Algebraic Topology Iv Epiphany Term Lecture

Home , Algebraic topology, Cap product, Cohomology ring, Cup product

ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES

MARK POWELL

This course is about cohomology of spaces, products on cohomology, and manifolds. Cohomology is useful primarily because it can be made into a ring using the cup product. We will study the cohomology ring and apply it to manifolds, especially of dimension 2, 3 and 4. For manifolds homology and cohomology can be related to each other in two different ways, using Poincaréduality and using the universal coefficient theorem. The interplay between these two relations can be extremely powerful.

Contents 1. CW complexes and manifolds 2 1.1. CW complexes 2 1.2. Manifolds 2 1.3. Smooth manifolds 3 1.4. Manifolds with boundary 4 2. Homology groups of manifolds 4 2.1. Recap of singular and cellular homology 4 2.2. Top dimensional homology of a manifold 6 2.3. Fundamental classes and degrees of maps 7 3. Cohomology 8 3.1. Hom groups 8 3.2. Algebra of cochain complexes 9 3.3. Cohomology of a topological space 11 3.4. CW cohomology 12 3.5. Understanding cohomology 13 3.6. Properties of cohomology 15 4. Relationships between homology and cohomology 17 4.1. Universal coefficient theorem 17 4.2. Poincaréduality 18 4.3. An application of both 19 5. Universal coefficient theorem 19 5.1. Ext groups 19 5.2. Examples and properties 23 5.3. The universal coefficient theorem for cohomology 25 6. Univeral coefficients for homology and the Künneththeorem 27 6.1. Tensor products 27 6.2. Tor groups 28 1 ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 2

6.3. Universal coefficient theorem for homology 28 6.4. The Künneththeorem for cohomology 29 7. Cup products 30 7.1. Definition of cup products 30 7.2. An example 32 7.3. The cohomology ring 34 7.4. More examples 36 8. Cap products 40 8.1. Definition of cap products 40 8.2. An example 42 8.3. Properties of cap product 43 9. Applications of cup and cup products combined with Poincaréduality 44 9.1. Degree one maps 45 9.2. Computations of cup products 46 10. Proof of F2 coefficient Poincaréduality 49 References 52

1. CW complexes and manifolds 1.1. CW complexes. Here is a better definition of CW complexes. It is important that the decomposition into cells is remembered as part of the data of a CW complex. Merely defining it as a space that “was built in a certain way” loses too much data that is needed when working with these spaces. Definition 1.1 (Finite dimensional CW complex). (a) A CW complex X of dimension −1 is ∅. Its topological realisation |X| is also the empty set. (b) Inductively define higher CW complexes. A CW complex of dimension ≤ n, Xn, consists of (i) A CW complex Xn−1 of dimension ≤ n − 1, with topological realisation |Xn−1|. n−1 n−1 (ii) A set of maps {ϕi : S → |X |}i∈In , the attaching maps. The topological realisation |Xn| of X is G |Xn| := |Xn−1 t Dn/ ∼

i∈In n−1 n n−1 where for all i ∈ In, x ∈ S = ∂Di , we identify x ∼ ϕi(x) ∈ |X |. (c) A (ﬁnite dimensional) CW structure on a topological space Y is a CW complex X and a homeomorphism f : |X| → Y . n n n−1 n ϕi n (d) The map ϕi : D → |X | such that S → D −→|X | equals ϕi is called a characteristic map. 1.2. Manifolds. An n-dimensional topological manifold M is a topological space that is Hausdorﬀ, second countable (countable basis of open sets), and locally ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 3

n homeomorphic to R . That is, for all x ∈ M, there is an open set Ux containing x n n and a homeomorphism ϕx : Ux → V ⊂ R where V is an open subset of R .

Example 1.2. n n+1 (i) The sphere S = {x ∈ R | kxk = 1}. (ii) Products of spheres Sn × Sm, of dimension n + m. (iii) The n-torus T n = S1 × · · · × S1, the product of n copies of the circle S1. n n+1 (iv) Real projective space RP = R r{0}/ ∼, where x ∼ λx for all λ ∈ Rr{0}. n n+1 (v) Similarly complex projective space CP = C r{0}/ ∼, where z ∼ λz for all λ ∈ Cr{0}. (vi) The surface Σg of genus g is a 2-dimensional manifold. (vii) The Klein bottle K. n n (viii) Quotients of S or R by a free, proper, continuous action of a group are manifolds. (ix) Products of manifolds are manifolds. For this last item, we will provide several examples of manifolds obtained as orbits spaces of free actions. First, the definitions of these adjectives associated with group actions. (a) An action of a group G on a manifold X is free if g·x = x implies that g = e ∈ G for every x ∈ X. (b) An action of a group G on a manifold X is proper if for all x, y ∈ X, there exist open sets U 3 x and V 3 y such that {g ∈ G | gU ∩ V 6= ∅} is finite. (c) A group action is continuous if the map x 7→ g · x is a continuous map X → X for all g ∈ G. Now the promised examples of manifolds defined by quotients by group actions. Example 1.3. ∼ 1 (1) The group Z acts on R by n · x := x + n. The quotient is R/Z = S . 2 2 (2) The group Z acts on R by (n, m) · (x, y) = (x + n, y + m). The quotient 2 2 ∼ 2 R /Z = T . n n ∼ n (3) The group Z/2 acts on S by 1 · x = −x. The quotient S /(Z/2) = RP . 1 2n+1 2n+1 n+1 (4) The group S acts on S as follows. Consider S ⊂ C as the 2 2 2πiθ 2πiθ set of points with |z0| + · · · |zn| = 1. Then e · (z0, . . . , zn) := (e · 2πiθ 2n+1 1 ∼ n z0, . . . , e · zn). The quotient is S /S = CP . 3 (5) The group Z/p acts on S as follows. Let p and q be coprime positive 3 2 2 2 ∼ integers. Consider S ⊂ C with |z| + |w| = 1. Write Z/p = Cp where Cp is the cyclic group generated by ξ = e2πi/p ∈ S1. Then ξj · (z, w) := (ξ · z, ξq · w).

3 The quotient is S /Cp = L(p, q), the lens space. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 4

1.3. Smooth manifolds. All the manifolds we talk about will be smooth.

n Deﬁnition 1.4. Let U ⊆ R be open. A function m f = (f1, . . . , fn): U → R is smooth is all partial derivatives of fi exist for all i. Deﬁnition 1.5. (i) An n-dimensional chart for a manifold M at x ∈ M is a homeomorphism n Φ: U → V where U 3 x is an open neighbourhood of x, and V ⊂ R is open. (ii) An n-dimensional atlas for M is a collection of n-dimensional charts {Φi : Ui → S Vi}i∈I with i Ui = M, where I is some indexing set. (iii) An atlas {Φi : Ui → Vi}i∈I is smooth if the transition map

−1 Φj ◦ Φi :Φi(Ui ∩ Uj) → Φj(Ui ∩ Uj) n is smooth as a map between open subsets of R . Deﬁnition 1.6. A smooth manifold is a topological manifold M with a choice of smooth atlas.

When we say “manifold” we will usually implicitly mean smooth manifold.

1.4. Manifolds with boundary. An n-dimensional topological manifold with boundary M is a topological space that is Hausdorﬀ, second countable, and locally n homeomorphic to either R or to the half space n n H := {x ∈ R | x1 ≥ 0}. We write n ∂M := {x ∈ M with Ux = H }. n For example, D , or a surface Σg,1. We will mostly consider closed manifolds. By deﬁnition a closed manifold is a compact manifold with empty boundary ∂M = ∅.

2. Homology groups of manifolds We start with a recollection of homology groups, then we discuss homology of manifolds.

2.1. Recap of singular and cellular homology. Homology theories are covariant functors

Hi : {Top spaces} → {Abelian groups} for each i, satisfying some axioms: homotopy invariance, long exact sequence of a pair, excision, and taking a ﬁxed value on a point: H0(pt) = Z. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 5

2.1.1. Singular homology. Singular homology sends

X 7→ Hi(X; Z) or Hi(X) for short, deﬁned as follows. The singular chain complex n i o Ci(X) := Z singular simplices ∆ → X is the free abelian group generated by the singular simplices. Recall that ∆i is the i-simplex i i+1 X {(t0, . . . , ti) ∈ R | 0 ≤ ti ≤ 1, ti = 1}. i=0 The vertices tj = 1 are naturally ordered by We denote the jth face inclusion i−1 i respecting the ordering of vertices by ιj : ∆ → ∆ . Then the boundary map

∂i : Ci(X) → Ci−1(X) is deﬁned by i i X j+1 i−1 ∂i(σ : ∆ → X) = (−1) (σ ◦ ιj : ∆ → X). j=0 The cycles are Zi(X) := ker(∂i : Ci(X) → Ci−1(X)) and the boundaries are

Bi(X) := im(∂i+1 : Ci+1(X) → Ci(X)). The ith singular homology of X is

Zi(X) Hi(X) := . Bi(X) 2.1.2. Cellular homology. Sometimes CW homology is easier for computing homology. This is a functor from spaces to abelian groups CW X 7→ Hi (X; Z) CW or Hi (X) for short. Deﬁne the CW chains by CW n o Ci (X) := Z i − cells of X and CW CW CW ∂i : Ci (X) → Ci−1 (X) is deﬁned by i X i−1 ∂i(D , ϕj) = nk (D , ϕk)

k∈Ii−1 where nk is the degree of the composition ϕ Si −→j Xi−1/Xi−2 → Di−1/∂Di−1 =∼ Si−1 where the second map is given by collapsing onto the kth (i−1)-cell. Here the degree i−1 i−1 ∼ i−1 i−1 ∼ of a map f : S → S is d ∈ Z such that Z = Hi−1(S ) → Hi−1(S ) = Z ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 6 is given by multiplication by d. There is a sign issue here that is important to get right. We will discuss it again later. Again, we have cycles

CW CW CW Zi (X) := ker(∂i : Ci (X) → Ci−1 (X)) and boundaries

CW CW CW Bi (X) := im(∂i+1 : Ci+1 (X) → Ci (X)). The ith cellular homology of X is

CW CW Zi (X) Hi (X) := CW . Bi (X) Theorem 2.1. For every ﬁnite dimensional CW complex X, we have that ∼ CW Hi(X) = Hi (X). 2.2. Top dimensional homology of a manifold. We will discuss the following theorem.

Theorem 2.2. Let M be a connected nonempty n-dimensional manifold. Then ( ∼ Z M closed and orientable Hn(M; Z) = 0 otherwise.

Also, Hm(M; Z) = 0 for m > n.

Recall that we also know for connected manifolds that H0(M; Z) = Z. In order to make sense of this theorem, we need to discuss what it means for a manifold to be orientable. For example, S2 and T 2 are orientable surfaces, whereas 2 RP and K are non-orientable; two sided surfaces are orientable and one-sided surfaces are non-orientable. First, note that this theorem can be applied to show that a given closed manifold is orientable or nonorientable.

2n 2n−1 Corollary 2.3. For n ≥ 1, the 2n-manifold RP is nonorientable, whereas RP is orientable. m CW m ∼ Proof. There is a cell structure on RP with Ci (RP ; Z) = Z for 0 ≤ i ≤ m and CW i i ∂i = 1+(−1) . The top dimensional homology is therefore ker(1+(−1) : Z → Z) which vanishes for i even and is Z for i odd. Then Theorem 2.2 implies the result. To given an algebraic topology deﬁnition of orientability, we need the next lemma. ∼ Lemma 2.4. Let n ≥ 1 and let M be a closed n-manifold. Then Hn(M,Mr{pt}) = Z for every point in M. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 7

Proof. We give the proof for n ≥ 2. The proof for n = 1 is similar but slightly different since removing a point from a neighbourhood of that point changes the num- ∼ n n ber of connected components. By excision, Hn(M,Mr{pt}) = Hn(R , R r{pt}). Then we may compute using the long exact sequence of a pair: n n n n n Hn(R ) → Hn(R , R r{pt}) → Hn−1(R r{pt}) → Hn−1(R ). n n Here Hn(R ) = Hn−1(R ) = 0, so n n ∼ n ∼ n−1 ∼ Hn(R , R r{pt}) = Hn−1(R r{pt}) = Hn−1(S ) = Z, n n−1 using the homotopy equivalence R r{pt}' S . Deﬁnition 2.5. An orientation on a n-dimensional connected manifold M is an assignment of a generator Op ∈ Hn(M,Mr{p}; Z) for every p ∈ M, such that: for n n every p, q ∈ M and for every R ⊂ M with p, q ∈ R , there is an element Op,q ∈ n n Hn(M,MrR ; Z) such that for r = p, q, under the maps Hn(M,MrR ; Z) → n Hn(M,Mr{r}; Z) induced by the inclusion MrR ⊆ Mr{r}, we have Op,q 7→ Or.

Definition 2.6. An n-manifold is orientable if an orientation {Op}p∈M exists. n n n 2n+1 All of S , CP , T , L(p, q), Σg and RP are orientable. Sketch proof of Theorem 2.2. First, every n-manifold admits a simplicial structure: a family of injective maps ni {ϕi : ∆ → X}i∈I with (X, {ϕi}) a simplicial complex, where simplices intersect in faces and the union of the images of the ϕi is X. There are some other conditions, we will not go into all the details. We say that a simplex c has order r if there are precisely r simplices of dimension k + 1 with c as a face. We will use the following lemma. Lemma 2.7. For every simplicial structure on a closed n-manifolds M: (i) The simplicial structure has finitely many simplices if and only if there are finitely many n simplices. (ii) If c is an (n − 1) dimensional simplex of M, then the order of c is two. Since this is just a sketch, we will not prove the lemma in detail. Now, ∼ CW ∼ CW CW Hn(M) = Hn (M) = ker(Cn (M) → Cn−1 (M)), thinking of the simplicial structure as a cellular structure. A generator of the cycles CW in Cn (M) is a sum of n-dimensional simplices with appropriate orientations. Start with one n-simplex σ, and choose an (n − 1)-dimensional face of it τ. The order of this face is two, so there is exactly one other n-simplex σ0 that also has τ as a face. Orient (choose the sign ± of) σ0 in an n-chain so that in ∂CW (σ + σ0), τ is cancelled. Since M is compact, there are finitely many n-simplices, so we can continue by induction. The appropriate choice of signs exists to produce a cycle if and only if M is orientable. This procedure in fact creates a generator of ∼ Hn(M) = Z, for M closed and orientable, but we will not show this precisely. If M has nonempty boundary, then the (n − 1)-simplices of the boundary have order one, so the induction stops there without creating an n-cycle. As mentioned ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 8 above, if M were nonorientable, then there is no appropriate choice of signs to form a cycle. So in these two cases, Hn(m) = 0. This completes our sketch proof of Theorem 2.2. 2.3. Fundamental classes and degrees of maps. Definition 2.8. If a connected nonempty n-manifold M is closed and orientable, then a choice of generator [M] ∈ Hn(M; Z) is called a fundamental class. There are two such choices corresponding to two possible orientations: an orientation determines a fundamental class and vice versa. Definition 2.9. Let f : M → N be a map between closed, oriented, connected n-manifolds with fundamental classes [M] and [N]. The degree of f is the integer d such that f∗([M]) = d[N] ∈ Hn(N; Z).

