## 4.23 Exact functors

In this section we define 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 4.14.11 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

$(A \times _{a, B, b} A)\times _{(\text{pr}_1, \text{pr}_2), A \times A, \Delta } A.$

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

Lemma 4.23.3. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor. Suppose all finite colimits exist in $\mathcal{A}$, see Lemma 4.18.7. The following are equivalent:

1. $F$ is right exact,

2. $F$ commutes with finite coproducts and coequalizers, and

3. $F$ transforms an initial object of $\mathcal{A}$ into an initial object of $\mathcal{B}$, and commutes with pushouts.

Proof. Dual to Lemma 4.23.2. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).