Proof that two spaces that are homotopic have the same de Rham cohomology 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.
 A: 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.
