I'm trying to see an example of the topological interpretation of group cohomology, with the free group $F(S)$ on a set $S$ of generators, with coefficients in $\mathbb{Z}$ (on which we act trivially), but I have a problem because I get a $0$ cohomology, which is not supposed to happen. Here's my reasoning :
Put $G=F(S)$
So then, a $K(G,1)$ is the bouquet of circles $\bigvee_{s\in S}S^1$, and a universal cover of this is the Cayley graph $\Gamma(G,S)$. It has a cellular decomposition that consists in $G$ $0$-cells (vertices are elements of $G$), $G\times S$ $1$-cells (edges are pairs $(g,s)$, going from $g$ to $gs$), and there are no higher cells.
Good thing, we automatically get $H^q(G,\mathbb{Z})=H_q(G,\mathbb{Z}) =0$ for $q>1$. Indeed, we have the following $G$-free resolution of $\mathbb{Z}$ given by the cellular decomposition : $\dots \to 0\to \mathbb{Z}\{G\times S\}\to \mathbb{Z}\{G\} \to\mathbb{Z}\to 0$ where $\mathbb{Z}\{X\}$ denotes the free abelian group on the set $X$; and the action on $\mathbb{Z}\{G\times S\}$ is given on the basis by $h\cdot (g,s) = (hg,s)$
Now the map $\mathbb{Z}\{G\} \to\mathbb{Z}$ is simply the augmentation map, easy to describe. The boundary $\partial : \mathbb{Z}\{G\times S\}\to \mathbb{Z}\{G\}$ is, up to sign, given on the basis by $\partial (g,s) = gs - g$.
For homology, taking co-invariants gives us $\mathbb{Z}\{G\}_G = \mathbb{Z}, \mathbb{Z}\{G\times S\}_G = \mathbb{Z}\{S\}$ by an easy computation, and the boundary becomes $0$ (easy to check as well), so that $H_0(G,\mathbb{Z}) = \mathbb{Z}$, and $H_1(G,\mathbb{Z})=\mathbb{Z}\{S\}$.
For cohomology, we take invariants. That's where my issue kicks in. For nonempty $S$, $G$ is infinite, and so there seems to be no invariants, should it be in $\mathbb{Z}\{G\}$, or in $\mathbb{Z}\{G\times S\}$. Indeed, for the first one, take $\displaystyle\sum_{g\in G}a_g g$ an invariant, then by acting with $h$ on it we get that for all $g,h$, $a_g=a_{h^{-1}g}$; which means that all the $a_g$ are equal, and so must be $0$ (there's infinitely many of them) - and the same type of argument applies for the other one: this would yield that the cohomology is $0$ in all degrees !
But of course it's not the case, I already know (by algebraic considerations) that $H^1(G,\mathbb{Z}) = \hom(G^{ab},\mathbb{Z}) = \mathbb{Z}^S$, which is far from $0$.
So where's my mistake ?
To compute cohomology, you don't take invariants of the chains on $\Gamma(G,S)$; you take invariants of the cochains. In other words, before taking invariants you first have to dualize your resolution (Hom it into your coefficient group $\mathbb{Z}$; or equivalently you can both dualize and take invariants at once by taking the $G$-equivariant Hom). So, your cochain groups will be a product, not a diret sum, of copies of $\mathbb{Z}$ and you are allowed to have cochains of infinite support.