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).
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.