Let $M$ be a manifold, $P$ a closed subset and sub-manifold of $M$, and $f : (M,\emptyset) \to (M,P)$ a proper smooth map (I note $f : (M,P) \to (N,Q)$ a smooth map $M \to N$ such that $f(P) \subset Q$). If $f$ is the the inclusion $j$, its 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...

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...

Let $M$ be a manifold, $P$ a closed subset and sub-manifold of $M$, and $f : (M,\emptyset) \to (M,P)$ a smooth map (I note $f : (M,P) \to (N,Q)$ a smooth map $M \to N$ such that $f(P) \subset Q$). Its pullback $f^* : \Omega_{dR,c}(M,P) \to \Omega_{dR,c}(M)$ 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 $\bar{f}^* : H_{dR,c}(M,P) \to H_{dR,c}(M)$.

But is $\bar{f}^*$ 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...

Let $M$ be a manifold, $P$ a closed subset and sub-manifold of $M$, and $f : (M,\emptyset) \to (M,P)$ a proper smooth map (I note $f : (M,P) \to (N,Q)$ a smooth map $M \to N$ such that $f(P) \subset Q$). If $f$ is the the inclusion $j$, its 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...