121B: ALGEBRAIC TOPOLOGY 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 tradi- tional 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.