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I know this is true, but how do I prove it? Specifically, I'm trying to calculate the de Rham cohomology of the 3-sphere by using the Mayer-Vietoris sequence and covering $S^3$ with two hemispherical sets $U, V$, such that $U$ intersects $V$ in a "2-band", an equatorial band homotopic to a 2-sphere.

JBC
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Julien Clancy
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  • What is your source for learning de Rham cohomology? – Jesse Madnick Apr 09 '12 at 22:37
  • My calculus/analysis class includes some topology (the teacher is a topologist) which we're doing without text. – Julien Clancy Apr 09 '12 at 22:39
  • Are you comfortable with using facts about chain complexes? Do you know what it means for two CHAIN COMPLEXES to be homotopic? – Alex Youcis Apr 09 '12 at 23:01
  • No, I don't. The only chain complex stuff we've covered is for the Mayer-Vietoris sequence. – Julien Clancy Apr 09 '12 at 23:50
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    If you look at Differential forms in Algebraic topology, by Bott and Tu, it seems they're proving what you want using only the definition of homotopy and pullback. You can look at Corollary 4.1.2 and 4.1.2.1. – M Turgeon Apr 10 '12 at 00:55
  • Another proof can be found in the beginnings of Chapter 15 of Introduction to Smooth Manifolds, by John M. Lee. I do not know of a proof that is particularly easy to swallow with no prior exposure to chain complexes. – JakeR Apr 10 '12 at 01:40
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    Just a remark: spaces cannot be homotopic. The correct notion for spaces is homotopy equivalence. – Cheerful Parsnip Jun 24 '12 at 13:36

1 Answers1

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Definition: two smooth maps $f,g:M\rightarrow N$ are smoothly homotopic if it exists $H:M\times\mathbb R\rightarrow N$ smooth such as $H(\cdot,t)=f$ for $t\le 0$, and $H(\cdot,t)=g$ for $t\ge 1$.

Theorem: if $f,g:M\rightarrow N$ are smoothly homotopic then their pullback are equal in de Rham cohomology (ie $f^*=g^*:H^*(N)\rightarrow H^*(M)$).

For a proof of this theorem, look at, for example, Bott&Tu

Lemma 1: if $f:M\rightarrow N$ is continuous then $f$ is (continuously) homotopic to a smooth map.
Lemma 2: if two smooth maps $f,g:M\rightarrow N$ are continuously homotopic then they are smoothly homotopic.

To prove these lemmas, you can embed your manifolds in $\mathbb R^n$ spaces. If I remember well, it's done in From Calculus to Cohomology by Ib Henning Madsen and Jørgen Tornehave.

We can start having fun:

Theorem: a continuous map $f:M\rightarrow N$ induces a pullback $f^*:H^*(N)\rightarrow H^*(M)$ in the de Rham cohomology.

Proof: by lemma 1, it exists $f'$ smooth which is continuously homotopic to $f$, and let $f^*=f'^*$. We have to check that $f^*$ is well defined : if $f''$ is another smooth map continuously homotopic to $f$, then $f'$ and $f''$ are continuously homotopic, and so, by lemma 2, smoothly homotopic, so $f'^*=f''^*$ by the theorem. $\blacksquare$

Corollary 1: if $f:M\rightarrow N$ and $g:N\rightarrow L$ are continuous between manifolds then $(g\circ f)^*=f^*\circ g^*$.

Corollary 2: if $f,g:M\rightarrow N$ are continuously homotopic then $f^*=g^*$.

Corollary 3: two manifolds with the same homotopy type have the same de Rham cohomology.
In particular, two homeomorphic manifolds have the same de Rham cohomology.

Remark: this result seems impressive to me, because it shows that the de Rham cohomology on a manifold doesn't depend on the differential structure.

JBC
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