3. Cohomology 3.1. Hom groups. We have to introduce groups of homomorphisms. Let A be a group, and let G be an abelian group. Deﬁne an abelian group Hom(A, G) := ({group homomorphisms A → G}, +, 0). The group structure is deﬁned by (f + g)(a) = f(a) + g(a) ∈ G. The identity is 0(a) = 0 ∈ G for all a ∈ A. Let f : A → B be a group homomorphism. There is a homomorphism induced by f: f ∗ : Hom(B,G) → Hom(A, G) (ϕ: B → G) 7→ (ϕ ◦ f : A → B → G). Let g : G → H be a homomorphism of abelian groups. Then g induces a homomorphism g∗ : Hom(A, G) → Hom(A, H) (ϕ: A → G) 7→ (g ◦ ϕ: A → G → H). Example 3.1. ∼ (1) Hom(Z,G) = G for all G, with the map ϕ 7→ ϕ(1) giving an isomorphism. ∼ (2) Hom(Z/n, G) = ker(·n: G → G). ∼ (3) Hom(Z/n, Z/m) = Z/ gcd(n, m).

Lemma 3.2. Let {Ai}i∈I be a sequence of abelian groups and let G be an abelian group. Then M ∼ Y Hom( Ai,G) = Hom(Ai,G) i∈I i∈I Q via the map f 7→ i∈I f|Ai . Note the change between direct sum and direct product. What is the difference? The elements of both are tuples (ai)i∈I , and the addition and the identity are the same, but the sets are different: in the direct product any tuples are allowed but in the direct sum only {(ai)i∈I | finitely many ai 6= eAi }. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 9

Lemma 3.3. Let 0 → A → B → C → 0 be a short exact sequence of abelian groups and let G be an abelian group. Suppose that C is free abelian. Then

0 → Hom(C,G) → Hom(B,G) → Hom(A, G) → 0 is also short exact.

The proof is an exercise on the problem sheet.

n ∼ n Lemma 3.4. Hom(Z , Z) = Z . L Q Proof. For I ﬁnite, I and I coincide. Therefore n n n n n ∼ M ∼ Y ∼ Y ∼ M n Hom(Z , Z) = Hom( Z, Z) = Hom(Z, Z) = Z = Z = Z . i=1 i=1 i=1 i=1

3.2. Algebra of cochain complexes. Later, we will apply the algebra of cochain complexes in this section to cochain complexes arising from spaces, in order to deﬁne and work with cohomology.

Deﬁnition 3.5. A cochain complex is a sequence of abelian groups

0 → D0 −→δ0 D1 −→δ1 D2 −→·δ2 · · → Dn −→·δn · · with δi+1 ◦ δi = 0 for all i. We call the δi the coboundary maps. The cohomology groups of a cochain complex are

n n+1 n ∗ ker(δn : D → D ) H (D ) := n−1 n . im(δn−1 : D → D )

n n+1 n−1 n Here ker(δn : D → D ) are the cocycles and im(δn−1 : D → D ) are the coboundaries.

∗ ∗ i Deﬁnition 3.6. A cochain map f : C → D consists of a homomorphism fi : C → Di for each i such that

fi Ci / Di

C D δi δi fi+1 Ci+1 / Di+1 commutes for all i, meaning that both routes round the square give the same outcome for every element of Ci.

Deﬁnition 3.7. A cochain homotopy between cochain maps f, g : Ci → Di consists i i−1 of homomorphisms hi : C → D for all i with fi − gi = hi+1 ◦ δi + δi−1 ◦ hi for ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 10 all i.

δi−1 δi δi+1 ··· / Ci−1 / Ci / Ci+1 / ···

hi hi+1 fi−1−gi−1 fi−gi fi+1−gi+1

} δi−1 } δi δi+1 ··· / Di−1 / Di / Di+1 / ··· Lemma 3.8. (i) Let f : C∗ → D∗ be a cochain map. Then n n f∗ : H (C) → H (D) [c] 7→ [fn(c)] is a well-deﬁned map. ∗ ∗ (ii) If f ∼ g : C → D are two cochain homotopic cochain maps then f∗ = n ∗ n ∗ g∗ : H (C ) → H (D ) for all n ∈ N0. The proof is essentially the same as the analogous fact for homology. Now we have a long exact sequence in cohomology from a short exact sequence of cochain complexes 0 → C∗ → B∗ → A∗ → 0. That is, we have a commuting diagram

f n gn 0 / Cn / Bn / An / 0

δn δn δn f n+1 gn+1 0 / Cn+1 / Bn+1 / An+1 / 0

We define a connecting homomorphism, that is a homomorphism d: Hn(A∗) → Hn+1(C∗). n n n Let a ∈ A be a cocycle, that is δn(a) = 0. Then a = g (b) for some b ∈ B . Note n+1 n n+1 that g (δn(b)) = δn(g (b)) = δn(a) = 0. Therefore there exists c ∈ C such n+1 that f (c) = δn(b). Define d([a]) = [c] ∈ Hn+1(C∗). It is a routine but slightly long check that this is well-defined, omitted here. This connecting homomorphism enables us to fit the cohomology groups together into a long exact sequence f ∗ g∗ g∗ · · · → Hn(C) −→ Hn(B) −→ Hn(A) −→d Hn+1(C) −→ Hn+1(B) → · · · We will apply this to obtain certain long exact sequences, in particular the long exact sequence of a pair and the Mayer-Vietoris sequence. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 11

Cochain complexes will, for us, most often arise from taking Hom of a chain complex. Let (C∗, ∂∗) be a chain complex, with ∂n : Cn → Cn−1. Let G be an abelian group. For n ∈ N0, deﬁne ∗ δn := ∂n+1 : Hom(Cn,G) → Hom(Cn+1,G).

We obtain a cochain complex Hom(C∗,G) given by

δ0 δ1 δ2 Hom(C0,G) −→ Hom(C1,G) −→ Hom(C2,G) −→· · ·

δn δn+1 → Hom(Cn,G) −→ Hom(Cn+1,G) −−−→· · ·

This is the cochain complex dual to (C∗, ∂∗). The cohomology of C∗ with coeﬃcients in G is n H (C; G) := ker(δn)/ im(δn−1). Lemma 3.9.

(i) Let f∗ : C∗ → D∗. Then ∗ f : Hom(Dn,G) → Hom(Cn,G) (ϕ: Dn → G) 7→ (ϕ ◦ f : Cn → G) ∗ n n is a cochain map, and therefore induces a map f : H (D∗) → H (C∗). (ii) Two chain homotopic chains map f, g : C∗ → D∗ induce cochain homotopic cochain maps ∗ ∗ f , g : Hom(D∗,G) → Hom(C∗,G), so that f ∗ = g∗ : Hn(D) → Hn(C).

Proof. For the proof of (i), just note that fn−1 ◦ ∂n = ∂n ◦ fn implies that ∗ ∗ ∗ ∗ ∗ ∗ δn−1 ◦ fn−1 = ∂n ◦ fn−1 = fn ◦ ∂n = fn ◦ δn−1. Then apply Lemma 3.8 (i) to see that cochain maps induce maps on cohomology. To prove (ii), start with ∂ ◦ h + h ◦ ∂ = f − g, and take the dual, to obtain h∗ ◦ δ + δ ◦ h∗ = f ∗ − g∗. Thus h∗ gives a cochain homotopy, so by Lemma 3.8 (ii), f ∗ and g∗ induce the same map on cohomology. 3.3. Cohomology of a topological space. Now we define the singular cohomology of a topological space. Let (X,A) be a pair of spaces. Usually we will have A = ∅, and then we write X = (X, ∅). Define the singular cochain group to be n C (X,A; G) := Hom(Cn(X,A),G) and ∗ δn := ∂n+1 : Hom(Cn(X,A),G) → Hom(Cn+1(X,A),G). We call n n+1 ker(δn : C (X,A; G) → C (X,A; G)) ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 12 the singular cocycles and n−1 n im(δn−1 : C (X,A; G) → C (X,A; G)) the singular coboundaries. The singular cochain complex is C0(X,A; G) −→δ0 C1(X,A; G) −→δ1 C2(X,A; G) −→·δ2 · · δ → Cn−1(X,A; G) −−−→n−1 Cn(X,A; G) −→δn Cn+1(X,A; G) → · · · We then define the singular cohomology n n ∗ H (X,A; G) := H (C (X,A; G)) = ker δn/ im δn−1. n When G = Z, we will often omit the coefficients and simply write H (X,A). 3.4. CW cohomology. Let X be a CW complex. Recall that there is the cellular chain complex CW CW (C∗ (X), ∂∗ ). Let G be an abelian group. Define the CW cochain groups to be i CW CCW (X; G) := Hom(Ci (X),G). The CW coboundary maps are given by CW CW ∗ CW CW δi := (∂i+1 ) : Hom(Ci (X),G) → Hom(Ci+1 (X),G). Then the CW cohomology is n CW CW HCW (X; G) := ker(δn )/ im(δn−1). Theorem 3.10. For every CW complex X, for every abelian group G, and for n ∼ n every n ∈ N0, we have H (X; G) = HCW (X; G). Example 3.11. Let us consider the cohomology of the sphere Sn. It has a CW structure with two cells: a 0-cell and an n-cell. The resulting CW chain complex is CW n C∗ (S ) = · · · → Z → 0 → · · · → 0 → Z ∼ with nonzero chain groups in degrees 0 and n. Since Hom(Z, Z) = Z and Hom(0, Z) = 0, the cochain complex is CW n C∗ (S ) = Z → 0 → · · · → 0 → Z → · · · where now the Z on the left is in degree 0 and the Z on the right is in degree n. Therefore the CW cohomology of Sn is ( i n ∼ Z i = 0, n HCW (S ; Z) = 0 otherwise. Example 3.12. Now we use general G coefficients, with G an abelian group. Let Σg be the surface of genus g. There is a CW structure with one 0-cell, 2g 1-cells, and one 2-cell. The associated CW chain complex is 0 M 0 Z −→ Z −→ Z. 2g ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 13 ∼ Since Hom(Z,G) = G, this gives rise to a cochain complex that also takes the form 0 M 0 G −→ G −→ G. 2g Then the cellular cohomology is  G i = 0, 2  i ∼ L2g HCW (Σg; G) = G i = 1 0 otherwise. 3.5. Understanding cohomology. Recall that n ∼ ∼ n C (X; G) = Hom(Cn(X),G) = {maps {∆ → X} to G} In particular, 0 ∼ ∼ 0 ∼ C (X; G) = Hom(C0(X),G) = Map({∆ → X},G) = {functions X → G}. Now, 1 ∼ ∼ 1 C (X; G) = Hom(C1(X),G) = Map({∆ → X},G) and the coboundary map is given by C0(X; G) → C1(X; G) (ϕ: X → G) 7→ (δϕ: {∆1 → X} → G) where δϕ(ψ) := ϕ(ψ(1)) = ϕ(ψ(0)). Therefore H0(X; G) = ker δ = {ϕ: X → G | ϕ(x) = ϕ(y) whenever there is ψ : ∆1 → X with ψ(0) = x, ψ(1) = y}

= {ϕ: {connected components of X} = π0(X) → G} Y = G

|π0(X)| An analogous computation holds for the zeroth CW cohomology of CW complexes. So zeroth cohomology is not too hard to understand: functions on X that are constant on path components. For further discussion of an intuitive idea behind cohomology, I recommend reading the beginning of Chapter 3 of Hatcher. As an alternative, we give an explicit generator for the ﬁrst cohomology of the circle S1.

Example 3.13. Consider the real line R. Consider the functions

αR : R → R x 7→ x 0 in C (R; R) (here the ﬁrst R is a topological space and the second R is an abelian group) and

αZ : R → Z x 7→ bxc ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 14

1 1 2 Let µ: ∆ → R be a singular 1-simplex. Recall that ∆ = {(1 − t, t) ∈ R | t ∈ [0, 1]}. Denote a point in ∆1 by the value of t. Then

δαR(µ) = αR(∂µ) = αR(µ(1)) − αR(µ(0)) = µ(1) − µ(0). Also ( |{n ∈ Z | µ(0) < n ≤ µ(1)}| if µ(0) ≤ µ(1) δαZ(µ) = αZ(∂µ) = −|{n ∈ Z | µ(1) < n ≤ µ(0)}| if µ(1) ≤ µ(0). This counts how many integers µ crosses from left to right, with sign, so that crossing an integer from right to left counts as −1.

1 1 Example 3.14. Now consider the circle S . Let p: R → S be the function 2πit 1 1 1 sending t → e ∈ S ⊂ C. Given a 1-simplex σ : ∆ → S , it lifts to a 1-simplex in C1(R; R) 1 σe : ∆ → R 1 1 such that p ◦ σe = σ : ∆ → S . Such a lift is not unique; any two lifts diﬀer by an 0 integer σe(x) − σe (x) ∈ Z. 1 1 Deﬁne an element of C (S ; R) by

θR(σ) := δαR(σe) = σe(1) − σe(0). 1 1 Define an element of C (S ; Z) by ( |{n ∈ Z | σe(0) < n ≤ σe(1)}| if σe(0) ≤ σe(1) θZ(σ) := δαZ(σe) = −|{n ∈ Z | σe(1) < n ≤ σe(0)}| if σe(1) ≤ σe(0). You should check that this is well-defined i.e. does not depend on the choice of lift 1 1 1 1 σe. We have that θR ∈ C (S ; R) and θZ ∈ C (S ; Z) Now, consider the 1-simplex µ: ∆1 → S1 defined by µ(t) = e2πit. Lift it, to obtain for example µe(t) = t. Then µe(1) − µg(0) = 1, so

θR(µ) = 1 and

θZ(µ) = 1.

In the next lemma, we will show that θR and θZ represent nontrivial cohomology 1 1 1 1 classes in H (S ; R) and H (S ; Z) respectively. Since they evaluate to 1 on µ = [S1], this will imply that they generate these cohomology groups. Lemma 3.15.

(1) The singular 1-cochains θR and θZ are cocycles, so represent elements θR ∈ 1 1 1 1 H (S ; R) and θZ ∈ H (S ; Z). 0 1 (2) [θR] and [θZ] are not coboundaries δc for any c ∈ C (S ; R), respectively 0 1 C (S ; Z). Proof. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 15

2 1 (1) We will show that θR is a cocycle, that is δ1θR = 0 ∈ C (S ; R). A similar 2 1 argument works for θZ. Let σ : ∆ → S . We want that (δ1θR)(σ) = 1 2πit 1 θR(∂σ) = 0. Recall the function p: R → S sending t → e ∈ S ⊂ C. 2 1 2 Then σ lifts to σe : ∆ → R and, with ιj : ∆ → ∆ the jth face inclusion, we have 2 X j θR(∂σ) = θR (−1) σ ◦ ιj j=0 2 X j = (δαR) (−1) σe ◦ ιj = αR(∂ ◦ ∂σe) = αR(0) = 0. j=0 0 1 (2) To show that θR and θZ are not coboundaries, let β : C (S ; R), and simi- 0 1 1 larly β ∈ C (S ; Z). We can think of β as a function β : S → R, respec- 1 1 1 2πit tively β : S → Z. Consider the 1-chain µ: ∆ → S sending t 7→ e . Then

δ0(β)(µ) = β(∂µ) = β(µ(1) − µ(0)) = β(1) − β(1) = 0.

