Michael Shulman
full morphism

In Cat and Gpd, in addition to the (eso, ff) factorization system that is enshrined in the notion of regular 2-category, there is also an (eso+full, faithful) factorization system. In Gpd, the (eso+full, faithful) factorization is the same as the comprehensive factorization, but in Cat they are different.

Mathieu Dupont has pointed out that given sufficient exactness, (eso+full, faithful) factorizations can be constructed from (eso,ff) factorizations. Here we give the argument.

eso+full and full morphisms

Recall that a morphism m:CDm:C\to D is said to be faithful if K(X,C)K(X,D)K(X,C)\to K(X,D) is faithful for any XX.

Definition

A morphism e:ABe:A\to B in a 2-category KK is eso+full if for any faithful m:CDm:C\to D, the following square is a pullback:

K(B,C) K(B,D) K(A,C) K(A,D)\array{K(B,C) & \to & K(B,D)\\ \downarrow & & \downarrow\\ K(A,C) & \to & K(A,D)}
Definition

A morphism ff is said to be full if we have fmef \cong m e where mm is ff and ee is eso+full.

Luckily, the terminology is consistent:

Lemma

In any 2-category, 1. ff is eso+full if and only if it is both eso and full. 1. ff is ff if and only if it is both faithful and full.

Proof

Since ffs are faithful, any eso+full is eso, and any eso+full clearly factors as an eso+full (itself) followed by an ff (the identity). Conversely, if ff is eso and fmef \cong m e where ee is eso+full and mm is ff, then mm is an equivalence, and hence ff, like ee, is eso+full.

Now, ffs are clearly faithful, and any ff clearly factors as an eso+full (the identity) followed by an ff (itself). Conversely, if ff is faithful and we have fmef\cong m e where ee is eso+full and mm is ff, then ee is also faithful, and hence an equivalence; thus ff, like mm, is ff.

Lemma

Any eso+full morphism is co-ff.

Proof

Let f:ABf:A\to B be eso+full; we want to show that K(B,C)K(A,C)K(B,C)\to K(A,C) is ff. It is faithful since ff is eso, so suppose that g,h:BCg,h:B \;\rightrightarrows\; C and that α:gfhf\alpha: g f \to h f; we want to show α=βf\alpha = \beta f for some β:gh\beta: g\to h. Let p:EBp:E\to B with γ:gphp\gamma: g p \to h p be the inserter of g,hg,h. Then we have a k:AEk:A\to E such that pkfp k\cong f and (modulo this isomorphism) γk=α\gamma k = \alpha. But since pp, being an inserter, is faithful, and ff is eso+full, we have q:BEq:B\to E with kqfk\cong q f and therefore α=γk=γqf\alpha = \gamma k = \gamma q f; thus β=γq\beta=\gamma q is our desired 2-cell.

Factorizations

Lemma

If A 1AA_1 \;\rightrightarrows\; A is a 2-congruence such that A 2A 1A^{\mathbf{2}} \to A_1 is eso, then its quotient f:ABf:A\to B (if it has one) is eso+full.

Proof

Suppose that m:CDm:C\to D is faithful and that mghfm g\cong h f; we want to show there is a k:BCk:B\to C with gkfg\cong k f and hmkh\cong m k (the rest follows by standard arguments). Since mgm g comes with an action by A 1AA_1 \;\rightrightarrows\; A, it suffices to lift this action to gg itself, and since mm is faithful it suffices to lift the 2-cell defining the action. This follows by the existence of a diagonal lift in the following rectangle:

A 2 C 2 A 1 (f/f) (m/m)\array{A^{\mathbf{2}} && \to && C^{\mathbf{2}} \\ \downarrow && && \downarrow\\ A_1 & \to & (f/f) & \to & (m/m)}

which exists since A 2A 1A^{\mathbf{2}} \to A_1 is eso (by assumption) and C 2(m/m)C^{\mathbf{2}}\to (m/m) is ff (since mm is faithful).

Lemma

In a regular 2-category, if f:ABf\colon A\to B is a map such that A 2(f/f)A^{\mathbf{2}} \to (f/f) is eso, then ff is full.

Proof

The (eso,ff) factorization of ff is constructed by taking the quotient of ker(f)ker(f). But by assumption, this kernel has the property of Lemma 3, so the eso part of the (eso,ff) factorization of ff is eso+full. Hence ff is full, by definition.

For comparison, recall that

  • A 2(f/f)A^{\mathbf{2}} \to (f/f) is always faithful,
  • A 2(f/f)A^{\mathbf{2}} \to (f/f) is full (equivalently, ff) iff ff is faithful, and
  • A 2(f/f)A^{\mathbf{2}} \to (f/f) is an equivalence iff ff is ff.

I do not know whether the converse of Lemma 4 is true in general, but it is at least true in an exact 2-category.

