Projective modules and $\hom(P,\cdot)$

Question

I'm working in the categories of $A$-modules. I'm trying to prove that $\hom(P,\cdot)$ is an exact functor $\iff$ $\exists\,Q$ such that $P\oplus Q$ is a free $A$-module.

I'm having trouble translating the property of "freeness" into the preservation of exact sequences (and vice-versa). They still seem like totally different things to me.

Any tips?

Answer 1

For the reverse, recall that $\hom(P,-)$ is always left exact. That is, if you have an exact sequence

$$0\to L\to M\to N\to 0$$

then applying $\hom(P,-)$ gives

$$0\to\hom(P,L)\to\hom(P,M)\to\hom(P,N),$$

but it's not always right exact. This is what you need to show, supposing $P\oplus Q$ is a free $A$-module for some $Q$.

What this amounts to showing is that given a morphism $f:P\to N$ and the surjection $M\to N\to 0$, there is a map $\hat f:P\to M$ which "lifts" $f$. But you immediately get a morphism $P\oplus Q\to N$ by taking $(f,0)$, and since $P\oplus Q$ is free, you can choose a basis, and lift their images to $M$ to get a map $P\oplus Q\to M$ (you should be more precise than this in your solution, but this is the idea). Now why does this give us a map $P\to M$ which lifts $f$?