Lemma 12.5.16. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ x \ar[r]^ f \ar[d]^\alpha & y \ar[r]^ g \ar[d]^\beta & z \ar[d]^\gamma \\ u \ar[r]^ k & v \ar[r]^ l & w } \]

be a commutative diagram.

If the first row is exact and $k$ is a monomorphism, then the induced sequence $\mathop{\mathrm{Ker}}(\alpha ) \to \mathop{\mathrm{Ker}}(\beta ) \to \mathop{\mathrm{Ker}}(\gamma )$ is exact.

If the second row is exact and $g$ is an epimorphism, then the induced sequence $\mathop{\mathrm{Coker}}(\alpha ) \to \mathop{\mathrm{Coker}}(\beta ) \to \mathop{\mathrm{Coker}}(\gamma )$ is exact.

**Proof.**
Suppose the first row is exact and $k$ is a monomorphism. Let $a:\mathop{\mathrm{Ker}}(\alpha )\to \mathop{\mathrm{Ker}}(\beta )$ and $b:\mathop{\mathrm{Ker}}(\beta )\to \mathop{\mathrm{Ker}}(\gamma )$ be the induced morphisms. Let $h:\mathop{\mathrm{Ker}}(\alpha )\to x$, $i:\mathop{\mathrm{Ker}}(\beta )\to y$ and $j:\mathop{\mathrm{Ker}}(\gamma )\to z$ be the canonical injections. As $j$ is a monomorphism we have $b\circ a=0$. Let $c:s\to \mathop{\mathrm{Ker}}(\beta )$ with $b\circ c=0$. Then, $g\circ i\circ c=j\circ b\circ c=0$. By Lemma 12.5.15 there are an object $t$, an epimorphism $d:t\to s$ and a morphism $e:t\to x$ with $i\circ c\circ d=f\circ e$. Then, $k\circ \alpha \circ e=\beta \circ f\circ e=\beta \circ i\circ c\circ d=0$. As $k$ is a monomorphism we get $\alpha \circ e=0$. So, there exists $m:t\to \mathop{\mathrm{Ker}}(\alpha )$ with $h\circ m=e$. It follows $i\circ a\circ m=f\circ h\circ m=f\circ e=i\circ c\circ d$. As $i$ is a monomorphism we get $a\circ m=c\circ d$. Thus, Lemma 12.5.15 implies (1), and then (2) follows by duality.
$\square$

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