Let $$ 0 \longrightarrow M_1 \overset{\phi_1}\longrightarrow M_2 \longrightarrow M_3 \longrightarrow 0 $$ be a short exact sequence of $R$-modules (we assume $R$ commutative). What are the conditions (if any) on $\phi_1$ which are sufficient to conclude $$ 0 \longrightarrow \mathrm{Hom}(M_3, N) \longrightarrow \mathrm{Hom}(M_2, N) \overset{\phi_1^*}\longrightarrow \mathrm{Hom}(M_1,N) \longrightarrow 0 $$ is a short exact sequence as well? In particular, when is $\phi_1^*$ surjective?
My educated guess is that it holds when $\phi_1$ is left-invertible, i.e. when the original sequence is split exact. In fact, in this case $M_2 \cong M_1 \oplus M_2/\phi_1(M_1)$, hence any $f : M_2 \to N$ can be 'split', by UP of direct sum, into $f_1 : M_1 \to N$ and $f_2 : M_2/\phi_1(M_1) \to N$. Then clearly any $f_1 \in \mathrm{Hom}(M_1,N)$ can be chosen to make $f$, and $\phi_1^* f = f_1$ is surjective.