If $0 \to \mathrm{Hom}(M, N') \to \mathrm{Hom}(M, N) \to \mathrm{Hom}(M, N'')$ is exact for all $M$, then $0 \to N' \to N \to N''$ is exact

Question

Let for all $A$-modules $M$, $$ 0 \rightarrow \operatorname{Hom}(M,N^{\prime}) \xrightarrow{\overline{f}} \operatorname{Hom}(M,N) \xrightarrow{\overline{g}} \operatorname{Hom}(M,N^{\prime\prime}) $$ be exact. Thus $\overline{f}$ is injective and $\ker(\overline{g}) =\operatorname{im}(\overline{f})$. We want to show that $0 \rightarrow N^{\prime} \xrightarrow{f} N \xrightarrow{g} N^{\prime\prime}$ is an exact sequence.

Let $M = N^{\prime} / \ker(f)$. If I can prove $N^{\prime} \cong M$, then $f$ will be one-one. But how to show that? Any hint will be appreciated.

Answer 1

Take $M=A$. Now, $\mbox{Hom}\left( A,N\right)\simeq N$, $\mbox{Hom}\left( A,N'\right)\simeq N'$, $\mbox{Hom}\left( A,N''\right)\simeq N''$. Now, by diagram chasing, you get that $0\to N'\to N\to N''$ is exact.