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Let $M$ be a manifold, $P$ a closed subset and sub-manifold of $M$, and $j : (M,\emptyset) \to (M,P)$ the injection (I note $f : (M,P) \to (N,Q)$ a smooth map $M \to N$ such that $f(P) \subset Q$). The pullback $j^* : \Omega_{dR,c}(M,P) \to \Omega_{dR,c}(M)$ is just $\alpha \mapsto \alpha$ for a form $\alpha$ null on $P$, and is clearly injective. It induced in de Rham cohomology with compact support $\bar{j}^* : H_{dR,c}(M,P) \to H_{dR,c}(M)$.

But is $\bar{j}^*$ injective too ?

If $\alpha \in Z_{dR,c}(M,P)$ is such that $[\alpha]_M = 0$, $\alpha = d\,\beta$ for $\beta \in \Omega_{dR,c}(M)$, but there is no reason why $\beta$ should be null on $P$. Is it always possible to find $\beta' \in \Omega_{dR,c}(M,P)$ such that $d\,\beta' = d\,\beta$ ? I have a big doubt...

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  • $\begingroup$ You probably need $f$ to be proper? But I don’t even believe that it holds even if $f$ is the inclusion. Think about the case when $M$ is the circle and $P$ are two points. $\endgroup$ – Thomas Rot Feb 10 '18 at 16:30
  • $\begingroup$ @Thomas, yes, I have edited, and in fact I am just interested in the case where f is the injection. $\endgroup$ – ychemama Feb 10 '18 at 16:56
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Not necessarily. For simplicity, let's assume a closed manifold $M$ and a closed submanifold $P$. Then all forms automatically have closed support and the de Rham cohomology is isomorphic to singular cohomology with real coefficients (my preferred way of thinking about cohomology). Then your map $\bar j^*$ is just the map $H^i(M,P)\to H^i(M)$ in the long exact sequence of the pair $$\cdots\to H^{i-1}(P)\to H^i(M,P)\to H^i(M)\to H^i(P)\to H^{i+1}(M,P)\to\cdots$$(coefficients suppressed). From there it's not difficult to come up with examples. For example, choose $M$ to be a large sphere but $P$ to be a submanifold in some middle dimension with non-trivial homology. Then you force $H^i(M,P)\cong H^{i-1}(P)\neq 0$ for some $i$ but the map $H^i(M,P)\to H^i(M)$ will be trivial but not injective.

You might have a look at something like Bott and Tu for the general theory.

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