Let $B$ be a cellular (simplicial, semi-simplicial etc) complex having $\mathbb{Z}^n$-symmetry (that is, there is a free action of $\mathbb{Z}^n$ on $B$, commuting with the boundary operators) and let $R$ be a commutative ring. In this case it is possible to establish an isomorphism (as $R$-modules) of the homology modules $H_*(B, R)$ and $H_*(B', R[\mathbb{Z}^n])$, where the "quotient" complex $B'$ is constructed in the following way:

1) In each dimension $d$, choose a representative cell in each orbit of $\mathbb{Z}^n$-action on $B$. Define the chain modules $C_d(B')$ as the free $R[\mathbb{Z}^n]$-modules on the basis of these representatives.

2) Let $e_l$ be a cell of dimension $d$ in $B$ and let $e$ be the corresponding representative in its $\mathbb{Z}^n$-orbit, with $l \in \mathbb{Z}^n$ bringing $e$ to $e_l$. Let also $f$ be the respective basis element in $C_d(B')$. Then the mapping of $e_l$ to $1l \cdot f$ extends to an $R$-module isomorphism $\alpha: C_d(B) \to C_d(B')$ (here $1l \in R[\mathbb{Z}^n]$ stands for the element of the group ring corresponding to $l \in \mathbb{Z}^n$).

3) Since the boundary operators $\partial$ in $B$ commute with the action of $\mathbb{Z}^n$, the operators $\alpha \circ \partial \circ \alpha^{-1}$ are $R[\mathbb{Z}^n]$-linear, and make $B'$ into a differential complex of $R[\mathbb{Z}^n]$-modules. Its homology modules are isomorphic (as $R$-modules) to those of $B$.

My question is: is there a more elegant and natural way (without choosing representatives) to construct this "quotient" complex $B'$ and the corresponding isomorphism of homology modules? Also, I have a feeling that this is a particular case of some more general theory, but do not see which one.