Proof the the general Poincare Lemma $H^*(X \times \mathbb{R}) \cong H^*(X )$

Question

I was wondering if someone could provide a reference or help me out with the proof of the Poincare lemma in the form: $$ H^*(X \times \mathbb{R}) \cong H^*(X) $$ Where $X$ is a smooth manifold and $H^*$ is the de Rham cohomology. I was attempting to prove it myself, using charts, but can't seem to figure out how to do this properly. It is mentioned in Bott and Tu as something that should be relatively easy using the fact that if $U,\varphi$ is a chart on $X$, then $(U \times \mathbb{R})$ is a chart on $X \times \mathbb{R}$, but I don't really know how pull this all together.

Answer 1

This is done, as you say, in some detail in the book by Bott and Tu on differential forms in algebraic topology.

Show that there is a contracting homotopy $K:\Omega(M\times\mathbb R)\to\Omega(M\times\mathbb R)$ and after that the proof is formally the same as what they write.

To do that, pick a form $\omega$ on the domain of $K$, restrict it to open subsets of the form $U\times\mathbb R$ with $U$ diffeomorphic to $\mathbb R^n$ and use what you have: this gives you what should be the restriction of $K(\omega)$ to $U\times\mathbb R$. Next show that these "restrictions" glue to a globally defined form on $M\times\mathbb R$.