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?