But θR(µ) = 1, and also θZ(µ) = 1, so these cannot be the coboundaries of any 0-chain. 1 1 ∼ The cohomology class θZ ∈ H (S ; Z) = Z is a generator. It evaluates to 1 on the fundamental class. For this reason we refer to it as the dual fundamental class 1 ∗ 1 1 and write θZ = [S ] ∈ H (S ; Z). Deﬁnition 3.16. Let M be a compact, oriented n-dimensional manifold. Call the canonical generator ∗ n [M] ∈ H (M, ∂M; Z) the dual fundamental class of M, with the property that h[M]∗, [M]i = 1. 3.6. Properties of cohomology. The following basic properties of cohomology are closely related to the homology versions. Let G be an abelian group. (1) Let f :(X,A) → (Y,B) be maps of pairs of spaces. Then for all n ∈ N0, f∗ : Cn(X,A) → Cn(Y,B) is a chain map, which induces a map on cohomology f ∗ : Hn(Y,B; G) → Hn(X,A; G). (2) The assigment (X,A) → Hn(X,A; G) deﬁnes a contravariant functor. (3) Let f, g :(X,A) → (Y,B) be homotopic maps of pairs. They induce chain homotopic chain maps, and therefore cochain homotopic cochain maps, whence equal maps on cohomology: f ∗ = g∗ : Hn(Y,B; G) → Hn(X,A; G). (4) Let f : X → Y be a homotopy equivalence. Then it follows from the previ- ous item that f ∗ : Hn(Y,B; G) → Hn(X,A; G) is an isomorphism of groups. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 16

(5) Let (X,A) be a pair of spaces. Then there is a long exact sequence: 0 → H0(X,A; G) → H0(X; G) → H0(A; G) ∗ −→δ H1(X,A) → H1(X; G) → H1(A; G) → · · · ∗ q∗ ∗ −→δ Hn(X,A) −→ Hn(X; G) −→i Hn(A; G) ∗ −→δ Hn+1(X,A) → Hn+1(X; G) → Hn+1(A; G) → · · ·

with i: A → X the inclusion map, q∗ : C∗(X) → C∗(X)/C∗(A) = C∗(X,A) the quotient map, and δ∗ the connecting homomorphism. We also have excision for cohomology. Theorem 3.17 (Excision for cohomology). Let X be a space, and let G be an abelian group. Let Z ⊂ A ⊂ X be subsets such that the closure of Z is contained in the interior of A. Then the map of pairs i:(XrZ,ArZ) → (X,A) induces an isomorphism ∗ n =∼ n i : H (X,A; G) −→ H (XrZ,ArZ; G) for every n ∈ N0. For example, let M be a manifold and let p ∈ M be an interior point. Let U be n an open subset containing p that is homeomorphic to R . Then with G coeﬃcients n ∼ n ∼ n H (M,Mr{p}) = H (Mr(MrU), (Mr{p})r(MrU)) = H (U, Ur{p}) ∼ n n n ∼ = H (R , R r{0}) = G. So as with homology, the local homology of a manifold does not depend on the point p, and is isomorphic to G in degree equal to the dimension of the manifold. We also have a Mayer-Vietoris sequence for cohomology. It is similar to the Mayer-Vietoris sequence for homology, but the arrows are reversed. We will deﬁne excisive triads after stating the theorem. Theorem 3.18 (Mayer-Vietoris sequence). Let (X, A, B) be an excisive triad. Let G be an abelian group, and suppose that X = A ∪ B. Then there is a natural long exact sequence ∗ ∗ ∗ ∗ · · · → Hn(X; G) −−−→i ⊕i Hn(A; G) ⊕ Hn(B; G) −−−−→i ⊕−i Hn(A ∩ B; G) −→d Hn+1(X; G) → · · · where d is the connecting homomorphism.

What is an excisive triad? Let (X,A1,A2) be a triad, i.e. A1,A2 ⊆ X and A1 ∪ A2 = X. Deﬁne m {A1,A2} X Cn (X) := ajσj ∈ Cn(X) | for each j, there is i ∈ {1, 2} with im(σj) ⊆ Ai . j=1 ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 17

Deﬁnition 3.19. The triad (X,A1,A2) is excisive if inclusion induces a chain homotopy equivalence {A1,A2} C∗ (X) → C∗(X).

Proposition 3.20. Let (X,A1,A2) be a triad. Suppose one of the following holds. (i) A1 ⊆ A2. (ii) A2 ⊆ A1. (iii) A1 = ∅. (iv) A2 = ∅. (v) A1 = A2. (vi) A1 and A2 are open in X. (vii) X is a CW complex and A1 and A2 are subcomplexes. (viii) X is a manifold and A1 and A2 are closed submanifolds, with A1 ∩ A2 a boundary component of A1 and A2. Then (X,A1,A2) is excisive. The idea is roughly that one can deﬁne a homotopy inverse to the chain map {A1,A2} C∗ (X) → C∗(X) by subdividing simplices in X so that the image of every simplex lies in A1, A2, or both. Now for excisive triads we can prove the Mayer- Vietoris theorem. Proof of Theorem 3.18. Suppose that (X, A, B) is an excisive triad with X = A∪B. Then i∗⊕−i∗ {A,B} 0 → C∗(A ∩ B) −−−−→ C∗(A) ⊕ C∗(B) → C∗ (X) → 0 is a short exact sequence. Therefore {A,B} 0 → Hom(C∗ (X),G) → Hom(C∗(A),G)⊕Hom(C∗(B),G) → Hom(C∗(X),G) → 0 {A,B} is exact, since C∗ (X) is free abelian. Now this short exact sequence of cochain complexes gives rise to a long exact sequence in cohomology. Note that n {A,B} ∼ n ∼ n H (Hom(C∗ (X),G)) = H (Hom(C∗(X),G)) = H (X; G) {A,B} since C∗ (X) and C∗(X) are chain homotopy equivalent. Therefore we obtain the long exact sequence ∗ ∗ ∗ ∗ · · · → Hn(X; G) −−−→i ⊕i Hn(A; G) ⊕ Hn(B; G) −−−−→i ⊕−i Hn(A ∩ B; G) −→d Hn+1(X; G) → · · · required.

4. Relationships between homology and cohomology There are two relationships between homology and cohomology. We state them here, to give you the idea, and then we explain them. We will study the first one, the universal coefficient theorem (UCT), in more detail shortly. The UCT is for any topological space. The second relationship, Poincaréduality, is only for manifolds. We will return to studying this in more detail later, because we need ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 18 the cap product to do so properly. But for context, and to have the two main relationships between homology and cohomology in one place, we state both now.

4.1. Universal coefficient theorem. Roughly, in the case G = Z the Ext group 1 Ext (Hn−1(X); Z) appearing in the next theorem is the torsion in Hn−1(X). For an abelian group P , the torsion subgroup TP is the subgroup of elements p ∈ P such that np = 0 for some n ∈ Z. We will study Ext groups in detail soon, so this rough description is to give you enough of an idea to parse the next theorem. Theorem 4.1 (Universal coefficient theorem). Let X be a topological space and let G be an abelian group. There is a split, natural, short exact sequence for each n ∈ N0 1 n 0 → Ext (Hn−1(X),G) → H (X; G) → Hom(Hn(X),G) → 0. n In the statement, split means that there is a map Hom(Hn(X),G) → H (X; G) which when post-composed with the map back to Hom(Hn(X),G) yields the identity on Hom(Hn(X),G). This implies that n ∼ 1 H (X; G) = Hom(Hn(X),G) ⊕ Ext (Hn−1(X),G). The naturality means that if there is a map X → Y , there is an associated map of short exact sequences induced by that map. But this map of short exact sequences does not respect the splitting, in general. Naturality will be rather unimportant in this course. As mentioned above, the idea for Z coefficients is that Hom(Hn(X), Z) sees the 1 “free part” of Hn(X), that is Hn(X)/T Hn(X), whereas Ext (Hn−1(X), Z) sees the torsion subgroup THn−1(X). 6 6 For example in Z ⊕ Z/7 ⊕ Z/3 ⊕ Z/9, Z might be called the “free part” (but beware this is not canonically defined) and the torsion subgroup is Z/7⊕Z/3⊕Z/9. This is a canonically defined subgroup. 4.2. Poincaréduality. Here is the second important relationship between homology and cohomology. Theorem 4.2 (Poincaréduality). Let M be a compact, orientable, n-dimensional manifold, possibly with ∂M 6= ∅. There are isomorphisms n−r ∼ H (M, ∂M; Z) = Hr(M; Z) and n−r ∼ H (M; Z) = Hr(M, ∂M; Z) for 0 ≤ r ≤ n. n−r ∼ If ∂M = ∅, so that M is closed, then we have H (M; Z) = Hr(M; Z). This also holds for other coefficient groups. In particular, using Z/2 coefficients, we can drop the assumption that M be orientable.

Theorem 4.3 (Z/2 Poincaréduality). Let M be a compact, n-dimensional manifold, possibly with ∂M 6= ∅. There are isomorphisms n−r ∼ H (M, ∂M; Z/2) = Hr(M; Z/2) ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 19 and n−r ∼ H (M; Z/2) = Hr(M, ∂M; Z/2) for 0 ≤ r ≤ n. 4.3. An application of both. Let M be a closed, connected, oriented 4-dimensional manifold with π1(M) = {1}, so that H1(M; Z) = 0. We try to describe the homology and cohomology of M. Since M is connected and is a closed, oriented 4-manifold, we know that ∼ 0 ∼ ∼ 4 H0(M; Z) = H (M; Z) = H4(M; Z) = H (M; Z). By Poincaréduality, we have 3 ∼ H (M; Z) = H1(M; Z) = 0. We have 1 ∼ 1 1 H (M; Z) = Hom(H1(M; Z), Z) ⊕ Ext (H0(M; Z), Z) = Hom(0, Z) ⊕ Ext (Z,Z). 1 1 Since Z is torsion-free, Ext (Z, Z) = 0, so H (M; Z) = 0. This implies by Poincaré duality that ∼ 1 H3(M; Z) = H (M; Z) = 0. Finally, ∼ 2 ∼ 1 H2(M; Z) = H (M; Z) = Hom(H2(M; Z), Z) ⊕ Ext (H1(M; Z), Z) ∼ 1 ∼ = Hom(H2(M; Z), Z) ⊕ Ext (0, Z) = Hom(H2(M; Z), Z). Since Hom(P, Z) is torsion free for every finitely generated abelian group P , this means that ∼ 2 ∼ n H2(M; Z) = H (M; Z) = Z for some n. So just the knowledge that a 4-manifold has H1(M; Z) = 0 enabled us to deduce quite a lot.

5. Universal coefficient theorem As promised, we proceed to explain and prove the universal coeﬃcient theorem.

5.1. Ext groups. Let F : Ab → Ab be a contravariant functor between abelian groups. Deﬁnition 5.1. (i) We say that the functor F is left exact if whenever A → B → C → 0 is exact, then 0 → F (C) → F (B) → F (A) is exact. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 20

(ii) We say that F is right exact if whenever 0 → A → B → C is exact, then F (C) → F (B) → F (A) → 0 is exact. Lemma 5.2. Let G be an abelian group. Then Hom(−,G) is left exact. This was already part of what was asked in Lemma 3.3, but with the extra assumption that C is free abelian. In the case that C is free abelian, the sequence 0 → A → B → C → 0 splits, so B =∼ A ⊕ C, and then Hom(B,G) =∼ Hom(A, G) ⊕ Hom(C,G). A bit more work shows this version. Let us now prove the more general Lemma 5.2, without restrictions on C. Proof. Let f g A −→ B −→ C → 0 be exact. Then consider the sequence of maps g∗ f ∗ 0 → Hom(C,G) −→ Hom(B,G) −→ Hom(A, G). We want to show that this is an exact sequence. Let ϕ ∈ ker g∗. That is, ϕ◦g : B → G is the zero map. Let c ∈ C. Then there is a b ∈ B such that g(b) = c. Then ϕ(c) = ϕ ◦ g(b) = 0, so ϕ = 0. Thus g∗ is injective. To show exactness at Hom(B,G), suppose that θ = g∗(ϕ) ∈ Hom(B,G). Then f ∗ ◦ g∗(ϕ)(a) = ϕ ◦ g ◦ f(a) = ϕ(0) = 0 for all a ∈ A. Therefore f ∗ ◦ g∗ = 0. Now let θ ∈ ker f ∗ ⊆ Hom(B,G). That is, θ ◦ f : A → G is the zero map. We want to show that θ = g∗(ϕ) for some ϕ: C → G. Define ϕ(c), for c ∈ C, by taking b ∈ B with g(b) = c and defining ϕ(c) = θ(b). Suppose that b0 ∈ B with g(b0) = c too. Then θ(b) − θ(b0) = θ(b − b0). Since b − b0 ∈ ker g, we have that b − b0 = f(a) for some a ∈ A. Then θ(b − b0) = θ(f(a)) = 0, so θ(b) = θ(b0). Therefore ϕ(c) is well-defined, and does not depend on the choice of b. This shows that the sequence is exact at Hom(B,G). Example 5.3. Consider the short exact sequence ·p 0 → Z −→ Z → Z/p → 0 for p > 1 an integer. Then taking Hom(−, Z) yields ·p 0 → Hom(Z/p, Z) → Hom(Z, Z) −→ Hom(Z, Z) → 0 ∼ Now Hom(Z/p, Z) = 0 and Hom(Z, Z) = Z, so this yields ·p 0 → 0 → Z −→→ Z → 0. This is not exact, since multiplication by p is not onto. So we see that Hom(−Z) is not right exact. To measure the failure of Hom(−,G) to be right exact, we define the Ext groups. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 21

Deﬁnition 5.4. Now we deﬁne the Ext groups of a pair of abelian groups H,G. A sequence of groups

f2 f1 f0 · · · → F2 −→ F1 −→ F0 −→ H → 0 with Fi a free abelian group, and ker fi = im fi+1 for all i, is called a free resolution of H. Then n n Ext (H,G) := H (F∗; G). That is replace H by 0, apply Hom(−,G) to the resulting chain complex, and then n take cohomology, to get H (Hom(F∗,G)). We shall be principally concerned with Ext1 and Ext0, since for abelian groups these are the only ones that are nonzero, as we will soon see.

