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

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).

• Hint: let $B \subset F$ be a basis, then $M \longmapsto \mathrm{Hom}(F,M)$ and $M \longmapsto M^B$ are isomorphic functors... Feb 1, 2021 at 9:26
• If you want a reference, this appears in the second chapter of Atiyah-Macdonald Feb 1, 2021 at 9:38
• off-topic: the image size/question ratio is $>> 1$, your paragraph already contains all the relevant info :) Feb 1, 2021 at 10:13
• @guidoar OK I will put the image at the bottom so people see the problem first.
– user614535
Feb 1, 2021 at 10:19

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.