121B: ALGEBRAIC TOPOLOGY 1. Introduction and Examples in The

1. Introduction and examples In the first week I am doing the definition of simplicial cohomology and some examples. 1.1. ∆-complex. Here is a formal definition of a ∆-complex.

Deﬁnition 1.1. A ∆-complex X is a sequence of sets X0,X1, ··· (Xn is the set of n-simplicies of X) and mappings ∂i : Xn → Xn−1 for i = 0, ··· , n (∂i(σ) is the ith face of σ given by deleting the ith vertex) so that:

∂i∂j = ∂j−1∂i if i < j. (If you delete the ith vertex then the jth vertex becomes the j − 1st vertex.)

Problem: Find a formula for the ith vertex of σ ∈ Xn. Answer: A ∆-complex X is an (ordered) simplicial complex if

(1) Every n-simplex σ ∈ Xn has n + 1 distinct vertices. (2) σ ∈ Xn is uniquely determined by its set of vertices. Under these conditions we can write

σ = [v0, v1, ··· , vn] The ith boundary map is given by

∂i[v0, v1, ··· , vn] = [v0, v1, ··· , vbi, ··· , vn] Hatcher uses this notation even if the vertices are all equal. This is meant to be an intuitive description of σ and its faces. 1.2. ∆-cohomology. If G is an additive group and X is a ∆ complex then an n-cochain n on X with coefficients in G is any mapping φ : Xn → G. These form a group ∆ (X; G) by pointwise addition: (φ + ψ)(x) = φ(x) + ψ(x). Let δ : ∆n(X; G) → ∆n+1(X; G) be given by P i δφ(x) = (−1) φ(∂ix). In Hatcher’s notation: n X i (δφ)[v0, v1, ··· , vn] = (−1) φ[v0, ··· , vbi, ··· , vn] i=0 Elements in the kernel of δ : ∆n(X; G) → ∆n+1(X; G) are called n-cocycles on X with coefficients in G and elements in the image are the n + 1 coboundaries. Zn(X; G) = {φ ∈ ∆n(X; G) | δφ = 0} Bn(X; G) = {φ ∈ ∆n(X; G) | φ = δψ for some ψ ∈ ∆n−1(X; G)} We have the inclusion of additive groups: Bn(X; G) ⊆ Zn(X; G) ⊆ ∆n(X; G). The nth cohomology group of X with coefficients in G is defined to be the quotient: Zn(X; G) Hn(X; G) := Bn(X; G) 1 Elements of Hn(X; G) are written [φ] = φ + Bn(X; G) 1.3. Cohomology of a sphere. The problem is to compute the cohomology of the sphere. A ∆-complex model for the n − 1 sphere is ∂∆n. I started by pointing out that the reduced homology of the n-simplex is trivial since ∆n is contractible. ∗ Theorem 1.2. H (∆n) = 0. We actually need the algebraic version of this. Namely, the augmented cellular chain complex is contractible. Lemma 1.3. An exact sequence of free abelian groups is chain contractible. You are supposed to do this as homework. You are given an exact sequence: ∂ ∂ ∂ 0 ← C−1 ←− C0 ←− C1 ←− C2 ← · · ·

where all groups Cn are free abelian. You need to find a chain contraction, i.e., a sequence of homomorphisms hn : Cn → Cn+1 so that h∂ + ∂h = id You may use the fact that every subgroup of a free abelian group is free. Corollary 1.4. The augmented chain complex of ∆n is chain contractible. I pointed out that the first and last terms of the cellular chain complex of ∆n are both Z: n n n 0 ← Z ← C0(∆ ) ← C1(∆ ) ← · · · ← Cn(∆ ) ← 0 | {z } =Z To finish we need the following lemma: ∗ Lemma 1.5. If C∗ is a contractible chain complex then the cochain complex C (C∗; G) is also contractible for any (additive) group G. ∗ The cochain complex C (C∗; G) has groups k C (C∗; G) := Hom(Ck,G) connected by a sequence of coboundary maps

k δ k+1 C (C∗; G) = Hom(Ck,G) −→ Hom(Ck+1,G) = C (C∗; G) given by the dual of (precomposition with) the boundary operator: δ(φ) = ∂∗(φ) = φ ◦ ∂

k+1 Ck+1 δ(φ) = φ∂ ∈ C (C∗; G) @ δ(φ) 6 ∂ @ δ ? @@R - k Ck G φ ∈ C (C ; G) φ ∗

∗ ∗ Proof. The chain contraction for C (C∗; G) is given by the dual h of the chain contraction h for C∗: h∗δ + δh∗ = (∂h + h∂)∗ = id∗ = id This uses the fact that Hom(−,G) is a contravariant functor. 2 Theorem 1.6. The cohomology of ∂∆n is given by Hk(∂∆n; G) = G for k = 0, n − 1 and Hk(∂∆; G) = 0 otherwise. Proof. We have the following exact sequence:

0 δ0 1 δn−2 n−1 n 0 → Hom(Z,G) → C (C∗; G) −→ C (C∗; G) → · · · −−→ C (C∗; G) → C (C∗; G) → 0 | {z } ∗ The middle terms (*) are the same as the cochain complex ∆∗(∂∆n; G). Since the entire sequence is exact, the subsequence (*) is exact except at the beginning and end where 0 n ∼ H (∂∆ ; G) = ker δ0 = Hom(Z,G) = G and n−1 n n ∼ H (∂∆ ; G) = coker δn−1 = Hom(Cn(∆ ),G) = Hom(Z,G) = G k n The exactness at other points gives H (∂∆ ; G) = 0 for k 6= 0, n − 1.

2. Universal coefficient theorem Hatcher proves the UCT twice: once conceptually and again computationally (the traditional method). I’ll just do the ﬁrst one. Here is the outline.

Lemma 2.1. Two free chain complexes C∗,D∗ are homotopy equivalent iﬀ they have the same (i.e., isomorphic) homologies. In HW1 you are doing the easiest case when the homology is trivial. L α ∼ L α Lemma 2.2. Homology commutes with direct sum: Hk( α C∗ ) = α Hk(C∗ )

In the case when each homology group Hk(C∗) if ﬁnitely generated, it is the direct sum of Z’s and Z/n’s. So, we only need to do two examples: C∗ = · · · ← 0 ← Z ← 0 ← · · · |{z} k k with homology Hk(C∗) = Z and cohomology H (C∗; G) = G and n D∗ = · · · ← 0 ← Z ←− Z ← 0 ← · · · |{z} |{z} k k+1

with homology Hk(C∗) = Zn and cohomology given by the following cochain complex: ∗ n C (D∗; G) = · · · → 0 → G −→ G → 0 → · · · |{z} |{z} k k+1 Which has cohomology: k H (D∗; G) = {g ∈ G | ng = 0} k+1 H (D∗; G) = G/nG

