) ( 1) σ Fn Fn 1 = − = − ◦ ◦ − Xi 0 Xj 0 + + = ( 1)i j σ Fi F j + ( 1)i j σ Fi F j − ◦ n ◦ n 1 − ◦ n ◦ n 1 0 j

Now we deﬁne the set of singular n-cycles by

Zn (X; R) := ker(∂ : Sn (X; R) Sn 1 (X; R)) → − = c Sn (X; R) ∂c = 0 { ∈ | } and the set of singular n-boundaries by

Bn (X; R) := im(∂ : Sn+ (X; R) Sn (X; R)) 1 → = c Sn (X; R) b Sn+ (X; R) such that c = ∂b . { ∈ | ∃ ∈ 1 }

Lemma 3.6 says Bn (X; R) Zn (X; R). ⊂

Deﬁnition 3.7. The quotient Zn (X; R) Hn (X; R) := Bn (X; R) is called n-th singular homology of X with coeﬃcients in R.

Remark 3.8. The n-th homology Hn (X; R) is an R-module.

Example 3.9. Assume that X = point . Then there is only one singular n-simplex, namely n { } the constant map σn : ∆ X. In other words, Sn (X; R) = R σn. Consequently, (i) = i = → · σn σn Fn σn 1 and ◦ −

n 0, n odd, i ∂σn = ( 1) σn 1 = σn 1, n even, n , 0, − −  − i=0  = X 0, n 0.  This implies   R σn, n odd or n = 0, Z (X; R) = · n , 0, n even and n 0,    82 3.1 Singular homology and R σn, n odd, Bn (X; R) = · 0, n even.  We conclude that   R, if n = 0, Hn (X; R) 0, otherwise.   As in homotopy theory we not only associate groups to spaces but also homomorphisms to maps. Let f : X Y be a continuous map. Then we obtain an R-module homomorphism → Sn ( f ) : Sn (X; R) Sn (Y; R) by setting → m m Sn ( f ) αi σi := αi ( f σi ). · · ◦ i=1 i=1 X X

Lemma 3.10. The diagram

Sn (f ) Sn (X; R) / Sn (Y; R)

∂ ∂

 Sn 1 (f ) Sn 1 (X; R) − / Sn 1 (Y; R) − − commutes for all n 0. ≥

Proof. We compute

∂(Sn ( f )(σ)) = ∂( f σ) n ◦ = ( 1)i ( f σ) Fi − ◦ ◦ n i=0 Xn = ( 1)i f (σ Fi ) − ◦ ◦ n i=0 X n = i i Sn 1 ( f ) ( 1) σ Fn − − ◦ i=0 X = Sn 1 ( f )(∂σ). −

This lemma implies Sn ( f )(Zn (X; R)) Zn (Y; R) and Sn ( f )(Bn (X; R)) Bn (Y; R). Hence ⊂ ⊂ we obtain a well-deﬁned R-module homomorphism Hn ( f ) : Hn (X; R) Hn (Y; R) where → Hn ( f )([x]) := [Sn ( f )(X)]. Here the square brackets denote the homology classes of the n-cycles. One sees directly from the deﬁnition that Hn ( ) has the functorial properties · = (i) Hn (idX ) idHn (X;R),

83 3 Homology Theory

(ii) Hn ( f g) = Hn ( f ) Hn (g). ◦ ◦ Exactly as for homotopy groups these functorial properties imply that a homeomorphism f : X Y induces an isomorphism Hn ( f ) : Hn (X; R) Hn (Y; R). Homeomorphic spaces have → → isomorphic homology groups.

3.2 Relative homology

For a topological space X and A X we call (X, A) a pair of spaces and set ⊂ C((X, A), (Y, B)) := f C(X,Y ) f (A) B . { ∈ | ⊂ }

We abbreviate Sn (X) = Sn (X; R) if R is understood. We observe that Sn (A) Sn (X) and ⊂ ∂(Sn (A)) Sn 1 (A). Writing Sn (X, A) = Sn (X, A; R) := Sn (X; R)/Sn (A; R), the map ∂ ⊂ − induces a well-deﬁned homomorphism ∂¯ such that

Sn (X) / Sn (X, A)

∂ ∂¯ Sn 1 (X) / Sn 1 (X, A) − − commutes. Since ∂ ∂ = 0 and since Sn (X) Sn (X, A) is onto we also have that ∂¯ ∂¯ = 0. Set ◦ → ◦

Zn (X, A) = Zn (X, A; R) := ker ∂¯ : Sn (X, A) Sn 1 (X, A) , → − Bn (X, A) = Bn (X, A; R) := im ∂¯ : Sn+ (X, A) Sn (X, A) . 1 → We deﬁne the relative singular homology of (X,A) by

Zn (X, A; R) Hn (X, A) = Hn (X, A; R) := . Bn (X, A; R)

Now consider the preimage of Zn (X, A) under Sn (X) Sn (X, A) and set → = Zn′ (X, A) : c Sn (X) ∂c Sn 1 (A) , { ∈ | ∈ − } B′ (X, A) := c Sn (X) b Sn+ (X) such that c + ∂b Sn (A) . n { ∈ | ∃ ∈ 1 ∈ } = = Since Zn (X, A) Zn′ (X, A)/Sn (A) and Bn (X, A) Bn′ (X, A)/Sn (A) we obtain

Zn′ (X, A)/Sn (A) Zn′ (X, A) Hn (X, A) = = . Bn′ (X, A)/Sn (A) Bn′ (X, A)

Remark 3.11. For A = we have the special cases ∅

Z ′ (X, ) = Zn (X), n ∅ B′ (X, ) = Bn (X), n ∅ Hn (X, ) = Hn (X). ∅

84 3.2 Relative homology

Example 3.12. Let X = S1 [0, 1] be the cylinder over S1 and A = S1 0 X. × ×{ } ⊂

To construct an element in the relative homology H1(X, A) take a 1-simplex

σ : ∆1 X, → (te + (1 t)e ) (cos (2πt), sin (2πt), 1). 1 − 0 7→

| > σ X

∆1 A

Since σ is a closed curve in X we ﬁnd for its boundary

∂σ = σ(e ) σ(e ) = 0. 1 − 0

85 3 Homology Theory

Therefore σ Z (X) Z (X, A) and, as we will see later, represents a nontrivial element in ∈ 1 ⊂ 1′ H1(X). For the 2-simplices σ1 and σ2 deﬁned as indicated in the following picture

σ > > X σ1

∆2

σ2

A we ﬁnd for the 2-chain σ + σ the boundary ∂(σ + σ ) = σ + a where a S (A). Hence, 1 2 1 2 ∈ 1 modulo a S (A), we have σ B (X, A) and therefore 0 = [σ] H (X, A). ∈ 1 ∈ 1 ∈ 1

Let (X, A) be a pair of spaces. The inclusion map i : A ֒ X induces a homomorphism → Hn (i) : Hn (A) Hn (X). Furthermore we have the inclusion map j : (X, ) (X, A) which → ∅ → induces the homomorphism Hn ( j) : Hn (X) = Hn (X, ) Hn (X, A). ∅ → We deﬁne the connecting homomorphism or boundary operator

∂ : Hn (X, A) Hn 1(A), (3.3) → − [c] [∂c], 7→ 2 = where c Zn′ (X, A). Note that ∂c Zn 1(A) since ∂ 0. The connecting homomorphism is ∈ ∈ − well deﬁned because replacing c by another representative c + ∂b + a where b Sn+ (X) and ∈ 1 a Sn (A) yields ∈ c + ∂b + a ∂(c + ∂b + a) = ∂c + ∂a. 7→ + = Since ∂a Bn 1(A) we get [∂c ∂a] [∂c] Hn 1(A). Since ∂ : Zn′ (X, A) Sn 1 (A) is a ∈ − ∈ − → − homomorphism, the connecting homomorphism is also a homomorphism.

86 3.3 The Eilenberg-Steenrod axioms and applications

Lemma 3.13. The connecting homomorphism is natural, i.e. the diagram

Hn (f ) Hn (X, A) / Hn (Y, B)

∂ ∂

 Hn 1 (f A) Hn 1(A) − | / Hn 1(B) − − commutes for every f C((X, A), (Y, B)). ∈

Proof. We compute

Hn 1( f A) ∂ αi σi = Hn 1( f A) ∂ αi σi − − | i | i X X j j = Hn 1( f A) αi ( 1) σj F − n 1 | i j − ◦ − X X = j j αi ( 1) ( f A) (σi Fn 1) i j − | ◦ ◦ − X X = j j αi ( 1) ( f σi ) Fn 1 i j − ◦ ◦ − X X = ∂ αi ( f σi ) i ◦ X = ∂ αi ( f σi ) i ◦ X = ∂ Hn ( f ) αi σi . i X

3.3 The Eilenberg-Steenrod axioms and applications

Now we list the most important properties of homology theory known as Eilenberg-Steenrod axioms. The ﬁrst axiom is exactly Example 3.9.

Dimension Axiom. R, n = 0 Hn ( point ; R) { } 0, otherwise ( The next axiom deals with homotopy invariance. Two continuous maps f , f : (X, A) (Y, B) 0 1 → of pairs of spaces are called homotopic (in symbols f f ) iﬀ there exists an H C(X [0, 1], Y ) 0 ≃ 1 ∈ ×

87 3 Homology Theory such that for all x X ∈

H(x, 0) = f0(x),

H(x, 1) = f1(x), H(A [0, 1]) B. × ⊂ Homotopy Axiom. Let f , f C((X, A), (Y, B)) be homotopic, f f . Then the induced maps on homology 0 1 ∈ 0 ≃ 1 coincide, i.e., Hn ( f ) = Hn ( f ) : Hn (X, A) Hn (Y, B) 0 1 → holds for all n.

Remark 3.14. Let (X, A) (Y, B), i.e., there exist f C((X, A), (Y, B)) and ≃ ∈ g C((Y, B), (X, A)) such that f g id(Y, B) and g f id(X, A). It then follows as be- ∈ ◦ ≃ ◦ ≃ fore that Hn (X, A; R) Hn (Y, B; R).

,Exactness Axiom. For any pair of spaces (X, A) and inclusion maps i : A ֒ X → j : (X, ) ֒ (X, A), the sequence ∅ →

Hn+1(j) Hn+ (X, A) o 1 ··· ∂

 Hn (i) Hn (j) Hn (A) / Hn (X) / Hn (X, A)

∂

Hn 1 (i) o − Hn 1(A) ··· − is exact and natural.

Notation: For A X we call ⊂ A˚ = U with U open in X U A [⊂ the interior of A and A¯ = B with B closed in X B X \⊂ the closure of A.

Excision Axiom. For every pair of spaces (X, A) and every U A with U¯ A˚ the homomorphism ⊂ ⊂

Hn ( j) : Hn (X U, A U) Hn (X, A) \ \ →

88 3.3 The Eilenberg-Steenrod axioms and applications induced by the inclusion map

(j : (X U, A U) ֒ (X, A \ \ → is an isomorphism. The proofs of axioms A2, A3, and A4 will be given later. Before that we wil show their usefulness by studying some basic examples.

Remark 3.15. (cf. Exercise 3.1) 1.) If X , is path-connected then H (X; R) R and generators are represented by every ∅ 0 0-simplex σ : ∆0 X. In other words, the isomorphism H (X; R) R is given by → 0 → α j σj α j . j 7→ j 2.)P If Xk , k K,P are the path-components of X then Hn (X; R) k K Hn (Xk ; R). ∈ ∈ L Theorem 3.16. For n 1 we have ≥ R, if m = 0 or m = n H (Sn; R) m 0, otherwise ( R, if m = n H (Dn, Sn 1; R) m − 0, otherwise (

Proof. a) For n 1 the sphere Sn is path-connected and therefore H (Sn; R) R. ≥ 0 For n = 0 we have that S0 = x, y with the discrete topology and therefore H (S0; R) R2. { } 0 b) We have the exact sequence H (S0) H (D1) H (D1, S0) 0 0 −→ 0 −→ 0 −→ R2 R ∈ ∈ (a, b) a + b 7−→ 1 1 0 Since the map (a, b) a + b is onto, the map H0(D ) H0(D , S ) must be zero. Hence 1 0 7→ → H0(D , S ) = 0. For n 2 we have the following exact sequence ≥ H (Sn 1) H (Dn) H (Dn, Sn 1) 0 0 − −→ 0 −→ 0 − −→

 id R R −→ n n 1 Again the second arrow has to be trivial and therefore H0(D , S − ) = 0. This settles the case m = 0.

89 3 Homology Theory c) We will now use the exact sequence

H (Dn ) H (Dn, Sn 1) H (Sn 1) H (Dn) 1 −→ 1 − −→ 0 − −→ 0 For n = 1 we have H (D1) H (D1, S0) H (S0) H (D1) 1 −→ 1 −→ 0 −→ 0 H ( point ) R2 R 1 { } ∈ ∈ = 0 (a, b) a + b 7−→ which implies H (D1, S0) ker((a, b) a + b) R. 1 7→ For n 2 we have ≥ 0 H (Dn ) H (Dn, Sn 1) H (Sn 1) H (Dn) 1 −→ 1 − −→ 0 − −→ 0 = 0

n n 1 from which we get H1(D , S − ) = 0. n n n = n d) Consider the pair of spaces (S , D ) where D /(+) x S x0 0 (x0 0) is the lower − { ∈ | ≥ ≤ } (resp. upper) hemisphere. −

Dn −

For n 1 we have ≥ n n n n n n H1(D ) H1(S ) H1(S , D ) H0(D ) H0(S ) − −→ −→ − −→ −→ = 0

Since the last arrow is an isomorphism, the connecting homomorphism n n n n n n H1(S , D ) H0(D ) must be zero. Therefore H1(S ) H1(S , D ) is onto. − n→ − n n n → − Since H1(D ) = 0 we get H1(S ) H1(S , D ). − − n = n 1 e) Put U : x S x0 < 2 . − { ∈ | }

90 3.3 The Eilenberg-Steenrod axioms and applications

Dn − U n −

By the excision axiom the inclusion ( Sn U n, Dn U n) ֒ (Sn, Dn) \ − \ − → induces an isomorphism

n n n n n n Hm (S U , D U ) Hm (S , D ) \ − \ − since U¯ n D˚ n. We also have that − ⊂ − n n n n n n 1 n n 1 (S U , D U ) (D+, S − ) (D , S − ) \ − \ − ≃ ≈ where the homeomorphism is given by a vertical projection. This gives us the isomorphism

n n n n n n 1 Hm (S U , D U ) Hm (D , S − ). \ − \ − In particular, R, for n = 1 H (Sn ) H (Sn, Dn ) H (Dn, Sn 1) = (3.4) 1 1 1 − 0, otherwise − ( This concludes the case m = 1. f) Finally we treat the case m 2 by induction. Observe that ≥ n n n n n Hm (D ) Hm (S ) Hm (S , D ) Hm 1 (D ) − −→ −→ − −→ − = = 0 0 and n n n 1 n 1 n Hm (D ) Hm (D , S − ) Hm 1 (S − ) Hm 1 (D ) −→ −→ − −→ − = = 0 0 This yields n n n n n 1 n 1 Hm (S ) Hm (S , D ) Hm (D , S − ) Hm 1 (S − ) (3.5) − − Induction over m concludes the proof.

91 3 Homology Theory

1 Remark 3.17. Let us describe a generator of H1(S ) Z geometrically. We use the isomor- 1 0 1 1 phisms in (3.4) and start with H1(D , S ). The map c : ∆ D , t cos(π(1 t)), is a singular → 7→ 0 2 − 1-simplex with ∂c = const1 const 1. Under the isomorphism H0(S ) R it maps to (1, 1) − − − which generates the kernel of the map R2 R given by (a, b) a + b. Hence c represents a 1 0 → 7→ generator of H1(D , S ). 1 0 1 0 The isomorphism H1(D+, S ) H1(D , S ) induced by vertical projection gives us a generator 1 0 → 1 1 iπ(1 t ) of H1(D+, S ), namely the homology class represented by c′ : ∆ D+, t e − . 1 0 1 1 → 7→ The isomorphism H1(D+, S ) H1(S , D ) is induced by the inclusion. Hence [c′] 1 1 → − ∈ H1(S , D ) is a generator. Now − 1 1 iπ(1 (s+1)t ) F : S [0, 1] S , (t, s) e − , × → 7→ is continuous and satisﬁes

F(t, 0) = c′(t), iπ(1 2t ) F(t, 1) = e − =: c′′(t), F(0, s) = 1 D1, − ∈ − iπs 1 F(1, s) = e− D . ∈ − 1 1 1 Thus c′ c′′ as maps (∆ , 0, 1 ) (S , D ). Using the homotopy axiom we compute ≃ { } → −

[c′] = [c′ id∆1 ] = H (c′)[id∆1 ] = H (c′′)[id∆1 ] = [c′′ id∆1 ] = [c′′]. ◦ 1 1 ◦ ( 1 1 ) = = Therefore [c′′] H1 S , D is a generator. Since ∂c′′ constc′′(1) constc′′(0) const 1 ∈ − 1 − − − const 1 = 0, the 1-simplex c′′ represents a homology class in H1(S ). Moreover, since inclusion − 1 1 1 1 1 induces an isomorphism H1(S ) H1(S , D ) we ﬁnd that [c′′] H (S ) is a generator. iπ→(s 2t ) − 2iπt ∈ Using the homotopy (t, s) e − we see that t e− also represents a generator of 1 7→ 7→ H1(S ). Finally, playing the same game with lower and upper hemispheres interchanged shows that t e2iπt represents a generator of H (S1) as well. 7→ 1

As a ﬁrst application of Theorem 3.16 we now prove Brouwer’s ﬁxed point theorem in all dimensions.

