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The proposition 2.9 of Atiyah and Macdonald syas that a sequence of $A$-modules

$$M'\xrightarrow u M \xrightarrow v M'' \rightarrow 0$$

is exact iff the dual sequence

$$0\rightarrow Hom (M'',N)\xrightarrow{\bar{v}} Hom(M,N)\xrightarrow{\bar{u}} Hom (M',N)$$

is exact for all $A$-modules $N$.

But I have trouble understanding the proof: Suppose the dual sequence is exact for all $A$-modules $N$, then since $\bar{v}$ is injective for all $N$, it follows that $v$ is surjective.

But I don't understand why it is true, please helps.

Michelle
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1 Answers1

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$v$ being surjective is a statement about the cokernel $C=\operatorname{Coker}(v)$ of $v$. So you should use the asssumption for $N=C$.

You get that $\operatorname{Hom}(M'',C) \to \operatorname{Hom}(M,C)$ is injective.

Now you should consider the projection $M'' \to C \in \operatorname{Hom}(M'',C)$. It gets mapped to zero by the above injective homomorphism, so the projection must be zero, which of course implies $C=0$.

MooS
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