Let $f\colon\thinspace G\to H$ be a surjective homomorphism of finitely generated groups. Are there any methods to decide whether $f$ factors through a free group? That is, does there exist a free group $F$ and homomorphisms $g\colon\thinspace G\to F$ and $h\colon\thinspace F\to H$ such that $f=h\circ g$?
I know of one necessary condition, given by cohomology. If $f$ factors through a free group then $f^*\colon\thinspace H^i(H;M)\to H^i(G;M)$ is zero for all $i>1$ and all $H$-modules $M$ (since free groups have cohomological dimension one).
This question was inspired by Tom Goodwillie's answer to my earlier question on cohomological dimension of group homomorphisms.
Best Answer
$f: G \to H$ factors through a free group iff there is a subgroup $N \le \operatorname{ker}(f)$ and a free group $F$ such that $G = N \rtimes F\;\;$ ($N$ normal ).
Proof: $(\Rightarrow)$ Let $G \xrightarrow{g} E \xrightarrow{} H$ be a factorization of $f$ with $E$ free and let $N$ be the kernel of $g$. Clearly $N \le \operatorname{ker}(f)$ and as a subgroup of a free group, $F := G/N$ is itself free. Thus we have an extensions $1 \to N \to G \to F \to 1$ that splits since $F$ is free.
$(\Leftarrow)$ Suppose $N\le \operatorname{ker}(f)$ and $G = N \rtimes F$ with $F$ free. Hence $N$ is normal in $G$ and because $N \le \operatorname{ker}(f)$, $f$ has a factorization $G \to G/N=F \to H$ through a free group.
Remark: That $f$ is surjective wasn't used. But if we know that $f$ is surjective, we can conclude that the rank of $F$ is greater or equal than the minimal number of generators of $H$.