Theorem 3.18 (Brouwer’s ﬁxed point theorem). Let f : Dn Dn, n 1, be a continuous → ≥ map. Then f has a ﬁxed point, i.e., there exists an x Dn such that f (x) = x. ∈

Proof. We assume that the map f does not have a ﬁxed point and then derive a contradiction. Let n 2 (the case n = 1 having been treated in Remark 1.6). ≥

92 3.3 The Eilenberg-Steenrod axioms and applications

Dn Now consider the continuous map g : Dn Sn 1 as in → − x the picture. Denote the inclusion map by ι : Sn 1 Dn g(x) − → f (x) • and note that g ι = id n 1 . S − ◦ n n 1 ∂D = S −

By the functorial properties of homology we obtain the following commutative diagram:

= n Hn 1(id) id n H (S 1; Z) Z − / H (S 1; Z) Z n 1 − ❙ n 1 − − ❙❙❙ ❦❦−❦5 ❙❙❙❙ ❦❦❦❦ ❙❙❙ ❦❦❦ Hn 1(ι) ❙❙ ❦❦ Hn 1 (g) − ❙❙ ❦❦❦ − ) n Hn 1(D ; Z) = 0 − Since the identity Z Z does not factor through 0 we run into a contradiction. →

Now we are in the position to answer the third question on page 3.

Theorem 3.19. For n , m the space Rn is not homeomorphic to Rm.

Proof. Let us assume there exists a homeomorphism f : Rn Rm. Then → f : Rn 0 Rm f (0) \{ } → \{ } is also a homeomorphism. Since Rn point Sn 1 we obtain an isomorphism on the level of \{ } ≃ − homology groups: n 1 n m m 1 Hj (S − ) Hj (R 0 ) Hj (R f (0) ) Hj (S − ). \{ } \{ } By Theorem 3.16 this is a contradiction for j = n 1 or j = m 1 unless n = m. − −

Proposition 3.20. For n 1 let s : Sn Sn be the reﬂection given by ≥ →

s(x , x ,..., xn ) = ( x , x ,..., xn ). 0 1 − 0 1 Then the map n n Hn (s) : Hn (S ) Hn (S ) → is given by Hn (s) = id. −

93 3 Homology Theory

Proof. The proof is given by induction. First consider the case n = 1. Let c : ∆1 S1 be a 1 → 1-simplex generating H1(S ) and let S1 (s)(c) be its image under the induced homomorphism on 1-chains.

S1 (s)

b b ∧ ∨

We need to show that H (s)[c] = [S (s)c] = [c]. We construct two 2-simplices. The ﬁrst one 1 1 − is given by

e2 b σ1

b ∧

b b e0 e1

Applying the boundary operator yields

∂σ = S (s)c const + c. 1 1 − The second 2-simplex is the constant map.

94 3.3 The Eilenberg-Steenrod axioms and applications

e2 b σ2

b b e0 e1

We apply the boundary operator again and we get

∂σ = const const + const = const 2 −

It follows that for the chain σ1 + σ2 that

∂(σ1 + σ2) = S1(s)c + c and hence 0 = [∂(σ1 + σ2)] = [S1 (s)c + c] = [S1 (s)c] + [c] . This yields the desired result in the case of n = 1. The induction step n 1 n follows from the following commutative diagram: − ⇒ n n n n n 1 n 1 Hn (S ) / Hn (S , D ) o Hn (D+, S − ) / Hn 1(S − ) − −

Hn (s) Hn (s) Hn (s) Hn 1(s) − n n n n n 1 n 1 Hn (S ) / Hn (S , D ) o Hn (D+, S − ) / Hn 1(S − ) − − where the horizontal isomorphisms are the ones in (3.5). Commutativity of the last square follows from Lemma 3.13 and that of the first and the second one from the fact that the horizontal isomorphisms are induced by inclusion maps (which commute with s). Note here that we have to choose the lower hemisphere with respect to a different coordinate than the one which is reflected by s so that s(Dn ) = Dn. − −

Remark 3.21. Let a : Sn Sn with a(x) = x for all x Sn be the antipodal map then n+1 → − ∈ Hn (a) = ( 1) . This follows from the fact that a is the composition of n + 1 reﬂections. −

95 3 Homology Theory

Deﬁnition 3.22. A vector ﬁeld on Sn is a map v : Sn Rn+1 such that v(x) x for all → ⊥ x Sn. ∈ x v(x) b

Theorem 3.23 (Combing a hedgehog). The n-dimensional sphere Sn admits a continuous vector field without zeros iff n is odd. In particular, every continuous vector field on S2 has a zero.

Loosely speaking, this means that every continuously combed hedgehog has a “bald” spot. Proof. If n is odd we simply set

v(x) := ( x1, x0, x3, x2,..., xn, xn 1) . − − − −

Now let n be even and let us assume that v : Sn Rn+1 is a → continuous vector ﬁeld without a zero. Then we can put w(x) := x v(x) n n v(x) . We deﬁne the continuous map F : S [0, 1] S by | | × → b F(x, t) := x cos(πt) + w(x) sin(πt) . ∧ w(x) x − Since F(x, 0) = x and F(x, 1) = x = a(x) with a(x) the − antipodal map we have found that a id. Hence Hn (a) = ≃ Hn (id) = 1.

n+1 This contradicts Hn (a) = ( 1) = 1. − −

3.4 The degree of a continuous map

For n 1 we consider a continuous map f : Sn Sn. Then the homomorphism ≥ → n n Hn ( f ) : Hn (S ; Z) Z Hn (S ; Z) Z → is given by multiplication with a number which we denote deg( f ) Z. ∈

96 3.4 The degree of a continuous map

Deﬁnition 3.24. The number deg( f ) is called the degree of the map f . The degree can be deﬁned in the same way for continuous maps f : (Dn, Sn 1) (Dn, Sn 1). − → −

Examples 3.25. 1.) For a reﬂection we have deg(s) = 1. − 2.) For the antipodal map we have seen that deg(a) = ( 1)n+1. −

Lemma 3.26. The degree of a function has the following properties

(i) deg(id) = 1;

(ii) deg(const) = 0;

(iii) deg( f g) = deg( f ) deg(g); ◦ (iv) If f g then deg( f ) = deg(g); ≃ (v) If the map f is a homotopy equivalence then deg( f ) = 1; ± n+1 n n+1 n (vi) For f : (D , S ) (D , S ) we have deg( f ) = deg( f n ). → S

Proof. The ﬁrst and third assertion follow directly from the functorial property of Hn ( f ). The fourth and ﬁfth statement follow from the homotopy axiom. The second assertion follows from the fact that the homomorphism induced by a constant map factors through Hn (pt) = 0. The last statement of the lemma follows fromm the commutativity of the following diagram:

n+1 n Hn+1(f ) n+1 n Hn+1(D , S ) / Hn+1(D , S )

 ∂ ∂ 

 n Hn+1(f S n ) n Hn (S ) | / Hn (S )

Theorem 3.27. (i) Every f C(Sn, Sn ) without fixed points satisfies deg( f ) = ( 1)n+1. ∈ − (ii) Every f C(Sn, Sn ) without an antipodal point, i.e., f (x) , x for all x Sn, satisfies ∈ − ∈ deg( f ) = 1.

(iii) For n even every f C(Sn, Sn ) has a ﬁxed point or an antipodal point. ∈

97 3 Homology Theory

Proof. (i) Let f C(Sn, Sn ) be without a ﬁxed point. Then the line segment joining f (x) ∈ and x does not contain the origin. −

x b

b 0

f (x) b b x −

Hence we can deﬁne the continuous map (1 t) f (x) tx F : Sn [0, 1] Sn, F(x, t) := − − . × → (1 t) f (x) tx | − − | Since F(x, 0) = f (x) and F(x, 1) = x = a(x) with a being the antipodal map the map F − is a homotopy for f a. Thus ≃ + deg( f ) = deg(a) = ( 1)n 1. − (ii) Now let f C(Sn, Sn ) be without antipodal points. Then we can deﬁne ∈ (1 t) f (x) + tx G(x, t) := − . (1 t) f (x) + tx | − |

b 0

b x

b − f (x)

Since G(x, 0) = f (x) and G(x, 1) = x we have that f id and deg( f ) = deg(id) = 1 ≃ follows.

(iii) Finally, assume that f C(Sn, Sn ) has neither ﬁxed points nor antipodal points. Then + ∈ deg( f ) = ( 1)n 1 by (i) and by deg( f ) = 1 by (ii). Thus n must be odd. −

98 3.4 The degree of a continuous map

Let µ : Sn Sn Sn for n 1 be a continuous map. We choose p Sn arbitrarily and deﬁne × → ≥ ∈ j : Sn Sn Sn, x (x, p), 1 → × 7→ j : Sn Sn Sn, x (p, x) . 2 → × 7→ We then get the following diagram:

2 n n (Hn (j1 ), Hn (j2)) n n Hn (µ) n Z Hn (S ) Hn (S ) / Hn (S S ) / Hn (S ) Z ⊕ × 6

(d1, d2) with dµ Z. ∈

Deﬁnition 3.28. The pair of numbers (d , d ) Z2 is called the bidegree of the map µ. 1 2 ∈

Remark 3.29. The bidegree does not depend on the choice of p. If one chooses another p′, then a path from p to p′ will yield a homotopy between the corresponding embedding maps jν and jν′ . Hence they induce the same homomorphisms on homology and therefore the same bidegrees.

Examples 3.30. 1.) Consider the case n = 1, S1 C and let the map µ be given by µ(z , z ) = ⊂ 1 2 z z . Choose p = 1 S1. Then µ j = id and thus 1 2 ∈ ◦ 1 d = deg(µ j ) = deg(id) = 1. 1 ◦ 1

Similarly, we get d2 = 1. Hence the bidegree by (d1, d2) = (1, 1). 2.) In the case of n = 3, S3 H we consider the map µ given by quaternionic multiplication, ⊂ µ(h1, h2) = h1h2. Similar reasoning shows that the bidegree is again given by (d1, d2) = (1, 1).

Remark 3.31. The quaternions H form a division algebra isomorphic to R4 as a vector space. The algebra H is associate and noncommutative. The standard vector space basis we be denoted by 1, i, j, k. Therefore any h H can be uniquely written as ∈

h = h0 + h1i + h2 j + h3 k .

Quaternionic multiplication is now determined by the relations i2 = j2 = k2 = ijk = 1. With − the help of the conjugation h∗ = h h i h j h k 0 − 1 − 2 − 3 we can deﬁne h := √h h. We regard S3 H as the set of unit-length quaternions. | | ∗ ⊂

99 3 Homology Theory

n n k Proposition 3.32. Let n = 1 or n = 3 and let k Z. The map fk : S S with fk : z z ∈ → 7→ given by complex multiplication in the case n = 1 and by quaternionic multiplication in the case n = 3 has degree k.

Proof. The proof is by induction on k. For k = 0 and k = 1 the statement is trivial because constant maps have degree 0 while the identity has degree 1. The case of k = 1 has already 1 − been shown for n = 1, since here z z− = z¯ is a reﬂection and hence deg( f 1) = 1. On the 7→ − − other hand for k = 1, n = 3 we note that S1 S2 S3 regarding C H. Now the following − ⊂ ⊂ ⊂ commutative diagram

1 2 3 H1(S ) H2(S ) H3(S )

= = = H1 (f 1 S1 ) deg(f 1 S1 ) 1 H3 (f 1) deg(f 1) − | − | − − − 1 2 3 H1(S ) H2(S ) H3(S ) implies that deg( f 1) = 1 in the quaternionic case too. The horizontal isomorphisms are − − obtained as the composition of the isomorphisms

k k+1 k k+1 k+1 k+1 Hk (S ) Hk+1(D , S ) Hk+1 (S , D ) Hk+1 (S ). − We used that the quaternionic multiplication restricted to the complex numbers C is just complex multiplication. Now we are ready to carry out the induction over k. We consider k > 0 and we show that the statement for k 1 implies that for k. Let µ be complex (resp. quaternionic) multiplication. Then − deg( fk ) = deg(µ ( fk 1, f1)) ◦ − deg( fk 1) = (1, 1) − · deg( f1) ! = deg( fk 1) + deg( f1) − = k 1 + 1 = k. − The case k < 0 is treated similarly.

Now we can show that the fundamental theorem of algebra 2.34 also holds for quaternionic polynomials.

Theorem 3.33 (Fundamental theorem of algebra for quaternions). Every quaternionic polynomial k k 1 p(z) = z + α1z − + ... + αk

100 3.4 The degree of a continuous map of positive degree k has a quaternionic root.

Proof. Suppose the polynomial p has no root. Then we can deﬁne the continuous map pˆ : S3 S3 withp ˆ(z) := p(z) . Now consider F(z, t) : S3 [0, 1] S3 with F(z, t) := p(tz) . → p(z) × → p(tz) We observe that F(z, 0) =| const| and F(z, 1) = pˆ(z), hence we havep ˆ(z) const and conse-| | ≃ quently deg(pˆ) = 0. On the other hand, we can put for z S3 and t > 0 ∈ k z t p t G(z, t) := tk p z | t | and observe that the expression z tk p = zk + tα zk 1 + ... + tk α t 1 − 3 3 extends continuously to t = 0. The map G : S [0, 1] S satisﬁes G(z, 0) = fk (z) and again × → G(z, 1) = pˆ(z). Thusp ˆ fk and hence deg(pˆ) = k. This contradicts k > 0. ≃

Now let U Sn be open, n 1. Consider a continuous map f : U Sn and a point p Sn ⊂ ≥ → ∈ such that f 1 (p) is compact. Then we have the following diagram (with deg ( f ) Z): − p ∈

n (i) n n 1 (ii) 1 Z Hn (S ; Z) / Hn (S , S f (p); Z) o Hn (U, U f (p); Z) (3.6) \ − \ − degp (f ) Hn (f )

 n (iii) n n Z Hn (S ; Z) / Hn (S , S p ; Z) \{ } We observe: Concerning (i): This homomorphism is induced by the inclusion

n n n n 1 .((S = (S , ) ֒ (S , S f − (p ∅ → \ Concerning (ii): The inclusion (U,U f 1 (p)) ֒ (Sn, Sn f 1 (p)) induces an isomorphism \ − → \ − by the excision axiom. Concerning (iii): Consider the exact homology sequence of (Sn, Sn p ) \{ } n n n n n n 0 = Hn (S p ) Hn (S ) Hn (S , S p ) Hn 1(S p ) Hn 1(S ) . \{ } → → \{ } → − \{ } → − 0 for n 2 = ≥ |Z {zfor n = }1   Hence in both cases n = 1 and n 2 the homomorphism (iii) is an isomorphism. ≥

101 3 Homology Theory

Deﬁnition 3.34. The number degp ( f ) is called the local degree of f over p.

= n n = Examples 3.35. 1.) If p < im( f ) then degp ( f ) 0 because Hn (S , S ) 0. 2.) If f : U Sn is the inclusion map then deg ( f ) = 1 for all p U. Namely, in this case the → p ∈ homomorphisms (i) and (iii) in (3.6) coincide and Hn ( f ) is the inverse of (ii). 3.) For a homeomorphism f : U f (U) Sn we have that deg ( f ) = 1 for all p f (U). → ⊂ p ± ∈ Namely, the homomorphisms (i) and Hn ( f ) in (3.6) are isomorphisms in this case, hence multiplication by deg ( f ) is an isomorphism, thus deg ( f ) = 1. p p ±

Proposition 3.36. Assume that f 1 (p) K V U where K is compact and V open. Then − ⊂ ⊂ ⊂ the degree degp ( f ) is given by:

n n n Hn (S ) / Hn (S , S K) o Hn (V, V K) \ \ degp (f ) Hn (f V ) | n n n Hn (S ) / Hn (S , S p ) \{ }

1 Hence we can replace f − (p) by a larger compact set in U and also U by a smaller open 1 neighborhood of f − (p). For this reason we call degp ( f ) local. Proof. The assertion follows from the commutativity of the following diagram:

H (Sn, Sn f 1(p)) o H (U, U f 1(p)) n − n − ❙ ♠♠6 O \ \O ❙❙❙ ♠♠♠ ❙❙❙Hn (f ) ♠♠♠ ❙❙❙ ♠♠♠ ❙❙❙❙ ♠♠ ❙) H (Sn ) H (Sn, Sn p ) o H (Sn ) n ◗ n n ◗◗◗ ❦❦❦5 \{ } ◗◗◗ ❦❦❦❦ ◗◗◗ ❦❦❦ ◗◗ ❦❦ Hn (f V ) ◗ ❦❦❦ | ( n n Hn (S , S K) o Hn (V, V K) \ \ where all but two arrows are induced by inclusions.

Remark 3.37. For f : Sn Sn we have that deg( f ) = deg ( f ) for all p Sn. Simply choose → p ∈ K = V = Sn in Proposition 3.36.

