Michael Shulman
regular 2-category


A 2-category KK 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.


  • 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

  • Every morphism ff factors as fmef\cong m e where mm is ff and ee is eso.

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


(Street’s Lemma) Let KK be a finitely complete 2-category where esos are stable under pullback, let e:ABe:A\to B be eso, and let n:BCn:B\to C be a map. 1. If the induced morphism ker(e)ker(ne)ker(e) \to ker(n e) is ff, then nn is faithful. 1. If ker(e)ker(ne)ker(e) \to ker(n e) is an equivalence, then nn is ff.


First note that ker(e)ker(ne)ker(e)\to ker(n e) being ff means that if a 1,a 2:YAa_1,a_2: Y \rightrightarrows A and δ 1,δ 2:ea 1ea 2\delta_1,\delta_2 : e a_1 \;\rightrightarrows\; e a_2 are such that nδ 1=nδ 2n \delta_1 = n \delta_2, then δ 1=δ 2\delta_1=\delta_2. Likewise, ker(e)ker(ne)ker(e)\to ker(n e) being an equivalence means that given any α:nea 1nea 2\alpha: n e a_1 \to n e a_2, there exists a unique δ:ea 1ea 2\delta: e a_1 \to e a_2 such that nδ=αn \delta = \alpha.

We first show that nn is faithful under the first hypothesis. Suppose we have b 1,b 2:XBb_1,b_2:X \rightrightarrows B and β 1,β 2:b 1b 2\beta_1,\beta_2:b_1\to b_2 with nβ 1=nβ 2n \beta_1 = n \beta_2. Take the pullback

Y r X (a 1,a 2) (b 1,b 2) A×A e×e B×B\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

β 1r,β 2r:b 1rb 2r\beta_1 r, \beta_2 r: b_1 r \;\rightrightarrows\; b_2 r

such that the composites

nea 1nb 1rnβ 1r=nβ 2rnb 2rnea 2n 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, nβ 1r=nβ 2rn \beta_1 r = n \beta_2 r implies β 1r=β 2r\beta_1 r = \beta_2 r. But rr is eso, since it is a pullback of the eso e×ee\times e, so this implies β 1=β 2\beta_1=\beta_2. Thus, nn is faithful.

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

(ne/ne) g n/n C 2 A×A e×e B×B n×n C×C\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 gg, being a pullback of e×ee\times e, is eso. We also have a commutative square

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

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


A 2-category is regular if and only if 1. it has finite limits, 1. esos are stable under pullback, 1. every morphism ff factors as fmef\cong m e where mm is ff and ee is eso, and 1. every eso is the quotient of its kernel.


First suppose KK is regular; we must show the last two conditions. Let f:ABf:A\to B be any morphism. By assumption, the kernel ker(f)ker(f) can be completed to an exact 2-fork ker(f)AeCker(f) \rightrightarrows A \overset{e}{\to} C. Since ee is the quotient of the 2-congruence ker(f)ker(f), it is eso, and since ff comes with an action by ker(f)ker(f), we have an induced map m:CBm:C\to B with fmef\cong m e. But since the 2-fork is exact, we also have ker(f)ker(e)ker(f)\simeq ker(e), so by Street’s Lemma, mm is ff.

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

Now suppose KK satisfies the conditions in the lemma. Let f:ABf:A\to B be any morphism; we must show that ker(f)ker(f) can be completed to an exact 2-fork. Factor f=mef = m e where mm is ff and ee is eso. Since mm is ff, we have ker(f)ker(e)ker(f)\simeq ker(e). But every eso is the quotient of its kernel, so the fork ker(f)AeCker(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.


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

If gmeg \cong m e where mm is ff and ee is eso, we call mm the image of gg. Taking images defines a left adjoint f:Sub(X)Sub(Y)\exists_f:Sub(X)\to Sub(Y) to f *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 KK is regular, so are:

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


If KK is regular, so are the fibrational slices Opf(X)Opf(X) and Fib(X)Fib(X).

Last revised on December 20, 2009 at 06:52:38. See the history of this page for a list of all contributions to it.