Let H be a finitely generated abelian group. Then by the classification of finitely generated abelian groups, k ∼ n M ni H = Z ⊕ Z/pi i=1 for some n, for some k, for some primes p1, . . . , pk, and for some integers n1, . . . , nk. There is a resolution of length one

k n k 0,Lk p i M i=1 i n M 0 → Z −−−−−−−−→ Z ⊕ Z → H → 0. i=1 i=1 The next lemma is a baby version of what is often called the fundamental lemma of homological algebra. It will be sufficient to prove that the groups Ext1(H,G) are well-defined for finitely generated abelian groups H. Lemma 5.5. Given two resolutions

0 → F1 → F0 → H → 0 and 0 0 0 → F1 → F0 → H → 0 of H, and a homomorphism ϕ: H → H, we have the following. (1) There is a chain map between the resolutions inducing ϕ. (2) Any two such chain maps are chain homotopic. Proof. We need to construct a map as shown in the next diagram.

f1 f0 F1 / F0 / H / 0

ϕ0 ϕ f 0 f 0 0 1 0 0 F1 / F0 / H / 0

Let x ∈ F0 be an element of a generating set. Note that ϕ(f0(x)) ∈ H. Choose 0 0 0 0 0 x ∈ F0 with f0(x ) = ϕ(f0(x)) ∈ H. Deﬁne ϕ0(x) = x . Do this for all the free ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 22

0 generators of F0, and then extend by linearity. This deﬁnes ϕ0 : F0 → F0. Now we want to deﬁne the map ϕ1 as shown in the next diagram.

f1 f0 F1 / F0 / H / 0

ϕ1 ϕ0 ϕ f 0 f 0 0 1 0 0 F1 / F0 / H / 0

Let y ∈ F1. Then f0 ◦ f1(y) = 0, so ϕ ◦ f0 ◦ f1(y) = 0. By commutativity, 0 0 0 0 0 f0 ◦ ϕ0 ◦ f1(y) = 0. By exactness, there exists y ∈ F1 with f1(y ) = ϕ0 ◦ f1(y). Deﬁne 0 ϕ1(y) = y . Do this for a generating set of F1, and extend by linearity. This deﬁnes ϕ1 : F1 → 0 0 F1. The map ϕ∗ : F∗ → F∗ is a chain map by construction. Now we want to show that any two such maps are chain homotopic.

f1 f0 F1 / F0 / H / 0

g 0 0 0 ϕ1 ϕ1 ϕ0 ϕ0 ϕ

f 0 f 0 0 1 0 0 F1 / F0 / H / 0

We want to define the chain homotopy g shown. Let x ∈ F0. Then 0 0 f0(ϕ0(x) − ϕ0(x)) = ϕ(f0(x)) − ϕ(f0(x)) = 0. 0 0 0 0 0 Therefore there exists a y ∈ F1 with f1(y ) = ϕ0(x) − ϕ0(x). Define g(x) = y0. 0 Do this for a generating set of F0, and extend by linearity. This defines g : F0 → F1. 0 0 It has the property that f1 ◦ g = ϕ0 − ϕ0, so satisfies the requirement for a chain homotopy at F0. Now, let y ∈ F1. We want to show that 0 ϕ1(y) − ϕ1(y) = g ◦ f1(y). 0 Since f1 is injective, it is enough to show that 0 0 0 f1(ϕ1(y) − ϕ1(y)) = f1 ◦ g ◦ f1(y). Now we compute: 0 0 0 0 0 0 f1(ϕ1(y) − ϕ1(y)) = f1 ◦ ϕ1(y) − f1 ◦ ϕ1(y) = ϕ0 ◦ f1(y) − ϕ0 ◦ f1(y) 0 0 = (ϕ0 − ϕ0)(f1(y)) = f1 ◦ g ◦ f1(y), as required. Therefore g is a chain homotopy. 0 0 Now let F∗,F∗ be two resolutions of H. With ϕ = Id, let ϕ∗ : F → F be a 0 chain map inducing ϕ, and let ψ∗ : F∗ → F∗ be a chain map inducing ψ = Id. Then ϕ∗ ◦ψ∗, Id and ψ∗ ◦ϕ∗, Id are two chain maps inducing Id on H. Then for both pairs, the two chain maps are chain homotopic. Therefore ϕ∗ ◦ ψ∗ ∼ Id and ψ∗ ◦ ϕ∗ ∼ Id, 0 so F∗ → H → 0 and F∗ → H → 0 are chain homotopy equivalent. It follows, since ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 23 applying Hom(−,G) to a chain equivalence yields a chain equivalence, that the Ext groups Ext1(H,G), Ext0(H,G) are well-defined, for every abelian group G. That is, for any two chain resolutions of H we obtain the same group Exti(H,G), i = 0, 1, up to canonical isomorphism. 5.2. Examples and properties. Now that we know the Ext groups Ext0 and Ext1 are well-defined, let us see some examples. Example 5.6. n (1) We compute Ext (Z, Z). There is a free resolution 0 → F1 = 0 → F0 = Z → Z → 0. So F∗ = 0 → Z → 0, supported in degree zero. Taking Hom(−, Z) yields 0 → Hom(Z, Z) → 0. So 0 ∼ ∼ Ext (Z, Z) = Hom(Z, Z) = Z and i Ext (Z, Z) = 0 for i > 0. n (2) We compute Ext (Z/p, Z). There is a free resolution ·p 0 → F1 = Z −→ F0 = Z → Z/p → 0. ·p So F∗ = 0 → Z −→ Z → 0, supported in degrees zero and one. Taking Hom(−, Z) yields ·p 0 → Hom(Z, Z) −→ Hom(Z, Z) → 0. So 0 Ext (Z/p, Z) = 0, 1 ∼ Ext (Z/p, Z) = Z/p, and i Ext (Z/p, Z) = 0 for i > 1. 1 ∼ The fact that Ext (Z/p, Z) = Z/p is nonzero corresponds to the failure of Hom(−, Z) to be right exact observed in Example 5.3. Next, we list some of the main important properties of the Ext groups. Proposition 5.7. Let G and H be abelian groups. (1) Ext0(H,G) =∼ Hom(H,G). (2) Exti(H,G) = 0 for i ≥ 2. 1 (3) Ext (H,G) = coker Hom(F0,G) → Hom(F1,G) , where 0 → F1 → F0 → H → 0 is a free resolution of length one. (4) If H is a free abelian group, then Ext1(H,G) = 0. 1 ∼ (5) Ext (Z/n, G) = G/nG. 1 (6) If H is a finitely generated abelian group, then Ext (H, Z) is the torsion subgroup of H, that is TH := {h ∈ H | ∃ n ∈ Nr{0} with nh = 0}. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 24

1 (7) Ext (H, Q) = 0 for any H. (8) Let H1,...,Hk and G1,...,Gk be abelian groups. Then k k 1 M ∼ M 1 Ext ( Hi,G) = Ext (Hi,G) i=1 i=1 and k k 1 M ∼ M 1 Ext (H, Gi) = Ext (H,Gi). i=1 i=1 We will not prove all of these in detail.

Proof. (1) Consider a resolution F1 → F0 → H → 0. This is exact. Therefore

0 → Hom(H,G) → Hom(F0,G) → Hom(F1,G) is exact since Hom(−,G) is left exact. Therefore 0 Hom(H,G) = ker(Hom(F0,G) → Hom(F1,G)) = Ext (H,G) where the last equality is by deﬁnition. (2) Since H is abelian, there is a resolution 0 → F1 → F0 → H → 0. Then i i Ext (H,G) = H (Hom(F0,G) → Hom(F1,G)) which, since the nonzero groups are in degrees zero and one, equals 0 for i ≥ 2. (3) We have 1 1 Ext (H,G) = H (Hom(F0,G) → Hom(F1,G)) = coker(Hom(F0,G) → Hom(F1,G)).

(4) Since H is free abelian, there is a free resolution with F0 = H and Fi = 0 for i ≥ 1. (5) We have 1 1 ·n ·n ∼ Ext (Z/n, G) = H (Hom(Z,G) −→ Hom(Z,G)) = coker(G −→ G) = G/nG. (6) Let H be a finitely generated abelian group. As noted above, by the classification of finitely generated abelian groups, k ∼ n M ni H = Z ⊕ Z/pi i=1

for some n, for some k, and for some primes p1, . . . , pk and some integers Lk ni n1, . . . , nk. The torsion subgroup is i=1 Z/pi There is a resolution of length one

k n k 0,Lk p i M i=1 i n M 0 → Z −−−−−−−−→ Z ⊕ Z → H → 0. i=1 i=1 Applying Hom(−, Z) to this yields k n k 0,Lk p i n M i=1 i M 0 → Z ⊕ Z −−−−−−−−→ Z → 0. i=1 i=1 ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 25

Taking H1 yields k 1 ∼ M ni Ext (H, Z) = Z/pi = TH, i=1 the torsion subgroup of H. (7) I am omitting the proof of this. Note that it uses that Q is divisible, and Q can be replaced with any divisible abelian group here. (8) I am also omitting a detailed proof of this. It is not too hard if one follows through the definitions. 5.3. The universal coefficient theorem for cohomology. Now we move on to the universal coefficient theorem, which was the reason for introducing the Ext groups. One of the maps in the theorem is the evaluation map, so let us define it.

Definition 5.8. Let (C∗, ∂) be a chain complex of free abelian groups, and let G be an abelian group. The evaluation map is n ev: H (C; G) → Hom(Hn(C),G) H (C) → G [ϕ: C → G] 7→ n n [c] 7→ h[ϕ], [c]i = ϕ(c). Here the pairing h[ϕ], [c]i = ϕ(c) is often called the Kronecker pairing. It simply means evaluate a representative cocycle of a homology class on a representative cycle of a homology class. In order to make the last “definition” really a definition, we should show that ev is well-defined. Lemma 5.9. The map ev is a well-defined group homomorphism. Proof. We will just show that it is well-defined, considering the effect of changes in representative cocycles and cycles. (ϕ + δψ)(c + ∂d) = ϕ(c) + δψ(c) + ϕ(∂d) + δψ(∂d) = ϕ(c) + ψ(∂c) + δϕ(d) + ψ(∂2d) = ϕ(c). Here ∂c = 0 because c is a cycle and δϕ = 0 because ϕ is a cocycle. Then ∂2 = 0, so the last term vanishes as well. The universal coefficient theorem for spaces will follow directly from the following purely algebraic statement.

Theorem 5.10. Let (C∗, ∂) be a chain complex of free abelian groups, and let G be an abelian group. For each n ∈ N0, there is a natural short exact sequence 1 n ev 0 → Ext (Hn−1(C),G) → H (C; G) −→ Hom(Hn(C),G) → 0 that splits, i.e. n ∼ 1 H (C; Γ) = Ext (Hn−1(C),G) ⊕ Hom(Hn(C),G). ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 26

Proof. Let (C∗, ∂) be a chain complex of free abelian groups, and let G be an abelian group. Write Zn := ker ∂n and Bn := im ∂n+1. Both are also free abelian groups. We have a short exact sequence of chain complexes. That is the rows are exact of the next diagram......

∂n 0 / Zn / Cn / Bn−1 / 0

0 ∂n 0

∂n−1 0 / Zn−1 / Cn−1 / Bn−2 / 0

...... Apply Hom(−,G) to obtain the next diagram. The rows are again exact because Bn is free abelian...... O O O

∗ ∂n 0 / Hom(Bn−1,G) / Hom(Cn,G) / Hom(Zn,G) / 0 O O O ∗ 0 ∂n 0

0 / Hom(Bn−2,G) / Hom(Cn−1,G) / Hom(Zn−1,G) / 0 O O O

...... By the snake lemma, we obtain a long exact sequence in cohomology. However the left and right vertical sequences have all coboundary maps trivial, so the cohomology is equal to the cochain groups. We therefore have a long exact sequence:

dn−1 n dn · · · → Hom(Zn−1,G) −−−→ Hom(Bn−1,G) → H (C; G) → Hom(Zn,G) −→ Hom(Bn,G) → · · ·

Here dn, dn−1 denotes the connecting homomorphism. For m = n, n − 1 we assert ∗ that dm = im, where im :: Bm → Zm is the inclusion. This is a straightforward check using the deﬁnition of the connecting homomorphism. We therefore have a short exact sequence n 0 → coker(dn−1) → H (C; G) → ker(dn) → 0. There is a resolution ∗ in−1=dn−1 0 → Bn−1 −−−−−−−→ Zn−1 → Hn−1(C) → 0. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 27

Therefore 1 coker(dn−1) = Ext (Hn−1(C),G). Similarly there is a resolution ∗ in=dn 0 → Bn −−−−→ Zn → Hn(C) → 0. Therefore 0 ∼ ker(dn) = Ext (Hn(C),G) = Hom(Hn(C),G). Putting this together yields a short exact sequence 1 n 0 → Ext (Hn−1(C),G) → H (C; G) → Hom(Hn(C),G) → 0 as desired. I will omit the proof that this is also a split sequence, and that it is natural, for time reasons. We end the discussion of the universal coefficient theorem by noting that with field coefficients, the relationship between homology and cohomology is much sim- pler.

Theorem 5.11. Let (X,A) be a pair of topological spaces. Let F be a ﬁeld. Then n ev: H (X,A; F) → HomF(Hn(X,A; F), F)

H (X,A; ) → ϕ 7→ n F F σ 7→ hϕ, σi is an isomorphism for every n ∈ N0.

6. Univeral coefficients for homology and the Kunneth¨ theorem 6.1. Tensor products. Let A and B be abelain groups. Then the tensor product is a quotient of the free abelian group generated by symbols of the form ai ⊗ bi, with ai in A and bi in B n X A ⊗ B = { ai ⊗ bi}/ ∼ i=1 where the relations ∼ are generated by (a + a0) ⊗ b = a ⊗ b + a0 ⊗ b and a ⊗ (b + b0) = a ⊗ b + a ⊗ b0. Here are some facts on tensor products. The proofs are omitted, and you should think about them to help you understand tensor products. (1) A ⊗ B =∼ B ⊗ A. (2) A ⊗ (B ⊗ C) =∼ (A ⊗ B) ⊗ C. ∼ (3) Z ⊗ A = A. ∼ (4) Z/k × A = A/kA. ∼ ∼ (5) Z/n ⊗ Z/m = Z/n/(mZ/m) = Z/ gcd(m, n). ∼ r ∼ r (6) If A is a ﬁnitely generated abelian group and A = Z ⊕TA, then Q⊗A = Q . ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 28

6.2. Tor groups. The Tor groups are to tensor product as Ext groups are to Hom. If 0 → A → B → C → 0 is a short exact sequence, then A ⊗ G → B ⊗ G → C ⊗ G → 0 is exact. So − ⊗ G is a covariant right exact functor from abelian groups to abelian groups. The failure to be left exact is measured by the Tor groups. Deﬁnition 6.1. Let A and B be abelian groups. Let

· · · → F2 → F1 → F0 → A → 0 be a free resolution of A. Then

Torn(A, B) := Hn(F∗ ⊗ B). That is, replace A by 0, tensor with B, then take homology.

