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*}

4.23 Exact functors

Definition 4.23.1. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.

  1. Suppose all finite limits exist in $\mathcal{A}$. We say $F$ is left exact if it commutes with all finite limits.

  2. Suppose all finite colimits exist in $\mathcal{A}$. We say $F$ is right exact if it commutes with all finite colimits.

  3. We say $F$ is exact if it is both left and right exact.

Lemma 4.23.2. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor. Suppose all finite limits exist in $\mathcal{A}$, see Lemma 4.18.4. The following are equivalent:

  1. $F$ is left exact,

  2. $F$ commutes with finite products and equalizers, and

  3. $F$ transforms a final object of $\mathcal{A}$ into a final object of $\mathcal{B}$, and commutes with fibre products.

Proof. Lemma 4.14.10 shows that (2) implies (1). Suppose (3) holds. The fibre product over the final object is the product. If $a, b : A \to B$ are morphisms of $\mathcal{A}$, then the equalizer of $a, b$ is

\[ (A \times _{a, B, b} A)\times _{(pr_1, pr_2), A \times A, \Delta } A. \]

Thus (3) implies (2). Finally (1) implies (3) because the empty limit is a final object, and fibre products are limits. $\square$


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