102 3.4 The degree of a continuous map

n n 1 n n 1 n 1 Lemma 3.38. Let f : (D+, S − ) (D+, S − ) be continuous and p D˚ + such that f − (p) n → ∈ ⊂ D˚ + is compact. Then

deg( f ) = deg f n . p |D˚ + 

Proof. We extend f to a continuous map F : (Sn, Dn ) (Sn, Dn) by mapping the circular arc n 1 − → − from a point x on the equator S − to the south pole to the corresponding arc from f (x) to the 1 n n south pole. Then by Remark 3.37 and Proposition 3.36 with K = f − (p), V = D˚ + and U = S we ﬁnd deg(F) = deg (F) = deg (F n ) = deg ( f n ) . (3.7) p p |D˚ + p |D˚ + On the other hand, the commutative diagram

n n 1 n n n (D+, S − ) / (S , D ) o S − f F F n n 1 n n n (D+, S − ) / (S , D ) o S − yields on the level of homology

n n 1 n n n Hn (D+, S − ) / Hn (S , D ) o Hn (S ) − deg(f ) H (F) deg(F) · n · n n 1 n n n Hn (D+, S − ) / Hn (S , D ) o Hn (S ) − Hence deg(F) = deg( f ). This together with (3.7) concludes the proof.

Proposition 3.39. Let f : U Sn Sn be continuous, let p Sn with f 1 (p) compact and ⊂ → ∈ − let g C(Sn, Sn ). Then ∈ deg ( f g) = deg ( f ) deg(g). p ◦ p ·

Proof. This follows from the commutative diagram:

n n n 1 1 Hn (S ) / Hn (S , S f − (p)) Hn (U,U f − (p)) O \ \ Hn (f ) degp (f ) = - n n n deg(g) Hn (g) 1 Hn (S , S p ) Hn (S ) \{ O } degp (f g) ◦ Hn (f g) ◦ n n n 1 1 1 1 Hn (S ) / Hn (S , S ( f g) (p)) Hn (g (U), g (U) ( f g) (p)) \ ◦ − − − \ ◦ −

103 3 Homology Theory

Remark 3.40. If Xi are the path-components of X then

Hm (X) Hm (Xi ), Mi see Exercise 3.1. The isomorphism is induced by the inclusion maps of the connected components into X. Similarly one sees that for A X and Ai = A Xi ⊂ ∩ Hm (X, A) Hm (Xi, Ai ) . Mi

Proposition 3.41 (Additivity of the degree). Let f : U Sn be a continuous map. Let n 1 = r → = p S be such that f − (p) is compact. Let U λ=1Uλ with Uλ open and put f λ : f Uλ . ∈ 1 ∪ | Suppose f λ− (p) are pairwise disjoint. Then

r = degp ( f ) degp ( f λ ) . = Xλ 1

Before proving the proposition we give some examples.

Example 3.42 U 1 Assume that f − (p) is a ﬁnite set, i.e. 1 b f (p) = p ,..., pr . Now choose Uλ such − { 1 } that pλ Uλ and pν < Uλ for ν , λ. It ∈ follows that

r = b degp ( f ) degp ( f λ ) . = Xλ 1 If the map f is a local homeomorphism, then = deg ( f ) 1 for all λ. b p ±

Uλ

1 1 k Example 3.43. Consider the map fk : S S with f (z) = z , k > 0, and set p = 1. We write → 1 f − (1) = k-th unit roots = ξ ,...,ξk k { } { 1 } and ﬁnd that fk ξ is a homeomorphism. Hence deg( fk, λ ) = 1. Since k > 0, |small neighborhood of λ ± the restriction of fk to a small neighborhood of ξλ is homotopic to an embedding of a (k times 1 = = larger) neighborhood of ξλ into S . Hence deg( fk, λ ) 1. We conclude deg1( fk ) k.

104 3.5 Homological algebra

1 Proof of Proposition 3.41. Choose open neighborhoods Vλ such that f λ− (p) Vλ Uλ with r ⊂ ⊂ Vλ Vµ = for λ , µ. Now put V = Vλ . Proposition 3.36 tells us deg ( f ) = deg ( f V ). ∩ ∅ ∪λ=1 p p | The commutative diagram

deg (f V ) p |

n 1 Hn (f V ) n n , n Hn (S ) / Hn (V, V f − (p)) | / Hn (S , S p ) o Hn (S ) \O O \{ } OO 1 . (1,...,1) (1,...,1) * . + . / .1/ . / r n r 1 λ Hn (fλ ) r n n r n ,= - Hn (S ) / = Hn (Vλ, Vλ f − (p)) ⊕ / = Hn (S , S p ) o = Hn (S ) λ 1 λ 1 \ λ λ 1 \{ } 2 λ 1 L L L L degp ( f1) . . . .* /+ . / . degp ( fr )/ . / , -

yields

degp ( f1) 1 . . deg ( f V ) = (1,..., 1) . . . = deg ( f1) + ... + deg ( fr ). p | · · . p p * + * + . degp ( fr )/ .1/ . / . / , - , - 3.5 Homological algebra

Before continuing with topological considerations we clarify some of the underlying algebra. Throughout this section let R be a commutative ring with unit element.

Deﬁnition 3.44. A complex of R-modules K is a sequence ∗

∂n+1 ∂n ... / Kn+1 / Kn / Kn 1 / ... −

of R-modules Kn together with homomorphisms ∂n such that

∂n ∂n+ = 0 (3.8) ◦ 1 for all n Z. We deﬁne the space of n-cycles ∈

Zn K := x Kn ∂n x = 0 = ker(∂n) , ∗ { ∈ | } and the space of n-boundaries

BnK := ∂n+1 x Kn+1 x Kn+1 = im(∂n+1). ∗ { ∈ | ∈ }

105 3 Homology Theory

By (3.8) BnK Zn K so that we can deﬁne the n-th homology by ⊂ ZnK Hn K := ∗ . ∗ Bn K ∗

Example 3.45. For

Sn (X, A; R) = Sn (X; R)/Sn (A; R), n 0 Kn = ≥ 0, n < 0   we obtain relative singular homology Hn K = Hn (X, A; R).  ∗

Deﬁnition 3.46. Given two complexes K and K ′ a chain map ϕ : K K ′ is a sequence of ∗ ∗ ∗ ∗ → ∗ homomorphisms ϕn : Kn K such that → n′

ϕn+1 Kn+1 / Kn′ +1

∂n+1 ∂n′ +1

 ϕn Kn / Kn′ commutes for all n Z. ∈

Example 3.47. For a continuous map f : (X, A) (Y, B) the homomorphisms ϕn = Sn ( f ) → constitute a chain map.

Given a general chain map ϕ we conclude from ∗

ϕn ∂n+ = ∂′ + ϕn+ ◦ 1 n 1 ◦ 1 that ϕn (ZnK ) ZnK ′ and also that ϕn (BnK ) Bn K ′. Hence ϕ induces homomorphisms ∗ ⊂ ∗ ∗ ⊂ ∗ ∗ Hn (ϕ ) : Hn K Hn K ′ by Hn (ϕ )([z]) = [ϕn z]. This construction is functorial in the sense ∗ ∗ → ∗ ∗ that = = Hn (ϕ ψ ) Hn (ϕ ) Hn (ψ ) and Hn (idK ) idHn K . ∗ ◦ ∗ ∗ ◦ ∗ ∗ ∗

ϕ ψ Deﬁnition 3.48. A sequence K ′ ∗ K ∗ K ′′ of chain maps is called exact ϕ···n −→ψn∗ −→ ∗ −→ ∗ −→ ··· iﬀ the sequence K Kn K is exact for every n Z. ··· −→ n′ −→ −→ n′′ −→ ··· ∈

106 3.5 Homological algebra

i p Proposition 3.49. If 0 K ′ ∗ K ∗ K ′′ 0 is an exact sequence of complexes then the ∗ Hn (i) → ∗H−→n (p) −→ ∗ → sequence HnK ′ HnK HnK ′′ is exact for every n Z. ∗ −→ ∗ −→ ∗ ∈

Proof. a) By assumption we have that p i = 0 and hence ∗ ◦ ∗ 0 = Hn (p i ) = Hn (p ) Hn (i ). ∗ ◦ ∗ ∗ ◦ ∗ Therefore im Hn (i ) ker Hn (p ). ∗ ⊂ ∗ b) It remains to show that ker Hn (p) im Hn (i). ⊂ Let z ZnK represent [z] ker Hn (p ). Hence pn z = ∂′′x′′ for some x′′ K ′′+ . Since ∈ ∗ ∈ ∗ ∈ n 1 pn+ is surjective we can choose x Kn+ with pn+ x = x . We compute 1 ∈ 1 1 ′′ pn (z ∂x) = ∂′′x′′ ∂′′pn+ x = ∂′′x′′ ∂′′x′′ = 0 . − − 1 − By exactness of the complex there exists y K such that z ∂x = in y . Now we get ′ ∈ n′ − ′ in 1∂′y′ = ∂in y′ = ∂[z ∂x] = ∂z ∂∂x = 0 . − − − Since in 1 is injective it follows that ∂′y′ = 0, i.e., y′ Zn K ′ represents an element in − ∈ ∗ homology. Finally we see

Hn (i )([y′]) = [in y′] = [z ∂x] = [z]. ∗ − This shows ker Hn (p ) im Hn (i ). ∗ ⊂ ∗

i p Deﬁnition 3.50. Let 0 K ′ ∗ K ∗ K ′′ 0 be an exact sequence of complexes. We → ∗ −→ ∗ −→ ∗ → construct the connecting homomorphism

∂ : HnK ′′ Hn 1K ′ ∗ ∗ → − ∗ as follows: Let z′′ Zn K ′′ represent an element [z′′] Hn K ′′. Since pn is surjective we can ∈ ∗ ∈ ∗ choose x Kn with pn x = z′′. For ∂x Kn 1 we observe ∈ ∈ −

pn 1∂x = ∂′′pn x = ∂′′z′′ = 0. − = By exactness there is a unique y′ Kn′ 1 with in 1 y′ ∂x. Moreover, ∈ − − in 2∂′y′ = ∂in 1 y′ = ∂∂x = 0. − −

Since in 2 is injective we have ∂′y′ = 0, i.e., y′ Zn 1K ′. Now put − ∈ − ∗ ∂ [z′′] := [y′]. ∗

107 3 Homology Theory

Lemma 3.51. The conncecting homomorphism ∂ : Hn K ′′ Hn 1K ′ is well deﬁned, i.e., ∗ ∗ → − ∗ independent of the choices made in its construction.

Proof. There are two choices in the construction of the connecting homomorphism: that of the preimage x with pn x = z′′ and that of the representing cycle z′′ itself. = = As to the choice of x let pn x pn x¯ z′′. For the corresponding elements y′, y¯′ Kn′ 1 we = = = ∈ − have and in 1 y′ ∂x and in 1 y¯′ ∂x¯. Since pn (x x¯) 0 there exists an ω′ Kn′ with − − − ∈ x x¯ = in (ω ). We compute − ′ in 1 (y′ y¯′) = ∂(x x¯) = ∂(inω′) = in 1∂ω′ , − − − − from which we conclude that y y¯ = ∂ω and therefore [y ] = [y ¯ ]. − ′ ′ ′ ′ As to the choice of the representing cycle z′′ it suﬃces to show that if z′′ is a boundary then [y ] = 0. Let z = ∂ ζ be a boundary. Since pn+ is onto we can choose ξ Kn+ with ′ ′′ ′′ ′′ 1 ∈ 1 pn+1ξ = ζ. Then x = ∂ξ is an admissible choice because

pn x = pn ∂ξ = ∂′′pn+1 ξ = ∂′′ζ = z′′.

Thus ∂x = ∂∂ξ = 0 and hence y′ = 0.

It is easy to see that the connecting homomorphism is indeed a homomorphism. Now we are ready to prove the Exactness Axiom.

i p Proposition 3.52. If 0 K ′ ∗ K ∗ K ′′ 0 is an exact sequence of complexes then the → ∗ −→ ∗ −→ ∗ → long homology sequence

∂ Hn (i ) Hn (p ) ∂ / Hn+1K ′′ ∗ / HnK ′ ∗ / Hn K ∗ / HnK ′′ ∗ / Hn 1K ′ / ··· ∗ ∗ ∗ ∗ − ∗ ··· is also exact.

Proof. In view of Proposition 3.49 it remains to show ker H(i) = im ∂ and ker ∂ = im H(p). ∗ ∗ a) im ∂ ker H(i): ∗ ⊂ Using the notation of Deﬁnition 3.50 we compute 1 H(i)∂ [z′′] = H(i)∂ [px] = H(i)[i− ∂x] = [∂x] = 0 . ∗ ∗ b) ker H(i) im ∂ : ⊂ ∗ Let H(i)[z′] = 0. Then we have [iz′] = 0 and therefore iz′ = ∂x. Put z′′ := px. Then ∂′′z′′ = ∂′′px = p∂x = piz′ = 0. Hence z′′ ZnK represents an element in homology. We ∈ ∗ compute 1 ∂ [z′′] = ∂ [px] = [i− ∂x] = [z′] . ∗ ∗ Hence [z′] im ∂ . ∈ ∗

108 3.5 Homological algebra c) im H(p) ker ∂ : ⊂ ∗ This follows from 1 ∂ H(p)[z] = ∂ [pz] = [i− ∂z ] = 0 . ∗ ∗ =0 d) ker ∂ im H(p): |{z} ∗ ⊂ Let ∂ [z′′] = 0. We write z′′ = px and compute ∗ 1 0 = ∂ [z′′] = ∂ [px] = [i− ∂x] . ∗ ∗ 1 1 This implies that i− ∂x is a boundary in K ′. Hence we have that i− ∂x = ∂′x′ and therefore

∂(x ix′) = ∂x i∂′x′ = 0 . − − This ﬁnally leads to

H(p)[x ix′] = [px pix′] = [px] = [z′′], − − hence [z ] im H(p). ′′ ∈

= = Example 3.53. For Kn′ Sn (A), Kn Sn (X) and

= = Sn (X) Kn′′ Sn (X, A) Sn (A) Proposition 3.52 now yields the Exactness Axiom for singular homology.

Example 3.54. Consider a triple of spaces (X, A, B) with B A X and set ⊂ ⊂ = = Kn′ Sn (A)/Sn (B) Sn (A, B) Kn = Sn (X)/Sn (B) = Sn (X, B) = = Kn′′ Sn (X)/Sn (A) Sn (X, A) .

Then the canonical sequence

0 / Kn′ / Kn / Kn′′ / 0 is again exact. By Proposition 3.52 we obtain the long exact homology sequence for a triple:

... / Hn+1(X, A) ❣❣❣ ∂ ❣❣❣❣❣ ❣❣❣∗❣❣ ❣❣❣❣❣ s❣❣❣ Hn (A, B) / Hn (X, B) / Hn (X, A) ❣❣❣❣ ∂ ❣❣❣❣❣ ❣❣∗❣❣ ❣❣❣❣❣ s❣❣❣ Hn 1(A, B) / ... −

109 3 Homology Theory

Proposition 3.55. The long homology sequence is natural, i.e., if the diagram of chain maps

i p 0 / K ′ ∗ / K ∗ / K ′′ / 0 ∗ ∗ ∗ ϕ′ ϕ ϕ′′ ∗ ∗ ∗ j q 0 / L′ ∗ / L ∗ / L′′ / 0 ∗ ∗ ∗ is commutative with exact rows then the diagram

Hn+1(p ) ∂ Hn (i ) Hn (p ) ∂ ... ∗ / Hn+1K ′′ ∗ / HnK ′ ∗ / Hn K ∗ / Hn K ′′ ∗ / ... ∗ ∗ ∗ ∗ Hn+1(ϕ′′) Hn (ϕ′ ) Hn (ϕ ) Hn (ϕ′′) ∗ ∗ ∗ ∗ Hn+1(q ) ∂ Hn (j ) Hn (q ) ∂ ... ∗ / Hn+1L′′ ∗ / Hn L′ ∗ / Hn L ∗ / Hn L′′ ∗ / ... ∗ ∗ ∗ ∗ is commutative as well.

Proof. By assumption we have ϕ i = j ϕ and ϕ p = q ϕ, hence ◦ ◦ ′ ′′ ◦ ◦ Hn (ϕ) Hn (i) = Hn ( j) Hn (ϕ′), ◦ ◦ Hn (ϕ′′) Hn (p) = Hn (q) Hn (ϕ). ◦ ◦ We are left to show that the diagram

∂ Hn K ′′ ∗ / Hn 1K ′ −

Hn (ϕ′′) Hn 1 (ϕ′) − ∂ Hn L′′ ∗ / Hn L′ commutes. We calculate

1 1 1 Hn 1(ϕ′)∂ [px] = Hn 1(ϕ′)[i− ∂x] = [ϕ′i− ∂x] = [j− ϕ∂x] − ∗ − 1 = [j− ∂ϕx] = ∂ [pϕx] = ∂ [ϕ′′px] = ∂ Hn (ϕ′′)[px], ∗ ∗ ∗ which proves the assertion.

Example 3.56. Let f C((X, A), (Y, B)). For K = Sn (A), Kn = Sn (X), K = Sn (X, A) ∈ n′ n′′ and ϕ = Sn ( f A), ϕn = Sn ( f ), and ϕ the induced homomorphism on Sn (X, A) we recover n′ | n′′ Lemma 3.13.