The proof that Torn is well-defined uses the fundamental lemma of homological algebra, that we showed for length one resolutions above in order to show that Ext is well defined. The next proposition is the analogue of the fact that Ext0(G, H) =∼ Hom(G, H). ∼ Proposition 6.2. Tor0(A, B) = A ⊗ B. We will not provide any proofs of these statements, since they are rather similar in character to the Ext group proofs we have already covered, and I think it best to move onto cup products as soon as possible. 6.3. Universal coefficient theorem for homology. Tensor product appears in two important places: in the universal coefficient theorem for homology and in the Künneththeorem for cohomology. Theorem 6.3 (Universal coefficient theorem for homology). Let X be a topological space and let G be an abelian group. For all n ∈ N0, there is a natural short exact sequence

0 → Hn(X) ⊗ G → Hn(X; G) → Tor1(Hn−1(X); G) → 0 that splits. ∼ Corollary 6.4. For every space X and for n ∈ N0, we have that Hn(X; Q) = Hn(X; Z) ⊗ Q.

Proof. Let H := Hn−1(X). There is a resolution of length one:

k n k 0,Lk p i M i=1 i n M 0 → Z −−−−−−−−→ Z ⊕ Z → H → 0. i=1 i=1 for some integers n, k, pi, ni. To compute Tor1(H,Q) we tensor with Q to obtain k n k 0,Lk p i M i=1 i n M 0 → Q −−−−−−−−→ Q ⊕ Q → 0 i=1 i=1 Since the pi are nonzero, the ﬁrst homology of this chain complex is trivial, so Tor1(H, Q) = 0. The corollary then follows from the universal coeﬃcient theorem exact sequence. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 29

3 Example 6.5. Let us consider the homology of RP . With Z coefficients, we 3 ∼ 2 ∼ 3 3 have H0(RP ; Z) = H3(RP ; Z) = Z, H2(RP ; Z) = 0 and H1(RP ; Z) = Z/2. We can compute the homology with Z/2 coefficients using the cellular chain complex CW 3 ∼ Ci (RP ) = Z for i = 0, 1, 2, 3 and boundary maps CW 3 ∼ 0 CW 3 ∼ 2 CW 3 ∼ 0 CW 3 ∼ 0 → C3 (RP ) = Z −→ C2 (RP ) = Z −→ C1 (RP ) = Z −→ C0 (RP ) = Z → 0. ∼ Tensor with Z/2 to obtain, using that Z ⊗ Z/2 = Z/2, the chain complex: 0 0 0 0 → Z/2 −→ Z/2 −→ Z/2 −→ Z/2 → 0. 3 ∼ CW 3 ∼ Therefore Hi(RP ; Z/2) = Hi (RP ; Z/2) = Z/2 for i = 0, 1, 2, 3, and is otherwise trivial. Now let us compute with the universal coefficient theorem for homology in- stead. What is Tor1(H, Z/2)? For H free abelian, Tor1(H, Z/2) = 0. However ∼ Tor1(Z/2, Z/2) = Z/2. To see this, note that Z/2 has a resolution 2 Z −→ Z → Z/2 → 0. Tensor with Z/2 to get 0 Z/2 Z/2 −→−−→ supported in degrees 0 and 1. The H1 of this is Tor1(Z/2, Z/2) = Z/2. Now, by the universal coefficient theorem for homology, this means that 3 3 ∼ Hi(RP ; Z/2) = Hi(RP ; Z) ⊗ Z/2 = Z/2 ∼ ∼ for i = 0, 1, 3, since Z ⊗ Z/2 = Z/2 for i = 0, 3, and since Z/2 ⊗ Z/2 = Z/2 for 3 3 i = 1. For H2(RP ; Z/2), we have H2(RP ; Z) = 0 so 3 ∼ 3 ∼ H2(RP ; Z/2) = Tor1(H1(RP ; Z), Z/2) = Tor1(Z/2, Z/2) = Z/2. 3 ∼ So once again we compute that Hi(RP ; Z/2) = Z/2 for i = 0, 1, 2, 3, and is otherwise trivial. 6.4. The Künneththeorem for cohomology. Now for another very important use of tensor product: the Künneththeorem that allows us to compute the cohomology of a product of space X × Y , given knowledge of the cohomology of X and of Y . Theorem 6.6. Let X and Y be topological spaces. Let p: X × Y → X and q : X × Y → Y be the projections. Suppose that the homology groups of X and Y are finitely generated. Then for all n ∈ N0, there is a short exact sequence M k ` n M k ` ∗ ∗ 0 → H (X; Z) ⊗ H (Y ; Z) → H (X × Y ; Z) Tor1(H (X; Z),H (Y ; Z)) → 0ϕ ⊗ ψ 7→ p (ϕ) ^ q (ψ) k+`=n k+`=n+1 which splits. For examples, see the problem sheet which focusses on applications of this theorem. In the Künneththeorem, we include a description of one of the maps in terms of cup products, the products that we are about to introduce. In problem classes we will also see how to use this information on the map, that ϕ ⊗ ψ 7→ p∗(ϕ) ^ q∗(ψ), to compute the cup products of certain product spaces. But first we have to introduce cup products. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 30

7. Cup products As advertised previously, one of the great advantages of cohomology is that it admits a beautiful multiplicative structure. Whereas homology associates to a topological space a sequence of abelian groups, in cohomology these abelian groups are collected together to produce a single ring. The multiplication in this ring is called the cup product, which we shall now define. In order to make cohomology into a ring, we will need to work with coefficients that are themselves a ring. So let R be a commutative ring. For example we will typically use R = Z, Q, R or Z/n for some n. This looks similar to before, since these are all abelian groups. Indeed any ring is an abelian group. But now we shall also use the ring structure on these abelian groups. 7.1. Definition of cup products. Here is our goal. For a space X, we want to define a map ^: Hi(X; R) × Hj(X; R) → Hi+j(X; R). To define such a map, we will define a map on cochains ^: Ci(X; R)×Cj(X; R) → Ci+j(X; R), and then we will check that this induces a well-defined map on cohomology groups. Let us recall and introduce some notation for singular simplices. A singular i- simplex of X is a continuous map ∆i → X. We write label the vertices 0, 1, . . . , i. i We write [v0, . . . , vj] for the face of ∆ spanned by the vertices v0, . . . , vj, with vk ∈ {0, . . . , i} for k = 0, . . . , j. We write i bj : [0, . . . , j] → [0, . . . , i] = ∆ for the front j face of ∆i. We write i cj :[i − j, . . . , i] → [0, . . . , i] = ∆ for the back j face of ∆i. Given a singular i + j-simplex σ : ∆i+j → X, we can consider its composition with bi and cj: i i+j σ ◦ bi : ∆ → ∆ → X and j i+j σ◦cj : ∆ → ∆ → X. Now we can define: ^: Ci(X; R) × Cj(X; R) → Ci+j(X; R) (ϕ, ψ) 7→ ϕ(σ ◦ bi)ψ(σ◦cj) = ϕ(σ|[0,...,i])ψ(σ|[i,...,i+j]). That is, we evaluate ϕ on the front i-face and ψ on the front j face. Note that in the last line we think of σ ◦ bi as the restriction of σ to a face, σ|[0,...,i]. Theorem 7.1. (1) Given two cocycles ϕ ∈ Ci(X; R) and ψ ∈ Cj(X; R) with δϕ = 0 and δψ = 0, then ϕ ^ ψ ∈ Ci+j(X; R) is also a cocycle, and therefore represents a cohomology class [ϕ ^ ψ] ∈ Hi+j(X; R). ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 31

(2) Cup product is well-deﬁned on cohomology. That is, for any θ ∈ Ci−1(X; R) and for any χ ∈ Cj−1(X; R), we have that

(ϕ + δθ) ^ (ψ + δχ) − ϕ ^ ψ

is a coboundary. Therefore

[(ϕ + δθ) ^ (ψ + δχ)] = [ϕ ^ ψ] ∈ Hi+j(X; R).

We need the following lemma to prove Theorem 7.1.

Lemma 7.2. Let ϕ ∈ Ci(X; R) and ψ ∈ Cj(X; R). Then

δ(ϕ ^ ψ) = δϕ ^ ψ + (−1)iϕ ^ δψ ∈ Ci+j+1(X; R).

Proof. Let σ : ∆i+j+1 → X. Then we have the following computations. Here bk indicates that the number is missed out. i+1 X (7.3) (δϕ ^ ψ)(σ) = (−1)kϕ(σ| ) · ψ(σ ) [0,...,bk,...,i+1] [i+1,...,i+j+1] k=1

i+j+1 X (7.4) (−1)i(ϕ ^ δψ)(σ) = (−1)kϕ(σ ) · ψ(σ ) [0,...,i] [i,...,bk,...,i+j+1] k=i The sum of the left hand sides of these two equations equals the right hand side of the equation we want to prove. In the sum of the right hand sides of these two equations, the last term of (7.3) cancels with the ﬁrst term of (7.4). The rest of the terms sum to

(7.5) δ(ϕ ^ ψ)(σ) = (ϕ ^ ψ)(∂σ) where ∂σ = Pi+j+1 σ . k=0 [0,...,bk,...,i+j+1] Proof of Theorem 7.1. First we prove that the cup product of two cocycles is a cocycle. Suppose that δϕ = 0 and δψ = 0, so that both are cocycles. Then

δ(ϕ ^ ψ) = δϕ ^ ψ + (−1)iϕ ^ δψ = 0 ^ ψ + (−1)iϕ ^ 0 = 0 + 0 = 0.

So ϕ ^ ψ is a cocycle. Next, by the lemma we have that

ϕ ^ δχ = ±δ(ϕ ^ χ) − (δϕ ^ χ) = ±δ(ϕ ^ χ).

Similarly δθ ^ ψ = ±δ(θ ^ ψ) ± (θ ^ δψ) = ±δ(θ ^ ψ). So both are coboundaries. Finally

δθ ^ δχ = ±δ(θ ^ δχ) ± (θ ^ δ2ψ) = ±δ(θ ^ δχ), ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 32 which is also a coboundary. We make no effort to get the signs right because they are irrelevant. Therefore (ϕ + δθ) ^ (ψ + δχ) − ϕ ^ ψ =ϕ ^ ψ + ϕ ^ δχ + δθ ^ ψ + δθ ^ δχ − ϕ ^ ψ = ± δ(ϕ ^ χ) ± δ(θ ^ ψ) ± δ(θ ^ δχ) =δ(±ϕ ^ χ ± θ ^ ψ ± θ ^ δχ) which is a coboundary, as desired. Therefore the cup products are well-defined on cohomology, and so define maps ^: Hi(X; R) × Hj(X; R) → Hi+j(X; R), as promised. 7.2. An example. Example 7.6. One of the most instructive examples of cup product is the torus T 2 = S1 × S1. We explicitly compute the cup product of the torus 1 1 1 1 1 1 2 1 1 ^: H (S × S ; Z) × H (S × S ; Z) → H (S × S ; Z). 1 1 1 ∼ 2 1 1 ∼ Recall that H (S × S ; Z) = Z ⊕ Z and H (S × S ; Z) = Z. So the cup product 2 corresponds, after having chosen generating cohomology classes, to a pairing Z × 2 Z → Z. Such a pairing can be represented by a 2 × 2 matrix A, in the sense that 2 2 the pairing of x = (x1, x2) ∈ Z and y = (y1, y2) ∈ Z is given by xAyT , as you may recall from bilinear forms or inner product spaces in linear algebra. We shall choose natural bases for the cohomology of the torus, and with respect to these bases the cup product is represented by 0 1 . −1 0 In order to compute the cup product of the torus, first we need to define some singular simplices and some singular cochains. In Figure 1, we draw the torus, we draw the projections p, q : S1 × S1 → S1 to the first and second factors respectively. We also show the images of two singular simplices with the property that −σ1 + σ2 represents a fundamental class, that is 2 1 1 ∼ [−σ1 + σ2] = [T ] ∈ H2(S × S ; Z) = Z. The figure indicates which 1-simplices correspond to the images of the 3 summands in the boundaries ∂σi. Then note that

∂(−σ1 + σ2) = −(σ1|[0,1] + σ1|[1,2] − σ1|[0,2]) + (σ2|[0,1] + σ2|[1,2] − σ2|[0,2])

= −σ1|[0,1] − σ1|[1,2] + σ1|[0,2]) + σ2|[0,1] + σ2|[1,2] − σ2|[0,2]

= −σ1|[0,1] − σ1|[1,2] + σ1|[0,2]) + σ1|[1,2] + σ1|[0,1] − σ1|[0,2] = 0.

Note that it is important here that, for example σ1|[0,1] shows up twice here with opposite signs in the formal sum. What emphatically does not happen is ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 33

Figure 1. A diagram of the torus. Identify the vertical edges with each other and the horizontal edges with each other to get the torus. The projections p, q : S1 × S1 → S1 are shown. The images of two singular 2-simplices are shown. At the bottom, a standard 2-simplex ∆2 is drawn. The orientation is anticlockwise, and the dot indicates the vertex 0. Going anticlockwise round the edges we come to vertex 1, then vertex 2, then back to vertex 0. The dots and the orientation arrows in the main diagram of T 2 indicate how 2 2 the singular 2-simplices σi : ∆ → T are deﬁned. This enables us to label the boundary 1-simplices, such as σ1|[0,1].

that σ1|[0,1] and σ1|[1,0] show up, both with a positive sign. In the singular chain complex, these two do not cancel. (This is a popular misapprehension, so I am trying to highlight it. In cellular homology, and in the simplicial theory, such cancellation does occur. But we are not working in those theories.) It is true that the sum σ1|[0,1] + σ1|[1,0] is the boundary of a singular 2-chain; I suggest this as an instructive exercise to write down such a 2-chain. Note that this exercise could just be about singular simplices in the interval [0, 1], and has nothing to do with the torus. Now, some more notation for the upcoming cup product computation. Let µ: ∆1 → S1 be a 1-simplex given by t 7→ e2πit, and let ν : ∆1 → S1 be the constant ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 34

1 1 1 1-simplex t 7→ 1 ∈ S . Let θ ∈ C (S ; Z) be a cocycle with θ(µ) = 1 and θ(ν) = 0, 1 1 ∼ so that [θ] ∈ H (S ; Z) = Z is a generator. For example, we could use the cocycle θZ constructed in Example 3.14. ∗ ∗ 1 1 1 The cohomology classes [p (θ)] and [q (θ)] ∈ H (S ×S ; Z) provide a generating 1 1 1 ∼ 2 2 ∗ ∗ set for H (S × S ; Z) = Z , and the dual class [T ] = [−σ1 + σ2] is a generator 2 1 1 ∼ for H (S × S ; Z) = Z. Now we can compute the cup product. We focus on the computation of p∗(θ) ^ q∗(θ), since this is the most exciting one. All other cup products p∗(θ) ^ p∗(θ), q∗(θ) ^ q∗(θ) and q∗(θ) ^ p∗(θ) can be computed similarly, but also we will soon be able to deduce them from the computation below and the graded commutativity property of the cup product. We compute the cup ∗ ∗ product p (θ) ^ q (θ) evaluated on σ1 and evaluated on σ2 separately. ∗ ∗ ∗ ∗ p (θ) ^ q (θ) (σ1) = p (θ)(σ1|[0,1]) · q (θ)(σ1|[1,2])

= θ(p ◦ σ1|[0,1]) · θ(q ◦ σ1|[1,2]) = θ(ν) · θ(ν) = 0 · 0 = 0.

