TITLE: Atiyah and Macdonald, Proposition 2.9
QUESTION [7 upvotes]: 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).

REPLY [4 votes]: I believe the most simple and direct way to prove it is by contradiction/(contrapositive). 
Assume that $v$ is not surjective, then $\mathrm{coker}(v)=\frac{N}{v(M)}\neq 0$. Therefore the two maps $\pi : N \rightarrow \frac{N}{v(M)}$ and $\mathbf{0}:N \rightarrow \frac{N}{v(M)}$ are unequal (where $\pi : x \mapsto x+v(M)$ and $\mathbf{0}:x \mapsto 0$).
But $v^*(\pi)=v^*(\mathbf{0})$, because $(\pi\circ v)(x)=0= (\mathbf{0}\circ v)(x)$ for all $x$.  
Therefore the map $v^*:\mathrm{Hom}\Big(N,\frac{N}{v(M)}\Big) \to\mathrm{Hom}\Big(M,\frac{N}{v(M)}\Big) $ is not injective. QED.