Requirements for Hom(-,A) to be exact.

[rasmhop]

I'm reading Algebra (GTM) by Serge Lang and I'm a bit confused about the requirements of theorem 6.3 in chapter 4 (Modules). Not the technical nature of them, but rather the motivation behind them so let me first present my take on the theorem. The theorem reads (text in square brackets are my comments to provide context):

Theorem 6.3: Let U, V, W be finite free modules over the commutative ring A, and let
$$0 \rightarrow W \xrightarrow{\lambda} V \xrightarrow{\varphi} U \rightarrow 0$$
be an exact sequence of A-homomorphisms. Then the induced sequence
$$0 \rightarrow \text{Hom}_A(U,A) \rightarrow \text{Hom}_A(V,A) \rightarrow \text{Hom}_A(W,A) \rightarrow 0$$
i.e.
$$0 \rightarrow U^\vee \rightarrow V^\vee \rightarrow W^\vee \rightarrow 0$$
is also exact [$U^\vee$ denotes the dual module].
Proof: This is a consequence of P2 [short exact sequences of modules with last module projective splits] because a free module is projective.

I can't really see where the finiteness of U, V, W enters into the proof. Actually only the fact that U is free (or more generally: projective) matters. Afraid my understanding was lacking I tried to carry out the proof in detail, but can't see the issue:
Theorem 6.3':
Let U, V, W be modules over the commutative ring A with U a projective (or free) module, and let
$$0 \rightarrow W \xrightarrow{\lambda} V \xrightarrow{\varphi} U \rightarrow 0$$
be an exact sequence of A-homomorphisms. Then the induced sequence
$$0 \rightarrow \text{Hom}_A(U,A) \rightarrow \text{Hom}_A(V,A) \rightarrow \text{Hom}_A(W,A) \rightarrow 0$$
is also exact.
Proof:
Since U is projective, the short exact sequence
$$0 \rightarrow W \xrightarrow{\lambda} V \xrightarrow{\varphi} U \rightarrow 0$$
splits, with an A-homomorphism $\psi : V \to W$ splitting $\lambda$ (i.e. $\psi \circ \lambda = 1_W$). In general $\text{Hom}_A(-,A)$ is a contravariant left-exact functor so we have the following exact sequence:
$$0 \rightarrow \text{Hom}_A(U,A) \xrightarrow{\varphi'} \text{Hom}_A(V,A) \xrightarrow{\lambda'} \text{Hom}_A(W,A)$$
Thus all we need is to show that $\lambda'$ is an epimorphism. Define: $\psi' : \text{Hom}_A(W,A) \to \text{Hom}_A(V,A)$ by $f \mapsto f \circ \psi$ for all $f : W \to A$. Then for all $f : W \to A$,
$$(\lambda' \circ \psi')(f) = (\lambda')(f \circ \psi) = f \circ \psi \circ \lambda = f \circ 1_W = f$$
so $\lambda'$ is a retraction and therefore epic which completes the proof.

I'm wondering why this slight generalization wasn't presented instead. The proof actually becomes easier as it's clear what to ignore, the original follows as an obvious corollary and this seems to be a theorem that may actually be useful in other contexts where the generality may be needed. I could have understood the concreteness if it was immediately used in the form presented, but as far as I can see it's pretty much standing alone, justified by its own merits not by the application in a larger theorem. The section is not even specifically about finite modules, but about free modules over commutative rings and their dual module. So as I see it my form is the correct level of generality for this section, but I suspect Lang had a reason for presenting it in this form so I'm hoping someone can enlighten me as to why Lang's version is more beautiful, simple, generalizable, or whatever the reason.

 

[fresh_42]

This is hard to answer as we cannot know what Lang had in mind. I cannot see a flaw in your generalization on a quick glimpse, which makes me wonder, too. It would have been more Bourbakish to choose the more general version, than the other one. Hence I'd rather seek for the loophole in your argumentation.

Lang's finte, free modules guarantee all splits, which means to gather all necessary conditions in one.

I have a problem with $\psi$. If $U$ is projective, then I get a homomorphism $U \longrightarrow V$ which splits $\varphi$, but no $\psi$. In addition I found the following theorem in my book:

$\operatorname{Hom}_A(-,I)$ is exact if and only if $I$ is injective.

Here we have $I=A$, such that $A$ is injective if each embedding $W \longrightarrow V$ becomes an epimorphism $\operatorname{Hom}_A(V,A) \twoheadrightarrow \operatorname{Hom}_A(W,A)$. And this is exactly what needs to be shown in your argumentation. But as I still don't see where you take your $\psi$ from, I assume that this is the point where the finite, free modules come into play!