I am looking at a statement that, for a short exact sequence of Abelian groups
$$0 \to A\mathop \to \limits^f B\mathop \to \limits^g C \to 0$$
if $C$ is a free Abelian group, then this sequence is split.
I cannot figured out why, can anybody help?
We want to find $h : C \to B$ such that $gh = \operatorname{id}_C$ holds. Since $C$ is free abelian, it is generated by a basis $S$. By exactness, $g$ is surjective, hence for any $x \in S$ we find $b_x \in B$, such that $g(b_x) = x$. Now define $h : C \to B$ to be the extension of the map $x \mapsto b_x$ (by the extension property). Now we have for any $x \in S$: $$gh(x) = g(b_x) = x,$$ which implies $gh = \operatorname{id}_C$.
Hint:
How do you define a homomorphism from a free abelian group (more generally a free $R$-module) into another group/module?
Deduce a homomorphism $s\colon C\to B$ such that $\;g\circ s=\operatorname{id}_C$, and prove that $$B\simeq A\oplus s(C).$$