Deﬁnition 3.57. Let ϕ, ψ : K K be chain maps. A chain homotopy between ϕ and ψ is a → ′ sequence of homomorphisms hn : Kn K ′ + → n 1

110 3.5 Homological algebra such that = + ϕn ψn ∂n′ +1hn hn 1∂n , − − for all n Z. The maps ϕ and ψ are called homotopic if such a homotopy exists. In this case ∈ we write ϕ ψ. ≃

Proposition 3.58. (i) The relation “ ” is an equivalence relation on the set of all chain ≃ maps K K . → ′ (ii) If ϕ ψ : K K and ϕ ψ : K K then ϕ ϕ ψ ψ : K K . ≃ → ′ ′ ≃ ′ ′ → ′′ ′ ≃ ′ → ′′

Proof. (i) First we show that “ ” is an equivalence relation. ≃ a) To show that ϕ ϕ choose h = 0. ≃ b) The statement ϕ ψ ψ ϕ follows from replacing h by h. ≃ ⇒ ≃ − c) If ϕ ψ and ψ χ then ψ χ . ≃h ≃k h≃+k

(ii) Assume that ϕ ψ and ϕ′ ψ′. Now we calculate ≃h h≃′

ϕ′ϕ ϕ′ψ = ϕ′(∂′h + h∂) = ∂ϕ′h + ϕ′h∂, −

hence ϕ′ϕ ϕ′ψ. Similarly we see ϕ≃′h

ϕ′ψ ψ′ψ = (∂′h′ + h′∂)ψ = ∂′h′ψ + h′ψ∂, −

hence ϕ′ψ ψ′ψ. Combining both equivalences we get the desired result, namely h≃′ψ

ϕ′ϕ ϕ′ψ ψ′ψ. ≃ ≃

Homotopic chain maps induce the same homomorphisms on homology:

Proposition 3.59. If ϕ ψ : K K then H(ϕ) = H(ψ) : HK HK . ≃ → ′ → ′

Proof. Let z ZnK. We compute ∈

H(ϕ)[z] = [ϕz] = [ψz + ∂′hz + h∂z] = [ψz] = H(ψ)[z] because ∂z = 0 and [∂′hz] = 0.

111 3 Homology Theory

Deﬁnition 3.60. A chain map ϕ : K K is called a homotopy equivalence if there exists a → ′ chain map ψ : K K such that ψϕ idK and ϕψ idK . If such a homotopy equivalence ′ → ≃ ≃ ′ exists we write K K . ≃ ′

Corollary 3.61. If K K ′ then HK HK ′. ≃ϕ H (ϕ)

3.6 Proof of the homotopy axiom

We put I := [0, 1] and deﬁne the aﬃne linear map

+ T j : ∆n 1 ∆n I n → × for j = 0,..., n by

j (ek, 0), k j, Tn (ek ) = ≤ (ek 1, 1), k > j.  − In the case n = 0 we have  T0 : ∆1 ∆0 I = e I , 0 → × { 0}× which can be visualized as

b (e0, 0) 0 T0

b b b (e0, 1) e0 e1

In the case n = 1 we ﬁnd

112 3.6 Proof of the homotopy axiom

e2 (e0, 1) (e1, 1)

b b b 0 T1

b b b b

e0 e1 (e0, 0) (e1, 0)

e2 (e0, 1) (e1, 1)

b b b 1 T1

b b b b

e0 e1 (e0, 0) (e1, 0)

Lemma 3.62. The composition of the operators T and the aﬃne linear map F (deﬁned in (3.1) on page 80) yields:

j+1 i = i j Tn Fn+1 (Fn id) Tn 1 ( j i), ◦ × ◦ − ≥ j i+1 i j T F + = (F id) T ( j < i), n ◦ n 1 n × ◦ n 1 i i i 1 i − T F + = T − F + (1 i n), n ◦ n 1 n ◦ n 1 ≤ ≤ 0 0 T F + = i , n ◦ n 1 1 n n+1 T F + = i , n ◦ n 1 0 n n with it : ∆ ∆ I being the inclusion map x (x, t). → × 7→

Proof. Let us prove the ﬁrst formula where j i. If k < i then ≥ j+1 i = j+1 = = i = i j Tn Fn+1 (ek ) Tn (ek ) (ek, 0) (Fn id)(ek, 0) (Fn id) Tn 1 (ek ). ◦ × × ◦ − If i k j then ≤ ≤ j+1 i j+1 i i j = + = + = = Tn Fn+1 (ek ) Tn (ek 1 ) (ek 1, 0) (Fn id)(ek, 0) (Fn id) Tn 1 (ek ). ◦ × × ◦ − If k > j then

j+1 i j+1 i i j = + = = = Tn Fn+1 (ek ) Tn (ek 1 ) (ek, 1) (Fn id)(ek 1, 1) (Fn id) Tn 1 (ek ). ◦ × − × ◦ − The proofs of the other formulas are similar exercises in index shifting.

113 3 Homology Theory

Now let σ : ∆n X be a singular n-simplex. Note that (σ id) T j C(∆n+1, X I). We → × ◦ n ∈ × deﬁne Pσ Sn+ (X I) by ∈ 1 × n j j Pσ := ( 1) (σ id) Tn . = − × ◦ Xj 0 We extend P linearly to chains and get a linear map

P : Sn (X) Sn+ (X I). → 1 × This homomorphism descends to relative chains

P : Sn (X, A) Sn+ (X I, A I). → 1 × × The operator P is called prism operator.

Lemma 3.63. Let (X, A) be a pair of spaces. The operator P is a chain homotopy for

S(iX ), S(iX ) : S(X, A) S(X I, A I) , 0 1 → × × where iX : X X I denotes the inclusion map x (x, t). t → × 7→

Proof. We have to show that S(iX ) S(iX ) = P∂ + ∂P. (3.9) 0 − 1 Let σ : ∆n X be a singular n-simplex. We compute, using Lemma 3.62: → n = i i P∂σ P ( 1) (σ Fn ) = − ◦ Xi 0 n n 1 = i − j i j ( 1) ( 1) (σ Fn id) Tn 1 = − = − ◦ × ◦ − Xi 0 Xj 0 + = ( 1)i j (σ id) (Fi id) T j − × ◦ n × ◦ n 1 0 j

114 3.6 Proof of the homotopy axiom

On the other hand, we ﬁnd n j j ∂Pσ = ∂ ( 1) (σ id) Tn = − × ◦ Xj 0 n n+1 = j i j i ( 1) ( 1) (σ id) Tn Fn+1 = − = − × ◦ ◦ Xj 0 Xi 0 i+j j i = ( 1) (σ id) T F + (i < j) − × ◦ n ◦ n 1 0 i< j n ≤X≤ n + i i = (σ id) Tn Fn+1 (i j) = × ◦ ◦ Xi 0 n+1 i 1 i = + (σ id) Tn− Fn+1 (i j 1) − = × ◦ ◦ Xi 1 + ( 1)i+j (σ id) T j Fi (i > j + 1) − × ◦ n ◦ n+1 1 j+1

Proposition 3.64. If f g : (X, A) (Y, B) then we have ≃ → S( f ) S(g) : S(X, A) S(Y, B) ≃ →

Proof. Let F be a homotopy for f and g, i.e., F C((X I, A I), (Y, B)) with ∈ × × f = F iX, g = F iX . ◦ 1 ◦ 0 Then by (3.9) we get S( f ) S(g) = S(F)S(iX ) S(F)S(iX ) = S(F)P∂ + S(F)∂P = S(F)P∂ + ∂S(F)P. − 1 − 0 Hence S(F)P is a chain homotopy for S( f ) and S(g).

115 3 Homology Theory

Corollary 3.65. The homotopy axiom holds for singular homology.

3.7 Proof of the excision axiom

We want to show that the inclusion (X U, A U) ֒ (X, A) induces an isomorphism \ \ →

Hn (X U, A U) Hn (X, A) \ \ provided U¯ A˚. ⊂ X

A U

Let X be a topological space and let = Ui i I be an open cover of X. A chain U { } ∈ n σ = α j σj Sn (X) is called -small iﬀ for each j there exists an i such that σj (∆ ) Ui . j ∈ U ⊂ We denote P

SU (X) := σ Sn (X) σ is -small n { ∈ | U } i Sn (ji ) ⊕ = im Sn (Ui ) Sn (X) i I −−−−−−−−−→ *M∈ + with ji : Ui X the inclusion map. For A X we put → , ⊂ - S (X) = nU SnU (X, A) : . SnU (A)

Theorem 3.66 (Small chain theorem). The inclusion S (X, A) Sn (X, A) induces an iso- nU → morphism in homology.

Before coming to the proof we show that the small chain theorem implies the excision axiom. To this extent let (X, A) be a pair of spaces and let U A be such that U¯ A˚. Now set U := A˚ ⊂ ⊂ 1

116 3.7 Proof of the excision axiom and U := X U¯ . Then = U , U forms an open cover of the space X. We compute 2 \ U { 1 2} ( ) = SnU X SnU (X, A) SnU (A) Sn (A˚) + Sn (X U¯ ) = \ Sn (A˚) + Sn (A U¯ ) \ Sn (X U¯ ) = \ (Sn (A˚) + Sn (A U¯ )) Sn (X U¯ ) \ ∩ \ Sn (X U¯ ) = \ . Sn (A U¯ ) \ Similarly we get

( ˚ ) + ( ¯ ) ( ¯ ) = Sn A U Sn X U = Sn X U SnU (X U, A U) \ \ \ . \ \ Sn (A˚ U) + Sn (A U¯ ) Sn (A U¯ ) \ \ \ In the following commutative diagram all arrows are induced by inclusions.

= S (X U, A U) / S (X, A) mU \ \ nU

Sm (X U, A U) / Sn (X, A) \ \ By the small chain theorem the vertical arrows induce isomorphisms on homology and the excision theorem is proved. It remains to prove Theorem 3.66. We deﬁne the barycenter of ∆n by

n = 1 Bn : + ej . n 1 = Xj 0

Example 3.67. For example, e2

B0 = e0, 1 B1 = (e0 + e1), b 2 B1 B2 1 B = (e + e + e ). e0 e1 e0 e1 2 3 0 1 2

n n+1 n+1 n+1 For an affine map σ : ∆ ∆ we define the affine map Cnσ : ∆ ∆ by → →

Bn+1, k = 0, (Cnσ)(ek ) = σ(ek 1), k 1.  − ≥  

117 3 Homology Theory

Example 3.68. Consider the case σ = F0 : ∆1 ∆2. 2 →

e b2

G b

b e0 e1

C1G e0 e1 e b2

b b e0 e1 Now we set aﬀ ∆n = ∆n = Sk ( ) : σ Sk ( ) σ α j σj and each σj is aﬃne . ∈ j X By linear extension we get a homomorphism

aﬀ n+1 aﬀ n+1 Cn : S (∆ ) S + (∆ ) . n → n 1

Lemma 3.69. The homomorphism Cn has the following properties:

(i) ∂C (c) = c ( α j )B where c = α j σj ; 0 − j 0 j (ii) ∂Cn (c) = c CPn 1∂c for n 1. P − − ≥

Proof. (i) It suffices to show the assertion for an affine simplex c. We then find

(C0c)(e0) = B0, (C0c)(e1) = c(e0) and hence

∂C (c)(e ) = (C (c))(e ) (C (c))(e ) = c(e ) B = (c 1B )(e ) 0 0 0 1 − 0 0 0 − 0 − 0 0

118 3.7 Proof of the excision axiom

as desired.

(ii) is left as an exercise.

Lemma 3.70. To each topological space X and each n N we can associate homomorphisms ∈ 0

Sdn : Sn (X) Sn (X), → Qn : Sn (X) Sn+ (X), → 1 such that

(i) Sd is a chain map, i.e., ∂ Sdn = Sdn 1 ∂; ∗ ◦ − ◦ (ii) Q is a chain homotopy between id and Sd , i.e., id Sdn = ∂ Qn + Qn 1 ∂; ∗ ∗ − ◦ − ◦ (iii) Sd and Q are natural, i.e., for every f C(X,Y ) the following diagrams commute: ∗ ∗ ∈

Sdn Qn Sn (X) / Sn (X) Sn (X) / Sn+1 (X)

Sn (f ) Sn (f ) Sn (f ) Sn+1 (f )

 Sdn Qn Sn (Y ) / Sn (Y ) Sn (Y ) / Sn+1 (Y )

n n (iv) If the map σ : ∆ ∆ is aﬃne then each simplex σj occuring in Sdn (σ) or in Qn (σ) → is aﬃne and n diam(σj ) diam(σ). ≤ n + 1

Proof. The construction of Sdn and Qn will be done recursively, the proof is by induction over n. We start by considering the case n = 0. We put Sd := id : S (X) S (X) and Q := 0. It 0 0 → 0 0 is obvious that the four assertions hold. In the case n 1 we assume that Sdn 1 and Qn 1 are already deﬁned and we set for a singular ≥ − − simplex σ : ∆n X: →

Sdn (σ) := Sn (σ)(Cn 1(Sdn 1(∂ (id∆n ) ))) − − Saff (∆n ) ∈ n aff ∆n S|{z}n 1 ( ) ∈ − |aff ∆{zn } Sn 1 ( ) ∈ − aff n | Sn (∆{z) } ∈ | {z }

119 3 Homology Theory and Qn (σ) = Sn+1 (σ)(Cn (id∆n Sdn (id∆n ) Qn 1∂ id∆n )). − − − aﬀ n Sn (∆ ) ∈ aﬀ ∆n | Sn+{z1 ( ) } ∈ We have to verify assertions (i)-(iv). We| check (iii): {z }

Sdn (Sn ( f )(σ)) = Sn (Sn ( f )(σ))(Cn 1(Sdn 1(∂(id∆n )))) − − = Sn ( f σ)(Cn 1(Sdn 1(∂(id∆n )))) ◦ − − = Sn ( f ) Sn (σ)(Cn 1(Sdn 1(∂(id∆n )))) ◦ − − = Sn ( f )(Sdn (σ)).

The computation for Qn is similar. n Next we check (i): First we consider the case that X = ∆ and σ = id∆n .

∂(Sdn (id∆n )) = ∂(Cn 1(Sdn 1(∂ id∆n ))) − − L. 3.69 = Sdn 1 (∂ id∆n ) Cn 2∂(Sdn 1 (∂ id∆n )) − − − − ind. hyp. = Sdn 1 (∂ id∆n ) Cn 2Sdn 2 ∂∂ id∆n − − − − =0 = Sdn 1 (∂ id∆n ). − |{z} For a general simplex σ : ∆n X we then ﬁnd →

∂(Sdn (σ)) = ∂(Sn (σ)(Sdn (id∆n )))

= Sn (σ)(∂(Sdn (id∆n )))

= Sn (σ)(Sdn 1 (∂ id∆n )) − (iii) = Sdn 1 (Sn (σ)(∂(id∆n ))) − = Sdn 1 (∂Sn (σ)(id∆n )) − = Sdn 1 (∂σ). − n Now we check (ii): Again we ﬁrst consider the case X = ∆ and σ = id∆n .

∂Q(id∆n ) = ∂Cn (id∆n Sdn (id∆n ) Qn 1∂ id∆n ) − − − L. 3.69 = id∆n Sdn (id∆n ) Qn 1∂ id∆n − − − Cn 1∂ id∆n Sdn (id∆n ) Qn 1∂ id∆n − − − − − ind. hyp. = id∆n Sdn (id∆n ) Qn 1∂ id∆n − − − Cn 1 ∂ id∆n ∂Sdn (id∆n ) (∂ id∆n Sdn 1 (∂ id∆n ) − − − − − − + Qn 2 ( ∂∂ id∆n )) − = 0 (i) = id∆n Sdn (id∆n ) Q|{z}n 1∂ id∆n . − − −

120 3.7 Proof of the excision axiom

The passage to general σ can now be done as before. Finally we check (iv): It is clear from the recursive definition of Sdn and of Qn that each simplex σj occuring in Sdn (σ) or in Qn (σ) is again affine. The diameter of an affine simplex is the maximal distance of any two of its vertices. We distinguish two cases:

1. The vertices p, q of σj of maximal distance lie on ∂σ.

p q

Then we ﬁnd by the induction hypothesis n 1 n d(p, q) − diam(face of σ) < diam(σ) . ≤ n n + 1

2. One vertex is Bn:

Bn pi

Then we ﬁnd

1 n ( ) = d pi, Bn pi + pj − n 1 = Xj 0 n 1 = ( ) + pi pj n 1 = − Xj 0 n 1 pi pj ≤ n + 1 | − | j=0 diam(σ) and =0 for j=i X ≤ n diam(σ) .| {z } ≤ n + 1

121 3 Homology Theory

Conclusion 3.71. We have that Sd id and therefore ≃ Sd Sd2 = Sd ∂ Q + Sd Q ∂ − ◦ ◦ ◦ ◦ = ∂ Sd Q + Sd Q ∂. ◦ ◦ ◦ ◦ We conclude that Sd2 Sd id. Iterating this procedure we ﬁnd that Sdr id for all r N. ≃ ≃ (r) ≃ ∈ Hence there exist homomorphisms Q : Sn (X) Sn+ (X) with n → 1 r = (r) + (r) Sdn id ∂ Qn Qn 1 ∂. − ◦ − ◦ r For σ aﬃne, we have for every σj occuring in Sd (σ) that n r diam(σj ) diam(σ) ≤ n + 1

Lemma 3.72. For σ : ∆n X continuous there exists a ε > 0 such that all ε-balls ∆n are →1 ∩ completely contained in σ (Ui ) for some Ui . − ∈U

n Proof. Assume the assertion were false. Then for εk = 1/k there exists a point pk ∆ such ∈ that n B 1 (pk ) = x ∆ x pk < εk k { ∈ | | − | } 1 n is not contained in any of the σ− (Ui ). After passing to a subsequence we have that pk p ∆ ∆n 1 1 → ∈ by compactness of . Choose i0 with p σ− (Ui0 ). Since σ− (Ui0 ) is open there exists a δ > 0 1 ∈ 1 such that Bδ (p) σ (Ui ). Now choose k so large, that pk p < δ/2 and εk = < δ/2. We ⊂ − 0 | − | k then ﬁnd 1 B 1 (pk ) Bδ (p) σ− (Ui0 ), k ⊂ ⊂ a contradiction.

