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\begin{document}

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\section*{regular 2-category}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{2category_theory}{}\paragraph*{{2-Category theory}}\label{2category_theory}

[[!include 2-category theory - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{factorizations}{Factorizations}\dotfill \pageref*{factorizations} \linebreak
\noindent\hyperlink{subobjects}{Subobjects}\dotfill \pageref*{subobjects} \linebreak
\noindent\hyperlink{preservation}{Preservation}\dotfill \pageref*{preservation} \linebreak
\noindent\hyperlink{regular_completion}{Regular completion}\dotfill \pageref*{regular_completion} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The notion of \emph{regular 2-catory} is the analog in [[2-category theory]] of the notion of \emph{[[regular category]]} in [[category theory]].

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\begin{defn}
\label{}\hypertarget{}{}
A [[2-category]] $K$ is called \textbf{regular} if

\begin{enumerate}%
\item It is finitely complete (has all [[finite limit|finite]] [[2-limits]] );


\item [[eso morphism|esos]] are stable under [[2-pullback]];


\item Every [[2-congruence]] which is a kernel can be completed to an exact [[2-fork]].



\end{enumerate}
\end{defn}
\begin{remark}
\label{}\hypertarget{}{}
In particular, the last condition implies that every 2-congruence which is a kernel has a [[quotient]].

\end{remark}
\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{itemize}%
\item [[nLab:Cat|Cat]] is regular.


\item A [[1-category]] is regular as a 2-category iff it is [[regular category|regular]] as a 1-category, since the esos in a 1-category are precisely the [[strong epimorphism|strong epis]].


\item Every finitely complete] [[(0,1)-category]] (that is, every [[meet-semilattice]]) is regular.



\end{itemize}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\hypertarget{factorizations}{}\subsubsection*{{Factorizations}}\label{factorizations}

In [[StreetCBS]] the last condition is replaced by

\begin{itemize}%
\item Every morphism $f$ factors as $f\cong m e$ where $m$ is [[ff morphism|ff]] and $e$ is eso.

\end{itemize}
We now show that this follows from our definition. First we need:

\begin{lemma}
\label{}\hypertarget{}{}
\textbf{(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.

\end{lemma}
\begin{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

\begin{displaymath}
\itexarray{&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}
\end{displaymath}
Then we have two 2-cells

\begin{displaymath}
\beta_1 r, \beta_2 r: b_1 r \;\rightrightarrows\; b_2 r
\end{displaymath}
such that the composites

\begin{displaymath}
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
\end{displaymath}
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:

\begin{displaymath}
\itexarray{(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}
\end{displaymath}
Then $g$, being a pullback of $e\times e$, is eso. We also have a commutative square

\begin{displaymath}
\itexarray{(e/e) & \to & (n e / n e)\\ \downarrow && \downarrow g \\
  B^{\mathbf{2}} & \to & (n/n).}
\end{displaymath}
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.

\end{proof}
\begin{theorem}
\label{StreetDefn}\hypertarget{StreetDefn}{}
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.

\end{theorem}
\begin{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.

\end{proof}
In [[StreetCBS]] it is claimed that the final condition in Theorem \ref{StreetDefn} follows from the other three, but there is a flaw in the proof.

\hypertarget{subobjects}{}\subsubsection*{{Subobjects}}\label{subobjects}

In a regular 2-category $K$, we call a ff $m:A\to X$ with codomain $X$ a \textbf{subobject} of $X$. We write $Sub(X)$ for the [[nLab:preorder|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 \textbf{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 [[nLab:Beck-Chevalley condition|Beck-Chevalley condition]] is satisfied for any pullback square, because esos are stable under pullback.

\hypertarget{preservation}{}\subsubsection*{{Preservation}}\label{preservation}

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

\begin{itemize}%
\item its [[nLab:opposite 2-category|2-cell dual]] $K^{co}$ (by the remarks about opposite [[2-congruence]]s).
\item the (2,1)-category $gpd(K)$ of [[groupoidal object]]s in $K$.
\item the (1,2)-category $pos(K)$ of [[posetal object]]s in $K$.
\item the 1-category $disc(K)$ of [[discrete object]]s in $K$.
\item and more generally the $n$-category $trunc_n(K)$ of $n$-[[truncated object|truncated objects]] in $K$.

\end{itemize}
The [[slice 2-category]] $K/X$ does not, in general, inherit regularity, but we have:

\begin{theorem}
\label{}\hypertarget{}{}
If $K$ is regular, so are the [[fibrational slice]]s $Opf(X)$ and $Fib(X)$.

\end{theorem}
\hypertarget{regular_completion}{}\subsubsection*{{Regular completion}}\label{regular_completion}

See at \emph{[[2-congruence]]} the section \emph{\href{http://ncatlab.org/nlab/show/2-congruence#Regularity}{Regularity}}.

\hypertarget{references}{}\subsection*{{References}}\label{references}

The above definitions and observations are originally due to

\begin{itemize}%
\item [[Michael Shulman]], \emph{[[michaelshulman:regular 2-category]]}

\end{itemize}
[[!redirects regular 2-categories]]

[[!redirects regular (2,1)-category]] [[!redirects regular (2,1)-categories]]



\end{document}