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Math 442: Final Project Cup Products and the Cohomology Ring Due Math 442: Final Project Cup Products and the Cohomology Ring Due Thursday, March 17 One reason for discussing the cohomology H∗(X; R) of a space X with co- efficients in a commutative R is that these groups assemble into a graded com- mutative ring. It is that ring structure that gives cohomology its impact. The point of this project is define and discuss this product, and to give some basic examples. The reference for this material is Chapter 3.2 of Hatcher. In follows, R is a general, but fixed, commutative ring. If X is a topological • space, C•(X) and C (X; R) will denote the singular chains and cochains on X. Thus an element in Cn(X; R) can be regarded as a function α : Sn(X) −→ R on the set Sn(X) of singular n-simplices of X. If σ : ∆n → X is a singular n-simplex, recall that we have the first p-face operator fpσ and the last q-face operator `qσ. For example, fpσ(t0, . , tp) = σ(t0, . , tp, 0,..., 0) is the induced singular p-simplex. Definition: Suppose α ∈ Cp(X; R) and β ∈ Cq(X; R). Define the cup product p+q α ` β ∈ C (X; R) by the formula (α ` β)(σ) = α(fpσ)β(`qσ). Problem 1. Show that the cup product has the following properties: 1. For all α1, α2, and β, (α1 + α2) ` β = α1 ` β + α2 ` β. 2. For all α, β, and γ, we have (α ` β) ` γ = α ` (β ` γ). 3. If f : X → Y is a continuous map and α and β are cochains on Y , then ] ] ] ] • • f (α ` β) = f α ∪ f β. Here f : C (Y ; R) → C (X; R) is the induced map on cochains. 4. If e : S0X → R is the cocycle so that e(σ) = 1 for all singular 0-simplices, then e ` α = α ` e = α Problem 2. The cup product descends to cohomology. Prove that if δ is the coboundary map on cochains, we have p δ(α ` β) = δα ` β + (−1) α ` δβ where p is the degree of α. Conclude that there is a natural associative graded ring structure on H∗(X; R). (You may have to define this notion.) Remark: On cohomology, the cup product has the following commutativity property: if x ∈ Hp(X; R) and y ∈ Hq(Y ; R), then pq x ` y = (−1) y ` x. This is certainly not true on the chain level and we won’t prove that here. See Hatcher p. 215. Problem 3. Here is a formal property of the cup product. Let X and Y be two spaces and X t Y their disjoint union. Prove that, as rings H∗(X t Y ; R) = H∗(X; R) × H∗(Y ; R). The ring structure on the right is (α1, α2) ` (β1, β2) = (α1 ` β1, α2 ` β2). Use this to calculate H∗(X ∨ Y ; R) where X ∨ Y is the one point union of two locally contractible spaces. Problem 4. Here are two specific examples. Assuming that you can use a ∆-set structure on a space and the resulting ∆-chains for computations. (You can.) 1. Let T be the torus. Find generators x, y of H1(T, Z) and z of H2(T, Z) so that x ` x = 0 = y ` y x ` y = z = −y ` x. 2. Let K be the Klein bottle. Find generators x, y of H1(T, Z/2Z) and z of H2(T, Z/2Z) so that x ` x = z = x ` y y ` y = 0. Problem 5. Show that the torus T and the space S1 ∨ S1 ∨ S2 have the same cohomology groups but different cohomology rings. Problem 6. Let Mg be the g-holed torus, g ≥ 1. Calculate the cohomology ∗ ring H (Mg, Z) and show that the cup product 1 1 2 ∼ (−) ` (−): H (Mq) × H (Mg) → H (Mg) = Z has the following perfect-pairing property: if x 6= 0, then there exists y so that x ` y is a generator. Hint: There is a map Mg → M1 ∨ ... ∨ M1, where the g copies of M1. See p. 228 of Hatcher. Problem 7. Consider a continuous map f : Mg → Mk from the g-holed torus to the k-holed torus and suppose f∗ : H2(Mg) → H2(Mk). is an isomorphism. Prove k ≤ g. Hint: This problem has two parts. If k ≤ g, you have to construct a map. If ∗ 1 1 k > g then f : H (Mk, Z) → H (Mg, Z) must have non-trivial kernel (why?) and this isn’t possible by the perfect pairing part of the previous problem. Problem 8. Show that the cup-product extends to relative cohomology in the following way: if A, B ⊆ X are subspaces, then there is a cup-product pairing p q p+q (−) ` (−): H (X, A; R) × H (X, B; R) → H (X, A ∪ B; R). Conclude that if X is the union of two contractible subspace A and B, then the cup product of two elements of positive degree in the cohomology of X must be zero. Show that a suspension of a space Y is an example of such a space..
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