$M' \to M \to M'' \to 0$ exact $\implies 0\to \text{Hom}(M'',N) \to \text{Hom}(M,N) \to \text{Hom}(M',N)$ is exact.

Let $M', M'', M, N$ be $A$-modules. If $$M' \stackrel{u}{\to} M \stackrel{v}{\to} M'' \to 0$$ is exact, then $$0\to \text{Hom}(M'',N) \stackrel{\bar{v}}{\to} \text{Hom}(M,N) \stackrel{\bar{u}}{\to} \text{Hom}(M',N)$$ is exact, where $\bar{u} (f) = f \circ u$ and similarly for $v$.

I've shown that $\ker (\bar{u}) \supseteq \text{im} (\bar{v})$ by starting with $v\circ u = 0$. How do I show the opposite inclusion?

Solution 1:

Let $f\in\text{Hom}(M,N)$ be in the kernel of $\overline{u}$. Therefore, for all $m'\in M'$, $(f\circ u)(m')=0$. Let $m_1,m_2\in M$ be such that $m_1-m_2\in M'$. Then $f(m_1-m_2)=0$, so $f(m_1)=f(m_2)$. Therefore, $f$ is constant on quotients $M/M'\simeq M''$.

Therefore, we can define a lift $\widetilde{f}\in\text{Hom}(M'',N)$ as follows. Let $m''\in M''$. Let $m$ be any preimage of $m''$ in $M$ (which exists because $v$ is surjective). Then define $\widetilde{f}(m'')=f(m)$. This is well-defined because any other preimage differs by an element of $M'$.

You should check that $\widetilde{f}$ is a homomorphism, but that is straight-forward. Also, it is easy to see that $\widetilde{f}\circ v=f$.

In general, the hard part of these types of proofs is constructing the correct object (which is usually done by defining it in the only way possible).