Homotopic smoothe maps induce isomorphic maps in (standard) de Rham cohomology. The same proof is supposed to work also for de Rham with compact support if the two maps are *proper* and the homotopy between them is *smooth and proper*. I have an explicit formula for the chain complex homotopy  

$$K(\alpha) = \int\_0^1 J\_t^\*(\iota\_{\partial t} \alpha) dt$$  

verifying $$J\_1^∗\,\alpha − J\_0^∗\,\alpha = dK(\alpha) + K(d\alpha)$$  

where $J\_t : x \mapsto (x,t)$, $\alpha \in \Omega^k(M \times R)$, and $K : \Omega^k(M \times R) \to \Omega^{k-1}(M)$ 


But how can I prove (clean proff not hand waving) that in fact that 
$$K : \Omega\_{\kappa}^k(M \times R) \to \Omega\_c^{k-1}(M)$$
where $\Omega\_{\kappa}^k(M \times R)$ is the vector space of $k$-forms on $M \times R$ whose support projected on $M$ is compact ? 


In Godbillon's book, he uses the local formula for $K$ on $U$ open domain of $M$ homeo to $R^n$, namely  

$$K(\beta) = (\int\_0^1 b dt) dx^{i\_1} \wedge \dots \wedge dx^{i\_{k-1}}$$
where $\beta = b(x,t) dt \wedge dx^{i\_1} \wedge \dots \wedge dx^{i\_{k-1}}$, and says that for compact support it is enough to take $U$ relatively compact... I don't understand...



Invariance of de Rham cohomology with compact support
You find a proof in 12.5 of [here](http://www.mat.univie.ac.at/~michor/dgbook.pdf).


