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...

properhomotopy $\endgroup$ – Denis Nardin Jan 28 '18 at 16:16