3 The ﬁrst group is naturally isomorphic to Hom(Zn,G). The second group is isomorphic to Zn ⊗ G but the isomorphism is not natural since Zn ⊗ G is covariant in both variables. 2.1. Reduction to the free resolution case. Lemma 2.2 is obvious is you know what the symbols mean. The direct sum of chain complexes is given by

(C∗ ⊕ D∗)n = Cn ⊕ Dn

⊕ C D ∂n = ∂n ⊕∂n ? ? (C∗ ⊕ D∗)n−1 = Cn−1 ⊕ Dn−1

L α L α Or, more generally, ( C )n = Cn with boundary map defined coordinatewise: ⊕ ∂ (··· , xα, xβ, ··· ) = (··· , ∂xα, ∂xβ, ··· ) The kernel and cokernels of ∂⊕ are also given coordinatewise: ⊕ C D α ker ∂ = ker ∂ ⊕ ker ∂ = ⊕Zn ⊕ C D α coker ∂ = coker ∂ ⊕ coker ∂ = ⊕Bn with homology M α M α M α ∼ M α Hn( C∗ ) = Zn / Bn = Hn(C∗ ) Lemma 2.1 has an obvious part and a nonobvious part. The obvious part, or at least the part that you are already supposed to know, is the following. (1) (⇒) Homotopy equivalent chain complexes have isomorphic homologies. (2) Next, C ' D implies that the cochain complexes are homotopy equivalent: ∗ ∗ C (C; G) = Hom(C∗,G) ' C (D; G) = Hom(D∗,G) ∗ ∼ ∗ (3) This implies that H (C∗; G) = H (D∗; G). Therefore, once we prove Lemma 2.1 we will know the following. Theorem 2.3. The homology of a free chain complex determines its cohomology with any coefficients. We broke the (nontrivial part of) Lemma 2.1 into two parts and proved the first part. Step 1. Every free chain complex C∗ is isomorphic to a direct sum of free chain complexes n C∗ of the following form:

n ∂n+1 (2.1) C∗ : · · · ← 0 ← Dn ←−−− Dn+1 ← Dn+2 ← 0 ··· n where D = C∗ is exact except possibly at Dn. If the n-homology is H then we have a long exact sequence:

∂n+1 (2.2) 0 ← H ←− Dn ←−−− Dn+1 ← Dn+2 ← 0 ··· The sequence (2.1) is called a free resolution of H. Step 2. Any two free resolutions of H are chain homotopy equivalent. To do step 1, we broke up any free chain complex

∂n ∂n+1 C∗ : · · · ← Cn−1 ←− Cn ←−−− Cn+1 ← · · · 4 at the point Cn. This uses the following short exact sequence:

∂n 0 ← Bn−1 ←− Cn ←-Zn ← 0 The exactness of this sequence follows from the deﬁnitions of the symbols. Now use the fact that Proposition 2.4. Every subgroup of a free abelian group is free abelian.

Thus Bn−1 is free which means that the above short exact sequence splits. I.e., there is a homomorphism s : Bn−1 → Cn so that ∂ns = idB. This makes Cn = sBn−1 ⊕ Zn. But, the image of ∂n+1 : Cn+1 → Cn lies in Zn and ∂n : Cn → Cn−1 is zero on Zn. Therefore the chain complex C∗ is isomorphic to the direct sum of the chain complexes:

∂n · · · ← Cn−1 ←− sBn−1 ← 0 ← · · ·

∂n+1 · · · ← 0 ← Zn ←−−− Cn+1 ← · · · The ﬁrst sequence is exact at degree n and above. The second sequence is zero in degrees < n. So, if we do this at every place where the homology of C∗ is nonzero, we get a decomposition of C∗ as desired. The traditional method is to do the above construction at every Cn, replacing each Cn with sBn−1 ⊕ Zn to get a direct sum of the free chain complexes:

∂n+1 · · · ← 0 ← Zn ←−−− sBn ← 0 ← · · ·

2.2. Free resolutions and Ext. It remains to explain the following important theorem. Theorem 2.5. Any abelian group has a free resolution and any two free resolutions are chain homotopy equivalent. Serge Lang, in his book Algebra, said “Take any book on homological algebra, and prove all the theorems without looking at the proofs in that book.” Homological algebra has advanced a lot in 40 years. However, it is probably good advice to prove the basic theorems yourself. You just need to know the exact statements of the theorems.

Proof. The existence of a free resolution is easy. Suppose that G is any abelian group. Then there exists a free abelian group F which maps onto G. (You could, e.g., take F to be the free group generated by all the elements of G.) The kernel of the epimorphism F → G is a subgroup R of F which is also free since all subgroups of free abelian groups are free. Then (2.3) 0 ← G ← F ← R ← 0 is a free resolution of G. This is called a free presentation of G. (A free resolution with Cn = 0 for n ≥ 2.) The following lemma shows the uniqueness of free resolutions up to homotopy: Given any two free resolutions C∗,D∗ of the same group G, the identity mapping G → G is covered by chain maps f∗ : C∗ → D∗ and g∗ : D∗ → C∗. The composition g∗f∗ : C∗ → C∗ must be homotopic to the identity chain map C∗ → C∗ by the lemma since both cover the identity map G → G. 5 Lemma 2.6. Suppose that G, H are abelian groups with free resolutions: ∂ 0 ← G ←− C0 ←− C1 ← · · · ∂0 0 ← H ←− D0 ←− D1 ← · · ·

Suppose that f : G → H is any homomorphism. Then there exist a chain map f∗ : C∗ → D∗ so that f0 = f : C0 → H. (We say that f∗ covers f.) Furthermore, any two such chain maps are chain homotopy equivalent.

Proof. Since C0 is free, the homomorphism f : C0 → H lifts to a homomorphism f0 : C0 → 0 D0. The homomorphism f0∂ : C1 → D0 has image in ker = im ∂ . So it lifts to a map f1 : C1 → D1, and so on. If there are two chain maps f∗, g∗ covering the same homomorphism f : G → H then the 0 difference f∗ − g∗ covers 0 : G → H. So, f0 − g0 has image in ker = im ∂ and therefore lifts to a homorphism h0 : C0 → D1, etc. The mappings hn give a chain homotopy h : f∗ ' g∗. I.e., 0 ∂n+1hn + hn−1∂n = fn − gn Now we can define the functor Ext = Ext1 . Z Definition 2.7. If G, H are abelian groups then Extn(H,G) is defined to be the nth coho- 1 Z mology of the cochain complex Hom(C∗,G) where C∗ is any free resolution of H.

