# Definition

A 2-category $K$ is called regular if

1. It is finitely complete,
2. esos are stable under pullback, and
3. Every 2-congruence which is a kernel can be completed to an exact 2-fork.

In particular, the last condition implies that every 2-congruence which is a kernel has a quotient.

# Examples

• Cat is regular.

• A 1-category is regular as a 2-category iff it is regular as a 1-category, since the esos in a 1-category are precisely the strong epics.

• Every finitely complete (0,1)-category (that is, every meet-semilattice) is regular.

# Factorizations

In StreetCBS the last condition is replaced by

• Every morphism $f$ factors as $f\cong me$ where $m$ is ff and $e$ is eso.

We now show that this follows from our definition. First we need:

###### Lemma

(Street’s Lemma) Let $K$ be a finitely complete 2-category where esos are stable under pullback, let $e:A\to B$ be eso, and let $n:B\to C$ be a map.

1. If the induced morphism $\mathrm{ker}\left(e\right)\to \mathrm{ker}\left(ne\right)$ is ff, then $n$ is faithful.
2. If $\mathrm{ker}\left(e\right)\to \mathrm{ker}\left(ne\right)$ is an equivalence, then $n$ is ff.
###### Proof

First note that $\mathrm{ker}\left(e\right)\to \mathrm{ker}\left(ne\right)$ being ff means that if ${a}_{1},{a}_{2}:Y⇉A$ and ${\delta }_{1},{\delta }_{2}:e{a}_{1}\phantom{\rule{thickmathspace}{0ex}}⇉\phantom{\rule{thickmathspace}{0ex}}e{a}_{2}$ are such that $n{\delta }_{1}=n{\delta }_{2}$, then ${\delta }_{1}={\delta }_{2}$. Likewise, $\mathrm{ker}\left(e\right)\to \mathrm{ker}\left(ne\right)$ being an equivalence means that given any $\alpha :ne{a}_{1}\to ne{a}_{2}$, there exists a unique $\delta :e{a}_{1}\to e{a}_{2}$ such that $n\delta =\alpha$.

We first show that $n$ is faithful under the first hypothesis. Suppose we have ${b}_{1},{b}_{2}:X⇉B$ and ${\beta }_{1},{\beta }_{2}:{b}_{1}\to {b}_{2}$ with $n{\beta }_{1}=n{\beta }_{2}$. Take the pullback

$\begin{array}{cccc}& Y& \stackrel{r}{\to }& X\\ \left({a}_{1},{a}_{2}\right)& ↓& & ↓& \left({b}_{1},{b}_{2}\right)\\ & A×A& \stackrel{e×e}{\to }& B×B\end{array}$\array{&Y & \overset{r}{\to} & X \\ (a_1,a_2) & \downarrow && \downarrow & (b_1,b_2)\\ & A\times A & \overset{e\times e}{\to} & B\times B}

Then we have two 2-cells

${\beta }_{1}r,{\beta }_{2}r:{b}_{1}r\phantom{\rule{thickmathspace}{0ex}}⇉\phantom{\rule{thickmathspace}{0ex}}{b}_{2}r$\beta_1 r, \beta_2 r: b_1 r \;\rightrightarrows\; b_2 r

such that the composites

$ne{a}_{1}\cong n{b}_{1}r\stackrel{n{\beta }_{1}r=n{\beta }_{2}r}{\to }n{b}_{2}r\cong ne{a}_{2}$n e a_1 \cong n b_1 r \overset{n \beta_1 r = n \beta_2 r}{\to} n b_2 r \cong n e a_2

are equal. By the hypothesis, $n{\beta }_{1}r=n{\beta }_{2}r$ implies ${\beta }_{1}r={\beta }_{2}r$. But $r$ is eso, since it is a pullback of the eso $e×e$, so this implies ${\beta }_{1}={\beta }_{2}$. Thus, $n$ is faithful.

