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Tag 0033

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. $\square$

    The code snippet corresponding to this tag is a part of the file categories.tex and is located in lines 2994–3039 (see updates for more information).

    \section{Exact functors}
    \label{section-exact-functor}
    
    
    \begin{definition}
    \label{definition-exact}
    Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
    \begin{enumerate}
    \item Suppose all finite limits exist in $\mathcal{A}$.
    We say $F$ is {\it left exact} if it commutes
    with all finite limits.
    \item Suppose all finite colimits exist in $\mathcal{A}$.
    We say $F$ is {\it right exact} if it commutes
    with all finite colimits.
    \item We say $F$ is {\it exact} if it is both left and right
    exact.
    \end{enumerate}
    \end{definition}
    
    \begin{lemma}
    \label{lemma-characterize-left-exact}
    Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
    Suppose all finite limits exist in $\mathcal{A}$,
    see Lemma \ref{lemma-finite-limits-exist}.
    The following are equivalent:
    \begin{enumerate}
    \item $F$ is left exact,
    \item $F$ commutes with finite products and equalizers, and
    \item $F$ transforms a final object of $\mathcal{A}$
    into a final object of $\mathcal{B}$, and commutes with fibre products.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Lemma \ref{lemma-limits-products-equalizers} 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.
    \end{proof}

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