Lecture 6: Homotopy Invariance of Homology

6 Homotopy invariance of homology We now know that the homology of a star-shaped region is trivial: in such a space, every cycle with augmentation 0 is a boundary. We will use that fact, which is a special case of homotopy invariance of homology, to prove the general result, which we state in somewhat stronger form: Theorem 6.1. A homotopy h : f0 ' f1 : X → Y determines a natural chain homotopy f0∗ ' f1∗ : S∗(X) → S∗(Y ). The proof uses naturality (a lot). For a start, notice that if k : g0 ' g1 : C∗ → D∗ is a chain homotopy, and j : D∗ → E∗ is another chain map, then the composites j ◦ kn : Cn → En+1 give a chain homootpy j ◦ g0 ' j ◦ g1 . So if we can produce a chain homotopy between the chain maps induced by the two inclusions i0 , i1 : X → X × I, we can get a chain homotopy k between f0∗ = h∗ ◦ i0∗ and f1∗ = h∗ ◦ i1∗ in the form h∗ ◦ k. So now we want to produce a natural chain homotopy, with components kn : Sn (X) → Sn+1 (X × I). The unit interval hosts a natural 1-simplex given by an identification ∆1 → I, and we should imagine k as being given by “multiplying” by that 1-chain. This “multiplication” is a special case of a chain map × : S∗ (X) × S∗ (Y ) → S∗ (X × Y ) , defined for any two spaces X and Y , with lots of good properties. It will ultimately be used to compute the homology of a product of two spaces in terms of the homology groups of the factors. Here’s the general result. Theorem 6.2. There exists a map × : Sp (X) × Sq (Y ) → Sp+q (X × Y ), the cross product, that is: • Natural, in the sense that if f : X → X 0 and g : Y → Y 0 , and a ∈ Sp (X) and b ∈ Sp (Y ) so that a × b ∈ Sp+q (X × Y ), then f∗ (a) × g∗ (b) = (f × g)∗ (a × b). • Bilinear, in the sense that (a + a0 ) × b = (a × b) + (a0 × b), and a × (b + b0 ) = a × b + a × b0 . • The Leibniz rule is satisfied, i.e., d(a × b) = (da) × b + (−1)p a × db. • Normalized, in the following sense. Let x ∈ X and y ∈ Y . Write jx : Y → X × Y for y 7→ (x, y), and write iy : X → X × Y for x 7→ (x, y). If b ∈ Sq (Y ), then c0x × b = (jx )∗ b ∈ Sq (X × Y ), and if a ∈ Sp (X), then a × c0y = (iy )∗ a ∈ Sp (X × Y ). The Leibniz rule contains the first occurence of the “topologist’s sign rule”; we’ll see these signs appearing often. Watch for when it appears in our proof. Proof. We’re going to use induction on p+q; the normalization axiom gives us the cases p+q = 0, 1. Let’s assume that we’ve constructed the cross-product in total dimension p + q − 1. We want to define σ × τ for σ ∈ Sp (X) and τ ∈ Sq (Y ). Note that there’s a universal example of a p-simplex, namely the identity map ιp : ∆p → ∆p . It’s universal in the sense any p-simplex σ : ∆p → X can be written as σ∗ (ιp ) where σ∗ : Sinp (∆p ) → Sinp (X) is the map induced by σ. To define σ × τ in general, then, it suffices to define ιp × ιq ∈ Sp+q (∆p × ∆q ); we can (and must) then take σ × τ = (σ × τ )∗ (ιp × ιq ). Our long list of axioms is useful in the induction. For one thing, if p = 0 or q = 0, normalization provides us with a choice. So now assume that both p and q are positive. We want the cross-product to satisfy the Leibnitz rule: d(ιp × ιq ) = (dιp ) × ιq + (−1)p ιp × dιq ∈ Sp+q−1 (∆p × ∆q ) Since d2 = 0, a necessary condition for ιp × ιq to exist is that d((dιp ) × ιq + (−1)p ιp × dιq ) = 0. Let’s compute what this is, using the Leibnitz rule in dimension p + q − 1 where we have it by the inductive assumption: d((dιp )×ιq +(−1)p ιp ×(dιq )) = (d2 ιp )×ιq +(−1)p−1 (dιp )×(dιq )+(−1)p (dιp )×(dιq )+(−1)q ιp ×(d2 ιq ) = 0 because d2 = 0. Note that this calculation would not have worked without the sign! The subspace ∆p ×∆q ⊆ Rp+1 ×Rq+1 is convex and nonempty, and hence star-shaped. Therefore we know that Hp+q−1 (∆p × ∆q ) = 0 (remember, p + q > 1), which means that every cycle is a boundary. In other words, our necessary condition is also sufficient! So, choose any element with the right boundary and declare it to be ιp × ιq . The induction is now complete provided we can check that this choice satisfies naturality, bilinearity, and the Leibniz rule. I leave this as a relaxing exercise for the listener. The essential point here is that the space supporting the universal pair of simplices – ∆p × ∆q – has trivial homology. Naturality transports the result of that fact to the general situation. The cross-product that this procedure constructs is not unique; it depends on a choice a choice of the chain ιp × ιq for each pair p, q with p + q > 1. The cone construction in the proof that star-shaped regions have vanishing homology provids us with a specific choice; but it turns out that any two choices are equivalent up to natural chain homotopy. We return to homotopy invariance. To define our chain homotopy hX : Sn (X) → Sn+1 (X × I), pick any 1-simplex ι : ∆1 → I such that d0 ι = c01 and d1 ι = c00 , and define hX σ = (−1)n σ × ι . Let’s compute: dhX σ = (−1)n d(σ × ι) = (−1)n (dσ) × ι + σ × (dι) 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