Here, to identify p ◦ σ1|[0,1] = ν = q ◦ σ1|[1,2] we look at Figure 1, and note that these 1-simplice indeed project to points. On the other hand, we have: ∗ ∗ ∗ ∗ p (θ) ^ q (θ) (σ2) = p (θ)(σ2|[0,1]) · q (θ)(σ2|[1,2])

= θ(p ◦ σ2|[0,1]) · θ(q ◦ σ2|[1,2]) = θ(µ) · θ(µ) = 1 · 1 = 1. Therefore ∗ ∗ 2 ∗ ∗ [p (θ) ^ q (θ)]([T ]) = p (θ) ^ q (θ) (−σ1 + σ2) = 0 + 1 = 1. So we have a nontrivial cup product. This corresponds to the top right entry of the matrix 0 1 −1 0 claimed earlier. The remaining entries can be calculated with analogous computations, or as mentioned above can be deduced from the fact, to be proven shortly, 2 2 ∼ 1 2 that ϕ ^ ψ = −ψ ^ ϕ ∈ H (T ; Z) = Z for ϕ, ψ ∈ H (T ; Z). 7.3. The cohomology ring. Let us return to the theory of cup products. Proposition 7.7. Let R be a commutative ring and let X be a topological space. (1) The cup product is R-bilinear and associative. (2) The conglomerate H∗(X; R) := L Hn(X,R), with the operations + n∈N0 and ^, forms a ring: M H∗(X; R) := Hn(X,R), +, ^ .

n∈N0 0 The unit of the ring is 1X , the cohomology class in H (X; R) represented by the constant map X → R that sends x 7→ 1 ∈ R for every x ∈ X. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 35

Proof. Let ϕ ∈ Ci(X; R), let ψ ∈ Cj(X; R) and let χ ∈ Ck(X; R). Let (σ : ∆i+j+k → X) ∈ Ci+j+k(X; R) be a singular simplex. We prove associativity:

(ϕ ^ ψ) ^ χ(σ) = (ϕ ^ ψ)(σ|[0,...,i+j]) · χ(σ|[i+j,...,i+j+k])

= ϕ(σ|[0,...,i]) · ψ((σ|[i,...,i+j])) · χ(σ|[i+j,...,i+j+k])

= ϕ(σ|[0,...,i]) · (ψ ^ χ)(σ|[i,...,i+j+k]) = ϕ ^ (ψ ^ χ)(σ). The R-bilinearity is easy and if left to the reader. n n Now let us check that 1X is a unit. Let [ϕ] ∈ H (X; R) and let σ : ∆ → X be an n-simplex in Cn(X; R). Then

(ϕ ^ 1X )(σ) = ϕ(σ) · 1X (σ|[n]) = ϕ(σ) · 1 = ϕ(σ). n Thus [ϕ ^ 1X ] = [ϕ] ∈ H (X; R). Similarly 1X ^ ϕ = ϕ. So 1X is an identity. ∗ Therefore H (X; R) is a ring as asserted. Theorem 7.8. Let f : X → Y be a continuous map of topological spaces. Then f ∗ : H∗(Y ; R) → H∗(X; R) is a ring homomorphism. Proof. Let ϕ ∈ Ck(Y ; R) and let ψ ∈ C`(Y ; R). Let σ : ∆k+` → X be a singular (k + `)-simplex. Then ∗ ∗ ∗ ∗ (f (ϕ) ^ f (ψ))(σ) = f (ϕ)(σ|[0,...,k]) · f (ψ)(σ|[k,...,k+`])

= ϕ(f ◦ σ|[0,...,k]) · ψ(f ◦ σ|[k,...,k+`]) = (ϕ ^ ψ)(f ◦ σ) = f ∗(ϕ ^ ψ)(σ) So f ∗(ϕ) ^ f ∗(ψ) = f ∗(ϕ ^ ψ). Since we already know that f ∗ is a homomorphism of the underlying abelian groups H∗(Y ; R) → H∗(X; R), it follows that f ∗ is a ring homomorphism as claimed. Recall that if f ∼ g : X → Y are homotopic maps, then f ∗ = g∗ : Hk(X; R) → k H (Y ; R) for all k ∈ N0. Corollary 7.9. Suppose that X ' Y are homotopy equivalent spaces. Then H∗(X; R) =∼ H∗(Y ; R) are isomorphic rings. Proof. Let F : X → Y and G: Y → X be maps realising the homotopy equivalence. Then f ◦ g ∼ Id and g ◦ f ∼ Id implies that g∗ ◦ f ∗ = Id and f ∗ ◦ g∗ = Id as ring homomorphisms. Therefore f ∗ and g∗ are inverse ring homomorphisms to one another. Now we sketch the proof that cup product is graded commutative on the level of cohomology. Note that this certainly does not hold in general on the chain level, only after passing to cohomology. Theorem 7.10 (Graded commutativity). Let [ϕ] ∈ Hk(X; R) and let [ψ] ∈ H`(X; R). Then [ϕ ^ ψ] = (−1)k`[ψ ^ ϕ] ∈ Hk+`(X; R). ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 36

Proof. We give a sketch of the proof. Please read [Hat] or [Fr] for the details. Define ρ: Cn(X) → Cn(X) σ 7→ εnσ where n(n+1)/2 εn := (−1) . Here σ(i) = σ(n − i), so σ is the composition [0, . . . , n] → [n, . . . , 0] −→σ X. The corresponding permutation of the vertices can be written as a product of n(n + 1)/2 transpositions, whence the definition of ε. We claim that ρ is a chain map and that ρ ∼ Id. We omit the proof of the claim, which is the main difficulty in the proof. See Page 211–2 of Hatcher, for example. Now, assuming the claim, we prove the theorem. We have ∗ ∗ ρ (ϕ) ^ ρ (ψ)(σ) = ϕ(εkσ|[k,...,0]) · ψ(ε`σ|[k+`,...,k]) ∗ ρ (ψ ^ ϕ) = εk+`ψ(σ|k+`,...,k) · ϕ(σ|[k,...,0]). It follows that ∗ ∗ ∗ εkε`(ρ (ϕ) ^ ρ (ψ)) = εk+`ρ (ψ ^ ϕ). It is trivial to check that k` εk+` = (−1) εkε`. Therefore ρ∗(ϕ) ^ ρ∗(ψ) = (−1)k`ρ∗(ψ ^ ϕ). Since ρ ∼ Id, on passing to cohomology this yields [ϕ] ^ [ψ] = (−1)k`[ψ ^ ϕ] as desired. 7.4. More examples.

Example 7.11. Consider Σg, the genus g orientable surface. We have that 1 ∼ 2g H (Σg) = Z = hα1, β1, . . . , αg, βgi, where αi and βi are classes dual to the standard 2g generators of ﬁrst homology H1(Σg). The cup product gives a map 2g 2g 2 ∼ ^: Z × Z → H (Σg; Z) = Z represented by the matrix M 0 1 −1 0 where the direct sum indicates block diagonal sum of matrices. Note that αi ^ 1·1 αi = (−1) αi ^ αi = −αi ^ αi for i = 1, . . . , g. Therefore αi ^ αi = 0, and similarly βi ^ βi = 0, for i = 1, . . . , g. We also have αi ^ βi = −βi ^ αi. So the matrix above had to be skew symmetric. Therefore the only computation that should be made directly, as we did for the torus above, is that αi ^ βj = δij. We leave this computation as an exercise in generalising the torus computation made above. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 37

2 Example 7.12. We compute the cup product of RP with F2 coeﬃcients. The interesting cup product is 1 2 1 2 2 2 ^ H (RP ; F2) × H (RP ; F2) → H (RP ; F2) which is a function F2 × F2 → F2. We have to compute the image of (1, 1). The answer is 1, so the cup product is nontrivial. We have to compute this though. See Figure 2 for a depiction of the image of some 2-simplices that sum to a 2 fundamental class over F2 for RP . That is, 2 2 [RP ]F2 := σ1 + σ2 + τ1 + τ2 ∈ C2(RP ; F2) 2 ∼ generates H2(RP ; F2) = F2.

2 Figure 2. A diagram of RP . Identify the edges with each other 2 to get RP . The images of four singular 2-simplices σ1, σ2, τ1 and τ2 are shown.

In the figure, which shows a disc, opposite points on the boundary circle are identified, as shown by the arrows. The convention for drawing σ1 and σ2 is the same as in our example of the torus above, and shown in Figure 1. The convention for drawing τ1 and τ2 is different, since these look like bigons, not triangles. 2 2 Remember that we are describing the images of maps ∆ → RP , not subsets 2 homeomorphic to ∆ . The τi are degenerate 2-simplices: one of the edges of each of the 2-simplices τi is mapped entirely to a single point, the vertex labelled with a ×. In both cases this edge is τi|[0,2]. Note that every singular 1-simplex appears exactly twice in ∂(σ1 + σ2 + τ1 + τ2), with the same orientation. Coming up with such a diagram is not so easy, as you will discover if you try to do it yourself for the Klein bottle. 1 2 2 We want to define a generator α ∈ H (RP ; F2) such that (α ^ α)([RP ]F2 ) = 1 2 1 ∈ F2. Let us explicitly define an element α of H (RP ; F2), then we will show it ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 38 has these properties. Consider the function

2 S → F2 ( 1 if (z > 0) or (z = 0 and y > 0) or (z = 0 = y and x = 1). (x, y, z) 7→ 0 else.

2 1 2 Note that for p ∈ S , f(−p) = 1 − f(p). Let σ : ∆ → RP be a 1-simplex. Let 2 2 2 2 2 π : S → RP be the projection that sends p ∈ S to [p] ∈ RP = S /{p ∼ −p}. 1 2 Pick a lift σe : ∆ → S , in the sense that π ◦ σe = σ. Then deﬁne a 1-cochain α by

α(σ) := f(σe(1)) − f(σe(0)) ∈ F2.

There are choices of lifts. However the end point σe(1) changing to its antipodal point would also change the endpoint σe(0) to its antipodal point, by continuity and the fact that σe is a lift. Since f(−p) = 1 − f(p), making both these changes does not affect α(σ). We note that α is a cocycle, as can be seen by lifting a 2-simplex 2 2 to S : α applied to the boundary of a 2-simplex in RP is computed by applying f to the image of the vertices 2-simplex under the lifted map to S2. Each vertex appears twice, since it is the vertex of two different edges, so in F2 all contributions vanish and so δα = 0. 1 2 πit 1 2 2 1 Let µ: ∆ → RP be defined by t 7→ e ⊂ S ⊂ S → RP , with S sitting inside S2 as the equator {z = 0}. Then µ is a cycle and

πi 0 α(µ) = f(e ) − f(e ) = 1 ∈ F2.

1 2 2 Therefore α is a generator of H (RP ; F2) and µ is a generator of H1(RP ; F2). Now that we have deﬁned all our cycles and cocycles, we can proceed to compute the cup product. We have:

α ^ α(σ1) = α(σ1|[0,1]) · α(σ1|[1,2]). To compute α on 1-simplices, think of the disc we have drawn in Figure 2 as the northern hemisphere of S2, that is the points {z ≥ 0}. Then f(p) = 1 everywhere in the interior of the disc, everywhere on the upper boundary arc, excluding the endpoints. f(p) = 0 on the lower boundary arc, excluding the endpoints. Finally f(p) = 1 for p the right hand vertex on the boundary, and f(p) = 0 for p the left hand vertex on the boundary. We assign to a 1-simplex the diﬀerence in the values of f at its endpoints. So any 1-simplex in Figure 2 with an endpoint at the left hand vertex has α evaluating to 1 on it, while the other 1-simplices have α evaluating to 0. This makes computing α now relatively straightforward. So we have

α ^ α(σ1) = α(σ1|[0,1]) · α(σ1|[1,2]) = 0 · 1 = 0

α ^ α(σ2) = α(σ2|[0,1]) · α(σ2|[1,2]) = 1 · 0 = 0

α ^ α(τ1) = α(τ1|[0,1]) · α(τ1|[1,2]) = 1 · 1 = 1

α ^ α(τ2) = α(τ2|[0,1]) · α(τ2|[1,2]) = 0 · 0 = 0. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 39

Therefore 2 (α ^ α)([RP ]F2 ) = (α ^ α)(σ1 + σ2 + τ1 + τ2) = (α ^ α)(σ1) + (α ^ α)(σ2) + (α ^ α)(τ1) + (α ^ α)(τ2) = 0 + 0 + 1 + 0 = 1. So we have a nontrivial cup product as asserted above. Example 7.13. We describe the cohomology ring of a sphere Sn. We have that Hi(Sn; R) =∼ R if i = 0, n, and is 0 otherwise. The cohomology ring is H∗(Sn; R) =∼ R[x]/(x2), the quotient of the polynomial ring R[x] by the ideal (x2). This is known as a truncated polynomial ring. Here x is the generator of Hn(Sn; R) =∼ R. So H∗(Sn; R) = {a + bx | a, b ∈ R}. Next we show that cup products in wedges of spheres are pretty boring. This will provide a good source of examples of spaces with a certain set of cohomology groups, and when we show that for certain other spaces, cup products are nontrivial, this will enable us to show that these spaces are not homotopy equivalent. Proposition 7.14. Consider a wedge of spheres Sn1 ∨ Sn2 ∨ · · · ∨ Snk .

Here ni ≥ 1 and i = 1, . . . , k. Then all cup products of degree at least one are zero. The analogue actually holds for general wedges, but we will not give the proof. n n n n Proof. Let pi : S 1 ∨ S 2 ∨ · · · ∨ S k → S i be the projection map, sending all spheres other than the ith sphere to the basepoint. For all n ∈ N0, we have an isomorphism ∼ ∗ ∗ n n1 n nk = n n1 nk p1 ⊕ · · · ⊕ pk : H (S ; R) ⊕ · · · ⊕ H (S ; R) −→ H (S ∨ · · · ∨ S ; R). The statement means that we need to show that

∗ ni ∗ ∗ nj ∗ ni+nj n1 nk pi ([S ] ) ^ pj ([S ] ) = 0 ∈ H (S ∨ · · · ∨ S ; R) for all i, j = 1, . . . , k. For i 6= j, let

n1 nk ni nj gij : S ∨ · · · ∨ S → S ∨ S be the projection, and let ni nj ni qi : S ∨ S → S and ni nj nj qj : S ∨ S → S also be projections. Note that qi ◦ gij = pi and qj ◦ gij = pj. We then compute:

∗ ni ∗ ∗ nj ∗ ∗ ∗ ni ∗ ∗ ∗ nj ∗ pi ([S ] ) ^ pj ([S ] ) = gij(qi ([S ] )) ^ gij(qj ([S ] )) ∗ ∗ ni ∗ ∗ nj ∗ = gij qi ([S ] ) ^ qj ([S ] ) ∗ = gij(0) = 0 ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 40

∗ ni ∗ ∗ nj ∗ ni+nj ni nj where we use that qi ([S ] ) ^ qj ([S ] ) ∈ H (S ∨ S ; R) = 0 since ni ≥ 1 ∗ ni ∗ ∗ nj ∗ for all i, so certainly qi ([S ] ) ^ qj ([S ] ) = 0. For i = j, we have

∗ ni ∗ ∗ ni ∗ ∗ ni ∗ ni ∗ ∗ pi ([S ] ) ^ pi ([S ] ) = pi ([S ] ^ [S ] ) = pi (0) = 0. n ∗ n ∗ 2n n Since [S i ] ^ [S i ] ∈ H i (S i ; R) = 0. In order to compute more interesting examples of cup products, it will be easiest to use Poincaréduality. In order to do this, we need to describe the chain level map that induces the Poincaréduality isomorphism. For this, we need cap product. So we study cap product next, then we can state Poincaréduality more precisely in terms of cap products, then we will use Poincaréduality to compute cup products, and from there derive interesting geometric conclusions about topological spaces. After that we will give a proof of the F2 coefficient version of Poincaréduality.

