Michael Shulman
balancedness in n-pretoposes

1-pretoposes are balanced

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

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

2-pretoposes are not balanced

For a similar statement in an n-pretopos for n>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:”

  1. We could replace “epic” by “cofaithful” and “regular monic” by “equifier,” or
  2. 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.”

(2,1)-pretoposes are balanced

Theorem

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 α:ff:AB in a 2-category is the equifier of α and 1 f. By an involution we mean an (invertible) 2-cell that is its own inverse.

Proof

Let m:AB be ff with B groupoidal, and consider first the (2,1)-congruence given by B+BB, 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×B). We have

(B+B)× B×B(B+B)(B× B×BB)+(B× B×BB)+(B× B×BB)+(B× B×BB)(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

B+A+A+B (B× B×BB)+(B× B×BB)+(B× B×BB)+(B× B×BB).\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×B)disc(Fib K(B×B)) is a 1-pretopos, this relation has a quotient, say [f,g]:B+BC. It is easy to verify that a,b:CB is then a (2,1)-congruence on B with af=ag=bf=bg=1 B. (This depends on B being groupoidal; otherwise it would be a homwise-discrete category but not necessarily a congruence.) Let q:BD be the quotient in K of this (2,1)-congruence. Then we have a 2-fork ϕ:qaqb such that ϕf=1 q and ϕg is an involution of q.

We claim that m is an identifier of ϕg. By construction of C, we have ϕgm=1 qm, so since m is ff, it suffices to show that for any x:XB with ϕgx=1 qx, x factors through m. But since C with ϕ is the kernel of q, the assumption ϕgx=1 qx=ϕfx implies that in fact fx=gx. If we write i,j:BB+B for the two inclusions, this means that ix:XB+B and jx:XB+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.

Corollary

In a (2,1)-pretopos,

  1. every ff is an equifier,
  2. every cofaithful ff is an equivalence, and
  3. every cofaithful morphism is eso.
Proof

Theorem 1 shows the first statement. Then any ff f:AB is an equifier of α,β:gh:BC, so in particular αf=βf; but if f is also cofaithful, this implies α=β, and thus their equifier f is an equivalence. Finally, if f is just cofaithful, we factor it as f=me where m is ff and e is eso; but then m is also cofaithful, hence an equivalence, and so f, like e, is eso.

(1,2)-pretoposes are balanced

Theorem

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

Proof

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

(B 0+B 1)×(B 0+B 1)=(B 0×B 0)+(B 0×B 1)+(B 1×B 0)+(B 1×B 1),(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

B 2 + B 2 + Y + B 2 (B 0×B 0) + (B 0×B 1) + (B 1×B 0) + (B 1×B 1).\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 YB 2 is defined to be the ff image of the “composition” morphism (1 B/m/1 B)B 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+BC be the quotient of this congruence. Analogously to the proof of Theorem 1, the fork defining this quotient gives a 2-cell ϕ:pq such that ϕm is an isomorphism. Thus, to show that m is the inverter of ϕ, it suffices to show that for any x:XB with ϕx invertible, x factors through m. Now ϕx is induced by the fork defining C together with the composite XBB 2. If ϕx is invertible, then its inverse is given by some map y:XY, which must lie over the diagonal XBB×B.

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

Corollary

In a (1,2)-pretopos,

  1. every ff is an inverter,
  2. every liberal ff is an equivalence, and
  3. every liberal morphism is eso.
Proof

Just like Corollary 1 but using Theorem 2 instead.

2-pretoposes are Cauchy balanced

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.

Corollary

In a 2-pretopos,

  1. every rff is an inverter,
  2. every liberal rff is an equivalence, and
  3. every liberal morphism is Cauchy surjective.

In particular, every liberal morphism is cofaithful.

Proof

The first statement is proven exactly like Theorem 2, 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 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 3 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 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.

Revised on November 18, 2009 19:53:57 by Mike Shulman (128.135.230.26)