Michael Shulman Cauchy factorization

Definitions

A morphism $m:A\to B$ in a 2-category $K$ is called ff and retract-closed or rff if $K(X,A)\to K(X,B)$ is fully faithful and closed under retracts for all $X$. Explicitly this means that (in addition to being ff) if $b:X\to B$ is a retract of $m a$ for some $a:X\to A$, then $b\cong m a'$ for some $a'\in A$.

A morphism $e:A\to B$ is called Cauchy surjective or cso if it is left orthogonal to all rffs; that is, if

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

is a pullback whenever $m:C\to D$ is rff.

Remarks

• Since rffs are stable under pullback, $e$ is cso if and only if whenever it factors through an rff $m:C\to B$, then $m$ is an equivalence.

• Every eso is cso, since every rff is ff. Conversely, in a (2,1)-category, every ff is rff, and thus every cso is eso.

• In $Cat$, the Cauchy surjective functors are those for which every object of $B$ is a retract of an object of $A$.

• Every equifier or inverter is rff. Therefore, every Cauchy surjective morphism is cofaithful and coconservative. In “Modulated bicategories” by Carboni, Johnson, Street, and Verity it is shown that conversely, every coconservative functor in $Cat$ is Cauchy surjective.

Factorization System

In $Cat$, (cso,rff) is a factorization system. We now show that the same is true in any regular 2-category.

Given $f:A\to B$, let $R_f$ be its retractor. This is a 2-categorical finite limit which comes with projections $p:R_f\to A$ and $q:R_f\to B$ and 2-cells $\rho:f p \to q$ and $\sigma:q \to f p$ such that $\rho \sigma = 1_q$, and which is universal with these properties.

Lemma

In any 2-category, if $q:R_f\to B$ is eso, then $f$ is Cauchy surjective.

Proof

Suppose that $q$ is eso and let $A\overset{g}{\to} C \overset{m}{\to} B$ be such that $f\cong m g$ and $m$ is rff; we want to show $m$ is an equivalence. But since $m$ is rff and $q$ is a retract of $f p \cong m g p$, it follows that $q$ factors through $m$; whence $m$ is an equivalence since $q$ is eso.

Now suppose that $K$ is regular, and for any $f:A\to B$, let $R_f \overset{j}{\to} C_f \overset{m}{\to} B$ be an (eso,ff) factorization of $q$. Since $f$ is a retract of $f 1_A$, there is a canonical $s:A\to R_f$ with $p s \cong 1_A$ and $f\cong q s \cong m j s$.

Lemma

In a regular 2-category, if $f:A\to B$ is ff and $C_f\subset A$ (hence $C_f\simeq A$), then $f$ is rff.

Proof

Suppose $f$ is ff and $C_f\simeq A$, and let $b:X\to B$ and $a:X\to A$ be such that $b$ is a retract of $a$. Then $b \cong q c$ and $a\cong p c$ for some $c:X\to R_f$. Then we have $j c:X\to C_f$ with $m j c \cong b$. But since $C_f\simeq A$ in $Sub(B)$, we have a $y:X\to A$ with $f y \cong b$ as well; hence $f$ is rff.

Theorem

In a regular 2-category $K$, 1. Every morphism $f$ factors as a Cauchy surjective morphism followed by an rff. 1. Therefore, (cso,rff) is a factorization system on $K$. 1. A morphism $f$ is cso if and only if $q:R_f\to B$ is eso. 1. Likewise, $f$ is rff if and only if it is ff and $C_f\simeq A$.

Proof

Given any $f:A\to B$, let $j$, $m$, and $s$ be as above, and define $e = j s:A\to C_f$. Then $m e \cong m j s\cong q s \cong f$. Is easy to check that $R_f \cong R_e$; thus $R_e\to C_f$ is eso, so by Lemma $e$ is Cauchy surjective.

Now let $R_m$ be the retractor of $m$, with $p':R_m\to C_f$, $q':R_m\to B$, and corresponding factorization $R_m \overset{j'}{\to} C_m \overset{m'}{\to} B$. Let

$$\array{P & \overset{u}{\to}& R_f\\ ^v \downarrow && \downarrow ^j\\ R_m & \underset{p'}{\to} & C_f}$$

be a pullback. Then $v$, like $j$, is eso, and thus so is $j' v:P \to C_m$. Now $m' j' v \cong q' v$ is a retract of $m p' v\cong m j u \cong q u$, and $q$ is a retract of $f p$, so $q u$ is a retract of $f p u$. Therefore, $m' j' v:P\to B$ is also a retract of $f p u$, so there is a morphism $w:P\to R_f$ with $m j w \cong q w \cong m' j' v$ (and $p w\cong p u$). Now in the square

$$\array{P & \overset{j w}{\to} & C_f\\ ^{j' v}\downarrow && \downarrow ^m\\ C_m & \underset{m'}{\to} & B,}$$

$m$ is ff and $j' v$ is eso, so we have $C_m\subset C_f$ in $Sub(B)$; thus by Lemma $m$ is rff.

Therefore, with $f\cong m e$ we have factored $f$ as a cso followed by an rff; this shows the first two statements. Moreover, $e$ satisfies the condition of Lemma , so if $f$ is cso and hence equivalent to $e$, so does it. Likewise, $m$ satisfies the condition of Lemma , so if $f$ is rff and hence equivalent to $m$, so does it.

Internal logic

From the characterization of csos and rffs in Theorem , we can conclude that csos and rffs have the expected descriptions in the internal logic.

Corollary

In a regular 2-category, a morphism $f:A\to B$ is Cauchy surjective if and only if the following sequent is valid in the internal logic:

$$y:B | \top \vdash (\exists x:A)(\exists \rho:hom_B(f(x),y))(\exists \sigma:hom_B(y,f(x)))(\rho \sigma = 1_y).$$

Proof

The context $y:B, x:A, \rho:hom_B(f(x),y), \sigma:hom_B(y,f(x)) | \rho \sigma = 1_y$ is easily seen to be precisely $R_f$.

Corollary

In a regular 2-category, a morphism $f:A\to B$ is rff if and only if it is ff and the following sequent is valid in the internal logic:

$$x:A,y:B,\rho:hom_B(f(x),y), \sigma:hom_B(y,f(x)) | \rho \sigma = 1_y \vdash (\exists x':A)(f(x')\cong y)$$

Proof

The validity of this sequent says that the pullback of $f:A\to B$ along $q:R_f \to B$ is eso. But since both $f$ and $q$ factor through the ff $m:C_f\to B$, this pullback is equivalent to the pullback of $e:A\to C_f$ along $j:R_f\to C_f$. Since $j$ is eso, this pullback being eso is equivalent to $e:A\to C_f$ being eso. And since $f$ is ff, so is $e$; thus this is equivalent to $e$ being an equivalence.