It is well-known that in a 1-pretopos,

- every monic is regular, and as a consequence
- a 1-pretopos is balanced (every monic epic is an isomorphism), and
- every epic is regular.

For a similar statement in an $n$-pretopos for $n\gt 1$, the natural first guess is to replace “monic” by “ff” and “regular epic” by “eso.” However, there are (at least) two reasonable replacements for “epic” and “regular monic:”

- We could replace “epic” by “cofaithful” and “regular monic” by “equifier,” or
- we could replace “epic” by “coconservative” (aka “liberal”) and “regular monic” by “inverter.”

Since esos in Cat are cofaithful and liberal but not co-ff, it wouldn’t work to replace “epic” by “co-ff.” Unfortunately, both ideas fail in Cat, where equifiers and inverters are not just ff but are closed under retracts. Likewise, the map from a category to its Cauchy completion is full, faithful, cofaithful, and liberal, but generally not an equivalence.

There are two ways to deal with this problem. One is to restrict to smaller values of $n$, for which there are no nontrivial retracts. The other is to go back and change “ff” to “ff and closed under retracts” and change “eso” to “surjective up to retracts.”

In a (2,1)-exact positive coherent 2-category, every ff with groupoidal codomain is an equifier (in fact, an identifier of an involution).

Recall that an *identifier* of a 2-cell $\alpha:f\to f:A\to B$ in a 2-category is the equifier of $\alpha$ and $1_f$. By an *involution* we mean an (invertible) 2-cell that is its own inverse.

Let $m:A\to B$ be ff with $B$ groupoidal, and consider first the (2,1)-congruence given by $B+B \;\rightrightarrows\; B$, where one copy of $B$ gives the identities, and the composition treats the other copy of $B$ as an involution. Its quotient is the copower of $B$ by the “walking involution.”

Now consider the following equivalence relation on $B+B$ in $disc(K/B\times B)$. We have

$(B+B)\times_{B\times B}(B+B) \simeq (B\times_{B\times B} B)+(B\times_{B\times B} B)+(B\times_{B\times B} B)+(B\times_{B\times B} B)$

and we define the relation to be given by

$\array{B+A+A+B \\ \downarrow\\
(B\times_{B\times B} B)+(B\times_{B\times B} B)+(B\times_{B\times B} B)+(B\times_{B\times B} B).}$

Since $disc(K/B\times B) \simeq disc(Fib_K(B\times B))$ is a 1-pretopos, this relation has a quotient, say $[f,g]:B+B \to C$. It is easy to verify that $a,b:C\;\rightrightarrows\; B$ is then a (2,1)-congruence on $B$ with $a f = a g = b f = b g = 1_B$. (This depends on $B$ being groupoidal; otherwise it would be a homwise-discrete category but not necessarily a congruence.) Let $q:B\to D$ be the quotient in $K$ of this (2,1)-congruence. Then we have a 2-fork $\phi:q a \to q b$ such that $\phi f = 1_q$ and $\phi g$ is an involution of $q$.

We claim that $m$ is an identifier of $\phi g$. By construction of $C$, we have $\phi g m = 1_{q m}$, so since $m$ is ff, it suffices to show that for any $x:X\to B$ with $\phi g x = 1_{q x}$, $x$ factors through $m$. But since $C$ with $\phi$ is the kernel of $q$, the assumption $\phi g x = 1_{q x} = \phi f x$ implies that in fact $f x = g x$. If we write $i,j:B\;\rightrightarrows\; B+B$ for the two inclusions, this means that $i x:X\to B+B$ and $j x:X\to B+B$ become equal in $C$, and therefore factor through the kernel pair of $[f,g]$, namely $B+A+A+B$. But this is evidently tantamount to saying that $x$ factors through $A$.

In a (2,1)-pretopos, 1. every ff is an equifier, 1. every cofaithful ff is an equivalence, and 1. every cofaithful morphism is eso.

Theorem shows the first statement. Then any ff $f:A\to B$ is an equifier of $\alpha,\beta:g\;\rightrightarrows\;h:B\;\rightrightarrows\; C$, so in particular $\alpha f = \beta f$; but if $f$ is also cofaithful, this implies $\alpha=\beta$, and thus their equifier $f$ is an equivalence. Finally, if $f$ is just cofaithful, we factor it as $f=m e$ where $m$ is ff and $e$ is eso; but then $m$ is also cofaithful, hence an equivalence, and so $f$, like $e$, is eso.

