---
tag: SCG4061G4W
kind: definition
name: Accessible category
msc: [18C35]
defines: FC8C0V4TFZ
uses: [A2DE66937D, 975PMY4FG8, BJS1CGTKQ5, 13ZV2F8JAR, VJXS8AK67E]
sources:
  - url: https://en.wikipedia.org/wiki/Accessible_category
---

## Definition

Let $\lambda$ be a [[regular cardinal|13ZV2F8JAR]].
A small category $\mathcal{I}$ is *$\lambda$-filtered* if every subcategory of $\mathcal{I}$ with fewer than $\lambda$ morphisms has a cocone in $\mathcal{I}$.
Let $\mathcal{C}$ be a locally small [[category|A2DE66937D]].
An object $A$ of $\mathcal{C}$ is *$\lambda$-presentable* if the [[hom functor|VJXS8AK67E]] $\mathcal{C}(A, -)$ preserves [[colimits|BJS1CGTKQ5]] of $\lambda$-filtered diagrams.
The category $\mathcal{C}$ is *$\lambda$-accessible* if both conditions hold.

1. Every diagram $\mathcal{I} \to \mathcal{C}$ with $\mathcal{I}$ small and $\lambda$-filtered has a colimit in $\mathcal{C}$.
2. There is a set $S$ of $\lambda$-presentable objects such that every object of $\mathcal{C}$ is a colimit of a $\lambda$-filtered diagram of objects of $S$.

The category $\mathcal{C}$ is *accessible* if it is $\lambda$-accessible for some regular cardinal $\lambda$.

## Remarks

For $\lambda = \aleph_0$, the $\lambda$-filtered categories are the [[filtered categories|975PMY4FG8]], and one says finitely presentable and finitely accessible.

Example: the category of sets is finitely accessible.
It has all colimits, so clause 1 holds.
A finite set $A$ is finitely presentable, since a map from $A$ into a filtered colimit factors through some stage, essentially uniquely.
Every set is the directed union, hence the filtered colimit, of its finite subsets, and the finite sets $\set{1, \dots, n}$ form a set $S$ up to isomorphism.

Non-example: let $\mathcal{M}$ be the category with one object $*$ and morphisms $1$ and $e$, where $e \circ e = e$.
The identity diagram of $\mathcal{M}$ is $\lambda$-filtered for every $\lambda$, with the cocone $e$.
It has no colimit: the cocone $e$ is the only one, and both $1$ and $e$ factor it through itself.
So clause 1 fails for every $\lambda$, and $\mathcal{M}$ is not accessible.