Theorem

Let KK be an nn-exact nn-category, where nn is 2, (2,1), (1,2), or 1. Then: 1. Every morphism ff factors as fmef\cong m e where ee is eso+full and mm is faithful. Thus, (eso+full, faithful) is a factorization system on KK. 1. A morphism f:ABf:A\to B is full iff A 2(f/f)A^{\mathbf{2}} \to (f/f) is eso. 1. Therefore, ff is eso+full iff it is the quotient of a 2-congruence A 1AA_1 \;\rightrightarrows\; A such that A 2A 1A^{\mathbf{2}} \to A_1 is eso.

Of course, the cases n=1n=1 and (1,2)(1,2) are fairly trivial, since in those cases every morphism in an nn-category is faithful and thus the only eso+full morphisms are the equivalences. However, we include them for completeness.

Proof

For the first statement, let f:ABf:A\to B be any morphism, and factor A 2(f/f)A^{\mathbf{2}} \to (f/f) as an eso followed by a ff to get A 2A 1(f/f)A^{\mathbf{2}} \to A_1 \to (f/f). Then A 1A_1 inherits the structure of a 2-congruence, which is an nn-congruence since (f/f)(f/f) is. Since KK is nn-exact, this nn-congruence has a quotient e:AQe:A\to Q, which is eso+full by Lemma 3, and clearly ff factors through ee as fmef\cong m e. Finally, since (e/e)=A 1(f/f)(e/e) = A_1 \to (f/f) is ff by definition, Street's Lemma implies that mm is faithful.

The “if” direction of the second statement is Lemma 4. Conversely, if fmef\cong m e where mm is ff and ee is eso+full, then ee must be obtained, up to equivalence, as the quotient of the nn-congruence A 1AA_1 \;\rightrightarrows\; A constructed above, for which A 2A 1A^{\mathbf{2}}\to A_1 is eso. But since mm is ff, we have (f/f)(e/e)=A 1(f/f)\simeq (e/e)= A_1, so A 2(f/f)A^{\mathbf{2}} \to (f/f) is eso as desired.

One direction of the third statement is Lemma 3. For the other, if ff is eso+full, then by the second statement A 2(f/f)A^{\mathbf{2}} \to (f/f) is eso, and because ff is eso it is the quotient of its kernel, namely (f/f)A(f/f) \;\rightrightarrows\; A.

Inspecting the proof shows that we don’t need the full strength of exactness, either; all we need is that any sub-congruence of a kernel has a quotient. This is a property in between regularity and exactness. For instance, it is satisfied by FinCatFinCat, which is not exact.

Corollary

In a (2,1)-exact (2,1)-category, eso+full morphisms are stable under pullback, and therefore so are full ones.

Proof

If p:ABp:A\to B is eso+full, then it is the quotient of its kernel, which in a (2,1)-category is A× BAAA\times_B A \to A. And since pp is eso+full, by Theorem 1, AA× BAA\to A\times_B A is eso. If

D A q p C f B\array{D & \to & A \\ ^q\downarrow && \downarrow ^p\\ C & \overset{f}{\to} & B}

is a pullback square, then the kernel D× CDD\times_C D is also the pullback of A× BAA\times_B A along ff. Thus, by properties of pullback squares, DD× CDD \to D\times_C D is a pullback of AA× BAA\to A\times_B A, and hence also eso. But qq, being eso (as a pullback of pp), is also the quotient of its kernel, so by Lemma 3 qq is eso+full. The second statement follows since ffs are certainly stable under pullback.

Corollary

Let KK be an nn-exact nn-category. Then: 1. Full morphisms are stable under composition. 1. A morphism is full iff it is (isomorphic to) a composite of eso+full and ff morphisms. 1. If gfg f is full and ff is eso, then gg is full.

Proof

Given f:ABf:A\to B and g:BCg:B\to C, we have two pullback squares

(f/f) (gf/gf) A×A B 2 (g/g) B×B.\array{ (f/f) & \to & (g f / g f) & \to & A\times A\\ \downarrow &&\downarrow && \downarrow\\ B ^{\mathbf{2}} & \to & (g/g) & \to & B\times B.}

Thus, if ff and gg are full, so that A 2(f/f)A^{\mathbf{2}} \to (f/f) and B 2(g/g)B^{\mathbf{2}} \to (g/g) are eso, then (f/f)(gf/gf)(f/f) \to (g f / g f) is also eso, hence so is A 2(gf/gf)A ^{\mathbf{2}} \to (g f / g f); thus gfg f is full. This proves the first statement, from which the second follows. For the third, if gfg f is full and ff is eso, then A 2(gf/gf)A ^{\mathbf{2}} \to (g f / g f) is eso, and so is (gf/gf)(g/g)(g f / g f) \to (g/g) since it is a pullback of f×ff\times f. Thus the composite A 2(g/g)A ^{\mathbf{2}} \to (g/g) is eso, and since it factors through B 2B ^{\mathbf{2}}, it follows that B 2(g/g)B ^{\mathbf{2}} \to (g/g) is also eso; hence gg is full.

Revised on July 21, 2010 13:08:28 by Jamie Vicary? (86.26.19.94)