 Solution

Put $\Gamma ={\pi }_1(X,x)$ and $R=\mathbb{Z}[\Gamma ]$. Give $\mathbb{Z}$ its trivial left $R$-module structure, through the augmentation $\epsilon :R\to \mathbb{Z}$, $\sum n_{g}g\mapsto \sum n_g$. The universal cover $\tilde{X}$ has the lifted cell structure, with $\Gamma$ acting by deck transformations.

For each cell of $X$, choose one lifted cell and orient every translate compatibly with its projection. The translates form one copy of the regular $R$-module. Thusis a free left $R$-module, and its boundary is $R$-linear. The augmentation $C_0(\tilde{X})\to \mathbb{Z}$ sends every vertex to one. The contractibility of $\tilde{X}$ says that its reduced homology vanishes, so the augmented complexis exact. It is therefore a free resolution of the trivial module. By the definition of the Ext functor,

An $R$-linear cochain with values in the trivial module has the same value on every translate of a lifted cell. It is therefore exactly an ordinary integral cellular cochain on $X$. The correspondence does not depend on which lift was initially chosen. It also respects the coboundary: applying a constant-on-orbits cochain to the lifted cellular boundary sums precisely the incidence coefficients of the boundary in the base. Henceas cochain complexes, not just as graded groups. Cellular cohomology computes the cohomology of a cell complex, givingAt $i=0$ both sides are $\mathbb{Z}$, by connectedness and ${\mathrm{Hom}}_R(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$. This is the cellular free resolution from a contractible universal cover description of group cohomology.