Let $G$ be a discrete, countable group and $X$ a finite, free, proper $G$-CW complex, such that the underlying $CW$-structure of $X$ is locally finite. Further, let $C_*(X,G)$ be the induced free-finite $\mathbb Z[G]$-chain complex, freely generated by the collection of all $G$-equivariant cells on $X$. We can form $2$ distinct cochain complexes, namely

1) $C^*_c(X,G) := \hom_{\mathbb Z[G]}(C_*(X,G),\mathbb Z[G])$,

2) $C^*(X,G) := \hom_{\mathbb Z}(C_*(X,G),\mathbb Z)$.

Denote the correspoding cohomology by $H^*_c(X,G)$ and $H^*(X,G)$, respectively.

Let $H \unlhd G$ be a finite index normal subgroup of $G$. Consider the quotient $X/H$, equipped with the induced free, proper $G/H$-equivariant CW structure. Denote by $\pi: X \to X/H$ the quotient map. Then, there exists a cochain map $T: C^*_c(X,G) \to C^*(X/H, G/H)$, defined at the $n$-th level as the following $\mathbb Z$-homomorphism:

\begin{align} \forall \alpha \in C(X,G)^n_c \; \; \forall [x] \in C(X/H,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}

The fact that $\alpha \in C(X,G)_c^n$ ensures that the right-hand sum is finite. Now my question is, roughly speaking, whether this map preserves any cohomology occuring in $H(X,G)^n_c$, or, more precisely:

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 is easily verified. However, does this assertion also hold when $G$ is infinite ?

*Remark:* I posted this same question, along with some more explanation, already on stackexchange some months ago, without getting any answer. However, I realized that this setting of equivariant cohomology might be better suited for the experts here at overflow.