By Theorem 2.5 we can choose any C∗. In particular, we can take a free presentation: f 0 ← H ← C0 ←− C1 ← 0

Then Hom(C∗,G) is the chain complex:

f # 0 → Hom(C0,G) −→ Hom(C1,G) → 0 I am using the symbol f # to indicate the dual of f given by pre-composition with f because f ∗ is overused. Since this cochain complex has only two terms, it only has cohomology in degrees 0, 1. In degree 0, the kernel of f # is obviously isomorphic to Hom(H,G) since any mapping C0 → G which is zero on C1 factors uniquely through the cokernel of f : C1 → C0. Proposition 2.8. Extn(H,G) = 0 for n ≥ 2 and Ext0 (H,G) = Hom(H,G). Z Z This means that the only interesting term is Ext1 which we call simply Ext Z Ext(H,G) := Ext1 (H,G) = coker f # Z With the following theorem we can compute Ext(H,G) for any ﬁnitely generated abelian group H. Theorem 2.9. For any abelian group G we have: ∼ (1) Ext(H1 ⊕ H2,G) = Ext(H1,G) ⊕ Ext(H2,G) (2) Ext(Z,G) = 0 ∼ (3) Ext(Zn,G) = G/nG

1A cochain complex is a sequence of abelian groups Cn and homomorphisms δ : Cn → Cn+1 so that δ2 = 0. The cohomology of C∗ is the kernel of δ modulo the image of δ. 6 2.3. UCT. Now that we have the deﬁnitions, we can state and prove the universal coeﬃcient theorem for cohomology.

Theorem 2.10. If C∗ is a free chain complex and G is an abelian group then we have a natural exact sequence: n 0 → Ext(Hn−1(C∗),G) → H (C∗; G) → Hom(Hn(C∗),G) → 0 Furthermore the sequence splits, but not naturally. L (k) (k) Proof. We know that C∗ is a direct sum of chain complexes C∗ = k C∗ where C∗ is a free resolution of Hk(C∗) shifted up in degree by k. So, Y Y Hn(C ; G) ∼= Hn(C(k); G) = Extn−k(H (C ),G) ∗ ∗ Z k ∗ k k Since Extn−k = 0 except when k = n, n − 1 we have a ﬁnite product (= direct sum): Z Hn(C ; G) ∼= Ext0 (H (C ),G) ⊕ Ext1 (H (C ),G) ∗ Z n ∗ Z n−1 ∗ = Hom(Hn(C∗),G) ⊕ Ext(Hn−1(C∗),G) A more detailed description of this isomorphism shows which part is natural. I am skipping this part. The important part is the naturality of the evaluation map n ev : H (C∗; G) → Hom(Hn(C∗),G)

This map has the following natural description. Let f ∈ Hom(Cn,G) be a cocycle represent- n ing a cohomology class [f] ∈ H (C∗; G) and let z ∈ Cn be a cycle representing the homology class [z] ∈ Hn(C∗). Then ev[f] is the homomorphism Hn(C∗) → G which sends [z] to f(z). If f 0 = f + δg and z0 = z + ∂y are any other choices for f, z then f 0(z0) = (f + g∂)(z + ∂y) = f(z) + f∂(y) + g∂z + g∂2y = f(z) since f∂ = δf = 0, ∂z = 0 and ∂2 = 0. So, ev is independent of all choices, i.e., it is well-deﬁned. “Naturality” of ev is obvious if you know what it means.

2.4. Problems. Problem 1. Find a free presentations of Zn and show that ∼ Ext(Zn,G) = G/nG Conclude that G/6G ∼= G/2G ⊕ G/3G. ∼ Problem 2. Since Z6 = Z2 ⊕ Z3 we get two free presentations of Z6. Find an explicit chain homotopy equivalence between these free resolutions. Problem 3. What does it mean that the evaluation map n ev : H (C∗; G) → Hom(Hn(C∗),G) is natural? Homework 2 This is the ﬁrst problem in HW2. Find a free presentation for the additive group 1 a Z[ p ] = { b ∈ Q | a, b ∈ Z and b is a power of p} 1 (p is prime) and use it to compute Ext(Z[ p ], Z).

7 3. Cohomology of spaces Using the universal coefficient theorem and our knowledge of homology, we can compute the cohomology of spaces. The general properties of cohomology follow from the corresponding property for homology by duality. The evaluation map is also a consequence of duality: n computation of Hn ⇒UCT computation of H n properties of Hn ⇒Hom properties of H 3.1. Singular cohomology. This is derived from singular homology. If X is a topological space then recall that C∗(X) is the singular chain complex of X. Cn(X) is the free abelian group generated by the set of all continuous mappings σ : ∆n → X. These are called singular n-simplices in X and ∂ : Cn(X) → Cn−1(X) is given by alternating sum of faces. Definition 3.1. The singular cochain complex of a spaces X with coefficients in G is the n n sequence of groups C (X; G) := Hom(Cn(X); G) with coboundary map δ : C (X; G) → Cn+1(X; G) given by δ = ∂#. n X δ(φ)σ = φ(∂σ) = φ(σ | [v0, ··· , vbi, ··· , vn]) i=0

3.1.1. UCT and computations. Since C∗(X) is a free chain complex, the UCT for free chain complexes (Theorem 2.10) implies the UCT for spaces. Theorem 3.2 (UCT for spaces). For any space X and any additive group G we have a natural exact sequence:

n h 0 → Ext(Hn−1(X),G) → H (X; G) −→ Hom(Hn(X),G) → 0 where h is the evaluation map. Furthermore this sequence splits, but not naturally. The evaluation map h is given by h[φ]([z]) = φ(z). In the special case when G = F is a ﬁeld (e.g., F = Zp, Q, R, C) there is a simpler theorem: Theorem 3.3 (UCT over a ﬁeld). The evaluation map gives a natural isomorphism:

n ≈ h : H (X; F ) −→ Hom(Hn(X; F ),F ) This does not follow from the UCT for free modules (= vector spaces) over the ﬁeld F n instead of free abelian groups and the fact that ExtF (V,W ) = 0 for all n > 0. 2 2 2 Example 3.4. Take the projective plane X = P = RP . We know that H0(P ) = 1 2 n 2 Z,H (P ) = Z2 and H (P ) = 0 for n ≥ 2. So, the UCT tells us that 0 2 ∼ 2 (0) H (P ; G) = Hom(H0(P ),G) = G. 1 2 ∼ 2 2 ∼ (1) H (P ; G) = Hom(H1(P ),G) ⊕ Ext(H0(P ); G) = Hom(Z2,G). This is the group of elements of G of order 2 (or 1). 2 2 ∼ 2 (2) H (P ; G) = Ext(H1(P ),G) = Ext(Z2,G) = G/2G. 0 1 2 n In the special case when G = Z2 this gives H = H = H = Z2 and H = 0 for n > 2. This 2 agrees with the second UCT since Hn(P ; Z2) = Z2 for n = 0, 1, 2. Problem Describe H0(X; G) for any X,G. 8 3.1.2. Properties of cohomology. If X is a ∆-complex or a CW-complex, the cellular chain complex ∆∗(X) has the same homology as the singular chain complex C∗(X). Therefore, they are homotopy equivalent: ∆∗(X) ' C∗(X). So, the cochain complexes are homotopy equivalent: ∗ ∗ ∆ (X; G) = Hom(∆∗(X),G) ' Hom(C∗(X); G) = C (X; G) This proves the following theorem. ∗ Theorem 3.5. If X is a ∆-complex or a CW-complex, its cellular cohomology H∆(X; G) is isomorphic to its singular cohomology H∗(X; G). Example 3.6. The cellular chain complex of P 2 using the CW decomposition P 2 = ∗∪e1∪e2 is given by: 0 2 0 ← Z ←− Z ←− Z ← 0. ∗ 2 2 The cellular cochain complex is ∆ (P ; G) = Hom(∆∗(P ),G): 0 → G −→0 G −→2 G → 0.

