# Applying hom-functor on short exact sequence

Let $N, M, M', M''$ be $R$-modules. Given homomorphisms

$f: M' \rightarrow M$

$g: M \rightarrow M''$

$\psi: N \rightarrow M$

with $\operatorname{im} f = \ker g$, $g \circ \psi = 0$ and $g$ surjective, $f$ injective I would like to show that there exists a homomorphism $\phi: N \rightarrow M'$ such that $f \circ \phi = \psi$. Trying to consider some factor modules did not lead me too far.

context: I want to show that if $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is a short exact sequence, then the following sequence is exact: $0 \rightarrow Hom(N, M') \rightarrow Hom(N,M) \rightarrow Hom(N,M'')$.

• This would be my idea to show that the hom sequence is exact in Hom(N,M). To do this we need to show that $im H(f) = ker H(g)$, when H(f) resp. H(g) is the image of f resp. g under the hom-functor. Thus we need to show that for any homomorphism $\psi \in H(N,M)$ the equivalence $\psi = f \circ \phi$ for some $\phi \in Hom(N,M')$ $\Leftrightarrow$ $g \circ \psi = 0$ holds. "$\Rightarrow$" follows immediately, I am struggling with "$\Leftarrow$".
– CHwC
Commented Apr 22, 2016 at 22:35
• However, I am definitely not sure about my thoughts, so please correct me if there is anything wrong. I am open to other suggestions.
– CHwC
Commented Apr 22, 2016 at 22:36

Since $f$ is injective, just define $\phi(n)=f^{-1}(\psi(n))$. To know that this definition makes sense, you need to know that $\psi(n)$ is always in the image of $f$. But the image of $f$ is the kernel of $g$, and $g(\psi(n))=0$ by assumption.
• How else do you imagine showing it? You have to actually define the function $\phi$ somehow, and defining a function means showing what it does to elements. You can give a "categorical" argument, but that's just hiding the elementwise definition in a proof that certain universal properties hold for your maps. Commented Apr 22, 2016 at 22:51