Corollary 3.73. Assume that σ : ∆n X is continuous. Then there exists an r(σ) N such r→ (r) ∈ that every simplex σj occuring in Sd (σ) or in Q (σ) for r r(σ) is completely contained r (r) ≥ in one of the Ui , i.e., Sd (σ), Q (σ) S (X). ∈ nU

Finally we can prove the small chain theorem.

Proof of Theorem 3.66. a) First we consider the case A = . ∅ i) We show injectivity: Assume that z Z U (X) with Hn ( j)([z] ) = 0. Then there ∈ n HnU

122 3.8 The Mayer-Vietoris sequence

exists an x Sn+ (X) with ∂x = z. Now we calculate ∈ 1 ∂ Sdr x = Sdr ∂x S (X) for large r ∈ nU+1 |{z} = Sdr z = z ∂Q(r) z Q(r) ∂z . − − =0 Hence |{z} z = ∂( Sdr x + Q(r) z )

S (X) for large r ∈ nU and therefore [z] = 0. HnU | {z } r ii) We show surjectivity: Let [z] Hn (X) be given. We know that Sd z S (X) for r ∈ ∈ nU large enough. We compute r = (r) (r) = [Sdn z] [z ∂Qn z Qn 1 ∂z ] [z]. − − − =0 Hn (j)(HnU (X)) ∈ |{z} |{z} b) Now we pass to general (X, A). Consider the commutative diagram with exact rows:

HnU (A) / HnU (X) / HnU (X, A) / HnU 1(A) / HnU 1(X) − − Hn (A) / Hn (X) / Hn (X, A) / Hn 1(A) / Hn 1(X) − − By part a) of the proof we know that the outer four arrows are isomorphisms. The Five Lemma (Exercise 3.11) implies that the map HnU (X, A) Hn (X, A) is also an isomorphism. →

3.8 The Mayer-Vietoris sequence

Let X be a topological space and let A X be a subset. Assume that U , U X are open with ⊂ 1 2 ⊂ U U = X, hence = U , U forms an open cover of X. Consider the exact sequence of 1 ∪ 2 U { 1 2} chain complexes

Sn (i1) Sn (i2 ) (Sn (j1 ), Sn (j2)) 0 / Sn (U U ) − / Sn (U ) Sn (U ) / S (X) / 0 1 ∩ 2 1 ⊕ 2 nU with the inclusion maps iν : U U Uν and jν : Uν X. We then get the following long 1 ∩ 2 → → exact homology sequence:

Hn (i1) Hn (i ) (Hn (j ), Hn (j )) ∂ − 2 1 2 Hn (U1 U2) / Hn (U1) Hn (U2) / HnU (X) Hn 1(U1 U2) ···→ ∩ ⊕ → − ∩ →···

123 3 Homology Theory

By the small chain theorem HnU (X) is canonically isomorphic to Hn (X). Using this isomorphism we can replace HnU (X) in the above exact homology sequence by Hn (X).

(Hn (j1), Hn (j2)) ∂ Hn (U1) Hn (U2) / HnU (X) / Hn 1(U1 U2) ···→ ⊕ ❱❱❱ −✐✐4 ∩ →··· ❱❱❱❱ ✐✐✐✐ ❱❱❱❱ ✐✐✐✐ ❱❱❱ ✐✐✐ MV (Hn (j1 ), Hn (j2)) ❱❱❱❱ ✐✐✐ ∂ ❱❱* ✐✐✐ Hn (X)

The same reasoning applies to relative homology and we obtain

Theorem 3.74 (Mayer-Vietoris sequence). Let X be a topological space, let A X and let ⊂ U , U X be open such that U U = X. Set Aν := A Uν and let 1 2 ⊂ 1 ∪ 2 ∩

iν :(U U , A A ) (Uν, Aν ), 1 ∩ 2 1 ∩ 2 → jν :(Uν, Aν ) (X, A), → be the inclusion maps, ν = 1, 2. Then the following sequence is exact and natural

Hn (i1 ) ∂MV Hn (i2 ) / Hn (U1 U2, A1 A2) − / Hn (U1, A1) Hn (U2, A2) ··· ∩ ∩ ❝❝❝❝ ⊕ ❝❝❝❝❝❝❝❝ ❝❝❝❝❝❝❝❝ ❝❝❝❝❝❝❝❝ (Hn (j1 ), Hn (j2)) ❝❝❝❝❝❝❝❝ q❝❝❝❝❝❝ Hn (X, A) / Hn 1(U1 U2, A1 A2) / ∂MV − ∩ ∩ ···

Example 3.75. We give a new computation of the homology of S1. To this extent we cover the circle as indicated in the picture.

U1 U2

We directly see that U U (0, 1) p 1 ≈ 2 ≈ ≃{ } and U U (0, 1) (0, 1) p , p 1 ∩ 2 ≈ ⊔ ≃{ 1 2} Consider the following part of the Mayer-Vietoris sequence.

124 3.8 The Mayer-Vietoris sequence

H (U U ) = H ( p , p ) / H (U ) H (U ) = H ( p ) H ( p ) / H (S1) n 1 2 n 1 2 n 1 n 2 n n ❝❝❝❝❝❝❝❝❝❝ n ∩ { } ⊕ { }❝❝⊕❝❝❝❝❝❝{❝❝} ❝❝❝❝❝❝❝❝❝❝ ❝❝❝❝❝❝❝❝❝❝ ❝❝❝❝❝❝❝❝❝ ∂MV q❝❝❝❝❝❝❝ Hn 1(U1 U2) = Hn 1( p1, p2 ) / Hn 1(U1) Hn 1(U2) = Hn 1( p ) Hn 1( p ) − ∩ − { } − ⊕ − − { } ⊕ − { } 1 If n 2 then all homologies of the point occuring in this diagram vanish. Hence Hn (S ) = 0 ≥ for all n 2. Since S1 is path-connected we have that H (S1) R. In the case n = 1 we ﬁnd ≥ 0 1 1 ( 1) 2 1 1 2 0 / H1 S / R / R 1 To compute H1(S ) we calculate

1 1 1 x H1(S ) ker = − x R R. 1 1 x ∈ ! ( ! )

Example 3.76. Now we consider the space X = G2 as given in the picture.

U1 U2

It is easy to see that U U S1 and U U p . Since G is path-connected we ﬁnd 1 ≈ 2 ≃ 1 ∩ 2 ≃ { } 2 H0(G2) R.

Hn (U1 U2) Hn (U1) Hn (U2) Hn (G2) Hn 1(U1 U2) ∩ → ⊕ → → − ∩ 1 1 Hn (S ) Hn (S ) Hn 1( p ) ⊕ − { }

In the case of n 2 we ﬁnd the| exact sequence{z } | {z } ≥ 0 / Hn (G2) / 0 and hence Hn (G2) = 0. For n = 1 we ﬁnd H ( p ) H (S1) H (S1) H (G ) H ( p ) H (S1) H (S1) . 1 { } → 1 ⊕ 1 → 1 2 → 0 { } → 0 ⊕ 0 =0 R2 R R2

The last map| {z in this} diagram| {z is given by} | {z } | {z } 1 1 2 H0( p ) R / H0(S ) H0(S ) R { } 1 ⊕; 1 and hence injective. Consequently we ﬁnd the isomorphism H (G ) H (U ) H (U ) R2 . 1 2 1 1 ⊕ 1 2

125 3 Homology Theory

3.9 Generalized Jordan curve theorem

In this section we will use the Mayer-Vietoris sequence to prove the Jordan curve theorem in arbitrary dimensions.

Lemma 3.77. Let X be a topological space and let Ui X be open with Ui Ui+ and ⊂ ⊂ 1 i NUi = X. Furthermore let ιn : Un ֒ X and ιn, m : Un ֒ Um for m n be the inclusion ∪ ∈ → → ≥ maps. Then we have:

(i) For each α Hk (X; R) there exists an n such that α im(Hk (ιn)) for all n n . ∈ 0 ∈ ≥ 0 (ii) For each αn Hk (Un; R) with Hk (ιn)(αn) = 0 there exists an m such that ∈ 0 Hk (ιn, m)(αn) = 0 for all m m . ≥ 0

Proof. (i) Let α Hk (X). We represent α by ∈ l k α j σj Zk (X; R), α j R, σj C(∆ , X). = ∈ ∈ ∈ Xj 1 k l k Note that σj (∆ ) X is a compact subset and therefore C := σj (∆ ) X is ⊂ j=1 ⊂ compact. Hence there exists an n such that for all n n we have C Un. We conclude 0 ≥ 0 ⊂ that S l α j σj Zk (Un; R) = ∈ Xj 1 for all n n and therefore ≥ 0

α = α j σj = Hn (ιn ) α j σj . H (X) H (U ) f g k f g k n X X l = (ii) Again represent αn Hk (Un) by j=1 α j σj . From Hn (ιn )(αn) 0 we know that there ∈ l exists a β Ck+ (X; R) such that α j σj = ∂ β. As before there exists a compact ∈ 1 P j=1 subset C′ X such that β Ck+1(C′; R). Since there exists an m0 with C′ Um for all ⊂ ∈ P ⊂ m m and thus β Ck+ (Um; R) we have that Hk (ιn, m)(αn) = 0. ≥ 0 ∈ 1

By an embedding we mean a continuous map f : X Y which is a homeomorphism onto its → image. In other words, f is continuous, open and injective. In the considerations which follow the domain will be compact and the target will be Hausdorﬀ so that f is automatically open.

Proposition 3.78. Let n N and let Y be a compact topological space such that for any ∈ embedding f : Y Sn → H (Sn f (Y ); R) H (point; R) . ∗ \ ∗

126 3.9 Generalized Jordan curve theorem

Then the space [0, 1] Y also has this property. ×

n n Proof. Let f : [0, 1] Y S be an embedding. Suppose 0 , α Hi (S f ([0, 1] Y )). Put × → ∈ \ × U := Sn f ([0, 1 ] Y ) and U := Sn f ([ 1, 1] Y ) . 0 \ 2 × 1 \ 2 × compact compact

Both U0 and U1 are open. We| also{z ﬁnd } | {z } U U = Sn f ([0, 1] Y ) and U U = Sn f ( 1 Y ) . 0 ∩ 1 \ × 0 ∪ 1 \ { 2 }× Y ≈

The Mayer-Vietoris sequence yields: | {z }

n 1 n 0 = Hi+ (S f ( Y )) / Hi (S f ([0, 1] Y )) 1 \ { 2 }× \ ×

n 1 n 1 Hi (S f ([0, ] Y )) Hi (S f ([ , 1] Y )) \ 2 × ⊕ \ 2 × ( Thus the inclusions Sn f ([0, 1] Y ) ֒ Sn f ([0, 1 ] Y ) and Sn f ([0, 1] Y ) ֒ Sn f ([ 1, 1] Y \ × → \ 2 × \ × → \ 2 × induce an injective homomorphism

n n 1 n 1 Hi (S f ([0, 1] Y )) Hi (S f ([0, ] Y ) Hi (S f ([ , 1] Y )) . \ × → \ 2 × ⊕ \ 2 × Hence , n 1 0 Hi (inclusion)(α) Hi (S f ([0, 2 ] Y )) or ∈ n \ 1 × 0 , Hi (inclusion)(α) Hi (S f ([ , 1] Y )) . ∈ \ 2 × By iterating this procedure we obtain a sequence of intervals Ik such that k I = [0, 1], Ik+ Ik, Ik = 2− 0 1 ⊂ | | and n 0 , Hi (ι ,k )(α) Hi (S f (Ik Y )) 0 ∈ \ × n n for all k. Here Vk := S f (Ik Y ) is open in S and ιk,l : Vk ֒ Vl for l k denotes the \ × → ≥ inclusion map. We ﬁnd

k N Ik = t ∈ ∩ { n} k NVk = S f ( t Y ) =: X ∪ ∈ \ { }× ,Now we apply the previous Lemma 3.77 for the inclusion ι : V ֒ X. Hence, for i 1 0 → ≥ n 0 , Hi (ι)(α) Hi (X) = Hi (S f ( t Y )) = 0 ∈ \ { }× Y ≈ giving a contradiction. The proof for i = 0 is similar (or in fact| {z the} same if one uses augmented homology).

127 3 Homology Theory

Corollary 3.79. If f : Dr Sn is an embedding then → H (Sn f (Dr )) H ( point ) ∗ \ ∗ { }

Proof. The proof is by induction on r. For r = 0 we ﬁnd

Sn f (D0) Rn point \ ≈ ≃{ } and hence H (Sn f (Dr )) H ( point ). ∗ \ ∗ { } For the induction step “r 1 r” we observe Dr W r = [0, 1] W r 1 [0, 1] Dr 1 and − ⇒ ≈ × − ≈ × − hence Proposition 3.78 applies.

Theorem 3.80. Let r < n and let f : Sr Sn be an embedding. Then → n r n r 1 H (S f (S )) H (S − − ). ∗ \ ∗

Proof. Again the proof is done by induction on r. For r = 0 we ﬁnd

n 0 n n n 1 S f (S ) = S p, q R 0 S − . \ \{ } ≈ \{ } ≃ r r r r r r 1 For the induction step “r 1 r” we write S = D+ D . Then D+ D = S − . We put n r − ⇒n r ∪ − ∩ − r U+ := S f (D+) and U := S f (D ). Both sets U+, U are open because f (D ) is compact. \ − \ − − ± In addition n r n r 1 U+ U = S f (S ) and U+ U = S f (S − ). ∩ − \ ∪ − \ Now we look at the Mayer-Vietoris sequence and use Corollary 3.79:

n r 1 n r Hi+1 (U+) Hi+1 (U ) / Hi+1 (S f (S − )) / Hi (S f (S )) ⊕ − \ \ H + (point) H + (point)=0 i 1 ⊕ i 1

| {z } Hi (U+) Hi (U ) ⊕ − = Hi (point) Hi (point) 0 if ⊕i 1 ≥ | {z } Thus for i 1 (and by similar reasoning also for i = 0) we get an isomorphism ≥ n r n r 1 n r n r 1 Hi (S f (S )) Hi+ (S f (S − )) Hi+ (S − ) Hi (S − − ). \ 1 \ 1 This proves the theorem.

128 3.9 Generalized Jordan curve theorem

n 1 n Theorem 3.81 (Jordan-Brouwer separation theorem). For any embedding f : S − S n n 1 → the complement S f (S − ) consists of exactly two path-components U and V. Both U and n \ n 1 V are open in S and ∂U = ∂V = f (S − ).

Proof. a) We know that H (Sn f (Sn 1)) H (S0) R2, hence Sn f (Sn 1) has exactly two 0 \ − 0 \ − path-components U and V. b) Since f (Sn 1) is compact the union U V = Sn f (Sn 1) is open and therefore both − ∪ \ − connected components U and V are open. c) Since U and V are open they contain no boundary point of U, thus ∂U f (Sn 1). Assume ⊂ − that there exists p Sn 1 with f (p) < ∂U. Then there exists W Sn open with f (p) W ∈ − ⊂ ∈ and W U = . Since f is continuous we can choose an open ball B Sn 1 with f (B) W. ∩ ∅ ⊂ − ⊂ Since Sn 1 B Dn 1 we ﬁnd that the space Y := Sn f (Sn 1 B) is path-connected because − \ ≈ − \ − \ of Corollary 3.79:

n n 1 H (Y ) = H (S f (S − B)) H (point) R . 0 0 \ \ 0 Moreover, we have

Y = U V f (B) U V W and U (V W ) = ∪ ∪ ⊂ ∪ ∪ ∩ ∪ ∅ and hence Y = (Y U) (Y (V W )) ∩ ⊔ ∩ ∪ U V ⊃

can be written as a disjoint union| of{z open} non-empty| {z subsets.} This contradicts Y being path-connected.

n 1 n Corollary 3.82 (Generalized Jordan curve theorem). For every embedding f : S − R n n 1 → the complement R f (S − ) consists of exactly two path components U and V. Both U and \ n 1 V are open, U is bounded, V is unbounded and ∂U = ∂V = f (S − ).