8. Cap products In this section we define cap products. Let R be a commutative ring. The cap product is an R-bilinear function k _: H (X; R) × H`(X; R) → H`−k(X; R) As with cup product, we define that cap product on the chain and cochain level, then we show that these maps induce well-defined maps on homology and cohomology. Then we show that these maps have the certain useful properties, in particular the cap and cup product interact in a useful way. As mentioned above, the Poincaré duality map is defined in terms of cap products.

8.1. Definition of cap products. Definition 8.1. Let ϕ ∈ Ck(X; R) be a singular cochain. Let σ : ∆` → X with k ≤ `. Then we define

ϕ _ σ = ϕ(σ|[0,...,k]) ⊗ σ|[k,...,`] ∈ R ⊗ Ck−`(X) = Ck−`(X; R).

Extend to all of C`(X; R) by linearity. This gives rise to a product k _: C (X; R) × C`(X; R) → C`−k(X; R) when ` ≥ k and if k > ` then we take the cap product to be the zero map by deﬁnition. To remember what is happening, you can think that the cochain is hungry, with appetite k, and the ` simplex is food. The cochain eats up k of the simplex, leaving an ` − k simplex left. The next lemma will be key for showing that the cap product descends to a well-deﬁned map on homology/cohomology.

k Lemma 8.2. Let ϕ ∈ C (X; R) and let σ ∈ C`(X; R). Then k ∂(ϕ _ σ) = (−1) (−δϕ _ σ + ϕ _ ∂σ) ∈ C`−k−1(X; R). ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 41

Proof. Assume that k ≤ `, or else this just says that 0 = 0. Let σ : ∆` → X be a singular `-simplex. We compute the three terms individually: (8.3) k ` X X ϕ _ ∂σ = (−1)iϕ(σ| )σ| + (−1)iϕ(σ| )σ| ; [0,...,bi,...,k+1] [k+1,...,`] [0,...,k] [k,...,bi,...,`] i=0 i=k+1 (8.4) δϕ _ σ = δϕ(σ|[0,...,k+1])σ|[k+1,...,`] = ϕ(∂σ|[0,...,k+1])σ|[k+1,...,`] k+1 X = (−1)iϕ(σ| )σ| ;(8.5) [0,...,bi,...,k+1] [k+1,...,`] i=0 (8.6) k k k (−1) ∂(ϕ _ σ) = (−1) ∂(ϕ(σ|[0,...,k])σ|[k,...,`]) = (−1) ϕ(σ|[0,...,k])∂σ|[k,...,`] ` X (8.7) = (−1)iϕ(σ| )σ| . [0,...,k] [k,...,bi,...,`] i=k When i = k + 1 in the (8.5) cancels with i = k in (8.7). Then we are left with (8.5) + (8.7) = (8.3). That is, ϕ _ ∂σ = δϕ _ σ + (−1)k∂(ϕ _ σ), which rearranges to the statement of the lemma. Lemma 8.8. The cap product induces a defined and well defined map k H (X; R) × H`(X; R) → H`−k(X; R) ([ϕ], [σ]) 7→ [ϕ _ σ] for all k, ` ∈ N0. k Proof. Let ϕ ∈ C (X; R) be a cocycle, so that δϕ = 0, and let σ ∈ C`(X; R) a cycle with ∂σ = 0. Then ∂(ϕ _ σ) = (−1)k(−δϕ _ σ + ϕ _ ∂σ) = (−1)k(−0 _ σ + ϕ _ 0) = 0. So ϕ _ σ is a cycle. Then we want to check that cap product is well-defined. [(ϕ + δχ) _ (σ + ∂τ)] = [ϕ _ σ + δχ _ σ + δχ _ ∂τ + ϕ _ ∂τ] =[ϕ _ σ ± χ _ ∂σ ± ∂(χ _ σ) ± χ _ ∂2τ ± ∂(χ _ ∂τ) ± δϕ _ ∂τ + ∂(ϕ _ τ)] =[ϕ _ σ ± χ _ 0 ± ∂(χ _ σ) ± χ _ 0 ± ∂(χ _ ∂τ) ± 0 _ ∂τ + ∂(ϕ _ τ)] =[ϕ _ σ + ∂(±χ _ σ ± χ _ ∂τ ± ϕ _ τ)] =[ϕ _ σ]. So the homology class produced does not depend on the cocycle representative for [ϕ], nor does it depend on the cycle representative for σ. Here are two basic lemmas on cap product: the identity of cohomology acts as identity on homology, and a special case where cap product is the same as evaluation. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 42

0 Lemma 8.9. Let σ ∈ Hk(X; R) and let 1X ∈ H (X; R) be the identity in the cohomology ring. Then 1X _ σ = σ. Proof.

1X _ σ = 1X (σ[0])σ|[0,...,k] = 1 · σ = σ. Lemma 8.10. Suppose that X is connected. Then k ∼ _: H (X; R) × Hk(X; R) → H0(X; R) = R ([ϕ], [σ]) 7→ hϕ, σi.

Proof. Let IX denote a 0-simplex that generates H0(X; R). We compute:

[ϕ _ σ] = [ϕ(σ|[0,...,k])σ[k]]

= hϕ, σiIX 7→ hϕ, σi ∈ R. 8.2. An example. Example 8.11. We compute some cap products in T 2 = S1 × S1. We use the same conventions as in Example 7.6. We repeat our diagram of the torus, and we use the same singular 2-simplices σ1 and σ2. We also show the projections p, q : S1 × S1 → S1. 1 1 1 1 Let θ ∈ C (S ; Z) represent a generator [θ] ∈ H (S ; Z). We have singular 1- 1 1 1 1 simplices µ, ν : ∆ → S , with [µ] ∈ H1(S ; Z) a generator and [ν] = 0 ∈ H1(S ; Z). 2 2 We also have that [−σ1 + σ2] = [T ] ∈ H2(T ; Z) is a fundamental class. 1 2 We compute the cap product of a generating set of H (T ; Z) with the fundamental class [T 2]. ∗ ∗ p (θ) _ σ1 = p (θ)(σ1|[0,1])σ1|[1,2]

= θ(p ◦ σ1|[0,1])σ1|[1,2]

= θ(ν)σ1|[1,2] = 0 On the other hand ∗ ∗ p (θ) _ σ2 = p (θ)(σ2|[0,1])σ2|[1,2]

= θ(p ◦ σ2|[0,1])σ2|[1,2] 1 = θ(µ)σ2|[1,2] = σ2|[1,2] = [1 × S ]. Butting these together we obtain ∗ 2 ∗ ∗ 1 [p (θ)] _ [T ] = −p (θ) _ σ1 + p (θ) _ σ2 = [1 × S ]. Similarly, we can compute that [q∗(θ)] _ [T 2] == −[S1 × 1]. The map 2 1 2 2 − _ [T ]: H (T ; Z) → H1(T ; Z) ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 43

Figure 3. A diagram of the torus. Identify the vertical edges with each other and the horizontal edges with each other to get the torus. The projections p, q : S1 × S1 → S1 are shown. The images of two singular 2-simplices are shown. At the bottom, a standard 2-simplex ∆2 is drawn. is in fact an isomorphism, with p∗(θ) 7→ [1 × S1] q∗(θ) 7→ −[S1 × 1]. This map is in fact the Poincar´eduality map, so it had to have been an isomorphism. 8.3. Properties of cap product. An important formula for the next section will be the following formula involving both the cup and the cap product. Theorem 8.12 (Cup-cap formula). Let X be a topological space, and let R be a k ` commutative ring. Let ϕ ∈ C (X; R), let ψ ∈ C (X; R) and let σ ∈ Cn(X; R). Then

ϕ _ (ψ _ σ) = (ψ ^ ϕ) _ σ ∈ Cn−k−`(X; R). Note that the order of the ϕ and the ψ switch. We could change them back, but then we would have to introduce a sign (−1)k`. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 44

Proof. Suppose that k + ` ≤ n, otherwise the statement is trivial since both sides are zero. Let σ : ∆n → X be a singular n-simplex. Then: ϕ _ (ψ _ σ) = ϕ _ ψ(σ|[0,...,`]) · σ|[`,...,n]

= ψ(σ|[0,...,`]) · ϕ(σ|[`,...,`+k]) · σ|[`+k,...,n]

= (ψ ^ φ)(σ|[0,...,`+k]) · σ|[`+k,...,n] = (ψ ^ ϕ) _ σ.

Extending this by linearity to general σ ∈ Cn(X; R) gives the result. We will also need to know the behaviour of cap product with respect to maps between spaces.

k Theorem 8.13. Let f : X → Y be a map of spaces. Let ϕ ∈ C (Y ; Z) and let σ ∈ Cn(X; Z). Then ∗ f∗(f (ϕ) _ σ) = ϕ _ f∗(σ). Proof. Let σ : ∆n → X. As usual we prove the theorem for this σ and extend by linearity. We compute: ∗ ∗ f∗(f (ϕ) _ σ) = f∗(f (ϕ)(σ|0,...,k)σ|[k,...,n]

= f∗ϕ(f ◦ σ|0,...,k)σ|[k,...,n]

= ϕ(f ◦ σ|0,...,k)f ◦ σ|[k,...,n]

= ϕ _ f∗(σ).

9. Applications of cup and cup products combined with Poincare´ duality One of the main reasons for introducing the cap product is its rôlein the formu- lation of the Poincaréduality theorem. In the definition of orientability and of fundamental classes we used Z coefficients. Replace Z in those definitions with any commutative ring R, to obtain the notion of R-orientable manifolds and an R-fundamental class, which is a generator for Hn(M; R), where M is a closed R-orientable n-manifold. Theorem 9.1 (PoincaréDuality). Let R be a commutative ring. Let M be a closed, R-orientable n-dimensional manifold. Let [M] ∈ Hn(M; R) be an R-fundamental class. Then for every r ∈ N0, the map ∼ r = PD : H (M; R) −→ Hn−r(M; R) [ϕ] 7→ [ϕ _ [M]] is an isomorphism.

We will give a proof (in the case R = F2 only) later. Example 9.2. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 45

(1) For Sn, we have n n n =∼ n ∼ − _ [S ]: H (S ; R) = R −→ H0(S ; R) = R is an isomorphism. (2) We also computes the cap product with the fundamental class of the torus: 0 1! −1 0 2 1 2 ∼ 2 2 ∼ 2 − _ [T ]: H (T ; Z) = Z −−−−−−−→ Z = H1(T ; Z) which is also an isomorphism. Now we start to discuss some applications of the whole theory. It has been quite a lot of work to develop homology, cohomology, cup and cap products, together with universal coeﬃcients and Poincar´eduality. So now we start to get some rewards for our labour. 9.1. Degree one maps. Proposition 9.3. Let f : M → N be a degree one map between n-dimensional manifolds. Then for all k ∈ N0 there is a direct summand of Hk(M; Z) isomorphic to Hk(N; Z). Proof. Consider the following diagram.

∗ n−k f n−k H (M; Z) o H (N; Z)

=∼ −_[M] =∼ −_[N]

f∗ Hk(M; Z) / Hk(N; Z). n−k We claim that it commutes. To see this let ϕ ∈ H (N; Z). Then ∗ f∗(f (ϕ) _ [M]) = ϕ _ f∗([M]) = ϕ _ [N]. The ﬁrst equality is the functoriality for cap product (Theorem 8.13), and the second is that f is degree one so f∗([M]) = [N]. Let ! f : Hk(N; Z) → Hk(M; Z) be deﬁned by the composition f ! := (− _ [M]) ◦ f ∗ ◦ (− _ [N])−1. By commuta- ! tivity of the diagram above we have that f∗ ◦ f = Id. Now consider the short exact sequence

f∗ 0 → ker f∗ → Hk(M; Z) −→ Hk(N; Z) → 0. ! ! The map f : Hk(N; Z) → Hk(M; Z) is a homomorphism and satisﬁes f∗ ◦ f = Id, so is a splitting. It follows that ∼ ! Hk(M; Z) = ker f∗ ⊕ f (Hk(N; Z)). In particular the existence of a degree one map M → N implies that there is a surjective homomorphism Hk(M; Z) → Hk(N; Z) for every k ∈ N0. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 46

Example 9.4. 3 3 3 3 ∼ (i) There is no degree one map S → RP . Here H1(S ) = 0 whereas H1(RP ) = 3 3 Z/2, so there can be no surjection H1(S ) → H1(RP ). 1 2 3 1 2 ∼ (ii) There is no degree one map S × S → RP . Here H1(S × S ) = Z and 3 ∼ H1(RP ) = Z/2, and Z/2 is not a direct summand of Z. (iii) There is no degree one map from a surface of genus one to a surface of genus two, since there can be no surjective homomorphism on first homology, which 2 4 would be a surjection Z → Z . 9.2. Computations of cup products. Recall that the complex projective plane 2 CP is by definition 3 2 C r{(0, 0, 0)} CP := . (z0, z1, z2) ∼ (λz0, λz1, λz2); λ ∈ Cr{0} This can be built, as we saw in the first exercise sheet, by a CW decomposition with three cells, one cell of dimension 0,2 and 4. That is 2 0 2 4 2 4 CP = e ∪ e ∪ e = S ∪η D . 1 ∗ 2 2 ∼ It is a 4-dimensional, oriented manifold. Let ϕ := [CP ] ∈ H (CP ; Z) = Z be a 2 2 2 2 generator. Let [CP ] ∈ H4(CP ; Z) be a fundamental class. Let ψ ∈∈ H (CP ; Z). Then 2 2 2 hϕ ^ ψ, [CP ]i = (ϕ ^ ψ) _ [CP ] = ψ _ (ϕ _ [CP ]) 2 2 2 2 Now, since the map (− _ [CP ]: H (CP ; Z) → H2(CP ; Z) is an isomorphism 2 2 by Poincaréduality, we have that ϕ _ [CP ] is a generator of H2(CP ; Z), so 2 1 ϕ _ [CP ] = ±[CP ]. 1 2 1 Next, note that Ext (H1(CP ; Z), Z) = Ext (0, Z) = 0, so 2 2 2 ev: H (CP ; Z) → Hom(H2(CP ; Z), Z) is an isomorphism by the universal coefficient theorem. We have that 2 2 2 hϕ ^ ψ, [CP ]i = ψ _ (ϕ _ [CP ]) = ev(ψ)(ϕ _ [CP ]) Taking ψ = ϕ, we have 2 2 1 ∗ 1 hϕ ^ ϕ, [CP ]i = ev(ϕ)(ϕ _ [CP ]) = [CP ] (±[CP ]) = ±1 2 ∗ It follows that ϕ ^ ϕ = ±[CP ] is a dual fundamental class. In particular the cup product 2 2 2 2 4 2 ^: H (CP ; Z) × H (CP ; Z) → H (CP ; Z) is nontrivial. In general, ψ = nϕ for some n ∈ Z, and so 2 2 1 ∗ 1 hϕ ^ ψ, [CP ]i = ev(nϕ)(ϕ _ [CP ]) = n · [CP ] (±[CP ]) = ±n. 2 4 2 Proposition 9.5. The spaces S ∨ S and CP are not homotopy equivalent. Note that the homology and cohomology of these spaces coincide, namely a Z in degrees 0,2, and 4. So one really needs cup products to distinguish the homotopy types of these spaces. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 47