If f : X → Y is a continuous mapping then we have a chain map f# : C∗(X) → C∗(Y ) given by left composition with f: f#(σ) = f ◦ σ : ∆ → Y . This induces a cochain map: f # : C∗(Y ; G) → C∗(X; G) given by right composition with f: f #(φ) = φ ◦ f. So, # f (φ)(σ) = φ(f ◦ σ) = φ(f#(σ)) The cochain map f # induces a map (homomorphism) in cohomology: f ∗ : H∗(Y ; G) → H∗(X; G) The key point is that it goes the wrong way (Y to X). So, cohomology is a contravariant functor. Theorem 3.7. If f, g : X → Y are homotopic then they induce the same map in cohomology: f ∗ = g∗ : H∗(Y ; G) → H∗(X; G)

Proof. We know that f#, g# : C∗(X) → C∗(Y ) are chain homotopic. Let h be a chain homotopy. Thus ∂h + h∂ = g# − f# Applying the functor Hom(−,G) to everything we get: h#δ + δh# = g# − f # Therefore, h# gives a homotopy f # ' g#. So, f #, g# induce the same map in cohomology. In case these symbols don’t mean anything to you, this follows from the previous equation by duality: (h#δ + δh#)(φ) = φ ◦ ∂ ◦ h + φ ◦ h ◦ ∂ = φ(∂h + h∂) # # = φ(g# − f#) = φ ◦ g + φ ◦ f = (g − f )(φ) # h ∂ φ (h δ)(φ) = φ ◦ ∂ ◦ h : Cn(X) −→ Cn+1(Y ) −→ Cn(Y ) −→ G HW2(b) Do problem 11 on p. 205. 9 3.2. Relative cohomology and long exact sequences. The deﬁnition of the relative cohomology is dual to relative homology and the long exact cohomology sequence of a pair is given by the same algebraic lemma, namely that a short exact sequence of cochain complexes n gives a long exact sequence in cohomology. (Some authors use the formal trick: C = C−n to convert a cochain complex into a chain complex.) If A is a subspace of X then we have the short exact sequence of free chain complexes:

j# 0 → C∗(A) −→ C∗(X) → C∗(X,A) → 0

It is important that C∗(X,A) is a free chain complex: Cn(X,A) is the free abelian group generated by singular n-simplices σ : ∆n → X whose images do not lie in A. When we hom this into a group G we get a short exact sequence of cochain complexes:

j# (3.1) 0 → C∗(X,A; G) → C∗(X; G) −→ C∗(A; G) → 0 The relative cohomology groups H∗(X,A; G) are the cohomology groups of the cochain complex C∗(X,A; G). 3.2.1. Interpretation of relative cohomology. Cohomology is nicer than homology in many ways. For example Cn(X,A) = Cn(X)/Cn(A) is a quotient group and elements of Cn(X,A) n n are cosets of Cn(A). But C (X,A; G) is naturally a subset of C (X; G). (It is the kernel of the restriction map j#.) Therefore, we have the following interpretation of relative cochains, cocycles and coboundaries. (1) A relative cochain φ ∈ Cn(X,A; G) is a function which assigns to each singular simplex σ : ∆n → X an element φ(σ) ∈ G so that φ(σ) = 0 if σ(∆n) ⊂ A. (2) φ is a relative cocycle if δφ = 0. I.e., Zn(X,A; G) = Cn(X,A; G) ∩ Zn(X; G) (3) φ is a relative coboundary if φ = δψ for some ψ ∈ Cn−1(X,A; G).

3.2.2. Computation of relative cohomology. Since C∗(X,A) is a free chain complex, its cohomology can be computed using the UCT. Theorem 3.8. There is a natural short exact sequence which splits unnaturally:

n h 0 → Ext(Hn−1(X,A),G) → H (X,A; G) −→ Hom(Hn(X,A),G) → 0 where h is the evaluation map:

h([φ])(z + Cn(A) + Bn(X)) = φ(z)

φ(z) ∈ G is well defined: φ = 0 on Cn(A) by definition of a relative cochain and φ = 0 on Bn(X) since δφ = 0. n n For example, if X is obtained from a space A by attaching r number of n-cells e1 , ··· , ek n−1 r along maps ηi : S → A then Hn(X,A) = Z and Hi(X,A) = 0 for i 6= n. So, n ∼ r r H (X,A; G) = Hom(Z ,G) = G 3.2.3. Long exact sequences. The short exact sequence of cochain complexes (3.1) gives a long exact sequence in cohomology, just as in the homology case: ∗ · · · → Hn(X,A; G) → Hn(X; G) −→ Hn(A; G) −→δ Hn+1(X,A; G) → · · · The mapping δ is given by diagram chasing, just like before. 10 3.2.4. Problems. (1) What is the definition of the connecting homomorphism δ in the long exact sequence? (2) Find an example of the previous example where δ : Hn−1(A; G) → Hn(X,A; G) is nonzero. (3) Show that the composition

n−1 δ n Ext(Hn−2(A),G) → H (A; G) −→ H (X,A; G) n n is zero when X = A ∪ e1 ∪ · · · ∪ er . We did (1) and (3) in class. Do (2) for homework.

11 4. Cup product If the coeﬃcient group is (the underlying additive group of) a ring R. Thus we have a multiplication map µ : R ⊗ R → R which is associative with unit. Then the cohomology groups Hn(X; R) form a graded ring L Hn(X; R). It is very important that this ring structure is deﬁned at the cochain level. I.e., L Cn(X; R) has an associative ring structure. The multiplication is call cup product.

4.1. Cup product at the cochain level. Here is my favorite formula for the cup product: φ ∪ ψ = µ ◦ (φ ⊗ ψ) ◦ ∆

k ` k+` Here φ ∈ C (X; R), ψ ∈ C (X; R) and φ ∪ ψ ∈ C (X; R) = Hom(Ck+`(X),R) is given by the composition

∆k,` φ⊗ψ µ Ck+`(X) −−→ Ck(X) ⊗ C`(X) −−→ R ⊗ R −→ R

4.1.1. The diagonal map ∆. The diagonal map is given by “front face” and “back face.”

∆k,`(σ) = fkσ ⊗ b`σ where fkσ is the front k-face of σ given by

fkσ = σ|[v0, v1, ··· , vk] and b`σ is the back `-face of σ given by

b`σ = σ|[vk, ··· , vk+`] Thus, the cup product is given by:

(φ ∪ ψ)(σ) = µ(φ ⊗ ψ)∆(σ) = µ(φ ⊗ ψ)(fkσ ⊗ b`σ) = φ(fkσ)ψ(b`σ)

= φ(σ|[v0, ··· , vk])ψ(σ|[vk, ··· , vk+`])

Lemma 4.1. We have the following identities of operators on Ci+j+k(X).

