Lemma 12.5.12. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]

be a commutative diagram.

If the diagram is cartesian, then the morphism $\mathop{\mathrm{Ker}}(f)\to \mathop{\mathrm{Ker}}(k)$ induced by $g$ is an isomorphism.

If the diagram is cocartesian, then the morphism $\mathop{\mathrm{Coker}}(f)\to \mathop{\mathrm{Coker}}(k)$ induced by $h$ is an isomorphism.

**Proof.**
Suppose the diagram is cartesian. Let $e:\mathop{\mathrm{Ker}}(f)\to \mathop{\mathrm{Ker}}(k)$ be induced by $g$. Let $i:\mathop{\mathrm{Ker}}(f)\to w$ and $j:\mathop{\mathrm{Ker}}(k)\to x$ be the canonical injections. There exists $t:\mathop{\mathrm{Ker}}(k)\to w$ with $f\circ t=0$ and $g\circ t=j$. Hence, there exists $u:\mathop{\mathrm{Ker}}(k)\to \mathop{\mathrm{Ker}}(f)$ with $i\circ u=t$. It follows $g\circ i\circ u\circ e=g\circ t\circ e=j\circ e=g\circ i$ and $f\circ i\circ u\circ e=0=f\circ i$, hence $i\circ u\circ e=i$. Since $i$ is a monomorphism this implies $u\circ e=\text{id}_{\mathop{\mathrm{Ker}}(f)}$. Furthermore, we have $j\circ e\circ u=g\circ i\circ u=g\circ t=j$. Since $j$ is a monomorphism this implies $e\circ u=\text{id}_{\mathop{\mathrm{Ker}}(k)}$. This proves (1). Now, (2) follows by duality.
$\square$

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