Projective Module equivalence

Question

I'm trying to prove that given an $A-module$: $P$

The functor $Hom(P,-)$ is exact $\implies$ $\exists Q (A-module): P\oplus Q$ is free

Can anyone guide me with some hints? It's the first time I'm seeing all of this.

Thanks

Answer 1

There exists a free module $F$ and a surjection $p:F\rightarrow P$, you have the exact sequence $0\rightarrow Ker(p)\rightarrow F\rightarrow P\rightarrow 0$, thus the sequence $0\rightarrow Hom(P,Ker(p))\rightarrow Hom(P,F)\rightarrow Hom(P,P)\rightarrow 0$ is exact, thus the morphism $Hom(P,F)\rightarrow Hom(P,P)$ is surjective. We deduce that there exists a morphism $h:P\rightarrow F$ such that $p\circ h=Id_P$, you can write $F=h(P)\oplus ker(p)$ and $h(P)\simeq P$.