Let $G$ be a group. Let $\Bbbk$ be a field of char. $0$. We denote with $C^{n}(G, \Bbbk)$ the set of maps $f\: : \: G^{n}\to \Bbbk$ and with $\partial\_{G}\: : \: C^{n-1}(G, \Bbbk)\to C^{n}(G, \Bbbk)$ the differential 
$$\partial\_{G}f(g\_{1},\ldots,g\_n)=f(g\_{2},\ldots,g\_n)+\sum\_{1\leq j\leq n-1}(-1)^{j-1}f(g\_1,\ldots,{g\_j}{g\_{j+1}},\ldots,g\_n)+(-1)^{n}f(g\_{1},\ldots,g\_{n-1})$$
We denote the cohomology of such a complex with $H(G, \Bbbk)$. It is the cohomology of $G$ with coefficients in $\Bbbk$. Let $F\_{2}$ be the free group on two generators $a\_{1}$, $a\_{2}$. It is well know that $$H^{n}(F\_{2}, \Bbbk)=\begin{cases} \Bbbk, & n=0,\\
\Bbbk^{2}, & n=1\\
0, & n>1
\end{cases}$$
In particular $H^{1}$ is generated by two group homomorphism $f\_{i}\: : \: G\to \Bbbk$ for $i=1,2$ defined by
$$
f\_{i}(a\_{j}):=\delta\_{ij}.
$$
Consider the maps $F\: : \: F\_{2}^{2}\to \Bbbk$ defined by $F(a,b):=f\_{1}(a)f\_{2}(b)$. In the language of simplicial set, this is the Alexander-Whitney product of $f\_{1},f\_{2}$. Here my questions:


1) $F$ is closed, hence it is exakt, do you know an explicit coboundary?


2)I am interested to the following: Let $F\_{n}$ be the free group on $n$ generators. Analogously we have $H^{n}(F\_{n}, \Bbbk)=0$ for $n>1$. Let $F\: : \: F\_{n}^{2}\to \Bbbk$ be a closed element. Is there some general formula for a $f\in C^{1}(F\_{n}, \Bbbk)$ such that $\partial\_{G}f=F$?



Coboundary for the Cohomology of free groups
There is a projective resolution $P\_\bullet$ of $\mathbb Z$ as an $F\_n$-module of the form $$0\to\mathbb ZF\_n\otimes V\to\mathbb ZF\_n\to\mathbb Z\to0$$ in which $V$ is the free $\mathbb Z$-module spanned by the generators $x\_1,\dots,x\_n$ of $F\_n$, and the differential is the $\mathbb ZF\_n$-linear map such that $d(1\otimes x\_i)=x\_i-1$. 


On the other hand, you have the bar resolution $B\_\bullet$ of $\mathbb Z$ as a $\mathbb ZF\_n$-module. The two complexes are homotopic, so that there are maps $\phi:P\_\bullet\to B\_\bullet$ and $\psi:B\_\bullet\to P\_\bullet$ such that $\phi\circ\psi$ and $\psi\circ\phi$ are homotopic to identity maps. If you make explicit an homotopy $s:\phi\circ\psi\simeq 1\_{B\_\bullet}$, you can do what you want: if $\alpha$ is a $p$-cocycle on the bar complex and $p>1$, then the boundary of $\alpha\circ s$ is $\alpha$.


