Michael Shulman
regular 2-category

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 fme 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:AB be eso, and let n:BC be a map.

  1. If the induced morphism ker(e)ker(ne) is ff, then n is faithful.
  2. If ker(e)ker(ne) is an equivalence, then n is ff.
Proof

First note that ker(e)ker(ne) being ff means that if a 1,a 2:YA and δ 1,δ 2:ea 1ea 2 are such that nδ 1=nδ 2, then δ 1=δ 2. Likewise, ker(e)ker(ne) being an equivalence means that given any α:nea 1nea 2, there exists a unique δ:ea 1ea 2 such that nδ=α.

We first show that n is faithful under the first hypothesis. Suppose we have b 1,b 2:XB and β 1,β 2:b 1b 2 with nβ 1=nβ 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 2r implies β 1r=β 2r. But r is eso, since it is a pullback of the eso e×e, so this implies β 1=β 2. Thus, n 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 g, being a pullback of e×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) is an equivalence. Since we have shown that n is faithful, the bottom map B 2(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,
  2. esos are stable under pullback,
  3. every morphism f factors as fme 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:AB be any morphism. By assumption, the kernel ker(f) can be completed to an exact 2-fork ker(f)AeC. 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:CB with fme. But since the 2-fork is exact, we also have ker(f)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:AB be any morphism; we must show that 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 ker(f)ker(e). But every eso is the quotient of its kernel, so the fork ker(f)AeC 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:AX 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)Sub(X) for any f:XY.

If gme where m is ff and e is eso, we call m the image of g. Taking images defines a left adjoint f:Sub(X)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:

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)