The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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.

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

  2. 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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