Right exactness of the Hom functor in the first argument for free modules Let $0 \rightarrow E_1 \stackrel{\sigma}{\rightarrow} E_2$ be an exact sequence of free $A$-modules, $A$ being a commutative ring. Let $F$ be a free $A$ module. Then is it true that the sequence $Hom_A(E_2,F) \rightarrow Hom_A(E_1,F) \rightarrow 0$ is exact? If yes, how can i show it?
 A: In Lang there is actually a stronger hypothesis. The theorem is that when you have $0\to U\to W\to V\to 0$ where all terms are finite rank free $A$-modules then $Hom_A(-,A)$ is exact. Note that the way it is written in the question allowed me to pick $\sigma$ as multiplication by $2$, but the cokernel is then $\mathbb{Z}/2\mathbb{Z}$ which is not a free $\mathbb{Z}$-module.
The key here is that any such short exact sequence with the given hypothesis actually splits and so you can reduce to the case of checking it works for the sequence $0\to U\to U\oplus V\to V\to 0$ where the first is just an inclusion of the free module and the second is the projection. (Note again that $\mathbb{Z}$ is not isomorphic to $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$, so the sequence I constructed for the counterexample does not split).
In that case what we want to check is that $Hom_A(U\oplus V, A)\to Hom_A(U,A)\to 0$ is exact. I.e. if you have a map $f: U\to A$ can you find some map $g: U\oplus V\to A$ such that composing with the inclusion $U\to U\oplus V\to A$ is the original? Sure. Just take $g=f\oplus 0$.
