A 2-category $K$ is called regular if
In particular, the last condition implies that every 2-congruence which is a kernel has a quotient.
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.
In StreetCBS the last condition is replaced by
We now show that this follows from our definition. First we need:
(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.
First note that $\mathrm{ker}(e)\to \mathrm{ker}(ne)$ being ff means that if ${a}_{1},{a}_{2}:Y\rightrightarrows A$ and ${\delta}_{1},{\delta}_{2}:e{a}_{1}\phantom{\rule{thickmathspace}{0ex}}\rightrightarrows \phantom{\rule{thickmathspace}{0ex}}e{a}_{2}$ are such that $n{\delta}_{1}=n{\delta}_{2}$, then ${\delta}_{1}={\delta}_{2}$. Likewise, $\mathrm{ker}(e)\to \mathrm{ker}(ne)$ 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 ${b}_{1},{b}_{2}:X\rightrightarrows B$ and ${\beta}_{1},{\beta}_{2}:{b}_{1}\to {b}_{2}$ with $n{\beta}_{1}=n{\beta}_{2}$. Take the pullback
Then we have two 2-cells
such that the composites
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:
Then $g$, being a pullback of $e\times e$, is eso. We also have a commutative square
By assumption, $(e/e)\to (ne/ne)$ is an equivalence. Since we have shown that $n$ is faithful, the bottom map ${B}^{2}\to (n/n)$ 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.
A 2-category is regular if and only if
First suppose $K$ is regular; we must show the last two conditions. Let $f:A\to B$ be any morphism. By assumption, the kernel $\mathrm{ker}(f)$ can be completed to an exact 2-fork $\mathrm{ker}(f)\rightrightarrows A\stackrel{e}{\to}C$. Since $e$ is the quotient of the 2-congruence $\mathrm{ker}(f)$, it is eso, and since $f$ comes with an action by $\mathrm{ker}(f)$, we have an induced map $m:C\to B$ with $f\cong me$. But since the 2-fork is exact, we also have $\mathrm{ker}(f)\simeq \mathrm{ker}(e)$, 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 $\mathrm{ker}(f)$ 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 $\mathrm{ker}(f)\simeq \mathrm{ker}(e)$. But every eso is the quotient of its kernel, so the fork $\mathrm{ker}(f)\rightrightarrows A\stackrel{e}{\to}C$ 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.
In a regular 2-category $K$, we call a ff $m:A\to X$ with codomain $X$ a subobject of $X$. We write $\mathrm{Sub}(X)$ 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 ${f}^{*}:\mathrm{Sub}(Y)\to \mathrm{Sub}(X)$ 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 ${\exists}_{f}:\mathrm{Sub}(X)\to \mathrm{Sub}(Y)$ to ${f}^{*}$ in any regular 2-category, and the Beck-Chevalley condition is satisfied for any pullback square, because esos are stable under pullback.
It is easy to check that if $K$ is regular, so are:
The slice 2-category $K/X$ does not, in general, inherit regularity, but we have:
If $K$ is regular, so are the fibrational slices $\mathrm{Opf}(X)$ and $\mathrm{Fib}(X)$.