$\newcommand{\Hom}{\operatorname{Hom}}$ For this problem, we let $0\rightarrow A\xrightarrow{i} B \xrightarrow{j}C\rightarrow 0$ be a short exact sequence of groups, and $G$ an abelian group. It's not too hard to see that this induces an exact sequence: $$0\rightarrow\Hom(G,A)\xrightarrow{i_*}\Hom(G,B)\xrightarrow{j_*}\Hom(G,C)$$ But I am then tasked with showing that if $G$ is free abelian, $j_*$ is surjective. The question suggests to me that such a condition must be necessary, but I can't think how I can use it to form a proof. Here's what I've come up with so far:
Given $\psi\in\Hom(G,C)$, we seek to construct $\phi\in\Hom(G,B)$ such that $j_*(\phi)=\psi$, or equivalently $j(\phi(g))=\psi(g)\; \forall g\in G$. Since $j$ is surjective by exactness, there exists $b\in B$ such that $j(b)=\psi(g)\; \forall g \in G$. Can we define $\phi$ by $\phi(g)=b$ where $j(b)=\psi(g)$? $j$ is not necessarily injective, so this may not be well-defined. Can we use the free-ness of $G$ at this point? I'm not sure how to proceed.
$\newcommand{\Hom}{\operatorname{Hom}}$ So, using the fact that $G$ free abelian (with basis $\{e_i\}_{i\in \mathcal{I}}$) means we can think of $\psi\in\Hom(G,C)$ as $\psi:e_i\mapsto c_i$ for some $c_i\in C$, each of which has a $j$-preimage $b_i\in B$, so we can define $\phi\in\Hom(G,B)$ by $\phi:e_i \mapsto b_i$, then $j\circ \phi=\psi$ as required.