The motivation behind this problem is that I need it to prove that an exact sequence, then if we passing to hom induces left exact sequence.
Here we assume that $\phi_2$ is surjective. This is an important assumption.
If $M_1,M_2,M_3$ are modules and $\phi_1 : M_1 \rightarrow M_3$,$\phi_2 : M_1 \rightarrow M_2$ such that $ker(\phi_2) \subset ker(\phi_1)$ then is there $\phi_3 : M_2 \rightarrow M_3$ making that diagram commute ?
I am thinking that $\phi_3(m_2) = \phi_1(\phi_2^{-1}(m_2))$. I proved that this is well defined. But, I am not sure why is it an A-module morphism?
Here is why it is well defined. I will add that for completeness sake. Suppose $x,y \in \phi_2^{-1}(m_2)$ then $\phi_2(x - y) = 0$ so we have that $\phi_1(x - y) = 0$. Hence $\phi_1(x) = \phi_1(y)$. I just want to know why it is an A-module homomorphism. The commutative part is easy.