Let $A$ be a ring and let $P$ be a projective $A$-module. Then, the exactness of the sequence:
$0\longrightarrow M_1 \overset{f}{\longrightarrow}M_2\overset{g}{\longrightarrow}M_3\longrightarrow 0 \tag{1}$
implies the exactness of the induced $\mathbb{Z}$-module sequence:
$0\longrightarrow \text{Hom}_A(P,\,M_1) \overset{f_\bullet}{\longrightarrow} \text{Hom}_A(P,\,M_2)\overset{g_\bullet}{\longrightarrow} \text{Hom}_A(P,\,M_3)\longrightarrow 0 \tag{2}$
Does the exactness of (2) imply that $g$ is an epimorphism?
Let $x\in M_3$. Suppose that there exists some morphism $\varphi\in\text{Hom}_A(P,\,M_3)$ such that $x\in\text{im}(\varphi)$; then, since $g_\bullet$ is an epimorphism, there exists some $\psi\in\text{Hom}_A(P,\,M_2)$ such that $\varphi=g_\bullet(\psi)=g\circ\psi$, so $x\in\text{im}(g)$. So it suffices to show that, for every $x\in M_3$ there exists some morphism in $\text{Hom}_A(P,\,M_3)$ such that $x$ is in its image. If $P$ is free this is obviously true; is it true if $P$ isn't?