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.



Homology modules and symmetry
Suppose a group $G$ acts `properly discontinuously' on a friendly topological space $B$, so that $G$ is acting freely on $B$, and, if we let $B^{\prime} = B/G$, then $B \rightarrow B^{\prime}$ is a covering map. It is a nice exercise in the standard lifting theorem of covering space theory to show that cellular structures on $B^{\prime}$ correspond to $G$--equivariant cellular structures on $B$. (Similarly for simplicial structures, etc.) 


If $M$ is a $R[G]$--module, $H\_\*(B^{\prime};M)$, the homology of $B^{\prime}$ with twisted coefficients $M$, is defined to be the homology of the complex $C\_\*(B) \otimes\_{R[G]} M$. There are two extreme cases. Let $M = R$. As $C\_\*(B) \otimes\_{R[G]} R = C\_\*(B^{\prime})$, we have $H\_\*(B^{\prime};R) = H\_\*(B^{\prime};R)$. (phew!). At the other extreme, let $M = R[G]$. As $C\_\*(B) \otimes\_{R[G]} R[G] = C\_\*(B)$, we have - yes - $H\_\*(B^{\prime};R[G]) = H\_\*(B;R)$.


