Let $G$ be an arbitrary countable group, $X$ a finite, free, proper $G$-CW complex and $C(G)_*$ the induced free, finite $G$-Chain complex. This means that $C(G)_*$ is a finite chain complex, whose members $C(G)_n$ are free, finitely generated $\mathbb ZG$-modules and whose boundary maps are $\mathbb ZG$-homomorphisms (i.e, $G$-equivariant homomorphisms). Now, there are essentialy 2 different ways of forming cochain complexes:
a) There is the $G$-cochain-complex $C(G)^* := \hom_{\mathbb Z}(C(G)_*,\mathbb Z)$ with the obvious $G$-action. Note that no member $C(G)^n$ needs to be free as a $\mathbb ZG$-module. Denote the corresponding homology by $H^*(X,G)$.
b) There is the $G$-subcomplex $C(G)^*_c \subseteq C(G)^*$ of compactly supported cochains, i.e, those cochains that are supported on some finitely generated $\mathbb Z$-submodule (I believe that one needs to assume that the underlying CW-structure on $X$ is locally finite in order for $C(G)^*_c$ to be a subcomplex). Denote the corresponding homology by $H^*_c(X,G)$.
One can show that $C(G)^*_c \cong \hom_{\mathbb ZG}(C_*(G),\mathbb ZG)$. Thus, $C(G)^*_c$ is in particular always again a free, finite $G$-(co)chain complex. Moreover, observe that trivially $C(G)^*_c = C(G)^*$ whenever $G$ is a finite group, since $\mathbb ZG$ is then itself a finitely generated abelian group.
For any normal subgroup $H \unlhd G$, we naturally obtain an induced free, finite $G/H$-chain complex $C(G/H)_*$, such that the quotient map $\pi: G \to G/H$ gives rise to a canonical chain map \begin{equation} \pi_*: C(G)_* \to C(G/H)_*.\end{equation} Evidently, $C(G/H)_*$ is the $G/H$-chain complex of a free, proper $G/H$-CW complex $X_H$ that is covered by $X$. Applying the contravariant functor $\hom_{\mathbb Z}(\;_\;,\mathbb Z)$ to the diagram, we obtain a canonical cochain map \begin{equation} \pi^*: C(G/H)^* \to C(G)^* \end{equation} (Both of these maps are only assumed to be ordinary homomorphisms at each level, we don't assume any equivariance with respect to some group action). Although there is (to me) no obvious cochain map from $C(G)^*$ to $C(G/H)^*$, there is always a cochain map $T: C(G)^*_c \to C(G/H)^*$ given for $n \in \mathbb N$ by \begin{align} \forall \alpha \in C(G)^n_c \; \; \forall [x] \in C(G/H)_n: T(\alpha)[x] := \sum_{\bar{x} \in \pi^{-1}([x])} \alpha(\bar{x}) = \sum_{h \in H} h.\alpha(x). \end{align} Since $\alpha$ is compactly supported, the right-hand sum will always be finite and one easily checks that $T$ indeed defines a cochain-homomorphism. With this setup in mind, I can now formulate my question.
Question: Suppose that $H \unlhd G$ is a normal subgroup with $[G:H] < \infty$. Assume that there is some $n \in \mathbb N$ and a cochain $\alpha \in H^n_c(X,G)$ such that $span_{\mathbb ZG}(\{\alpha\}) \cong \mathbb ZG \subseteq H^n_c(X,G)$. Is it true that for $T(\alpha) \in H^n(X_H,G/H)$, we have \begin{equation} span_{\mathbb Z[G/H]}(\{T(\alpha)\}) \cong \mathbb Z[G/H] ? \end{equation} Whenever $G$ is finite, this assertion holds true. This follows since in that case, we have $H^n(X,G) = H^n_c(X,G)$ and we can consider the $G/H$-span of the element $(\pi^n \circ T)(\alpha) \in H^n(X,G)$. However, does this assertion also hold when $G$ is infinite ?