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:CD is said to be faithful if K(X,C)K(X,D) is faithful for any X.


A morphism e:AB in a 2-category K is eso+full if for any faithful m:CD, 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)}

A morphism f is said to be full if we have fme where m is ff and e is eso+full.

Luckily, the terminology is consistent:


In any 2-category,

  1. f is eso+full if and only if it is both eso and full.
  2. f is ff if and only if it is both faithful and full.

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 f is eso and fme where e is eso+full and m is ff, then m is an equivalence, and hence f, like e, 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 f is faithful and we have fme where e is eso+full and m is ff, then e is also faithful, and hence an equivalence; thus f, like m, is ff.


Any eso+full morphism is co-ff.


Let f:AB be eso+full; we want to show that K(B,C)K(A,C) is ff. It is faithful since f is eso, so suppose that g,h:BC and that α:gfhf; we want to show α=βf for some β:gh. Let p:EB with γ:gphp be the inserter of g,h. Then we have a k:AE such that pkf and (modulo this isomorphism) γk=α. But since p, being an inserter, is faithful, and f is eso+full, we have q:BE with kqf and therefore α=γk=γqf; thus β=γq is our desired 2-cell.



If A 1A is a 2-congruence such that A 2A 1 is eso, then its quotient f:AB (if it has one) is eso+full.


Suppose that m:CD is faithful and that mghf; we want to show there is a k:BC with gkf and hmk (the rest follows by standard arguments). Since mg comes with an action by A 1A, it suffices to lift this action to g itself, and since m 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 1 is eso (by assumption) and C 2(m/m) is ff (since m is faithful).


In a regular 2-category, if f:AB is a map such that A 2(f/f) is eso, then f is full.


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

For comparison, recall that

  • A 2(f/f) is always faithful,
  • A 2(f/f) is full (equivalently, ff) iff f is faithful, and
  • A 2(f/f) is an equivalence iff f 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.


Let K be an n-exact n-category, where n is 2, (2,1), (1,2), or 1. Then:

  1. Every morphism f factors as fme where e is eso+full and m is faithful. Thus, (eso+full, faithful) is a factorization system on K.
  2. A morphism f:AB is full iff A 2(f/f) is eso.
  3. Therefore, f is eso+full iff it is the quotient of a 2-congruence A 1A such that A 2A 1 is eso.

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


For the first statement, let f:AB be any morphism, and factor A 2(f/f) as an eso followed by a ff to get A 2A 1(f/f). Then A 1 inherits the structure of a 2-congruence, which is an n-congruence since (f/f) is. Since K is n-exact, this n-congruence has a quotient e:AQ, which is eso+full by Lemma 3, and clearly f factors through e as fme. Finally, since (e/e)=A 1(f/f) is ff by definition, Street's Lemma implies that m is faithful.

The “if” direction of the second statement is Lemma 4. Conversely, if fme where m is ff and e is eso+full, then e must be obtained, up to equivalence, as the quotient of the n-congruence A 1A constructed above, for which A 2A 1 is eso. But since m is ff, we have (f/f)(e/e)=A 1, so A 2(f/f) is eso as desired.

One direction of the third statement is Lemma 3. For the other, if f is eso+full, then by the second statement A 2(f/f) is eso, and because f is eso it is the quotient of its kernel, namely (f/f)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 FinCat, which is not exact.


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


If p:AB is eso+full, then it is the quotient of its kernel, which in a (2,1)-category is A× BAA. And since p is eso+full, by Theorem 1, AA× BA 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× CD is also the pullback of A× BA along f. Thus, by properties of pullback squares, DD× CD is a pullback of AA× BA, and hence also eso. But q, being eso (as a pullback of p), is also the quotient of its kernel, so by Lemma 3 q is eso+full. The second statement follows since ffs are certainly stable under pullback.


Let K be an n-exact n-category. Then:

  1. Full morphisms are stable under composition.
  2. A morphism is full iff it is (isomorphic to) a composite of eso+full and ff morphisms.
  3. If gf is full and f is eso, then g is full.

Given f:AB and g:BC, 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 f and g are full, so that A 2(f/f) and B 2(g/g) are eso, then (f/f)(gf/gf) is also eso, hence so is A 2(gf/gf); thus gf is full. This proves the first statement, from which the second follows. For the third, if gf is full and f is eso, then A 2(gf/gf) is eso, and so is (gf/gf)(g/g) since it is a pullback of f×f. Thus the composite A 2(g/g) is eso, and since it factors through B 2, it follows that B 2(g/g) is also eso; hence g is full.

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