# Will there exist A-module morphism making the diagram commute?

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.

• How did you prove it's well defined? You can't assert that $\phi_2$ is injective, so how are you defining $\phi_2^{-1}$? – quasi Mar 16 '17 at 4:19
• @quasi: $\phi^{-1}_2(m_2)$ can refer to the entire preimage of $m_2$. This is fine as long as $\phi_2$ is surjective, because it's just a coset for the kernel of $\phi_2$, so after applying $\phi_1$ you get a single element. But if $\phi_2$ is not surjective then $\phi_2^{-1}(m_2)$ may be empty. – Qiaochu Yuan Mar 16 '17 at 4:20
• I could be wrong, but my sense is that you should look for a counterexample, not a proof. – quasi Mar 16 '17 at 4:22
• @Qiaochu Yuan Oh yeah right so under my assumption we have that $\phi_2$ is surjective. – user111750 Mar 16 '17 at 4:28
• If $\phi_2$ is surjective this is straightforward; it follows e.g. from the isomorphism theorems. – Qiaochu Yuan Mar 16 '17 at 4:30

Here is the answer. Suppose $\psi : M_1 \rightarrow M_3$ and $\phi : M_2 \rightarrow M_3$ with $\phi$ surjective and $ker(\phi) \subset ker(\psi)$. Then define $\eta : M_1 / ker(\phi) \rightarrow M_3$ as follows $\eta([x]) = \psi(x)$ this is well defined by the fact $ker(\phi) \subset ker(\psi)$