Atiyah and Macdonald, Proposition 2.9

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).

Solution 1:

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.

Solution 2:

Let $P=N/v(M)$, then the image of the natural quotient map $q :N \rightarrow N/v(M)$ is zero, so $q$ is in fact zero, and $N=v(M)$.

Solution 3:

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.