2 4 2 Proof. Suppose that there is a homotopy equivalence f : S ∨ S → CP . Then we have a commutative diagram as follows:

2 2 2 2 ^ 4 2 H (CP ) × H (CP ) / H (CP )

f ∗×f ∗ f ∗

^ H2(S2 ∨ S4) × H2(S2 ∨ S4) / H4(S2 ∨ S4) This commutes by functoriality of the cup product. Since f is a homotopy equiva- ∼ 2 2 lence, the vertical maps are isomorphisms. Start with (1, 1) ∈ Z × Z = H (CP ) × 2 2 4 2 4 2 4 H (CP ). This maps to ±1 ∈ H (CP ) and therefore to ±1 in H (S ∨ S ). On the other hand (1, 1) maps to ±(1, 1) ∈ H2(S2 ∨ S4) × H2(S2 ∨ S4). Since the cup product of wedges of spheres vanishes, this maps to 0 ∈ H4(S2 ∨ S4). This violates f ∗(ϕ) ^ f ∗(ϕ) = f ∗(ϕ ^ ϕ), i.e. the commutativity of the diagram. So not such homotopy equivalence f can exist. 2 2 4 4 3 2 Recall that CP = S ∪η D . The 4-cell D is attached by a map η : S → S . This is a famous map called the Hopf map. If this map were null homotopic, then 2 4 2 4 we would have that S ∪η D ' S ∨ S . Since we just showed this is not the case, we must have that the map η : S3 → S2 is nontrivial. The set of (based) homotopy classes of maps from S3 to S2 form a group, a higher dimensional analogue of the 2 2 fundamental group, called π3(S ). We have just shown that π3(S ) is nontrivial. 2 ∼ In fact π3(S ) = Z, and η is a generator, but we shall not prove that here. Example 9.6. We compute the cup products of S2 × S2. 2 2 2 ∼ 2 2 2 2 ∼ 2 4 2 2 ∼ ^: H (S × S ) = Z × H (S × S ) = Z → H (S × S ) = Z 2 2 2 This can be represented by a 2×2 matrix. Let pi : S ×S → S be the projection to 2 2 ∗ ∗ the ith factor and let θ ∈ H (S ) be a generator. Then p1(θ), p2(θ) are generators 2 2 2 ∼ 2 of H (S × S ) = Z . We have that ∗ ∗ ∗ ∗ pi (θ) ^ pi (θ) = pi (θ ^ θ) = pi (0) = 0. So the matrix looks like 0 x . x 0 We need to ﬁnd x. We know that the oﬀ-diagonal entries are equal by the symmetry 2 2 2 ∼ 2 of the cup product. Let ϕ = (1, 0) and let ψ = (0, 1) in H (S × S ) = Z . Then (ϕ ^ ψ) _ [S2 × S2] = ψ _ (ϕ _ [S2 × S2]) = ev(ψ)(ϕ _ [S2 × S2]) = m for some m ∈ Z, that we shall determine. This m is such that ϕ _ [S2 × S2] = n · [S2 × pt] + m · [pt ×S2]. Note that 0 = (ϕ ^ ϕ) _ [S2 × S2] == ev(ϕ)(ϕ _ [S2 × S2] = n. 2 2 2 2 2 So ϕ _ [S × S ] = m · [pt ×S ]. Since ϕ is a generator of Z , and − _ [S × 2 2 2 2 2 2 2 S ]: H (S × S ) → H2(S × S ) is an isomorphism, then m · [pt ×S ] must also ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 48

2 be a generator. For the vector (m, 0) to be part of a generating set of Z , there 2 must be a (y, z) ∈ Z with m y 0 z having determinant ±1. But the determinant is my, and for this to be ±1 we must have m = ±1. So ϕ ^ ψ = ±[S2 × S2]∗ ∗ and the matrix for the cup product, with respect to the chosen basis pi (θ), is 0 ±1 . ±1 0 We could change our choice of fundamental class to its negative, if we want to remove minus signs that might appear here. 2 2 2 2 Proposition 9.7. The spaces S ×S and CP #CP are not homotopy equivalent. 2 Proof. Both spaces are closed, orientable 4-manifolds with homology Z, 0, Z , 0Z. With respect to the usual bases, the cup product on S2 × S2 is represented by 0 1 1 0 while the cup product on 2 is represented by . Suppose that 1 0 CP 0 1 2 2 2 2 2 2 2 there is a homotopy equivalence S ×S → CP #CP . Let β = (1, 0) ∈ H (CP #CP ) 2 2 ∗ be a generator: then β ^ β = ±[CP #CP ] . Therefore ∗ ∗ ∗ ∗ 2 2 ∗ 2 2 ∗ f (β) ^ f (β) = f (β ^ β) = f (±[CP #CP ] ) = ±[S × S ] . This is an odd multiple ±1 of the dual fundamental class. Suppose that f ∗(β) = 2 2 2 ∼ 2 ∗ ∗ (a, b) for some (a, b) in H (S × S ) = Z . Then f (β) ^ f (β) is computed by 0 1 a a a b = b a = 2ab. 1 0 b b This is even for every (a, b), so can never equal an odd multiple of the dual fun- 2 2 2 2 damental class. This contradiction yields that S × S and CP #CP are not homotopy equivalent. n Example 9.8. Here is an application of cup products on CP . We want to show m n that there is no retract r : CP → CP , when m > n ≥ 1. For a subspace X ⊆ Y , with i: X → Y the inclusion map, a retract is a continuous map r : Y → X with r ◦ i = Id: X → X. m n Assume for a contradiction that there is a retract r : CP → CP . Then r◦i = Id implies that

∗ ∗ ∗ 2 n r∗ 2 m i∗ 2 n i ◦ r = Id : H (CP ) −→ H (CP ) −→ H (CP ). Since i∗ and Id are isomorphisms on second cohomology, we deduce that so is r∗. 1 ∗ 2 n ∼ ∗ 1 ∗ Let x := [CP ]n ∈ H (CP ) = Z be a generator. Then r (x) = ±[CP ]m is a 2 m m 2m n generator of H (CP ). Therefore, since x ∈ H (CP ) = 0, we have ∗ ∗ m ∗ m 1 ∗ m m ∗ 2m m 0 = r (0) = r (x ) = r (x) = (±[CP ] ) = ±[CP ] 6= 0 ∈ H (CP ). This is a contradiction, so no retraction r exists, as desired. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 49

10. Proof of F2 coefficient Poincare´ duality

We shall outline a proof of the Poincaréduality theorem with F2 coefficients, using simplicial structures. This gives a nice intuition of the idea behind Poincaré duality, without having to go into cohomology with compact supports, as in e.g. Hatcher. For the full proof we refer to [Hat] or [Br].

Theorem 10.1. Let M be a closed n-dimensional manifold. For every k ∈ N0, there is an isomorphism

n−k =∼ PD : H (M; F2) −→ Hk(M; F2). Before we give a detailed sketch of the proof we introduce some notation, as well as the notion of barycentric subdivision. Recall that the m-simplex is m m+1 X ∆ := {(t0, . . . , tm) ∈ R | ti = 1}. i Deﬁnition 10.2. (1) The barycentre of ∆m is 1 1 B := ( ,..., ) ∈ ∆m. m m + 1 m + 1 m (2) Let P0,...,Pk ∈ ∆ be a collection of points. Then we deﬁne a map k m [P0,...,Pk]: ∆ → ∆ Pk (t0, . . . , tk) 7→ i=0 tiPi m (3) Let hP0,...,Pki ⊆ ∆ denote the image of the map [P0,...,Pk].

(4) Let 0 ≤ n1 < ··· < nr ≤ m. Then hvn1 , . . . , vnr i an r-dimensional face of ∆m. r m (5) The image of Br under [vn1 , . . . , vnr ]: ∆ → ∆ is the barycentre of the face. Given a face τ we write τ for the barycentre of τ. (6) The set of singular simplices k m [σ0, . . . , σk]: ∆ → ∆ | σ0 ⊂ σ1 ⊂ · · · ⊂ σk ⊂ σ is the barycentric subdivision of ∆m. Note that the barycentric subdivision of a face is equal to the barycentric subdivision of ∆m restricted to a face. Let M be a closed, oriented n-manifold. Then M admits a simplicial structure n {ϕi : ∆ i → M}i∈I . By deﬁnition, this is a collection of maps with: n (1) ϕi : ∆ i → M injective. (2) For all x ∈ M, there exists a unique i with x in the interior of the image of ϕi. (3) Every face of a simplex is again a simplex. (4) If ϕi and ϕj intersect, they do so in a face. n Given P0,...,Pk ∈ ϕi(∆ i ), deﬁne ϕ −1 −1 k ni i [P0,...,Pk] := ϕi ◦ [ϕi (P0), . . . , ϕi (Pk)]: ∆ → ∆ −→ M. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 50

Figure 4. Barycentric subdivision of a 2-simplex.

Proof of Theorem 10.1. Here is a detailed sketch of a proof. Let M be a closed, n oriented n-manifold. Choose a simplicial structure {ϕi : ∆ i → M}i∈I for M. We have a barycentric subdivision of this simplicial structure, obtained by applying the barycentric subdivision deﬁned above to every simplex in M. This makes sense by the observation made above that the barycentric subdivision of a face is equal to the barycentric subdivision of ∆m restricted to a face. Now let σ be a k simplex of M. Then deﬁne the dual cell to σ as the union

d [ σ := hσ0, σ1, ··· , σsi. σ=σ0⊂σ1⊂···⊂σs The boundary of the dual cell is

d [ ∂σ := hσ1, ··· , σsi. σ⊂σ1⊂···⊂σs There are two claims in this proof, whose proofs we shall omit. These omissions are why this is just a sketch of the proof. Claim. (σd, ∂σd) =∼ (Dn−k,Sn−k) ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 51

Figure 5. A triangulation of a hexagon and its barycentric subdivision are shown. A 0-simplex σ and a 1-simplex τ are chosen, and their dual cells are depicted.

The proof of this claim uses crucially that M is a manifold. It is not too hard to construct topological spaces with a simplicial structure where this claim would fail. Claim. Suppose that σ and τ are simplices. Then τ is a face of σ if and only if σd ⊂ τ d.

We have two CW complex structures on M. Let Sk(M) be the set of k-simplices in {ϕi}. Then the CW complex X is the space M with CW structure having as k-cells the elements of Sk(M). On the other hand the CW complex Y is the space M with k-cells d {σ | σ an (n − k)-simplex in Sn−k(M)}. This is a CW structure by the two claims above. Given a k-cell e in Y , its dual ed is the simplex σ = ed with σd = e. Now let σ ∈ Sk(M) and deﬁne CW PD(σ): Cn−k(Y ) → F2 to be the map determined by sending σd to 1 and all other (n − k)-cells to 0. Claim. The map CW CW PD : Ck (X) ⊗ F2 → Hom(Cn−k(Y ), F2)

Pm Pm i=1 σi ⊗ ai 7→ i=1 aiPD(σi) induces isomorphisms

CW =∼ n−k Hk (X; F2) −→ HCW (Y ; F2) for every k ∈ N0. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 52

By the claim, ∼ CW =∼ n−k ∼ k Hk(M; F2) = Hk (X; F2) −→ HCW (Y ; F2) = H (M; F2) since singular and CW homology are isomorphic. This completes the proof of Poincar´eduality modulo the claims. We indicate the proof of this last claim. The inverse to the map is CW CW Hom(Cn−k(Y ), F2) → Ck (X) ⊗ F2 f 7→ P σ ⊗ f(σd). σ∈Sk(M) You should check that this indeed an inverse. Therefore the maps are isomorphisms between the chain groups. To see that they induce a chain isomorphism, we need to see that the following diagram commutes.

CW PD CW Ck (X) ⊗ F2 / Hom(Cn−k(Y ), F2)

∂ ∂∗ CW PD CW Ck−1 (X) ⊗ F2 / Hom(Cn−(k−1)(Y ), F2) Then the map PD is chain map and an isomorphism on chain groups, so is a chain isomorphism and so induces an isomorphism between homology and cohomology. To see that the diagram commutes, let σ ∈ Sk(M). We want to show that ∗ CW PD(∂σ) = ∂ (PD(σ)) ∈ Hom(Cn−(k−1)(Y ), F2). d That is, for every τ ∈ Sk−1(M), we have an (n − (k − 1))-cell τ , and we want that d d (10.3) PD(∂σ)(τ ) = PD(σ)(∂τ ) ∈ F2. d Let y1, . . . , ym be the (n−k)-cells in the boundary of τ , which is an (n−k+1)-cell. The only possible (n − k)-cells in τ d are in its boundary. Then we have: PD(∂σ)(τ d) = 1 ⇔τ is a face of σ d ⇔ there is an i with yi = σ d ⇔ there is an i with yi = σ ⇔PD(σ)(∂τ d) = 1. This proves (10.3), which completes the proof of the claim, and therefore completes our sketch proof of F2 coeﬃcient Poincar´eduality.

References [Ad] J. F. Adams, Algebraic Topology; a student’s guide. [Br] G. E. Bredon, Topology and Geometry, Graduate Texts in Mathematics 139, Springer- Verlag New York, (1993). [DK] J. Davis and P. Kirk, Lecture notes on algebraic topology. [Fr] S. Friedl, Algebraic topology lecture notes. [Hat] A. Hatcher,Algebraic Topology, Cambridge University Press (2002). [May] J. P. May, A concise course in algebraic topology. [Sp] E. H. Spanier, Algebraic Topology, McGraw-Hill, 1966. ALGEBRAIC TOPOLOGY IV k EPIPHANY TERM LECTURE NOTES 53

Department of Mathematical Sciences, Durham University, United Kingdom E-mail address: [email protected]