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.
$f$ 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.