How to prove that $G$ is semidirect product?

Question

If $ N\triangleleft G $ and $ G / N $ is a free group, I want to show that $ G $ is a semi-direct product of $ N $ by $ G / N $, but I don't see how to prove it, I know if $ N $ is normal to $ G $, then $ G $ is a semi-direct product of $ N $ and $ G / N $ if and only if there is a homomorphism $s: G / N \to G$ such that $v\circ s: G/N \to G/N = id$, where $v: G\to G/N$ is the natural projection. but I don't know where to put the hypothesis that $ G / N $ is a free group.

Answer 1

Basically, free groups are projective, so every surjection onto a free group has a section. Explicitly:

If $G/N$ is a free group, say it is (isomorphic to) a free group on the set $X$. For each $x\in X$, let $g_x\in G$ be such that $\pi(g_x)=x$, where $\pi\colon G\to G/N$ is the canonical projection.. By the universal property of the free group, there exists a group homomorphism $f\colon G/N\to G$ such that $f(x)=g_x$ for each $x\in X$. Since $\pi\circ f$ is the identity on $X$, the universal property of the free group tells us that $\pi\circ f=\mathrm{id}_{G/N}$, and so it follows that $f$ is one-to-one. Therefore, $G$ contains a subgroup isomorphic to $G/N$.

Now verify that $f(G/N)\cap N=\{e\}$ and that $G$ is generated by $N$ and $f(G/N)$.