(1) fifi+j = fi: the front i-face of the front i + j-face is the front i-face. (2) bkbj+k = bk: the back face of the back face is the back face. (3) bjfi+j = fjbj+k: this is the middle face mjσ = σ|[vi, ··· , vi+j].

In terms of the diagonal maps deﬁned on Ci+j+k(X) these three formulas give one formula:

(∆ij ⊗ id) ◦ ∆i+j,k = (id ⊗ ∆jk) ◦ ∆i,j+k When we apply these operators to σ we get:

fi(fi+jσ) ⊗ bj(fi+jσ) ⊗ bkσ = fiσ ⊗ fj(bj+kσ) ⊗ bk(bj+kσ) When we sum over all i, j, k we get: (∆ ⊗ id)∆ = (id ⊗ ∆)∆ i.e., ∆ is coassociative. 12 4.1.2. Properties of cup product. The coassociativity of ∆ (and the associativity of µ) imply that cup product is associative at the cochain level. Theorem 4.2. If α ∈ Ci(X; R), β ∈ Cj(X; R), γ ∈ Ck(X; R) then (α ∪ β) ∪ γ = α ∪ (β ∪ γ) Proof. Associativity of (α ∪ β) ∪ γ = µ((α ∪ β) ⊗ γ)∆ = µ(µ ⊗ id)(α ⊗ β ⊗ γ)(∆ ⊗ id)∆ α ∪ (β ∪ γ) = µ(α ⊗ (β ∪ γ))∆ = µ(id ⊗ µ)(α ⊗ β ⊗ γ)(id ⊗ ∆)∆ But µ(µ⊗id) = µ(id⊗µ) by associativity of µ and (∆⊗id)∆ = (id⊗∆)∆ by coassociativity of ∆

Problem Show that the augmentation map : C0(X) → R (given by composing the usual augmentation map : C0(X) → Z with the unique ring homomorphism Z → R) is the unit for cup product, i.e. ∪ φ = φ = φ ∪ Problem Show that cup product is natural, i.e., given any continuous mapping f : X → Y and φ ∈ Ck(Y ; R), ψ ∈ C`(Y ; R) we have: f #(φ ∪ ψ) = f #(φ) ∪ f #(ψ) In other words, the following diagram commutes. f #⊗f # Ck(X; R) ⊗ C`(X; R) Ck(Y ; R) ⊗ C`(Y ; R)

∪X ∪Y ? ? f # Ck|`(X; R) Ck+`(Y ; R) By looking at this diagram and the deﬁnition of ∪ we realized that we need to show:

φ(ffkσ)ψ(fb`σ) = φ(fk(fσ))ψ(b`(fσ))

I.e., we need to know that ffkσ = fk(fσ) and fb`σ = b`(fσ). To show the ﬁrst one we looked at another diagram:

fk ∆k - ∆k+` @ @ fσ @ σ @ fkσ @ @ @@R ? @@R f XY- We realized that the front k-face of a singular k + ` simplex σ : ∆k+` → X is actually k k+` a composition: fkσ = σ ◦ fk where fk : ∆ → ∆ is the inclusion of the front k-face. Therefore: f(fkσ) = f ◦ (σ ◦ fk) = (f ◦ σ) ◦ fk = fk(fσ).

13 4.1.3. Derivation. The main property of cup product is that δ is a graded derivation: Theorem 4.3. δ(φ ∪ ψ) = δφ ∪ ψ + (−1)kφ ∪ δψ if φ ∈ Ck(X; R). Proof. Apply both sides to any σ : ∆k+`+1 → X. k+`+1 k+`+1 X i X i (4.1) δ(φ ∪ ψ)σ = (φ ∪ ψ)∂σ = (−1) (φ ∪ ψ)∂iσ = (−1) φ(fk∂iσ)ψ(b`∂iσ) i=0 i=0 This expression has k + ` + 2 terms. k+1 X i (4.2) (δφ ∪ ψ)σ = δφ(fk+1σ)ψ(b`σ) = (−1) φ(∂ifk+1σ)ψ(b`σ) i=0

But ∂ifk+1σ = fk∂iσ and b`σ = σ|[vk+1, ··· , vk+`+1] = b`∂iσ for i ≤ k. So, the terms with i ≤ k in (4.2) are equal to the ﬁrst k + 1 terms in (4.1). `+1 k k X k+j (4.3) (−1) (φ ∪ δψ)σ = (−1) φ(fkσ)δψ(b`+1σ) = (−1) φ(fkσ)ψ(∂jb`+1σ) j=0

Here ∂jb`+1σ = b`∂j+kσ and, for j ≥ 1, fkσ = fk∂j+kσ. So, the terms with j ≥ 1 in (4.3) are equal to the last ` + 1 terms in (4.1). Finally, the last term of (4.2) cancels the ﬁrst terms of (4.3). They are

φ(fkσ)ψ(b`σ) with opposite signs. 4.2. Cup product in cohomology. Theorem 4.3 implies that cup product induces a multiplication in cohomology. (1) If φ, ψ are both cocycles then φ ∪ ψ is a cocycle. (2) If φ is a coboundary φ = δα and ψ is a cocycle then δ(α ∪ ψ) = δα ∪ ψ ± α ∪ δψ = φ ∪ ψ So, φ ∪ ψ is a coboundary. (3) Similarly, if φ is a cocycle and ψ is a coboundary, then φ ∪ ψ is a coboundary. Corollary 4.4. Cup product induces a multiplication in cohomology: ∪ : Hk(X; R) ⊗ H`(X; R) → Hk+`(X; R) by [φ] ∪ [ψ] = [φ ∪ ψ]. Furthermore, this multiplication is associative with unit. So, M Hn(X; R) is a graded ring for any associative ring R. The word graded ring means that all elements are sums of homogeneous elements and the product of a homogeneous element of degree k with a homogeneous element of degree ` is a homogeneous element of degree k + `. E.g., polynomial rings are graded but power series rings are not.

