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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.

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    What is your source for learning de Rham cohomology?2012-04-09
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    My calculus/analysis class includes some topology (the teacher is a topologist) which we're doing without text.2012-04-09
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    Are you comfortable with using facts about chain complexes? Do you know what it means for two CHAIN COMPLEXES to be homotopic?2012-04-09
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    No, I don't. The only chain complex stuff we've covered is for the Mayer-Vietoris sequence.2012-04-09
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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 4.1.2.1.2012-04-10
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    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.2012-04-10
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    Just a remark: spaces cannot be *homotopic.* The correct notion for spaces is *homotopy equivalence.*2012-06-24

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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.