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

Definition

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:

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

Definition

A morphism $f$ is said to be full if we have $f\cong me$ where $m$ is ff and $e$ is eso+full.

Luckily, the terminology is consistent:

Lemma

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.

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 $f$ is eso and $f\cong me$ 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 $f\cong me$ where $e$ is eso+full and $m$ is ff, then $e$ is also faithful, and hence an equivalence; thus $f$, like $m$, is ff.

Lemma

Any eso+full morphism is co-ff.

Proof

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 $\alpha :gf\to hf$; we want to show $\alpha =\beta f$ for some $\beta :g\to h$. Let $p:E\to B$ with $\gamma :gp\to hp$ be the inserter of $g,h$. Then we have a $k:A\to E$ such that $pk\cong f$ 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 $k\cong qf$ and therefore $\alpha =\gamma k=\gamma qf$; thus $\beta =\gamma q$ is our desired 2-cell.

Factorizations

Lemma

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

Proof

Suppose that $m:C\to D$ is faithful and that $mg\cong hf$; we want to show there is a $k:B\to C$ with $g\cong kf$ and $h\cong mk$ (the rest follows by standard arguments). Since $mg$ 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:

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

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

Theorem

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 $f\cong me$ where $e$ is eso+full and $m$ is faithful. Thus, (eso+full, faithful) is a factorization system on $K$.
2. A morphism $f:A\to B$ is full iff $A^2\to (f/f)$ is eso.
3. Therefore, $f$ is eso+full iff it is the quotient of a 2-congruence $A_1\rightrightarrows A$ such that $A^2\to A_1$ is eso.

For comparison to the second statement, recall that

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

Of course, the cases $n=1$ and $(\mathrm{1,2})$ 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.

Proof

For the first statement, let $f:A\to B$ be any morphism, and factor $A^2\to (f/f)$ as an eso followed by a ff to get $A^2\to A_1\to (f/f)$. 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 3, and clearly $f$ factors through $e$ as $f\cong me$. Finally, since $(e/e)=A_1\to (f/f)$ is ff by definition, Street's Lemma implies that $m$ is faithful.

For the second statement, if $A^2\to (f/f)$ is eso then the above construction produces the same factorization as the usual (eso,ff) factorization of $f$. Therefore, the first factor is eso+full and the second factor is ff, so $f$ is full. Conversely, if $f\cong me$ 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 $A^2\to A_1$ is eso. But since $m$ is ff, we have $(f/f)\simeq (e/e)=A_1$, so $A^2\to (f/f)$ 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 $A^2\to (f/f)$ is eso, and because $f$ is eso it is the quotient of its kernel, namely $(f/f)\rightrightarrows A$.

Corollary

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

Proof

If $p:A\to B$ 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 1, $A\to A{\times }_{B}A$ is eso. If

$$\begin{array}{ccc}D& \to & A\\ {}^q\downarrow & & {\downarrow }^p\\ C& \stackrel{f}{\to }& B\end{array}$$\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 3 $q$ is eso+full. The second statement follows since ffs are certainly stable under pullback.

Corollary

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.

Proof

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

$$\begin{array}{ccccc}(f/f)& \to & (gf/gf)& \to & A\times A\\ \downarrow & & \downarrow & & \downarrow \\ B^2& \to & (g/g)& \to & B\times B.\end{array}$$\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 $A^2\to (f/f)$ and $B^2\to (g/g)$ are eso, then $(f/f)\to (gf/gf)$ is also eso, hence so is $A^2\to (gf/gf)$; 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 $A^2\to (gf/gf)$ is eso, and so is $(gf/gf)\to (g/g)$ since it is a pullback of $f\times f$. Thus the composite $A^2\to (g/g)$ is eso, and since it factors through $B^2$, it follows that $B^2\to (g/g)$ is also eso; hence $g$ is full.