Well-definedness of a coboundary map between a reduced $L^2$ de Rham cohomology group and a relative cohomology group I'm working right now with this paper of Carron.
And I think I'm stuck at a relatively simple question. On page 11 he is defining a coboundary map $b : H^k_{2, \text{reduced}}(M - K) \to H^{k+1}(K, \partial K)$, while $K$ is a smooth, compact submanifold of a complete manifold $M$ and $H^{k+1}(K,\partial K)$ is the relative cohomology group of K and $H^k_{2, \text{reduced}}(M - K)$ is the reduced $L^2$ de Rham cohomology.
The coboundary map is defined as follows:
For $c \in H^k_{2, \text{reduced}}(M - K)$ take a smooth representative $\alpha$ of $c$, choose a smooth extension $\bar\alpha \in C^\infty(\Lambda^kT^*M)$. $d\bar\alpha$ is a closed form and zero on $M - K$.
Now we define $b(c) = [d\bar\alpha] \in H^{k+1}(K, \partial K)$.
But how can I show that this doesn't depend on my choice of the smooth representative of $c$?
 A: Let $\beta$ be another smooth representative of $c$, and let $\bar\beta$ be a smooth extension of $\beta$. We need to prove that $[d\bar\alpha]-[d\bar\beta]=0\in H^{k+1}(K,\partial K)$. Таke some $\gamma\in H_{k+1}(K,\partial K)$ and $\Gamma\in\gamma$. That is, $\Gamma$ is a $(k+1)$-chain in $K$ and $\partial\Gamma\subset\partial K$. Then 
$$
\int_{\Gamma} (d\bar\alpha-d\bar\beta) =\int_{\partial\Gamma}(\bar\alpha-\bar\beta) 
$$
Since $\alpha,\beta\in c$, it follows that $\alpha-\beta=d\omega$, where $\omega$ is a smooth $(k-1)$-form on $M\setminus K$. Let $\partial K_\epsilon$ be the geodesic shift of $\partial K$ by $\epsilon$ along the exterior normal to $\partial K$, and let $\partial\Gamma_\epsilon\subset \partial K_\epsilon$ be the shift of $\partial\Gamma$. Then $\partial\Gamma_\epsilon\subset M\setminus K$ for all $\epsilon>0$,  and we have
$$
\int_{\partial\Gamma}(\bar\alpha-\bar\beta)=\lim_{\epsilon\to0} 
\int_{\partial\Gamma_\epsilon}(\alpha-\beta)
=\lim_{\epsilon\to0} 
\int_{\partial\Gamma_\epsilon}d\omega=
\lim_{\epsilon\to0} 
\int_{\partial(\partial\Gamma_\epsilon)}\omega=0,
$$ 
because $\partial(\partial\Gamma_\epsilon)=$ shift by $\epsilon$ of $\partial^2\Gamma=\emptyset$. Thus, $\langle[d\bar\alpha],\gamma\rangle=\langle[d\bar\beta],\gamma\rangle$ for every $\gamma\in H_{k+1}(K,\partial K)$. Now $[d\bar\alpha]=[d\bar\beta]$ by the relative de Rham theorem.
