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.

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

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

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

A morphism $f$ is said to be **full** if we have $f \cong m e$ 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. 1. $f$ 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 $f \cong m e$ 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 $f\cong m e$ 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 $f:A\to B$ be eso+full; we want to show that $K(B,C)\to K(A,C)$ is ff. It is faithful since $f$ is eso, so suppose that $g,h:B \;\rightrightarrows\; C$ and that $\alpha: g f \to h f$; we want to show $\alpha = \beta f$ for some $\beta: g\to h$. Let $p:E\to B$ with $\gamma: g p \to h p$ be the inserter of $g,h$. Then we have a $k:A\to E$ such that $p k\cong f$ and (modulo this isomorphism) $\gamma k = \alpha$. But since $p$, being an inserter, is faithful, and $f$ is eso+full, we have $q:B\to E$ with $k\cong q f$ and therefore $\alpha = \gamma k = \gamma q f$; thus $\beta=\gamma q$ is our desired 2-cell.

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

Suppose that $m:C\to D$ is faithful and that $m g\cong h f$; we want to show there is a $k:B\to C$ with $g\cong k f$ and $h\cong m k$ (the rest follows by standard arguments). Since $m g$ comes with an action by $A_1 \;\rightrightarrows\; A$, 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:

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

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

In a regular 2-category, if $f\colon A\to B$ is a map such that $A^{\mathbf{2}} \to (f/f)$ 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 , so the eso part of the (eso,ff) factorization of $f$ is eso+full. Hence $f$ is full, by definition.

For comparison, recall that

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

I do not know whether the converse of Lemma 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 $f\cong m e$ where $e$ is eso+full and $m$ is faithful. Thus, (eso+full, faithful) is a factorization system on $K$. 1. A morphism $f:A\to B$ is full iff $A^{\mathbf{2}} \to (f/f)$ is eso. 1. Therefore, $f$ is eso+full iff it is the quotient of a 2-congruence $A_1 \;\rightrightarrows\; A$ such that $A^{\mathbf{2}} \to A_1$ 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 $f:A\to B$ be any morphism, and factor $A^{\mathbf{2}} \to (f/f)$ as an eso followed by a ff to get $A^{\mathbf{2}} \to A_1 \to (f/f)$. 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 $e:A\to Q$, which is eso+full by Lemma , and clearly $f$ factors through $e$ as $f\cong m e$. Finally, since $(e/e) = A_1 \to (f/f)$ is ff by definition, Street's Lemma implies that $m$ is faithful.

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

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

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

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

is a pullback square, then the kernel $D\times_C D$ is also the pullback of $A\times_B A$ along $f$. Thus, by properties of pullback squares, $D \to D\times_C D$ is a pullback of $A\to A\times_B A$, and hence also eso. But $q$, being eso (as a pullback of $p$), is also the quotient of its kernel, so by Lemma $q$ 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. 1. A morphism is full iff it is (isomorphic to) a composite of eso+full and ff morphisms. 1. If $g f$ is full and $f$ is eso, then $g$ is full.

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

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

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

Last revised on July 21, 2010 at 13:08:28. See the history of this page for a list of all contributions to it.