Proof. We use the stereographic projection to identify Rn with a subset of Sn,

n n n 1 n n S = R , f : S − R S . ∪ {∞} → ⊂ By the Jordan-Brouwer separation theorem the two path-components U˜ and V˜ of Sn f (Sn 1) \ − are both open and ∂U˜ = ∂V˜ = f (Sn 1). Let V˜ be the component containing . Then U = U˜ − ∞ and V = V˜ are the two path-components of Rn f (Sn 1). Clearly, V is unbounded and \ {∞} \ − ∂U = ∂U˜ = ∂V = ∂V˜ = f (Sn 1). If U were unbounded then ∂U which is not the case. − ∞ ∈ Hence U is bounded.

129 3 Homology Theory

Remark 3.83. Consider an embedding f : Sn 1 Sn. If i 1 we ﬁnd for the homology of the − → ≥ components of the complement

0 H(U V ) = Hi (U) Hi (V ) Hi (S ) = 0 ⊔ ⊕ and hence U and V have the same homology as a point. This makes us suspect that U, V D˚ n. ≈ For n = 2 this is indeed true, but it is false for n 3. The Alexander horned sphere is an ≥ example of an embedding of S2 into S3 where one component of the complement is not even simply connected:

The red circle in the picture is a non-contractible loop giving rise to a non-trivial element in the fundamental group. A very nice video illustrating this embedding of S2 can be found at http://www.youtube.com/watch?v=d1Vjsm9pQlc.

3.10 CW-complexes

We now describe a type of topological spaces for which there is a particularly eﬃcient way to compute their homology. These space are obtained by gluing together balls of various dimensions.

Definition 3.84. A finite CW-complex is a pair (X, ) where X is a Hausdorff space, X = n N n, n (X) and < with the following properties: X ∈ 0 X X ⊂ P |X| ∞

(i) `X = σ σ. ∈X n = n 1 (ii) Set X` : σ m σ X. For every σ n we haveσ ¯ σ X − . ∪ m∈Xn ⊂ ∈ X \ ⊂ ≤

130 3.10 CW-complexes

(iii) For every σ n there exists a surjective continuous map ∈ X n ϕσ : D σ¯ X → ⊂ n such that ϕσ n : D˚ σ is a homeomorphism. |D˚ →

Deﬁnition 3.85. An element σ n is called an n-cell. The map ϕσ is called the characteris- n ∈ X n 1 n 1 tic map of σ and X is called the n-skeleton of (X, ). The map ϕσ n 1= n : S − X − X S − ∂D → is called the attaching map of σ.

X 2

b b X 0

X 1

b b

Example 3.86. Consider X = Sn for n 1. Then the choice ≥

0 = e1 , X {{ }} n n = σn = S e , X { \{ 1}} m = , otherwise, X ∅ turns the n-sphere into a CW-complex.

131 3 Homology Theory

e1 Dn b Sn

ϕσ n

The attaching map to the n-cell is the constant map Sn 1 e . We have − →{ 1} 0 1 2 n 1 X = X = X = ... = X − = e , { 1} Sn = X n = X n+1 = ...

Example 3.87. If a space X has the structure of a CW-complex there are in general many diﬀerent ways to write X as a CW-complex, i.e., there are many diﬀerent for the same X. Let X us look again at X = Sn. We start with the case n = 0. Here the CW-structure is unique:

e1 , e1 , m = 0 m = {{ } {− }} X  , m > 0 . ∅ n 1 n  For n > 0 we use that S − S and deﬁne recursively ⊂ n 1 S − m , m n 1 n X ≤ − S = ˚ n ˚ n = m :  D+, D , m n X { − }  , m > n ∅  

n + S ◦ Dn

n 1 S −

◦ − Dn

132 3.10 CW-complexes

Therefore we have in this case

X 0 = S0, X 1 = S1,..., X n = Sn .

Example 3.88. Real projective space RPn. We deﬁne the real projective space as

+ + RPn = 1-dimensional real vector subspace of Rn 1 = Rn 1 0 / { } \{ } ∼ where x = (x ,..., xn ) y = (y ,..., yn ) 0 ∼ 0 iﬀ there exists a t , 0 such that x = ty. We consider the canonical projection map

n+1 n π : R 0 RP , x [x ,..., xn]. \{ } → 7→ 0 n n n The restriction ψ := π n : S RP is continuous and surjective. Thus RP is compact. S → Clearly ψ(x) = ψ(y) iﬀ x = y. We set ± n σk := [x ,..., xn] RP xk+ = ... = xn = 0, xk , 0 . 0 ∈ | 1 Then n k σ¯ k = [x ,..., xn] RP xk+ = ... = xn = 0 RP . 0 ∈ | 1 ≈ k For the characteristic map ϕk : D σ¯ k we can take →

2 ξ [ξ ,...,ξk, 1 ξ , 0,..., 0] . 7→ 1 − | | q It is clear that the map ϕk is continuous. Now we check that it is also surjective. Let [x , xk, 0] ′ ∈ σ¯ k where x′ = (x0,..., xk 1). Without loss of generality we assume that xk 0. Set − ≥ ξ := x′ Dk . We compute 2 2 √ x + xk ∈ | ′ | | |

k Next we show that ϕk k is injective. Let ϕk (ξ) = ϕk (η) for ξ, η D˚ . This leads to the two |D˚ ∈ equations: ξ = tη and 1 ξ 2 = t 1 η 2 − | | − | | for some t , 0. Squaring and adding both equationsq we ﬁndq

ξ 2 + 1 ξ 2 = t2 η 2 + t2(1 η 2) | | − | | | | − | |

133 3 Homology Theory leading to t2 = 1 and consequently t = 1. Since ξ , η < 1 it follows that 1 ξ 2 > 0 and ± | | | | − | | 1 η 2 > 0 and therefore t > 0. Hence t = 1 and thus ξ = η. − | | k p The map ϕk : D σ¯ k is closed, therefore the restriction p →

ϕk ˚ : D˚ σk Dk → is bijective, continuous and closed, hence a homeomorphism. We ﬁnd:

X 0 = point X 1 X 2 X n . { } ⊂ ⊂ ⊂···⊂ RP1 RP2 =RPn ≈ ≈ |{z} |{z}k k 1 |{z} Finally let us discuss the gluing map. For ξ ∂D = S − , i.e. ξ = 1, the gluing map is given k 1∈ k 1 k 1 | | by ϕk (ξ) = [ξ, 0, 0] and hence ϕk = ψ : S X RP . − → − ≈ −

Example 3.89. Complex projective space CPn. For the complex projective space the discussion is analogous to the real case with complex parameters instead of real parameters,

+ + CPn = 1-dimensional complex vector subspace of Cn 1 = Cn 1 0 / { } \{ } ∼ with x y iﬀ x = ty for some t C, t , 0. We ﬁnd ∼ ∈ 1, m even and 0 m 2n m = ≤ ≤ |X | 0, otherwise  and  0 1 2 3 4 5 2n 1 2n X = point = X X = X X = X X − = X . { } ⊂ ⊂ ⊂···⊂ CP1 CP2 =CPn ≈ ≈ |{z} |{z} |{z} Example 3.90. Quaternionic projective space HPn. Similarly, for the quaternionic projective space

HPn = 1-dimensional quaternionic vector subspace of Hn+1 = Hn+1 0 / { } \{ } ∼ with x y iff x = ty for some t H, t , 0, we find that ∼ ∈ 1, m divisible by 4 and 0 m 4n, m = ≤ ≤ |X | 0, otherwise,  and  0 1  2 3 4 4n 3 4n X = point = X = X = X X = ... X − = = X . { } ⊂ ⊂ ··· HP1 =HPn ≈ |{z} |{z} Remark 3.91. Every compact differentiable manifold can be triangulated and is consequently a finite CW-complex.

134 3.11 Homology of CW-complexes

3.11 Homology of CW-complexes

Throughout this section we assume that (X, ) is a ﬁnite CW-complex. Our goal is to prove X Theorem 3.100 which will provide us with an eﬃcient way to compute the homology of X.

Lemma 3.92. The map

σ n Hi (ϕσ ) n n 1 ⊕ ∈X n n 1 Hi (D , S − ) / Hi (X , X − ) σ n L∈X is an isomorphism.

Once we have this lemma, Theorem 3.16 implies

Corollary 3.93. We have the isomorphisms

n = n n 1 R|X |, for i n Hi (X , X − ) 0, otherwise   

Proof of Lemma 3.92. We set D˙ n := Dn 0 and \{ } n n 1 n n X˙ := X − ϕσ (D˙ ) = X ϕσ (0) σ n . ∪ \{ | ∈ X } σ n [∈X n 1 ˙ n x . The inclusion S − ֒ D is a homotopy equivalence with homotopy inverse x x → 7→ | | Dn

We deﬁne n n n 1 n 1 n n Y := D , Y − := S − , Y˙ := D˙

σ n σ n σ n a∈X a∈X a∈X The inclusions yield homotopy equivalences Y n 1 Y˙ n and X n 1 X˙ n. − → − →

135 3 Homology Theory

X 2

b X 1

Now consider:

Φ = : σ n ϕσ n n 1 n n ∪ ∈X n n _ n n 1 (Y ,Y − ) / (Y , Y˙ ) / (X , X˙ )o ? (X , X − )

Both inclusions induce isomorphisms on Hi . Due to the diagram

n 1 n n n 1 n 1 n Hi (Y − ) / Hi (Y ) / Hi (Y ,Y − ) / Hi 1 (Y − ) / Hi 1 (Y ) − − = = n n n n n n Hi (Y˙ ) / Hi (Y ) / Hi (Y ,Y˙ ) / Hi 1 (Y˙ ) / Hi 1 (Y ) − − n n 1 n n and the Five Lemma the map Hi (Y ,Y − ) Hi (Y , Y˙ ) is also an isomorphism. Similarly, n n 1 →n n n n 1 ( − we get that the inclusion (X , X − ) ֒ (X , X˙ ) induces an isomorphism Hi (X , X n n → → Hi (X , X˙ ). The inclusions

n n 1 n n 1 n n ,( ˙Y Y − , Y˙ Y − ) ֒ (Y ,Y) \ \ → n n 1 n n 1 n n ,( ˙X X − , X˙ X − ) ֒ (X , X) \ \ → induce isomorphisms on homology by the excision axiom. In addition, we have

n n 1 n n 1 n n 1 n n 1 (Y Y − ,Y˙ Y − ) (X X − , X˙ X − ) \ \ ≈ \ \ Hence we ﬁnd

n n 1 n n 1 Hi (D , S − ) Hi (Y ,Y − ) σ n M∈X n n Hi (Y ,Y˙ ) n n 1 n n 1 Hi (Y Y − , Y˙ Y − ) \ \ n n 1 n n 1 Hi (X X − , X˙ X − ) n \ n \ Hi (X , X˙ ) n n 1 Hi (X , X − ) .

n n 1 Set Kn (X, ) := Hn (X , X ) R n . We deﬁne a homomorphism X − |X |

∂n : Kn (X, ) Kn 1 (X, ) X → − X

136 3.11 Homology of CW-complexes by the following diagram:

n n 1 ∂ n 1 n / Hn (X , X − ) / Hn 1(X − ) / Hn 1(X ) / ··· − − ··· = ∂n * n 2 n 1 n 1 n 2 / Hn 1(X − ) / Hn 1(X − ) / Hn 1(X − , X − ) / ··· − − − ···

Lemma 3.94. The sequence of groups Kn (X, ) together with ∂n forms a complex. X

The pair (K (X, ),∂ ) is called the cellular complex of (X, ). ∗ X ∗ X Proof. The diagram

n n 1 ∂ n 1 Hn (X , X − ) / Hn 1(X − ) − = ∂n * n 1 n 1 n 2 ∂ n 2 Hn 1(X − ) / Hn 1(X − , X − ) / Hn 1(X − ) − − 2 − =0 = ∂n 1 − * n 2 n 2 n 3 Hn 1(X − ) / Hn 2(X − , X − ) − − shows ∂n 1 ∂n = 0 which is the claim. − ◦

p q Lemma 3.95. For p q n or n > p q we have Hn (X , X ) = 0. ≥ ≥ ≥

Proof. The proof is by induction on p q. The assertion is certainly true for p q = 0. To analyze − − + the situation p q > 0 we look at the exact homology sequence for the triple (X p, X q 1, X q ): −

q+1 q p q p q+1 ind. hyp. Hn (X , X ) / Hn (X , X ) / Hn (X , X ) = 0.

q+1 q Since either q n or q < p < n we have n , q + 1. Thus Hn (X , X ) = 0 by Lemma 3.92 ≥ p q and hence Hn (X , X ) = 0.

p Corollary 3.96. For n > p we have that Hn (X ) = 0.

p 0 Proof. Lemma 3.95 with q = 0 says Hn (X , X ) = 0. The claim now follows from the exact sequence 0 p p 0 0 = Hn (X ) Hn (X ) Hn (X , X ) = 0. −→ −→

137 3 Homology Theory

q Corollary 3.97. For q n we have Hn (X, X ) = 0. ≥

Proof. Choose p q so large that X p = X and use Lemma 3.95. ≥

Corollary 3.98. For r > n the inclusion X r ֒ X induces an isomorphism → r Hn (X) Hn (X ) .

Proof. The assertion follows from Corollary 3.97 and the exact sequence

r r r 0 = Hn+ (X, X ) Hn (X ) Hn (X) Hn (X, X ) = 0. 1 −→ −→ −→

Lemma 3.99. For r > n and r q the inclusion induces an isomorphism ≥ q r q Hn (X, X ) Hn (X , X ) .

r r Proof. Since r n + 1, Corollary 3.97 gives us Hn+ (X, X ) = Hn (X, X ) = 0. The assertion ≥ 1 now follows from the exact homology sequence of the triple (X, X r, X q ):

r r q q r Hn+ (X, X ) Hn (X , X ) Hn (X, X ) Hn (X, X ). 1 −→ −→ −→

Theorem 3.100. We have the following isomorphism:

HnK (X, ) Hn (X) ∗ X

138 3.11 Homology of CW-complexes

Proof. Consider the commutative diagram with exact columns and rows

n+1 n n 2 Hn+1(X , X ) Hn 1(X − ) = 0 ❘❘❘ − ❘❘❘ ∂n ∂ ❘❘❘ ❘❘❘ ❘) n 1 n i n n 1 n 1 0 = Hn (X − ) / Hn (X ) ∗ / Hn (X , X − ) / Hn 1(X − ) ❘❘❘ − ❘❘❘∂n 1 ❘❘❘− ❘❘❘ ❘) n+1 n 1 n 2 Hn (X ) Hn 1(X − , X − ) −

 n+1 n Hn (X , X ) = 0

Now we compute

n+1 Hn (X) Hn (X ) H (X n) n n+1 n ∂Hn+1(X , X ) n i Hn (X ) ∗ n+1 n i ∂Hn+1(X , X ) ∗ n n 1 n 1 ker(Hn (X , X − ) Hn 1(X − )) − n→+1 n ∂n Hn+1(X , X ) ker(∂n 1) − n+1 n ∂n Hn+1(X , X ) = HnK (X, ) ∗ X and the theorem is proved.

In the following examples we set αn := n . |X |

Example 3.101. We consider X = Sn for n 2. The CW-decomposition of Sn from Exam- ≥ ple 3.86 has α0 = αn = 1, α j = 0 otherwise. Hence we have to compute the homology of the complex

R 0 0 R 0 ←− ←− ··· ←− ←− ←− ←− ··· Since all arrows are 0 the homology coincides with the complex, hence

= n R, j 0 or n, Hj (S ) 0, otherwise,   which conﬁrms our earlier ﬁndings. 

139 3 Homology Theory

Example 3.102. Now look at X = CPn. Then we have for the CW-decomposition from Exam- ple 3.89 α0 = α2 = ... = α2n = 1, α j = 0 otherwise. Again all arrows in the complex

R 0 0 R 0 ←− ←− ··· ←− ←− ←− ←− ··· must be zero, hence the homology is given by = n R, j 0, 2,..., 2n, Hj (CP ) 0, otherwise.   Example 3.103. Consider X = HPn. For the CW-decomposition from Example 3.90 we have

α0 = α4 = α8 = ... = α4n = 1, α j = 0 otherwise and all arrows in

0 R 0 0 0 R 0 R 0 ←− ←− ←− ←− ←− ←− ··· ←− ←− ←− ←− ··· must be zero. We ﬁnd for the homology = n R, j 0, 4, 8,..., 4n, Hj (HP ) 0, otherwise.   3.12 Betti numbers and the Euler number

Throughout this section let R be a ﬁeld.

Deﬁnition 3.104. The dimension bj (X; R) := dimR Hj (X; R) is called j-th Betti number of the space X (over the ﬁeld R).

Now let (X, ) be a ﬁnite CW-complex. Denote the number of j-cells in (X, ) by α j . Then X X we get the following estimate for the Betti numbers:

bj (X; R) = dimR Hj (X; R) ker(∂j : Kj (X, ) Kj 1 (X, )) = dimR X → − X im(∂j+ : Kj+ (X, ) Kj (X, )) 1 1 X → X dimR ker(∂j : Kj (X, ) Kj 1 (X, )) ≤ X → − X dimR Kj (X, ) ≤ X = α j .

We conclude that bj (X; R) α j . In particular, the Betti numbers are ﬁnite, bj (X; R) < . ≤ ∞

140 3.12 Betti numbers and the Euler number

Deﬁnition 3.105. For a ﬁnite CW-complex (X, ) we call X

∞ i χ(X, ) = ( 1) αi X = − Xi 0 the Euler number or Euler-Poincaré characteristic.

