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I'm trying to read through Atiyah and MacDonald's Introduction to Commutative Algebra.

Proposition 2.9 says a sequence of $A$-modules and homomorphisms $$ M'\stackrel{u}{\to} M\stackrel{v}{\to} M''\to 0 $$ is exact iff for all $A$-modules $N$, the sequence

$$ 0\to\operatorname{Hom}(M'',N)\stackrel{\bar{v}}{\to}\operatorname{Hom}(M,N)\stackrel{\bar{u}}{\to}\operatorname{Hom}(M',N) $$ is exact. Here $\bar{u}(f)=f\circ u$ for $f\colon M\to N$, and similarly for $\bar{v}$.

I can prove most of this, but the one thing I'm stuck showing $\ker{\bar{u}}\subseteq\text{Im}(\bar{v})$. I take $f\colon M\to N$, and suppose $\bar{u}(f)=f\circ u$ is the $0$ homomorphism from $M'$ to $N$. I don't know how to show $f$ is in the image of $\bar{v}$. Why is this? Thanks.

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up vote 5 down vote accepted

$f$ vanishes on $\operatorname{im} u = \ker v$, so there exists a (unique) $f_*\colon M/\ker v \to N$ sending $x + \ker v$ to $f(x)$. Now, $v$ induces an isomorphism $v_*\colon M/\ker v \to M''$ sending $x + \ker v$ to $v(x)$. I think $f_* \circ v_*^{-1}$ is an element of $\operatorname{Hom}(M'', N)$ that does the job.

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This is much clearer if you draw out the diagrams. I might do that later. – Dylan Moreland Jan 23 '12 at 5:17
Thanks Dylan. I see that the maps are well-defined, and so for any $x\in M$, $$\bar{v}(f_*\circ v_*^{-1})(x)=(f_*v_*^{-1})(v(x))=f_*(x+\ker v)=f(x).$$ Perfect, thanks! :) – Vika Jan 23 '12 at 5:32
+1 Nice answer. – user38268 Jan 23 '12 at 15:29

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