The following simple claim is used without proof in Proposition 2.9 of Atiyah and MacDonald (p.23). Although I believe I can prove it with a fairly involved argument, the claim is treated by the authors as a triviality so I feel like there should be a simple argument. Is there such an argument?
Let $M, N$ be $A$-modules. Let $v\colon M \to N$ be a homomorphism. Suppose that for all $A$-modules $P$, the dual maps $v^*\colon\operatorname{Hom}(N,P) \to \operatorname{Hom}(M,P)$ are injective. Then $v$ is surjective.
(Note that the claim is trivial for free modules, but I am looking for a general proof).
Best Answer
Consider the exact sequence $M \to N \to C \to 0$, where $C$ is the cokernel of $v$. Then for any $A$-module $P$ we have an exact sequence $\DeclareMathOperator{\h}{Hom} 0 \to \h(C,P) \to \h(N,P) \to \h(M,P)$. By assumption the last map is injective, so we have $\h(C,P) = 0$. Since this holds for all $P$ it holds especially for $P = C$, so we conclude $C=0$ and hence that $v$ is surjective.