Cfrgtkky

Let $M^{prime}xrightarrow{f}Mxrightarrow{g}M^{primeprime}rightarrow 0 DeclareMathOperator{Hom}{Hom}$ be a sequence of $A$-modules and homomorphisms. I want to show that the sequence is exact if and only if for all $A$-modules $N$, the sequence $$0rightarrow Hom(M^{primeprime},N)xrightarrow{overline{g}} Hom(M,N)xrightarrow{overline{f}} Hom(M^{prime},N)$$ is exact.

I have shown that $M^{prime}xrightarrow{f}Mxrightarrow{g}M^{primeprime}rightarrow 0$ is exact implies $0rightarrow Hom(M^{primeprime},N)xrightarrow{overline{g}} Hom(M,N)xrightarrow{overline{f}} Hom(M^{prime},N)$ is exact for all $A$-modules $N$.

Now suppose for all $A$-modules $N$, $0rightarrow Hom(M^{primeprime},N)xrightarrow{overline{g}} Hom(M,N)xrightarrow{overline{f}} Hom(M^{prime},N)$ is exact. I want to show that $M^{prime}xrightarrow{f}Mxrightarrow{g}M^{primeprime}rightarrow 0$ is exact. That is I have to show $g$ is onto and $ker(g)=operatorname{Image}(f)$. In Atiyah Macdonald's Commutative Algebra, it is written that since $overline{g}$ is injective for all $A$-modules $N$, therefore $g$ is onto. But I could not understand it. Any explanation will be appreciated.

Hint:

Take for $N$ the cokernel of $g$, i.e. consider the (surjective) canonical map
$$M''longrightarrow M''/g(M)$$
and deduce from the hypothesis that $M''=g(M)$.