Lecture 7: Homology Cross Product

But dι = c01 − c00 ∈ S0 (I), which means that we can continue (remembering that |∂σ| = n − 1): = −hX dσ + (σ × c01 − σ × c00 ) = −hX dσ + (ι1∗ σ − ι0∗ σ) , using the normalization axiom of the cross-product. This is the result. 7 Homology cross product In the last lecture we proved homotopy invariance of homology using the construction of a chain level bilinear cross-product × : Sp (X) × Sq (Y ) → Sp+q (X × Y ) that satisfied the Leibniz formula d(a × b) = (da) × b + (−1)p a × (db) What else does this map give us? Let’s abstract a little bit. Suppose we have three chain complexes A∗ , B∗ , and C∗ , and suppose we have maps × : Ap × Bq → Cp+q that satisfy bilinearity and the Leibniz formula. What does this induce in homology? Lemma 7.1. These data determine a bilinear map × : Hp (A) × Hq (B) → Hp+q (C). Proof. Let a ∈ Zp (A) and b ∈ Zq (B). We want to define [a] × [b] ∈ Hp+q (C). We hope that [a] × [b] = [a × b]. We need to check that a × b is a cycle. By Leibniz, d(a × b) = da × b + (−1)p a × db, which vanishes becauxe a, b are cycles. Now we need to check that homology class depends only on the homology classes we started with. So pick other cycles a0 and b0 in the same homology classes. We want [a × b] = [a0 × b0 ]. In other words, we need to show that a × b differs from a0 × b0 by a boundary. We can write a0 = a + da and b0 = b + db, and compute, using bilinearity: a0 × b0 = (a + da) + (b + db) = a × b + a × db + (da) × b + (da) × (db) We need to deal with the last three terms here. But since da = 0, d(a × b) = (−1)p a × (db) . Since db = 0, d((a) × b) = (da) × b . And since d2 b = 0, d(a × b) = (da) × (db) . This means that a0 × b0 and a × b differ by
d (−1)p (a × b) + a × b + a × db , and so are homologous. The last step is to check bilinearity, which is left to the listener. This gives the following result. Theorem 7.2. There is a map × : Hp (X) × Hq (Y ) → Hp+q (X × Y ) that is natural, bilinear, and normalized. We will see that this map is also uniquely defined by these conditions, unlike the chain-level cross product. I just want to mention an explicit choice of ιp × ιq . This is called the Eilenberg-Zilber chain. You’re highly encouraged to think about this yourself. It comes from a triangulation of the prism. The simplices in this triangulation are indexed by order preserving injections ω : [p + q] → [p] × [q] Injectivity forces ω(0) = (0, 0) and ω(p + q) = (p, q). Each such map determines an affine map ∆p+q → ∆p × ∆q of the same name. These will be the singular simplices making up ιp × ιq . To specify the coefficients, think of ω as a staircase in the rectangle [0, p] × [0, q]. Let A(ω) denote the area under that staircase. Then the Eilenberg-Zilber chain is given by X ι p × ιq = (−1)A(ω) ω 3 3 2 3 2 1 2 1 1 0 0 0 This chain is due to Eilenberg and Mac Lane; the description appears in a paper [4] by Eilenberg and Moore. It’s very pretty, but it’s combinatorially annoying to check that this satisfies the conditions of the theorem. It provides an explicit chain map βX,Y : S∗ (X) × S∗ (Y ) → S∗ (X × Y ) that satisfies many good properties on the nose and not just up to chain homotopy. For example, it’s associative – βX,Y ×1 / S∗ (X × Y ) × S∗ (Z) S∗ (X) × S∗ (Y ) × S∗ (Z)  1×βY,Z S∗ (X) × S∗ (Y × Z) βX,Y ×Z  βX×Y,Z / S∗ (X × Y × Z) commutes – and commutative – S∗ (X) × S∗ (Y )  βX,Y T S∗ (Y ) × S∗ (X) βY,X / S∗ (X × Y )  S∗ (T ) / S∗ (X × Y ) commutes, where on spaces T (x, y) = (y, x), and on chain complexes T (a, b) = (−1)pq (b, a) when a has degree p and b has degree q. We will see that these properties hold up to chain homotopy for any choice of chain-level cross product. 8 Relative homology An ultimate goal of algebraic topology is to find means to compute the set of homotopy classes of maps from one space to another. This is important because many geometrical problems can be rephrased as such a computation. It’s a lot more modest than wanting to characterize, somehow, all continuous maps from X to Y ; but the very fact that it still contains a great deal of interesting information means that it is still a very challenging problem. Homology is in a certain sense the best “additive” approximation to this problem; and its additivity makes it much more computable. To justify this, we want to describe the sense in which homology is “additive.” Here are two related aspects of this claim.