Note that the sum in this deﬁnition is ﬁnite. It ends at the highest dimension occuring in the cell decomposition.

Proposition 3.106. We have the following relation between Euler and Betti numbers:

∞ i χ(X, ) = ( 1) bi (X; R). X = − Xi 0

In particular, the Euler number does not depend on the cell decomposition because the Betti numbers don’t. On the other hand, the Euler number does not depend on the coeﬃcient ﬁeld because the αi ’s don’t. We will henceforth write χ(X) instead of χ(X, ). X In order to prove the proposition we use the following

Lemma 3.107. Let

d1 d2 dn 0 o V o V o o Vn o 0 o 0 o 0 1 ··· ··· be a complex of ﬁnite-dimensional R-vector spaces. Then

∞ j ∞ j ( 1) dim HjV = ( 1) dim Vj . = − ∗ = − Xj 0 Xj 0

Proof of Lemma 3.107. To show the lemma we compute

j j ( 1) dim Vj = ( 1) (dim(d j Vj ) + dim ker(d j )) − − Xj Xj j = ( 1) (dim ker(d j ) dim(d j+1Vj+1)) j − − X ker(d j ) = ( 1) j dim d + V + j − j 1 j 1 X !

141 3 Homology Theory

j = ( 1) dim HjV . ∗ j − X

Proof of Proposition 3.106. The proposition follows from Lemma 3.107 with Vj = Kj (X, ) X Rα j .

Example 3.108. For X = Sn we have

b0 = bn = 1, bj = 0 otherwise and therefore 2, n even χ(Sn ) = 0, n odd .  The case n = 2 contains Euler’s classical formula for polyhedra as a special case. It says that for the alternating sum of the number of vertices, edges and faces of a polyhedron is always equal to 2. In particular, for platonic solids we have the following list:

α0 α1 α2 Tetrahedron 4 6 4 Cube 8 12 6 Octahedron 6 12 8 Dodecahedron 20 30 12 Icosahedron 12 30 20

Examples 3.109. We compute the Euler numbers of the projective spaces. n n 1.) For X = CP we have b0 = b2 = ... = b2n = 1 and bj = 0 otherwise. Hence χ(CP ) = n+1. n n 2.) For X = HP we have b0 = b4 = ... = b4n = 1 and bj = 0 otherwise. Hence χ(HP ) = n+1. 3.) In the case of X = RPn we do not know the Betti numbers yet. So we use Proposition 3.106 to compute the Euler number. For the cell decomposition described in Example 3.88 we have α0 = ... = αn = 1, α j = 0 otherwise.

142 3.13 Incidence numbers

Hence 1, n even χ(RPn ) = 0, n odd   3.13 Incidence numbers 

We return to a general commutative ring R with unit 1. Throughout this section let (X, ) be X a ﬁnite CW-complex. In order to compute the cellular homology of (X, ) we need a better X understanding of the homomorphism ∂n+ : Kn+ (X, ) Kn (X, ). In the case of n = 0 we 1 1 X → X ﬁnd ∂ K (X, ) 1 / K (X, ) 1 X 0 X

1 0 ∂ 0 H1(X , X ) / H0(X )

1 0 H1(ϕτ )(H1(D , S )) R τ 1 σ 0 L∈X L∈X Hence ∂ is given by the (α α )-matrix (∂τ ) where τ and σ . The entries ∂τ R 1 1 × 0 σ ∈ X1 ∈ X0 σ ∈ of this matrix are easily computed. Namely, recall that D1 = [ 1, 1] and that a generator of − H (D1, S0) is represented by c : [0, 1] [ 1, 1], t 2t 1. Then 1 → − 7→ −

∂ H (ϕτ )([c]) = ∂[ϕτ c] = ϕτ (c(1)) ϕτ (c(0)) = ϕτ (1) ϕτ ( 1). 1 1 ◦ − − − τ Thus if ϕτ ( 1) = ϕτ (1) then ∂ = 0 for all σ. If ϕτ ( 1) , ϕτ (1) then − σ −

1, for σ = ϕτ (1), τ = = ∂σ  1, for σ ϕτ ( 1), − − 0, otherwise.   Example 3.110. We compute the homology of the following CW-complex consisting of two 0-cells and three 1-cells:

τ2 τ1 σ1 b b σ2 τ3

τ1 = τ1 = Clearly, ∂σ1 ∂σ2 0. Depending on how the characteristic maps parametrize the 1-cells τ2 and τ3 we get ∂τ2 = ∂τ3 = 1 and ∂τ2 = ∂τ3 = 1, σ1 σ1 σ2 σ2 −

143 3 Homology Theory or possibly different signs which will not affect the homology however. Thus the cellular complex is 0 1 1 0 1 1 o 0 o R2 .*o − − /+ R3 o 0 o ··· , - ··· The image of the matrix is (x, x) x R = R (1, 1) and hence { − | ∈ } · − H (X; R) R2/R (1, 1) R 0 · − and 0 1 1 H (X; R) ker = (x, y, y) x, y R R2. 1 0 1 1 { − | ∈ } − − ! Returning to our general CW-complex (X, ) we find for n 1: X ≥ ∂n+1 Kn+ (X, ) / Kn (X, ) 1 X X

n+1 n n n 1 Hn+1(X , X ) Hn (X , X − )

n+1 n n n 1 Hn+1(ϕτ )(Hn+1(D , S )) Hn (ϕσ )(Hn (D , S − )) τ + σ n ∈Xn 1 ∈X L L τ (∂σ ) Rαn+1 / Rαn τ Hence ∂n+ is given by the (αn+ αn)-matrix (∂ ) where τ n+ and σ n. The entries 1 1 × σ ∈ X 1 ∈ X (∂τ ) R of this matrix are called the incidence numbers. We want to see how we can compute σ ∈ them. n Fix σ n, τ n+ , and p D˚ . From the commutative diagram ∈ X ∈ X 1 ∈ ∂n+1

+ n+1 n ∂ n n n 1 prσ n n 1 Hn+1(X , X ) / Hn (X ) / Hn (X , X − ) / Hn (ϕσ )Hn (D , S − ) O O O H + (ϕτ ) Hn (ϕτ n ) n 1 |S 1 1 n n deg (ϕ ϕτ :ϕ (σ) D S ) n+1 n ∂ n p σ− ◦ τ− → ⊂ n n n 1 Hn+1(D , S ) / Hn (S ) / Hn (S ) Hn (D , S − )

 n n 1 Hn (S , S ϕτ− (ϕσ (p))) \ O 

1 1 1 Hn (ϕ (σ),ϕ (σ) ϕ (ϕσ (p))) τ− τ− \ τ− Hn (ϕτ ) Hn (σ, σ ϕσ (p) ) \{ } 1 Hn (ϕσ− ) n n n n Hn (D˚ , D˚ p ) / Hn (D , D p ) \{ } \{ }

144 3.13 Incidence numbers we conclude that τ 1 1 n n ∂ = deg (ϕ− ϕτ : ϕ− (σ) D S ). σ p σ ◦ τ → ⊂ n n n 1 We used the canonical isomorphism Hn (S ) Hn (D , S − ). We have intepreted the incidence numbers as certain local mapping degrees. In applications they can often be computed by counting preimages.

Example 3.111. Look at X = RPn. We want to compute the homology of

∂ ∂ ∂ ∂ 0 o R o 1 R o 2 R o 3 o n R o 0 o ··· ··· τ The operator ∂j+1 is given by the (1 1)-matrix (∂σ ) with σ j and τ j+1. The gluing map j j j × ∈ X ∈ X j ϕτ S j : S X RP is given by the canonical projection. The image point ϕσ (p) X has | → ≈ j 1 1 ∈ j exactly two preimages under ϕτ which we call x, x S . Put Φ := ϕ ϕτ : ϕ (σ) D . − ∈ σ− ◦ τ− → From the additivity of the local degree we obtain Φ 1 j = Φ j + Φ j degp ( : ϕτ− (σ) D ) degp ( U (x) : U(x) D ) degp ( U ( x) : U( x) D ) → → − − → where U(x) is a small neighborhood of x. W.l.o.g. we assume U( x) = U(x). Since Φ is − − U (x) a homeomorphism onto its image we have

deg (Φ : U(x) D j ) = 1 =: ε. p U (x) → ± Let a : S j S j be the antipodal map. Then Φ = Φ a and hence → ◦ Φ j = Φ j degp ( U ( x) : U( x) D ) degp ( a : U( x) D ) − − → ◦ − → = deg (Φ : U(x) Dn) deg(a) p → · = ε ( 1) j+1. · − j+1 We conclude that ∂j+ = ε (1 + ( 1) ). For n even we obtain the complex 1 · − 0 2 0 2 0 o R o R o ± o R o ± R o 0 ··· whereas for n odd we ﬁnd 0 2 0 0 o R o R o o ± R o R o 0. ··· For R = Z/2Z all arrows are zero so that = n Z/2Z, j 0,..., n, Hj (RP ; Z/2Z) = 0, otherwise.  For R = Z the homology is computed to   Z, j = 0, Z/2Z, j = 1, 3, n 1 or n 2 resp.,  − − H (RPn; Z) = Z, j = n odd, j   = 0, j n even,  0, otherwise.    145 3 Homology Theory

3.14 Homotopy versus homology

The ﬁnal goal of this chapter is to compare homotopy groups and homology groups. We start by examining the fundamental group of CW-complexes.

Proposition 3.112. Let (X, ) be a ﬁnite CW-complex and let x X 0. Then the inclusion X 0 ∈ map j : X 2 ֒ X induces an isomorphism → j : π (X 2; x ) π (X; x ) . # 1 0 → 1 0

Proof. We have to show that attaching a k cell for k 3 does not alter π1 in the sense that the − ≥ 2 inclusion induces an isomorphism on π1. We assume that X is path-connected, since otherwise 2 we may replace X by the path-component that contains x0. Let Y be a path-connected ﬁnite CW-complex and let Y˜ be obtained from Y by attaching a k-cell. k k k 1 k 1 More precisely, Y˜ = Y ϕ D = Y D / where x ϕ(x) for all x S . Here ϕ : S Y ∪ ⊔ ∼ ∼ ∈ − − → is a continuous map. We have to show that the inclusion map induces an isomorphism

j : π (Y; x ) π(Y˜; x ) # 1 0 → 0 if k 3. Let Dk ( 1 ) D˚ k be the closed k-dimensional subball of radius 1 . Cover Y˜ by the two ≥ 2 ⊂ 2 open subsets U1, U2 where

U = D˚ k, U = Y˜ Dk ( 1 ) Y 1 2 \ 2 ≃

Y U2

146 3.14 Homotopy versus homology

We then ﬁnd U U = D˚ k Dk ( 1 ) Sk 1. To apply the Seifert-van-Kampen Theorem 2.68 1 ∩ 2 \ 2 ≃ − we calculate

π (U ) = 1 , 1 1 { } π1(U2) π1(Y ), k 1 π (U U ) π (S − ) = 1 , (here we use k 1 2) 1 1 ∩ 2 1 { } − ≥ and we ﬁnd π (U ) ⋆π (U ) π (Y˜ ) 1 1 1 2 π (Y ) 1 im π (U U ) 1 1 1 ∩ 2 the isomorphisms being induced by inclusions.

Example 3.113. Consider complex-projective space X = CPn. We use the cell decomposition from Example 3.89:

0 1 2 3 4 5 2n 1 2n X = point = X X = X X = X ... X − X { } ⊂ ⊂ ⊂ ⊂ ⊂ CP1 CP2 =CPn ≈ ≈ We are now able to calculate the fundamental|{z} group|{z} |{z}

π (CPn ) π (X 2) π (CP1) = π (S2) = 1 . 1 1 1 1 { } Hence complex-projective space CPn is simply connected.

Example 3.114. Similarly, for X = HPn we use the cell decomposition from Example 3.90:

0 1 2 3 4 4n 3 4n X = point = X = X = X X = ... X − = ... X { } ⊂ ⊂ ⊂ HP1 =HPn ≈ For the fundamental group we ﬁnd |{z} |{z}

π (HPn ) = π (X 2) = π ( point ) = 1 . 1 1 1 { } { } Thus HPn is also simply connected.

Remark 3.115. Proposition 3.112 can be generalized as follows: For a ﬁnite CW-complex the n+1 n+1 ( inclusion map j : X ֒ X always induces an isomorphism j : πn (X ; x ) πn (X; x → # 0 → 0 where x X 0. 0 ∈

Now we relate homotopy and homology groups. Recall that for the n-dimensional cube n n n n n n 1 n n W = [0, 1] we have (W ,∂W ) (D , S − ). Fix a generator αn Hn (W ,∂W ; Z) n n 1 ≈ ∈ n Hn (D , S − ; Z) Z. The elements of πn (X; x0) are homotopy classes relative to ∂W of n n maps f : W X with f (∂W ) = x . Then Hn ( f )(αn) Hn (X, x ; Z). The long exact → { 0} ∈ { 0} homology sequence of the pair (X, x ) yields for n 1 { 0} ≥ 0 0 = Hn ( x0 ; Z) / Hn (X; Z) / Hn (X, x0 ; Z) / Hn 1( x0 ; Z). { } { } − { }

147 3 Homology Theory

Namely, if n 2 then Hn 1( x0 ; Z) = 0. For n = 1 the arrow emanating from H0( x0 ; Z) ≥ − { } { } is injective so that the incoming arrow must again be zero. In either case the inclusion j : (X, ) ֒ (X, x ) induces an isomorphism Hn ( j) : Hn (X; Z) Hn (X, x ; Z). Now set ∅ → { 0} −→ { 0} 1 h([ f ]) := Hn ( j)− Hn ( f )(αn) Hn (X; Z) . ∈ Due to homotopy invariance the expression h([ f ]) only depends on the homotopy class of the map f . Hence we have constructed a well-deﬁned map

h : πn (X; x ) Hn (X; Z), n 1. 0 → ≥

Proposition 3.116. The map h : πn (X; x ) Hn (X; Z) is a homomorphism. 0 →

Proof. Consider the map s : W n = [0, 1 ] [0, 1]n 1 W n given by 1 1 2 × − →

(x ,..., xn ) (2x , x ,..., xn ) 1 7→ 1 2 and the map s : W n = [ 1, 1] [0, 1]n 1 W n. 2 2 2 × − →

(x ,..., xn ) (2x 1, x ,..., xn ) . 1 7→ 1 − 2

n n n n W W1 W2 W s1

For [ f ], [ f ] πn (X; x ) we have [ f ] + [ f ] = [g] with 1 2 ∈ 0 1 2 f (s (x)), x 1, g(x) = 1 1 1 ≤ 2 ( ( )) 1  f2 s2 x , x1 2 .  ≥ + n  n n We represent αn by c1 c2 Sn (W ; Z), where cν S (Wν ; Z) and cν represents the generator 1 n ∈ n ∈ Hn (sν )− (αn) of Hn (Wν ,∂Wν ; Z).

148 3.14 Homotopy versus homology

n = 1 n = 2

σ1 σ3 c1 = σ1 + σ2 c2 = σ3 + σ4 σ2 σ4

c1 c2

Now the proposition follows from

h([ f ] [ f ]) = h([g]) 1 · 2 1 = Hn ( j)− Hn (g)(αn) 1 = Hn ( j)− Hn (g)([c1 + c2]) 1 1 = Hn ( j)− Hn (g)([c1]) + Hn ( j)− Hn (g)([c2]) 1 1 = Hn ( j)− Hn ( f s )([c ]) + Hn ( j)− Hn ( f s )([c ]) 1 ◦ 1 1 2 ◦ 2 2 1 1 = Hn ( j)− Hn ( f1)Hn (s1)([c1]) + Hn ( j)− Hn ( f2)Hn (s2)([c1]) 1 1 = Hn ( j)− Hn ( f1)(αn) + Hn ( j)− Hn ( f2)(αn)

= h( f1) + h( f2).

Deﬁnition 3.117. The map h : πn (X; x ) Hn (X; Z) is called Hurewicz homomorphism. 0 →

Proposition 3.118. The Hurewicz homomorphism h is natural, i.e., for every f C(X,Y ) ∈ with f (x0) = y0 the following diagram commutes:

h πn (X; x0) / Hn (X; Z)

f# Hn (f ) h πn (Y; y0) / Hn (Y; Z)

Proof. The inclusion maps

( jX : (X, ) ֒ (X, x ∅ → { 0} ( jY : (Y, ) ֒ (Y, y ∅ → { 0}

149 3 Homology Theory satisfy

( jY f )(x) = jY ( f (x)) = ( f (x), y ) ◦ 0 ( f jX )(x) = f ((x, x )) = ( f (x), f (x )) = ( f (x), y ) ◦ 0 0 0 and therefore jY f = f jX . On the level of homology groups we ﬁnd ◦ ◦

Hn ( jY ) Hn ( f ) = Hn ( f ) Hn ( jX ) ◦ ◦ and consequently 1 1 Hn ( f ) Hn ( jX )− = Hn ( jY )− Hn ( f ) . ◦ ◦ Thus

1 (Hn ( f ) h)([σ]) = Hn ( f ) Hn ( jX )− Hn (σ)(αn) ◦ ◦ ◦ 1 = Hn ( jY )− Hn ( f ) Hn (σ)(αn) ◦ ◦ 1 = Hn ( jY )− Hn ( f σ)(αn) ◦ ◦ = h([ f σ]) ◦ = (h f )([σ]) ◦ # as claimed.