14 4.2.1. Note on tensor product. The coeﬃcient ring R will be suppressed from the notation. Also, we will from now on, assume that R is commutative. I actually only deﬁned cup product on the Cartesian product Ck(X) × C`(X) −→∪ Ck+`(X)

However, the cup product is biadditive, i.e., (φ1 +φ2)∪ψ = φ1 ∪ψ+φ2 ∪ψ and φ∪(ψ1 +ψ2) = φ ∪ ψ1 + φ ∪ ψ2. So, it induces a map on the tensor product Ck(X) ⊗ C`(X) −→∪ Ck+`(X) It is also R-bilinear in the sense that it also satisfies (rφ) ∪ ψ = r(φ ∪ ψ) = φ ∪ (rψ). So, we get: k ` ∪ k+` C (X) ⊗R C (X) −→ C (X) 4.2.2. Cup product on relative cohomology. If A ⊆ B ⊆ X then the cup product gives a map: k ` ∪ k+` C (X,A) ⊗R C (X,B) −→ C (X,B) We analyzed why this is true. (1) If φ ∈ Ck(X,A) then φ(σ) = 0 whenever σ(∆k) ⊆ A. (2) If ψ ∈ C`(X,B) then ψ(σ) = 0 whenever σ(∆`) ⊆ B. ` (3) So, (φ ∪ ψ)σ = φ(fkσ)ψ(b`σ) = 0 if σ(∆ ) ⊆ B. It would also work to put Ck+`(X,A) instead of Ck+`(X,B). I said that it was usually “better” to use B instead of A. In cohomology we get: k ` ∪ k+` H (X,A) ⊗R H (X,B) −→ H (X,B) 4.2.3. Cross product. Definition 4.5. For a Cartesian product X × Y the cross product k ` k+` × : H (X; R) ⊗R H (Y ; R) → H (X × Y ; R) is defined by ∗ ∗ α × β = pX (α) ∪ pY (β) ∗ ∗ where pX , pY are the maps in cohomology induced by the projection maps pX : X × Y → X and pY : X × Y → Y . The cross product is defined at the cochain level by # # φ × ψ = pX (φ) ∪ pY (ψ)

15 5. Kunneth¨ formula The K¨unneth formula gives the homology and cohomology of a product of spaces. In my approach to this I used universal objects which are called “acyclic models” and the “Yoneda element.” I used the algebraic part of the K¨unnethformula to prove the geometric part.

5.1. Logic. To dispel the appearance of a circular argument, I explained the logic of the proof. The proof is divided into two parts which are independent. The ﬁrst part is a geometric theorem called the Eilenberg-Zilber Theorem. Theorem 5.1 (Thm A: Eilenberg-Zilber). The singular chain complex of a product space X ×Y is naturally homotopy equivalent to the tensor product of the singular chain complexes of the factors:

C∗(X × Y ) ' C∗(X) ⊗ C∗(Y ) The second part (Theorem B) computes the homology and cohomology of a tensor product of free chain complexes. This is done in two parts B1, B2. We only need the ﬁrst part to prove part A (Thm B1 ⇒ Thm A).

Theorem 5.2 (Thm B1). Suppose that C∗,D∗ are free chain complexes so that the homology of D∗ is free abelian. Then the homology of C∗ ⊗ C∗ is isomorphic to the tensor product of the homologies of C∗,D∗: ∼ M Hn(C∗ ⊗ D∗) = Hk(C∗) ⊗ H`(D∗) k+`=n We will discuss the general formula (Theorem B2) later.

5.2. Tensor product of chain complexes. We went over the deﬁnition of the tensor product of chain complexes and proved Theorem B1.

C D Definition 5.3. If C∗,D∗ are chain complexes with differentials ∂ , ∂ then the chain complex C∗ ⊗ D∗ is defined as follows. The groups in the chain complex are: M (C∗ ⊗ D∗)n := k+`=n ⊗ The boundary map ∂ : Ck ⊗ D` → (C∗ ⊗ D∗)k+`−1 is given by ∂⊗(x ⊗ y) = ∂C x ⊗ y + (−1)kx ⊗ ∂Dy The sign (−1)k insures that (∂⊗)2 = 0.

0 0 0 0 Lemma 5.4. If C∗ ' C∗ and D∗ ' D∗ then C∗ ⊗ D∗ ' C∗ ⊗ D∗.

Proof of Theorem B1. If H∗(D∗) is free abelian then D∗ ' H∗(D∗) are homotopic free chain complexes (the boundary map being zero in H∗(D∗)). Therefore, by the lemma above, C∗ ⊗ D∗ ' C∗ ⊗ H∗(D∗) with homology ∼ ∼ M ∼ M Hn(C∗ ⊗ D∗) = Hn(C∗ ⊗ H∗(D∗)) = Hk(C∗; H`(D∗)) = Hk(C∗) ⊗ H`(D∗) k+`=n k+`=n 16 5.3. Skew-commutativity of cup product. The ﬁrst theorem we proved using Theorem B1 and the universal-Yoneda element argument was the skew commutativity of cup product. Theorem 5.5. If R is commutative and α ∈ Hk(X; R), β ∈ H`(X; R) then α ∪ β = (−1)k`β ∪ α Proof. I made the deﬁnition: α ∪0 β = (−1)k`β ∪ α We want to show that ∪0 = ∪. To show this I went to the cocycles. Suppose that φ, ψ are cocycles representing α, β and σ : ∆k+` → X. Then

(φ ∪ ψ)(σ) = φ(fkσ)ψ(b`σ) 0 k` k` (φ ∪ ψ)(σ) = (−1) (ψ ∪ φ)(σ) = (−1) φ(bkσ)ψ(f`σ) Next, I eliminate all three variables (φ, ψ, σ) by using my favorite formula:

φ ∪ ψ = µ(φ ⊗ ψ)∆k.` 0 0 φ ∪ ψ = µ(φ ⊗ ψ)∆k.`

where ∆k,`σ = fkσ ⊗ b`σ and 0 k` ∆k,`σ = (−1) bkσ ⊗ f`σ So, it suﬃces to show that ∆ ' ∆0. This follows from the following two lemmas. Lemma 5.6. ∆ and ∆0 are natural augmented chain maps

C∗(X) → C∗(X) ⊗ C∗(X) The fact that they are natural and augmented is obvious. (They are augmented since 0 = ( ⊗ )∆0,0 = ( ⊗ )∆0,0.) A calculation shows they are chain maps.

Lemma 5.7. Any two natural augmented chain maps C∗(X) → C∗(X)⊗C∗(X) are naturally chain homotopic.

