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 $f_{*}(\phi) = \phi \circ f$ for all $\phi \in Hom_A(M,D)$ and $g_{*} = \psi \circ g$ for all $\psi \in Hom_A(M'',D)$.
I am having trouble showing $\ker(f_{*}) \subset Im(g_{*})$. I will lists the steps I have taken up until where I got stuck:
Let $\phi \in \ker(f_{*}) \Rightarrow f_{*}( \phi) = 0 $ for all $ \phi \in Hom_A(M,D)$
$\Rightarrow \phi \circ f = 0 \Rightarrow Im(f) \subset \ker(\phi)$ By exactness then we have $\ker(g) \subset \ker(\phi)$ and this is where I am stuck.
How do we conclude from the last statement that there exists an induced map $\psi : M'' \rightarrow L$ such that $\psi \circ g = \phi$?
I have in my notes that $\psi = (\phi/\ker(g) ):M'' \rightarrow N$ but I dont really understand this notation. What is it about the containment of the kernel of g that induces the map?