Let $M$ be abelian group and $$ A^{'}\stackrel{f}{\rightarrow} A \stackrel{g}{\rightarrow} A^{''} \rightarrow 0 $$ an exact sequence of abelian groups. I want to show that $$0\rightarrow Hom(A^{''},M)\stackrel{g^*}{\rightarrow}Hom(A,M)\stackrel{f^*}{\rightarrow}Hom(A^{'},M)$$ is also an exact sequence where $f^*(\phi)$ denotes $\phi \circ f$ and $g^*$ the same.
I am having a hard time proving that $im(g^*)=\ker(f^*)$. For every $\phi \in \ker(f^*)$ I tried to construct a group homomorphism $\sigma: A^{\prime \prime} \rightarrow M$ with $\phi=\sigma \circ g$ using the axiom of choice, but I don't think I get a group homomorphism this way. Thank you