## Tag `0033`

## 4.23. Exact functors

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

- Suppose all finite limits exist in $\mathcal{A}$. We say $F$ is
left exactif it commutes with all finite limits.- Suppose all finite colimits exist in $\mathcal{A}$. We say $F$ is
right exactif it commutes with all finite colimits.- We say $F$ is
exactif 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:

- $F$ is left exact,
- $F$ commutes with finite products and equalizers, and
- $F$ 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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