Suppose that f, g : C∗(X) → C∗(X) ⊗ C∗(X) are two natural augmented chain maps. Then, the homotopy that we want is a sequence of maps X hn : Cn(X) → (C∗(X) ⊗ C∗(X))n+1 satisfying two condition ⊗ (1) ∂ hn + hn−1∂ = gn − fn X (2) hn is natural, i.e., for any map f : X → Y we get a commuting square. n X Since Cn(X) is freely generated by σ : ∆ → X, it suﬃces to ﬁnd an equation for hn (σ). This X is the point where I used the universal argument. The mapping hn exists in the universal case when X = ∆n. ∆n n n n h : C∗(∆ ) → C∗(∆ ) ⊗ C∗(∆ ) n n n n This is because ∆ is contractible and thus C∗(∆ ) and C∗(∆ ) ⊗ C∗(∆ ) are acyclic (by B1 (Theorem 5.2). Thus, they are both free resolutions of Z and any two free resolutions of the same abelian group are chain homotopy equivalent by augmented chain maps which ∆n are unique up to homotopy. Furthermore, the maps hm in the homotopy are constructed inductively on m ≥ 0. 17 X The logic is that, by induction on n, the maps hn−1 exist for all X and are natural. Then, ∆n hn exists. We apply this map to the Yoneda element n n n id∆n : ∆ → ∆ ∈ Cn(∆ ) X Naturality of hn means that we want the following diagram to commute: ∆n n hn - n n id∆n ∈ Cn(∆ )(C∗(∆ ) ⊗ C∗(∆ ))n+1

σ# (σ⊗σ)#

? X ? hn - σ ∈ Cn(X)(C∗(X) ⊗ C∗(X))n+1

X This forces the following formula which we took as the deﬁnition of fn (σ): X ∆n fn (σ) := (σ ⊗ σ)∗(fn (id∆n ) X In other words, fn (σ) is determined for all X and all σ by the value of hn in the universal case on the Yoneda element.

18 5.4. Homotopy invariance. The second example of acyclic models and the Yoneda element was the homotopy invariance of homology and cohomology. Theorem 5.8. Homotopic mappings f ' g : X → Y induce chain homotopic maps on the singular chain complexes:

f# ' g# : C∗ → D∗ By an easy argument which I explained before, this theorem implies that homology and cohomology are homotopy invariant: Corollary 5.9. Homotopic maps f ' g : X → Y induce the same map in homology and ∗ ∗ cohomology with arbitrary coefficients. (f∗ = g∗ and f = g .) Thus it suffices to prove Theorem 5.8. 5.4.1. Eliminate Y . We want to eliminate X and Y . I.e., instead of proving the theorem for all X and all Y it suffices to prove it for a particular case where it is obvious. The first step is to eliminate Y as an independent variable.

Problem. It suﬃces to consider the special case when Y = X × I, f = j0 and g = j1 where jt : X → X × I is the mapping jt(x) = (x, t). Why?

Answer. Suppose that h : X × I → Y is a homotopy from f to g. Then f = h ◦ j0 and g = h ◦ j1. Therefore, f# = h# ◦ (j0)#

(j0)# h# f# : C∗(X) −−−→ C∗(X × I) −→ C∗(Y )

and g# = h# ◦ (j1)#. Now suppose that J is a chain homotopy from (j0)# to (j1)#. Then ∂J + J∂ = (j1)# − (j0)# and h#J is a homotopy from f# to g#. The reason is that h# is a chain map:

∂(h#J) + (h#J)∂ = h#∂J + h#J∂ = h#(∂J + J∂) = h#((j1)# − (j0)#) = g# − f# 5.4.2. Universal example. We want to construct a chain homotopy between the two chain maps (j0)#, (j1)# : C∗(X) → C∗(X ×I) for all spaces X. This will consist of homomorphisms X hn : Cn(X) → Cn+1(X × I) for all spaces X and all n ≥ 0 satisfying the equation: X X (5.1) ∂hn + hn−1∂ = (j1)# − (j0)# n First of all, since Cn(X) is freely generated by the singular n simplices σ : ∆ → X, the X X homomorphism hn is determined by the elements hn (σ) ∈ Cn+1(X ×I) which can be chosen X arbitrarily. However, we also need hn to satisfy (5.1) and we want it to be natural. In particular we want the following diagram to commute for all σ : ∆n → X. ∆n n hn - n id∆n ∈ Cn(∆ ) Cn+1(∆ × I)

σ# (σ×idI )#

? X ? hn - σ ∈ Cn(X) Cn+1(X × I)

n n Here ∆ is the universal object and id∆n ∈ Cn(∆ ) is the Yoneda element. 19 X We assume by induction on n that hn−1 has been constructed for all X so that it is natural n n n and satisfies (5.1). Since ∆ and ∆ × I are contractible, the chain complexes C∗(∆ ) and n ∆n X C∗(∆ × I) acyclic. Therefore, hn exists satisfying (5.1). Choose one. Then hn can be defined for all X by X ∆n hn (σ) := (σ × idI )#hn (id∆n ) It is easy to see that this is natural and satisfies (5.1). This completes the proof of Theorem 5.8.

5.5. Eilenberg-Zilber. The last example of acyclic models is the Eilenberg-Zilber Theorem:

C∗(X) ⊗ C∗(Y ) ' C∗(X × Y ) We need to construct natural chain maps

f : C∗(X) ⊗ C∗(Y ) → C∗(X × Y ) and g going the other way so that the compositions f ◦ g, g ◦ f are homotopic to the identity. First we construct f.

5.5.1. Construction of f. We want to construct homomorphisms

X,Y fk,` : Ck(X) ⊗ C`(Y ) → Ck+`(X × Y ) for all X,Y and all k, ` ≥ 0 satisfying two properties: (1) f is a chain map: ∂f = f∂. (2) f is natural. k The group Ck(X) ⊗ C`(Y ) is freely generated by the elements σ ⊗ τ where σ : ∆ → X and τ : ∆` → Y . The universal example is given by X = ∆k and Y = ∆`. The Yoneda element is k ` id∆k ⊗ id∆` ∈ Ck(∆ ) ⊗ C`(∆ ) Naturality of f requires that the following diagram commute. k ` f ∆ ,∆ k ` k.` - k ` id∆k ⊗ id∆` ∈ Ck(∆ ) ⊗ C`(∆ ) Ck+`(∆ × ∆ )

σ#⊗τ# (σ×τ)#

? ? f X,Y ? k,` - σ ⊗ τ ∈ Ck(X) ⊗ C`(Y ) Ck+`(X × Y )

k ` k ` ∆k,∆` Since C∗(∆ ) ⊗ C∗(∆ ) and C∗(∆ × ∆ ) are acyclic, the chain map f exists and is unique up to homotopy. Furthermore the construction of this chain map is by induction X,Y on degree. Therefore, we may assume by induction on κ + ` that fk,` exists in all smaller ∆k,∆` X,Y degree sums for all X,Y and we may choose fk,` so that ∂f = f∂. Then fk,` can be deﬁned for all X,Y by

X,Y ∆k,∆` fk,` (σ ⊗ τ) := (σ × τ)#fk,` (id∆k ⊗ id∆` ) 20 5.5.2. Construction of g. Now we want to construct a natural chain map g : C∗(X × Y ) → C∗(X) ⊗ C∗(Y ). This is a sequence of homomorphisms

gn : Cn(X × Y ) → (C∗(X) ⊗ C∗(Y ))n

⊗ n satisfying ∂ gn = gn−1∂. Here the universal example is X = Y = ∆ and the Yoneda element is the diagonal map ∆ : ∆n → ∆n ×∆n. The reason that this is the Yoneda element is the following. n A generator of Cn(X × Y ) is a singular n simplex σ : ∆ → X × Y . But this is given by the pair of maps σX = pX ◦ σ and σY = pY ◦ σ since

σ(t) = (σX (t), σY (t))

The pair of maps σX , σY sends the diagonal ∆ to σ = (σX , σY ).

