Let $A$ be a commutative ring with identity. Let D,M,M',M'' be $A$-modules and suppose that 0\rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0 is an exact sequence. Label the maps f:M'\rightarrow M and g: M\rightarrow M''.
Consider the induced sequence 0 \rightarrow Hom_A(M'',D) \rightarrow Hom_A(M,D) \rightarrow Hom_A(M',D) and label the map f_{*} : Hom_A(M,D) \rightarrow Hom_A(M',D) given by $f_{*}(\phi) = \phi(f)$ for all $\phi \in Hom_A(M,D)$. I know the induced sequence is exact but I am missing a step in the proof.
How do we prove that the map $f_{*}$ is surjective? Or does $f_{*}$ need not be surjective for the sequence to be exact since we do not have the map Hom_A(M',D) \rightarrow 0?
I don't know if this is the place in module theory where you use baers criterion but I am having trouble filling in this step.