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...

  • $\begingroup$ deRham cohomology with compact support is not invariant under homotopy, as the cases of the identity and constant map on $\mathbb{R}^n$ shows. Where did you get this information? It is invariant under proper homotopy $\endgroup$ – Denis Nardin Jan 28 '18 at 16:16
  • $\begingroup$ @Denis, You're right, I should indeed have precise it. I have edited... $\endgroup$ – ychemama Jan 28 '18 at 16:32

You find a proof in 12.5 of here.

  • $\begingroup$ thx a lot. I have already used many times your very detailed pdf, but forgot to check it first for this question. $\endgroup$ – ychemama Jan 29 '18 at 17:14

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