# Definition

A 2-category $K$ is called regular if 1. It is finitely complete, 1. esos are stable under pullback, and 1. 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 $f\cong m e$ 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 $ker(e) \to ker(n e)$ is ff, then $n$ is faithful. 1. If $ker(e) \to ker(n e)$ is an equivalence, then $n$ is ff.

###### Proof

First note that $ker(e)\to ker(n e)$ being ff means that if $a_1,a_2: Y \rightrightarrows A$ and $\delta_1,\delta_2 : e a_1 \;\rightrightarrows\; e a_2$ are such that $n \delta_1 = n \delta_2$, then $\delta_1=\delta_2$. Likewise, $ker(e)\to ker(n e)$ being an equivalence means that given any $\alpha: n e a_1 \to n e a_2$, 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 $b_1,b_2:X \rightrightarrows B$ and $\beta_1,\beta_2:b_1\to b_2$ with $n \beta_1 = n \beta_2$. Take the pullback

$\array{&Y & \overset{r}{\to} & X \\ (a_1,a_2) & \downarrow && \downarrow & (b_1,b_2)\\ & A\times A & \overset{e\times e}{\to} & B\times B}$

Then we have two 2-cells

$\beta_1 r, \beta_2 r: b_1 r \;\rightrightarrows\; b_2 r$

such that the composites

$n e a_1 \cong n b_1 r \overset{n \beta_1 r = n \beta_2 r}{\to} n b_2 r \cong n e a_2$

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 $e\times e$, so this implies $\beta_1=\beta_2$. Thus, $n$ is faithful.

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

$\array{(n e / n e) & \overset{g}{\to} & n / n & \to & C^{\mathbf{2}}\\ \downarrow && \downarrow && \downarrow \\ A\times A & \overset{e\times e}{\to} & B\times B & \overset{n\times n}{\to} & C\times C}$

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

$\array{(e/e) & \to & (n e / n e)\\ \downarrow && \downarrow g \\ B^{\mathbf{2}} & \to & (n/n).}$

By assumption, $(e/e) \to (n e / n e)$ is an equivalence. Since we have shown that $n$ is faithful, the bottom map $B^{\mathbf{2}} \to (n/n)$ 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, 1. esos are stable under pullback, 1. every morphism $f$ factors as $f\cong m e$ where $m$ is ff and $e$ is eso, and 1. 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 $ker(f)$ can be completed to an exact 2-fork $ker(f) \rightrightarrows A \overset{e}{\to} C$. Since $e$ is the quotient of the 2-congruence $ker(f)$, it is eso, and since $f$ comes with an action by $ker(f)$, we have an induced map $m:C\to B$ with $f\cong m e$. But since the 2-fork is exact, we also have $ker(f)\simeq ker(e)$, 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 $ker(f)$ can be completed to an exact 2-fork. Factor $f = m e$ where $m$ is ff and $e$ is eso. Since $m$ is ff, we have $ker(f)\simeq ker(e)$. But every eso is the quotient of its kernel, so the fork $ker(f) \rightrightarrows A \overset{e} \to C$ 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 $Sub(X)$ 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 $f^*:Sub(Y)\to Sub(X)$ for any $f:X\to Y$.

If $g \cong m e$ where $m$ is ff and $e$ is eso, we call $m$ the image of $g$. Taking images defines a left adjoint $\exists_f:Sub(X)\to Sub(Y)$ 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 $K^{co}$ (by the remarks about opposite 2-congruences).
• the (2,1)-category $gpd(K)$ of groupoidal objects in $K$.
• the (1,2)-category $pos(K)$ of posetal objects in $K$.
• the 1-category $disc(K)$ of discrete objects in $K$.
• and more generally the $n$-category $trunc_n(K)$ 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 $Opf(X)$ and $Fib(X)$.

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