Coboundary for the Cohomology of free groups

Question

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 $\partial_{G}f=F$?

Answer 1

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 $\alpha$.