If $F$ is free, then the functor $M \mapsto \operatorname{Hom}_A(F,M)$ is exact.

Question

This is from Serge Lang's Algebra 3rd Edition (see the picture down below if you feel confused). My problem is, how to prove that if $F$ is free, then $M \mapsto \operatorname{Hom}_A(F,M)$ is exact. Indeed I know I can prove this using the fact that free modules are projective. But this statement is inside the declaration of projective module, so it's not a good idea to use properties of projective modules. Besides, it's chapter 3 into the book and many advanced concepts are not introduced. So I my thought is to prove it using properties of free modules. I was thinking about working on the basis of a module but don't know where to start.

I searched the internet but didn't find a suitable one. By suitable I mean the proof can be inserted before the introduction of projective modules.

You can find some discussions of exact functor and projective modules here (turn to 10.4 and 10.5).

Answer 1

Note that $\hom(F,-)$ is always left exact, so the desired result reduces to showing that $\hom(F,-)$ preserves epimorphisms. Fix $p \colon M \to N$ an epi, and let us prove that

$$ p_\ast \colon \hom(F,M) \to \hom(F,N) $$

is surjective. Given $g \colon F \to N$, we need $h \colon F \to M$ such that $ph = g$. For a given basis $B$ of $F$; one can define $h$ such that $h(b) \in p^{-1}(g(b))$ for each $b \in B$, guaranteeing the former to hold. This makes sense because each preimage is non-empty, here we use that $p$ is surjective. The existence of $h$ and its well definedness stem from the fact that $F$ is free.