Theorem 3.119 (Hurewicz). Let X be a topological space, x X and n 2. Assume that 0 ∈ ≥

π0(X; x0) = π1(X; x0) = ... = πn 1 (X; x0) = 1 . − { } We then have H1(X; Z) = ... = Hn 1(X; Z) = 0 − and the map h : πn (X; Z) Hn (X; Z) → is an isomorphism.

For a proof of this theorem see e.g. [2, Sec. 4.2].

Remark 3.120. For n = 1 this theorem cannot be true as it stands because H1(X; Z) is always abelian while π1(X; x0) is not in general. However, h induces a homomorphism

h¯ : π (X; x )/ π (X x ),π (X x ) H (X; Z). 1 0 [ 1 ; 0 1 ; 0 ] → 1 Already Poincaré showed that if π (X, x ) = 1 (i.e., X is path-connected) then the map h¯ is an 0 0 { } isomorphism.

150 3.14 Homotopy versus homology

Example 3.121. Consider X = Sn for n 2. We already know that Sn is 1-connected. Now ≥ apply the Hurewicz isomorphism with n = 2: = n n Z, n 2, π2(S ) H2(S ; Z) = 0, n 3.  ≥ For n 3 we ﬁnd due to Hurewicz:  ≥  = n n Z, n 3, π3(S ) H3(S ; Z) = 0, n 4.  ≥ n n By induction we then deduce that π1(Sn ) = ... = πn 1(S ) = 0 and πn (S ) Z. −

Example 3.122. We know that X = CPn is 1-connected. With the help of the Hurewicz isomorphism we calculate n n π2(CP ) H2(CP ; Z) Z .

Example 3.123. We also know that X = HPn is 1-connected. We apply the Hurewicz isomorphism three times and we get

n n π2(HP ) H2(HP ; Z) = 0, n n π3(HP ) H3(HP ; Z) = 0, n n π4(HP ) H4(HP ; Z) Z.

Example 3.124. Now let us analyze the space X = RPn for n 2. The map ψ : Sn RPn is a ≥ → two-fold covering and by Theorem 2.102

ψ♯ π (Sn ) π (RPn ) π ( p, q ) π (Sn ) 1 −−→ 1 → 0 { } → 0 n n is exact. Since S is simply connected π1(RP ) π0( p, q ) is an isomorphism of pointed n → { } sets, thus π1(RP ) has exactly two elements. There is only one group with two elements, hence n n n π1(RP ) Z/2Z. By Corollary 2.105 ψ# : πk (S ) πk (RP ) is an isomorphism for all k 2. n n → n ≥ We conclude that π2(RP ) = ... = πn 1(RP ) = 0 and πn (RP ) Z. −

Remark 3.125. Under the assumptions of the theorem of Hurewicz 3.119 not much can be said about h : πk (X, x0) Hk (X; Z) for k > n. For example, consider the Hopf ﬁbration 3 2 1 → 2 3 S S with ﬁber S . By Theorem 2.102 it induces an isomorphism π3(S ) π3(S ) Z. → 2 2 2 But H3(S ; Z) = 0 and hence h : π3(S ) H3(S ; Z) is not injective. 2 → 2 2 On the other hand, for the 2-torus T we have again by Corollary 2.105 π2(T ) π2(R ) = 0 while one can compute H (T2; Z) Z. Thus h : π (T2) H (T2; Z) is not surjective. 2 2 → 2

n Remark 3.126. We have seen that Hk (S ; Z) = 0 whenever k > n. But in general this is not 2 n true for the higher homotopy groups of the sphere, e.g. π3(S ) Z. The computation of πk (S ) for k > n is a diﬃcult problem and many of these groups are not known to date.

151 3 Homology Theory

3.15 Exercises

3.1. Let X be a topological space. Show: a) If X is path-connected then H0(X; R) R b) If Xk , k K, are the path-components of X then ∈

Hn (X; R) Hn (Xk ; R) k K M∈

3.2. Let Yk = 1, ..., k be equipped with the discrete topology. Compute Hn (Yk ; R) without { } using Exercise 3.1 directly with the help of the Eilenberg-Steenrod axioms. Hint: Consider the pair (Yk,Yk 1). −

3.3. Let X be a topological space. The augmented boundary operator is deﬁned by

∂# : S (X; R) R, 0 →

# ∂ αi σi = αi,

0 X X where σi C(∆ , X). ∈ a) Verify ∂# ∂ = 0. ◦ b) Compute the augmented homology

ker(∂# : S (X; R) R) H#(X; R) := 0 → 0 im(∂ : S (X; R) S (X; R)) 1 → 0 for X = point . { }

n n n n 1 3.4. a) Show that homeomorphisms σ : ∆ D represent generators of Hn (D , S ; R). → − n b) Describe generators of Hn (S ; Z). Make a drawing for n = 2.

3.5 (Topological invariance of the dimension). Let U Rn and V Rm be open and ⊂ ⊂ nonempty. Show: If U and V are homeomorphic then n = m. Hint: For p U and q V consider the pairs (U,U p ) and (V, V q ). ∈ ∈ \{ } \{ }

3.6. Let p be a complex polynomial without zeros on the unit circle S1 C. Show: The degree ⊂ of the map p(z) pˆ : S1 S1, pˆ(z) = , → p(z) | | coincides with the number of zeros of p in the interior of the unit disk (counted with multiplicities).

152 3.15 Exercises

3.7 (Homotopy invariance of the local mapping degree). Let V Sn be open and F : V n = n ⊂ 1 × [0, 1] S continuous. We put ft (x) : F(x, t). Let p S such that t [0,1] ft− (p) is → ∈ ∈ compact. Show: = S degp ( f1) degp ( f0).

3.8. Let (X, A) be a pair such that A is closed and a strong deformation retract of an open neighborhood U. Show that Hn (X, A) = Hn (X/A) for n , 0.

1 1 1 3.9. Let Z = S [0, 1] be the cylinder. Compute Hn (Z, S 0 S 1 ) for all n. Sketch × ×{ } ∪ ×{ } generators of the nontrivial homology groups. Hint: Use the homology sequence of the triple

(Z, S1 0 S1 1 , S1 0 ). ×{ } ∪ ×{ } ×{ }

th 3.10. Let (X, A) be a pair. Describe the 0 singular relative homology group H0(X, A).

3.11 (Five lemma). Let the rows in the following commutative diagram of abelian groups be exact: f1 f2 f3 f4 A1 / A2 / A3 / A4 / A5

ϕ1 ϕ2 ϕ3 ϕ4 ϕ5

 g1 g2 g3 g4 B1 / B2 / B3 / B4 / B5

Show that if ϕ1, ϕ2, ϕ4, and ϕ5 are isomorphisms then so is ϕ3. Hint: ϕ3 is injective if ϕ1 is surjective and ϕ2, ϕ4 are injective; ϕ3 is surjective if ϕ5 is injective and ϕ2, ϕ4 are surjective.

3.12. Suppose fn+1 fn Gn+1 Gn Gn 1 ···→ −−−−−−−−−−→ −−−−−−−−→ − →··· is a long exact sequence of abelian groups and

f n′ +1 fn′ G G G n′ +1 −−−−−−−−−−→ n′ −−−−−−−−→ n′ 1 ···→ − →··· is a subsequence, i.e., G Gn and f = f n . Prove that the subsequence is exact if and only n′ ⊂ n′ |G′n if the quotient sequence

Gn+1/Gn′ +1 Gn/Gn′ Gn 1/Gn′ 1 ···→ → → − − →··· is exact.

3.13. Let n N and m Z. Show that there exists f C(Sn, Sn ) with deg( f ) = m. ∈ ∈ ∈

153 3 Homology Theory

n n n n n n n 1 n n 3.14. Let f C(S , S ) with f (D+ ) D+ and f (D ) D . We identify S − with D+ D ∈ ⊂ − ⊂ − ∩ − and therefore have f (Sn 1) Sn 1. Show − ⊂ −

deg( f ) = deg( f n 1 ). |S −

3.15. Show that H1(R, Q; Z) is a free abelian group and ﬁnd a basis as a Z-module.

3.16. Let M be an n-dimensional manifold, n 3. Let p M. Show that the inclusion map ≥ ∈ M p ֒ M induces an isomorphism \{ } →

Hj (M p ; R) Hj (M; R) \{ } for all j 1, ..., n 2 . ∈ { − }

3.17. a) Show that for disjoint closed subsets A, B R2 we have ⊂ H (R2 (A B)) H (R2 A) H (R2 B). 1 \ ∪ 1 \ ⊕ 1 \ 2 2 b) Let p , ..., pn R be pairwise distinct. Compute H (R p ,..., pn ) and sketch generators. 1 ∈ 1 \{ 1 }

3.18. Use the Mayer-Vietoris sequence to compute a) the homology of the 2-torus; b) the homology of surfaces Fg of genus g 2. ≥ Sketch generators.

3.19. Let (X, ) be a ﬁnite CW-complex. Show that X is compact. X

3.20. Describe a CW-decomposition for the surfaces of genus g 1. ≥

3.21. Show: a) Each nonempty CW-complex has at least one 0-cell. b) Each CW-complex consisting of exactly two cells is homeomorphic to a sphere.

3.22. Let n 2 and k 1. Let X be the topological space obtained from k copies of Sn by ≥ ≥ identifying them all at one point. More formally,

k X = j Sn { }× ∼ j=1 [ where ( j, x) ( j , x ) iﬀ x = x = e . Compute the homology of X. ∼ ′ ′ ′ 1

154 3.15 Exercises

3.23. Find a CW-decomposition of the 2-torus with exactly one 0-cell, two 1-cells, and one 2-cell and use it to compute the homology.

155

Sources list for pictures

titlepage http://www.sxc.hu p. 4 taken from [3] p. 5 based on http://commons.wikimedia.org/wiki/File:Hilbert_curve.png p. 6, 2. and 4. picture http://www.sxc.hu p. 33 based on http://www.sxc.hu p. 130 http://www.math.ohio-state.edu/ fiedorow/ ∼

157

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[8] Warner, F. W.: Foundations of diﬀerentiable manifolds and Lie groups, Springer-Verlag, New York, 1983

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Index

Symbols h, Hurewicz homomorphism ...... 148 Hn ( f ), map induced on homology . . . . . 83 a,antipodal map...... 95 HnK , n-thhomology...... 106 ¯ ∗ A, closure of A ...... 88 Hn (ϕ ), map induced on homology . . . 106 ˚ ∗ A, interior of A ...... 88 Hn (X, A; R), relative n-th singular bj (X; R),Bettinumber...... 140 homology ...... 84 Bn,barycenter...... 117 Hn (X; R), n-th singular homology . . . . . 82 n BnK , n-boundaries...... 106 HP , quaternionic projective space . . . . 134 ∗ Bn (X, A; R), relative singular n-boundaries Kn (X, ), cellular complex...... 136 X 84 (S), normal subgroup generated by . 41 N S Bn (X; R), singular n-boundaries ...... 82 Ω (X; x0, x1),setofpaths...... 21 B(x, r), open ball of radius r centered at x 7 Ω (X; x0),setofloops...... 21 χ(X, ),Eulernumber ...... 141 P,prismoperator...... 114 X Cn, barycentric subdivision ...... 117 ϕσ,attachingmap...... 131 n CP , complex projective space . . . . 77, 134 π1(X; x0), fundamental group ...... 22 CX, cone over X ...... 14 πn (X; x0), n-th homotopy group ...... 58 C((X, A), (Y, B)), set of continuous maps Qn, chain homotopy for subdivision map between pairs of spaces ...... 84 119 C(X,Y ), set of continuous maps...... 17 RPn, real projective space . . . . . 73, 77, 133 deg,degree...... 30, 97 Sdn, subdivision chain map ...... 119 aﬀ ∆n degp,localdegree...... 102 Sk ( ),aﬃnechains...... 118 n D , closed n-dimensional unit ball ...... 6 Sn ( f ), map induced on singular chains. .83 n D ,hemisphere ...... 90 Sn (X, A; R), relative singular n-chains . . 84 n± ∆ , standard simplex ...... 79 Sn (X; R), singular n-chains...... 80 ∂, boundary of singular chain ...... 81 SnU (X, A), small singular n-chains. . . . .116 ∂, connecting homomorphism for ΣX, suspension of X ...... 14 homology groups ...... 86, 107 σ(i), i-th face of singular simplex...... 81 i ∂, connecting homomorphism for Tn ...... 112 n homotopy groups ...... 67 Rn , standard topology of R ...... 7 T ∂¯, boundary of relative singular chain . . . 84 X , topology of X ...... 7 T εx0 ,constantloop...... 21 U( f, p),windingnumber...... 75 Exp, covering map of S1 ...... 28 W n, n-dimensional cube...... 13 n f#, map induced on fundamental group . 24 X , n-skeleton ...... 131 Fg, surface of genus g ...... 6 ZnK , n-cycles...... 105 i ∗ Fn,facemap...... 80 Zn (X, A; R), relative singular n-cycles . . 84 H,quaternions...... 99 Zn (X; R), singular n-cycles...... 82

161 Index

,homotopic...... 17, 111 connected...... 39 ≃ A, homotopic relative to A...... 17 1-connected...... 39 ≃ ,homeomorphic...... 9 connecting homomorphism ...... 86, 107 ≈ G/H, homogeneous space ...... 64 continuous map...... 8 M#N, connected sum ...... 51 contractible ...... 19 G1 ⋆ G2,freeproduct...... 43 covering...... 73 ω ⋆ η, concatenation ...... 21 cycle...... 105 cylinder...... 14 A D Alexander horned sphere ...... 130 antipodalmap...... 95 deformation retract...... 19 attachingmap...... 131 degree...... 30, 97 augmented boundary operator...... 152 dimensionaxiom ...... 87 augmented homology ...... 152 discrete topology ...... 8

B E barycenter ...... 117 Eilenberg-Steenrod axioms ...... 87 basespace...... 62 embedding...... 126 Bettinumber...... 140 Eulernumber ...... 141 bidegree...... 99 Euler’s formula for polyhedra ...... 142 Borsuk-Ulam theorem ...... 33 Euler-Poincaré characteristic ...... 141 boundary...... 105 exact sequence of complexes ...... 106 boundary of a singular simplex ...... 81 exactness axiom ...... 88 boundary operator ...... 86 excisionaxiom ...... 88 augmented...... 152 F Brouwer’s fixed point theorem . . . . 37 f., 92 face of a singular simplex...... 81 C fiber...... 62 fiberbundle...... 63 cell...... 131 finiteCW-complex...... 130 cellular complex ...... 137 finitely presentable group ...... 55 chainhomotopy...... 110 freeproduct...... 42 chainmap ...... 106 fundamental group ...... 22 characteristic map ...... 131 chord...... 76 H closedsubset...... 8 closure ...... 88 ham-sandwich theorem ...... 34 coarsetopology...... 8 Hausdorffspace ...... 10 combspace...... 19, 75 Heine-Borel theorem ...... 9 compactsubset...... 9 Hilbertcurve...... 4 complex...... 105 HLP...... 61 complex projective space...... 77, 134 homeomorphic ...... 9 cone...... 14 homeomorphism...... 9

162 Index homogeneous space ...... 64 open subset of Rn ...... 7 homology...... 106 open subset of a topological space ...... 8 augmented...... 152 relative singular ...... 84 P singular...... 82 pairofspaces...... 84 homotopic chain maps ...... 111 path...... 21 homotopic relative to A...... 17 path-connected ...... 39 homotopy axiom...... 88 plane-filling curve ...... 4 homotopy equivalence ...... 18, 112 pointedset...... 61 homotopy equivalent ...... 18 presentation of a group ...... 55 homotopy group...... 58 prismoperator...... 114 homotopy lifting property ...... 61 projective space homotopy relative to A ...... 17 complex...... 77, 134 Hopffibration...... 77 quaternionic ...... 134 Hurewicz homomorphism ...... 149 real ...... 73, 77, 133 I Q incidence numbers ...... 144 induced topology of metric space ...... 8 quaternionic projective space ...... 134 induced topology of subspace ...... 8 quaternions ...... 99 interior...... 88 R L real projective space ...... 73, 77, 133 lift...... 30 relative singular homology...... 84 localdegree...... 102 retract...... 19 long exact homology sequence of a pair of retraction ...... 19 spaces...... 88 S long exact homotopy sequence of a Serre fibration...... 69 Seifert-van Kampen theorem ...... 47 loop...... 21 Serrefibration...... 62 simply-connected ...... 39 M singular n-boundary...... 82 Mayer-Vietoris sequence ...... 124 singular n-chain...... 80 Möbiusstrip ...... 76 singular n-cycle ...... 82 singular n-simplex...... 80 N singular homology ...... 82 skeleton ...... 131 natural...... 87 smallchain ...... 116 normal...... 40 small chain theorem ...... 116 normal subgroup generated by a set . . . . .41 split exact sequence ...... 78 O standard simplex...... 79 standard topology of Rn ...... 7 open (for the product topology) ...... 13 strong deformation retract ...... 19

163 Index subspace topology ...... 8 of product topology ...... 13 surface of genus g ...... 52 of quotient topology ...... 11 suspension...... 14 V T vectorﬁeld...... 96 totalspace ...... 62 W U weakﬁbration...... 62 universal property windingnumber...... 75

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