In Godbillon's book "Topologie Algébrique", chapter "de Rham cohomology with compact support", he states thm 3.1:

Let $M$ be a smooth manifold, $N$ a closed (as subset) submanifold, the injections $i : M\setminus N \to M$ and $j : (M,\emptyset) \to (M,N)$, then the homomorphism $i_* : H_c(M\setminus N) \to H_c(M)$ (which is extension by $0$ of a form with compact support on $M\setminus N$) factors through $H_c(M\setminus N) \xrightarrow{i'_*} H_c(M,N) \xrightarrow{j^*} H_c(M)$, i.e. $i_* = j^* \circ i'_*$.

Thm 3.1: $H_c(M\setminus N) \xrightarrow{i'_*} H_c(M,N)$ is an isomorphism.

His proof uses a version (not the usual one) of the "Tubular Neighborough theorem" : $M$ and $N$ as before, there exists a closed submanifold $T$ of $M$ having the same dimension, such as 1) $N \subset \mathring{T}$, 2) there is a proper smooth retraction $r : T \to N$, 3) with the injection $k : N \to T$, $k \circ r$ is smoothly and properly homotopic to $Id_T$. So here $k^* : H_c(T) \to H_c(N)$ is an isomorphism.

Then for the surjectivity of $i'_*$ he goes : let $\alpha \in Z_c(M,N)$. $\alpha$ being null on $N$ there is a form $\beta \in \Omega_c(T)$ such that $\alpha = d\,\beta$ (me : *because $k^*$ is an isomorphism, but $\beta$ is not null on $N$ a priori*). Let $\gamma \in \Omega_c(M)$ such that $\gamma = \beta$ on $T$ (me : *because here, with $l : T \to M$ the injection, $l^* : H_c(M) \to H_c(T)$ is surjective*). So $\alpha' = \alpha - d\;\gamma$ is a closed form with compact support, cohomolog to $\alpha$, null on $T$ (* so $Supp(\alpha') \subset M \setminus N$*) and $\alpha' \in Z_c(M\setminus N)$ completing the proof.

I am OK with its "notation shortcuts", but we're looking for a form in $Z_c(M\setminus N)$ whose image by $i'_*$ (which is extension by $0$) is cohomologous to $\alpha$ *relative to $N$*. He shows that $\alpha'$ is cohomologous to $\alpha$ in $\Omega_c(M)$, but not that it is cohomologous relative to $N$, i.e. in $\Omega_c(M,N)$, since there is no reason why $\beta$ or $\gamma$ should be null on $N$.

Is Godbillon's proof flawed, or am I missing something obvious ?