Lecture 25: Künneth and Eilenberg-Zilber

25 Künneth and Eilenberg-Zilber We want to compute the homology of a product. Long ago, in Lecture 7, we constructed a bilinear map Sp(X) × Sq (Y ) → Sp+q (X × Y ), called the cross product. So we get a linear map Sp(X) ⊗ Sq (Y ) → Sp+q (X × Y ), and it satisfies the Leibniz formula, i.e., d(x × y) = dx × y + (−1)px × dy. The method we used works with any coefficient ring, not just the integers. Definition 25.1. Let C∗ , D∗ be two chain complexes. Their tensor product is the chain complex with M (C∗ ⊗ D∗ )n = Cp ⊗ Dq . p+q=n The differential (C∗ ⊗D∗ )n → (C∗ ⊗D∗ )n−1 sends Cp ⊗Dq into the submodule Cp−1 ⊗Dq by x ⊗ y 7→ dx ⊗ y + (−1)p x ⊗ dy . L Cp ⊗Dq−1 So the cross product is a map of chain complexes S∗ (X) ⊗ S∗ (Y ) → S∗ (X × Y ). There are two questions: (1) Is this map an isomorphism in homology? (2) How is the homology of a tensor product of chain complexes related to the tensor product of their homologies? It’s easy to see what happens in dimension zero, because π0 (X) × π0 (Y ) = π0 (X × Y ) implies ∼ = that H0 (X) ⊗ H0 (Y ) − → H0 (X × Y ). Let’s dispose of the purely algebraic question (2) first. Theorem 25.2. Let R be a PID and C∗ , D∗ be chain complexes of R-modules. Assume that Cn is a free R-module for all n. There is a short exact sequence M M 0→ Hp (C) ⊗ Hq (D) → Hn (C∗ ⊗ D∗ ) → TorR 1 (Hp (C), Hq (D)) → 0 p+q=n p+q=n−1 natural in these data, that splits (but not naturally). Proof. This is exactly the same as the proof for the UCT. It’s a good idea to work through this on your own. Corollary 25.3. Let R be a PID and assume Cn0 and Cn are R free for all n. If C∗0 → C∗ and D∗0 → D∗ are homology isomorphisms then so is C∗0 ⊗ D∗0 → C∗ ⊗ D∗ . Our attack on question (1) is via the method of “acyclic models.” This is really a special case of the Fundamental Theorem of Homological Algebra, Theorem 22.1. Definition 25.4. Let C be a category, and fix a set M of objects in C, to be called the “models.” A functor F : C → Ab is M-free if it is the free abelian group generated by a coproduct of corepresentable functors. That is, F is a direct sum of functors of the form ZC(M, −) where M ∈ M. Example 25.5. Since we are interested in the singular homology of a product of two spaces, it may be sensible to take as C the category of ordered pairs of spaces, C = Top2 , and for M the set of pairs of simplicies, M = {(∆p , ∆q ) : p, q ≥ 0}. Then Sn (X × Y ) = Z[Top(∆n × X) × Top(∆n , Y )] = ZTop2 ((∆n , ∆n ), (X, Y )) . is M-free. Example 25.6. With the same category and models, M (S∗ (X) ⊗ S∗ (Y ))n = Sp (X) ⊗ Sq (Y ) , p+q=n is M-free, since the tensor product has as free basis the set a a Top2 ((∆p , ∆q ), (X, Y )) . Sinp (X) × Sinq (Y ) = p+q=n p+q=n Definition 25.7. A natural transformation of functors θ : F → G is an M-epimorphism if θM : F (M ) → G(M ) is a surjection of abelian groups for every M ∈ M. A sequence of natural transformations is a composable pair G0 → G → G00 with trivial composition. Let K be the objectwise kernel of G → G00 . There is a factorization G0 → K. The sequence is M-exact if G0 → K is a M-epimorphism. Equivalently, G0 (M ) → G(M ) → G00 (M ) is exact for all M ∈ M. Example 25.8. We claim that · · · → Sn (X × Y ) → Sn−1 (X × Y ) → · · · → S0 (X × Y ) → H0 (X × Y ) → 0 is M-exact. Just plug in (∆p , ∆q ): you get an exact sequence, since ∆p × ∆q is contractible. Example 25.9. The sequence · · · → (S∗ (X) ⊗ S∗ (Y ))n → (S∗ (X) ⊗ S∗ (Y ))n−1 → · · · → S0 (X) ⊗ S0 (Y ) → H0 (X) ⊗ H0 (Y ) → 0 . is also M-exact, by Corollary 25.3. The terms “M-free” and “M-exact” relate to each other in the expected way: Lemma 25.10. Let C be a category with a set of models M and let F, G, G0 : C → Ab be functors. Suppose that F is M-free, let G0 → G be a M-epimorphism, and let f : F → G be any natural transformation. Then there is a lifting: >G f f F 0  /G Proof. Clearly we may assume that F (X) = ZC(M, X). Suppose that X = M ∈ M. We get: G0 (M ) fM ZC(M, M ) fM 8  / G(M ) Consider 1M ∈ ZC(M, M ). Its image fM (1M ) ∈ G(M ) is hit by some element in cM ∈ G0 (M ), since G0 → G is an M-epimorphism. Define f M (1M ) = cM . Now we exploit naturality! Any ϕ : M → X produces a commutative diagram C(M, M )  fM ϕ∗ C(M, X) fX / G0 (M )  ϕ∗ / G0 (X) Chase 1M around the diagram, to see what the value of f X (ϕ) must be: f X (ϕ) = f X (ϕ∗ (1M )) = ϕ∗ (f M (1M )) = ϕ∗ (cM ) . Now extend linearly. You should check that this does define a natural transformation. This is precisely the condition required to prove the Fundamental Theorem of Homological Algebra. So we have the Theorem 25.11 (Acyclic Models). Let M be a set of models in a category C. Let θ : F → G be a natural transformation of functors from C to Ab. Let F∗ and G∗ be functors from C to chain complexes, with augmentations F0 → F and G0 → G. Assume that Fn is M-free for all n, and that G∗ → G → 0 is an M-exact sequence. Then there is a unique chain homotopy class of chain maps F∗ → G∗ covering θ. Corollary 25.12. Suppose furthermore that θ is a natural isomorphism. If each Gn is M-free and F∗ → F → 0 is an M-exact sequence, then any natural chain map F∗ → G∗ covering θ is a natural chain homotopy equivalence. Applying this to our category Top2 with models as before, we get the following theorem that completes work we did in Lecture 7. Theorem 25.13 (Eilenberg-Zilber theorem). There are unique chain homotopy classes of natural chain maps: S∗ (X) ⊗ S∗ (Y )  S∗ (X × Y ) covering the usual isomorphism H0 (X) ⊗ H0 (Y ) ∼ = H0 (X × Y ) , and they are natural chain homotopy inverses. Corollary 25.14. There is a canonical natural isomorphism H(S∗ (X) ⊗ S∗ (Y )) ∼ = H∗ (X × Y ). Combining this theorem with the algebraic Künneth theorem, we get: Theorem 25.15 (Künneth theorem). Take coefficients in a PID R. There is a short exact sequence M M 0→ Hp (X) ⊗R Hq (Y ) → Hn (X × Y ) → TorR 1 (Hp (X), Hq (Y )) → 0 p+q=n p+q=n−1 natural in X, Y . It splits as R-modules, but not naturally. Example 25.16. If R = k is a field, every module is free, so the Tor term vanishes, and you get a Künneth isomorphism: ∼ = × : H∗ (X; k) ⊗k H∗ (Y ; k) − → H∗ (X × Y ; k) This is rather spectacular. For example, what is H∗ (RP3 × RP3 ; k), where k is a field? Well, if k has characteristic different from 2, RP3 has the same homology as S 3 , so the product has the same homology as S 3 × S 3 : the dimensions are 1, 0, 0, 2, 0, 0, 1. If char k = 2, on the other hand, the cohomology modules are either 0 or k, and we need to form the graded tensor product: k k k k k k k k k k k k k k k k so the dimensions of the homology of the product are 1, 2, 3, 4, 3, 2, 1. The palindromic character of this sequence will be explained by Poincaré duality. Let’s look also at what happens over the integers. Then we have the table of tensor products Z Z/2Z Z Z Z/2Z Z/2Z Z/2Z Z/2Z 0 0 0 Z Z Z/2Z 0 Z 0 Z 0 Z/2Z 0 0 0 Z There is only one nonzero Tor group, namely 3 3 TorZ 1 (H1 (RP ), H1 (RP )) = Z/2Z. Putting this together, we get the groups Z H0 H1 Z/2Z ⊕ Z/2Z H2 Z/2Z H3 Z ⊕ Z ⊕ Z/2Z H4 Z/2Z ⊕ Z/2Z H5 0 H6 Z The failure of perfect symmetry here is interesting, and will also be explained by Poincaré duality.