Now suppose the (stronger) second hypothesis, and form the pair of pullbacks:

$\begin{array}{ccccc}\left(ne/ne\right)& \stackrel{g}{\to }& n/n& \to & {C}^{2}\\ ↓& & ↓& & ↓\\ A×A& \stackrel{e×e}{\to }& B×B& \stackrel{n×n}{\to }& C×C\end{array}$\array{(n e / n e) & \overset{g}{\to} & n / n & \to & C^{\mathbf{2}}\\ \downarrow && \downarrow && \downarrow \\ A\times A & \overset{e\times e}{\to} & B\times B & \overset{n\times n}{\to} & C\times C}

Then $g$, being a pullback of $e×e$, is eso. We also have a commutative square

$\begin{array}{ccc}\left(e/e\right)& \to & \left(ne/ne\right)\\ ↓& & ↓g\\ {B}^{2}& \to & \left(n/n\right).\end{array}$\array{(e/e) & \to & (n e / n e)\\ \downarrow && \downarrow g \\ B^{\mathbf{2}} & \to & (n/n).}

By assumption, $\left(e/e\right)\to \left(ne/ne\right)$ is an equivalence. Since we have shown that $n$ is faithful, the bottom map ${B}^{2}\to \left(n/n\right)$ is ff, so since the eso $g$ factors through it, it must be an equivalence as well. But this says precisely that $n$ is ff.

###### Theorem

A 2-category is regular if and only if

1. it has finite limits,
2. esos are stable under pullback,
3. every morphism $f$ factors as $f\cong me$ where $m$ is ff and $e$ is eso, and
4. every eso is the quotient of its kernel.
###### Proof

First suppose $K$ is regular; we must show the last two conditions. Let $f:A\to B$ be any morphism. By assumption, the kernel $\mathrm{ker}\left(f\right)$ can be completed to an exact 2-fork $\mathrm{ker}\left(f\right)⇉A\stackrel{e}{\to }C$. Since $e$ is the quotient of the 2-congruence $\mathrm{ker}\left(f\right)$, it is eso, and since $f$ comes with an action by $\mathrm{ker}\left(f\right)$, we have an induced map $m:C\to B$ with $f\cong me$. But since the 2-fork is exact, we also have $\mathrm{ker}\left(f\right)\simeq \mathrm{ker}\left(e\right)$, so by Street’s Lemma, $m$ is ff.

Now suppose that in the previous paragraph $f$ were already eso. Then since it factors through the ff $m$, $m$ must be an equivalence; thus $f$ is equivalent to $e$ and hence is a quotient of its kernel.

Now suppose $K$ satisfies the conditions in the lemma. Let $f:A\to B$ be any morphism; we must show that $\mathrm{ker}\left(f\right)$ can be completed to an exact 2-fork. Factor $f=me$ where $m$ is ff and $e$ is eso. Since $m$ is ff, we have $\mathrm{ker}\left(f\right)\simeq \mathrm{ker}\left(e\right)$. But every eso is the quotient of its kernel, so the fork $\mathrm{ker}\left(f\right)⇉A\stackrel{e}{\to }C$ is exact.

In StreetCBS it is claimed that the final condition in Theorem 1 follows from the other three, but there is a flaw in the proof.

# Subobjects

In a regular 2-category $K$, we call a ff $m:A\to X$ with codomain $X$ a subobject of $X$. We write $\mathrm{Sub}\left(X\right)$ for the preorder of subobjects of $X$, as a full sub-2-category of the slice 2-category $K/X$. Since $K$ is finitely complete and pullbacks preserve ffs, we have pullback functors ${f}^{*}:\mathrm{Sub}\left(Y\right)\to \mathrm{Sub}\left(X\right)$ for any $f:X\to Y$.

If $g\cong me$ where $m$ is ff and $e$ is eso, we call $m$ the image of $g$. Taking images defines a left adjoint ${\exists }_{f}:\mathrm{Sub}\left(X\right)\to \mathrm{Sub}\left(Y\right)$ to ${f}^{*}$ in any regular 2-category, and the Beck-Chevalley condition is satisfied for any pullback square, because esos are stable under pullback.

# Preservation

It is easy to check that if $K$ is regular, so are:

• its 2-cell dual ${K}^{\mathrm{co}}$ (by the remarks about opposite 2-congruences).
• the (2,1)-category $\mathrm{gpd}\left(K\right)$ of groupoidal objects in $K$.
• the (1,2)-category $\mathrm{pos}\left(K\right)$ of posetal objects in $K$.
• the 1-category $\mathrm{disc}\left(K\right)$ of discrete objects in $K$.
• and more generally the $n$-category ${\mathrm{trunc}}_{n}\left(K\right)$ of $n$-truncated objects in $K$.

The slice 2-category $K/X$ does not, in general, inherit regularity, but we have:

###### Theorem

If $K$ is regular, so are the fibrational slices $\mathrm{Opf}\left(X\right)$ and $\mathrm{Fib}\left(X\right)$.

Revised on December 20, 2009 06:52:38 by Mike Shulman (173.8.161.189)