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For example, I know that $H^k(S^n \times I^m) = H^k(S^n)$ (where $I$ is an open interval) because there is a deformation retraction from $S^n \times I^m$ to $S^n$.

Why are the cohomology groups invariant under deformation retractions?

user157227
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  • de Rham cohomology is only defined on smooth manifolds; $I^m$ is not a smooth manifold. On smooth manifolds, it agrees with something more general called singular cohomology (with real coefficients), and that is invariant not only under deformation retractions but under arbitrary homotopy equivalences. This is a long story and there isn't a really short way to summarize it. Crack open a book on differential forms or algebraic topology if you want to learn it. – Qiaochu Yuan Mar 19 '15 at 04:04
  • @Qiaochu I am taking $I$ to be an open interval. I will edit my post for clarity. – user157227 Mar 19 '15 at 04:09
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    In any case, I don't know what kind of explanation you're asking for given that you say you don't know much algebraic topology. Is there something you want to know that you don't expect to get from a book on differential forms and / or algebraic topology? – Qiaochu Yuan Mar 19 '15 at 04:12
  • @Qiaochu If you are interested, I have found an answer that avoids singular cohomology. – user157227 May 17 '15 at 20:16

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I will outline a proof that the cohomology groups are preserved under homotopy equivalence. Every function is $C^{\infty}$.

  1. Let $f : M \to N$. Then $f^*$ (pullback map) induces a map between $H^k_{de}(N) \to H^k_{de}(M)$.

  2. Prove a generalization of Poincare's lemma that states that if $f,g : M \to N$ are homotopic then $f(\omega) - g(\omega)$ is an exact form. Therefore, they induce the same map from $H^k_{de}(N) \to H^k_{de}(M)$. (The proof of this is almost exactly the same as the proof of Poincare's lemma.)

  3. Let $M$ and $N$ be homotopic manifolds. Let $f :M\to N$ and $g: N \to M$ be functions so that $f\circ g$ and $g\circ f$ are homotopic to the identity map. Then $(f\circ g)^* = g^*\circ f^*$ induces the identity map on $H^k_{de}$ and similarly for $(g\circ f)^*$. Therefore $f^*$ and $g^*$ are inverses of each other and give a bijection between $H^k_{de}(M)$ and $H^k_{de}(N)$.

user157227
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