In a (1,2)-exact positive coherent 2-category, every ff with posetal codomain is an inverter.

Let $m:A\to B$ be ff, and consider the 2-congruence on $B+B$ defined as follows. We have

$(B_0+B_1)\times (B_0+B_1) = (B_0\times B_0)+(B_0\times B_1)+(B_1\times B_0)+(B_1\times B_1),$

(adding subscripts to distinguish the two copies of $B$) and the congruence is given by

$\array{
B ^{\mathbf{2}} &+& B ^{\mathbf{2}} &+& Y &+& B ^{\mathbf{2}}\\
&&& \downarrow\\
(B_0\times B_0) &+& (B_0\times B_1) &+& (B_1\times B_0) &+& (B_1\times B_1).}$

Here $Y\hookrightarrow B ^{\mathbf{2}}$ is defined to be the ff image of the “composition” morphism $(1_B/m/1_B) \to B ^{\mathbf{2}}$; in other words it is “the object of arrows in $B$ which factor through some element of $A$.” The composition is easy to define making this into a 2-congruence, and if $B$ is posetal, then it is a (1,2)-congruence.

Let $[p,q]:B+B\to C$ be the quotient of this congruence. Analogously to the proof of Theorem , the fork defining this quotient gives a 2-cell $\phi:p\to q$ such that $\phi m$ is an isomorphism. Thus, to show that $m$ is the inverter of $\phi$, it suffices to show that for any $x:X\to B$ with $\phi x$ invertible, $x$ factors through $m$. Now $\phi x$ is induced by the fork defining $C$ together with the composite $X\to B \to B ^{\mathbf{2}}$. If $\phi x$ is invertible, then its inverse is given by some map $y:X\to Y$, which must lie over the diagonal $X\to B \to B\times B$.

Now, pulling back the eso $(1/m/1)\to Y$ along $y$ we obtain an eso $r:Z\to X$ with a morphism $s:Z\to A$ such that the identity 2-cell of $x r$ is the composite of 2-cells $x r \to m s \to x r$; in other words, $x r$ is a retract of $m s$. But since $B$ is posetal, this means $x r \cong m s$, and then the fact that $r$ is eso implies that $x$ factors through $m$, as desired.

In a (1,2)-pretopos, 1. every ff is an inverter, 1. every liberal ff is an equivalence, and 1. every liberal morphism is eso.

Now, in any regular 2-category, in addition to the (eso,ff) factorization system we also have a Cauchy factorization system consisting of the cso (Cauchy surjective) and rff (ff and retract-closed) morphisms. Moreover, every cso is cofaithful and liberal. In “Modulated bicategories” by Carboni, Johnson, Street, and Verity (CJSV), it is shown that the liberal functors in Cat are precisely the Cauchy surjective ones; we now show that the same is true in any 2-pretopos.

In a 2-pretopos, 1. every rff is an inverter, 1. every liberal rff is an equivalence, and 1. every liberal morphism is Cauchy surjective.

In particular, every liberal morphism is cofaithful.

The first statement is proven exactly like Theorem , except that at the last step we use the assumption that $m$ is retract-closed rather than the assumption that $B$ is posetal. The other statements follow as before, recalling that Cauchy surjective morphisms are cofaithful.

In (CJSV) a 2-category is defined to be *co-conservational* (*liberational*?) if

- it has finite colimits,
- it has (liberal, strong conservative) factorizations,
- strong conservative morphisms are preserved by copowers with ${\mathbf{2}}$, and
- strong conservative morphisms are stable under pushout,

and *faithfully co-conservational* if moreover

- every liberal morphism is cofaithful.

Here a *strong conservative* morphism is one that is right orthogonal to all liberals. Theorem shows that in a 2-pretopos, strong conservatives coincide with rffs, since liberals coincide with csos; thus any 2-pretopos satisfies the second condition to be co-conservational. It need not have finite colimits, but this can be remedied with some infinitary structure. The construction of copowers in a 2-pretopos can be used to show that rffs are preserved by copowers with ${\mathbf{2}}$. And we have also seen that since every liberal is cso, it is cofaithful.

However, even $Cat$ fails the final condition, as shown in (CJSV, Prop. 3.4). This can be remedied by passing to the 2-category $Cat_{cc}$ of Cauchy-complete categories. I do not yet know whether a similar construction is possible in any 2-pretopos.

Last revised on November 18, 2009 at 19:53:57. See the history of this page for a list of all contributions to it.