∆n,∆n n n gn - n n ∆ ∈ Cn(∆ × ∆ )(C∗(∆ ) ⊗ C∗(∆ ))n

(σX ×σY )# (σX )#⊗(σY )#

? ? X,Y ? gn - σ = (σX , σY ) ∈ Cn(X × Y )(C∗(X) ⊗ C∗(Y ))n)

X,Y As before, we assume by induction on n that a natural map gn−1 has been constructed for ∆n,∆n all X,Y . Then the chain complexes at the top are acyclic, the map gn exists and we X,Y may deﬁne gn by X,Y ∆n,∆n gn (σX , σY ) := ((σX )# ⊗ (σY )#) gn (∆)

5.5.3. The homotopies. What I didn’t have time to do in class was the homotopies. I’ll show that fg : C∗(X × Y ) → C∗(X × Y ) is homotopic to the identity map. The other way (gf ' id) is similar. The precise statement is the following.

Lemma 5.10. Any two natural chain maps C∗(X × Y ) → C∗(X × Y ) are homotopic. Proof. We want to construct a natural chain homotopy

X,Y hn : Cn(X × Y ) → Cn+1(X × Y )

n n hn−1 exists for all X,Y by induction and hn exists for the universal example X = ∆ ,Y = ∆ . ∆n,∆n X,Y The value of hn on the Yoneda element ∆ determines hn :

X,Y ∆n,∆n hn (σX , σY ) := (σX × σY )#hn (∆).

5.6. The Yoneda element in general. The Yoneda element is, by definition, always equal to the identity map of the universal object (or it correspond to the identity map). (What n n about ∆ ∈ Cn(∆ ×∆ )?) The theorem that we saw several examples of is called the Yoneda Lemma. It is considered to be trivial. There are two variations of the Yoneda Lemma, one in the category of sets and one in the category of abelian groups. 21 5.6.1. Set theoretic Yoneda Lemma. Definition 5.11. Suppose the C is any category and U is any fixed object of C.(U will be the universal object.) For any other object X of C let F (X) = MorC(U, X) be the set of all morphisms from U to X in the category C. Then idU ∈ F (U) is the Yoneda element for F . A functor F 0 : C → Sets is called representable if there is a natural isomorphism F 0(X) ∼= 0 MorC(U, X). The element of F (U) corresponding to the identity map on U is called the Yoneda element of F 0.

There is a similar deﬁnition for contravariant functors: F (X) = MorC(X,U) is the representable contravariant functor with universal object U and Yoneda element idU ∈ F (U). Theorem 5.12 (Yoneda Lemma). Suppose that G : C → Sets is any functor and F is as above. Then, a natural map φX : F (X) → G(X) is given uniquely by

φX (σ : U → X) := σ∗(x)

where x = φU (idU ) is an arbitrary ﬁxed element of G(U). This theorem is usually stated as follows: There is a bijection between the set of natural transformations φ : F → G and the set of elements of G(U). 5.6.2. Two examples. Here are two examples in the homotopy category. First, the categories. Let H be the category whose objects are all nice spaces (those homotopy equivalent to ∆- complexes) and whose morphisms are homotopy classes of maps [f]: X → Y . Thus:

MorH (X,Y ) = [X,Y ] where [X,Y ] is the standard notation for the set of homotopy classes of maps X → Y . Let H0 be the category of pointed nice topological spaces (X, x0) and pointed homotopy classes of maps [f]:(X, x0) → (Y, y0). This means that f(x0) = y0 and f ' g if the homotopy h is also pointed (h(x0, t) = y0 for all t ∈ I). So,

MorH0 (X,Y ) = [X,Y ]0

The base point is understood and indicated by the subscript 0. n Example 5.13. Take C = H0 with universal object U = (S , ∗). Then πn(X, x0) := n n [(S , ∗), (X, x0)]0. The Yoneda element is [idSn ] ∈ πn(S ). The Yoneda Lemma says, e.g., that the set of all natural maps πn(X) → πm(X) are in 1-1 corespondence with the element n of G(U) = πm(S ). Example 5.14. Take C = H with some universal object U. Then F (X) = [X,U] is a representable contravariant functor with Yoneda element [idU ] ∈ F (U). If G(X) = [X,V ] is another representable functor then the Yoneda Lemma says that the set of natural transformations F (X) → G(X) is in 1-1 correspondence with the elements of G(U) = [U, V ]. An important special case of this example is singular cohomology: Hn(X; G) ∼= [X,K(G, n)] The Yoneda Lemma says that the set of all natural maps Hn(X; G) → Hm(X; H) is in 1-1 correspondence with the elements of Hm(K(G, n)). Here is an example. Cup square is a natural map Hn(X; R) → H2n(X; R). By the Yoneda Lemma, it corresponds to some element of H2n(K(R, n); R). What is it? 22 5.6.3. Algebraic Yoneda Lemma. The version of the Yoneda Lemma that we have been using is the following.

Lemma 5.15. Suppose that F : C → Sets is a representable functor F (X) = MorC(U, X). For each X in C let C(X) is the free abelian group generated by F (X). Then C is a functor C → Ab and, for any other functor G : C → Ab, there is a 1-1 correspondence between the set of natural transformations φ : C → G and the elements of G(U). We say that C is free on the universal object U.

Proof. Any natural map φX : C(X) → G(X) is given uniquely by

φX (f : U → X) = f∗(x) where x = φU (idU ) is an arbitrary element of G(U).

The main example is the singular chain complex C∗(X). For each n ≥ 0, Cn(X) is free on the universal object ∆n. The Eilenberg-Zilber Theorem is an example of the following general theorem.

Theorem 5.16 (Acyclic Model Theorem). Suppose that C∗(X),D∗(X) are augmented chain complexes deﬁned naturally for all objects X of some category C. Suppose that Cn(X),Dn(X) are free with universal objects Un,Vn. If C∗(Un),C∗(Vn),D∗(Un),D∗(Vn) are acyclic for all n then C∗(X) ' D∗(X).

This also holds in the case when Cn(X),Dn(X) are direct sums of free functors whose universal objects are acyclic in both C∗ and D∗. Universal objects are also called models. The Acyclic Model Theorem follows from the following lemma. This is also the lemma that we actually used several times.

Lemma 5.17. Suppose that C∗(X),D∗(X) are augmented chain complexes deﬁned naturally for all objects X of some category C. Suppose that Cn(X) are free on models Un. If D∗(Un) are acyclic for all n then there is a natural augmented chain map C∗(X) → D∗(X) and any two such chain maps are naturally chain homotopic.

Again, this also holds in the case when Cn(X) is a direct sum of free functors on models Un,i as long as D∗(Un,i) are all acyclic.