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 nn-pretopos for n>1n\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:”

  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 nn, 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\alpha:f\to f:A\to B in a 2-category is the equifier of α\alpha and 1 f1_f. By an involution we mean an (invertible) 2-cell that is its own inverse.

Proof

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

Now consider the following equivalence relation on B+BB+B in disc(K/B×B)disc(K/B\times 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))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+BC[f,g]:B+B \to C. It is easy to verify that a,b:CBa,b:C\;\rightrightarrows\; B is then a (2,1)-congruence on BB with af=ag=bf=bg=1 Ba f = a g = b f = b g = 1_B. (This depends on BB being groupoidal; otherwise it would be a homwise-discrete category but not necessarily a congruence.) Let q:BDq:B\to D be the quotient in KK of this (2,1)-congruence. Then we have a 2-fork ϕ:qaqb\phi:q a \to q b such that ϕf=1 q\phi f = 1_q and ϕg\phi g is an involution of qq.

We claim that mm is an identifier of ϕg\phi g. By construction of CC, we have ϕgm=1 qm\phi g m = 1_{q m}, so since mm is ff, it suffices to show that for any x:XBx:X\to B with ϕgx=1 qx\phi g x = 1_{q x}, xx factors through mm. But since CC with ϕ\phi is the kernel of qq, the assumption ϕgx=1 qx=ϕfx\phi g x = 1_{q x} = \phi f x implies that in fact fx=gxf x = g x. If we write i,j:BB+Bi,j:B\;\rightrightarrows\; B+B for the two inclusions, this means that ix:XB+Bi x:X\to B+B and jx:XB+Bj x:X\to B+B become equal in CC, and therefore factor through the kernel pair of [f,g][f,g], namely B+A+A+BB+A+A+B. But this is evidently tantamount to saying that xx factors through AA.

Corollary

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.

Proof

Theorem 1 shows the first statement. Then any ff f:ABf:A\to B is an equifier of α,β:gh:BC\alpha,\beta:g\;\rightrightarrows\;h:B\;\rightrightarrows\; C, so in particular αf=βf\alpha f = \beta f; but if ff is also cofaithful, this implies α=β\alpha=\beta, and thus their equifier ff is an equivalence. Finally, if ff is just cofaithful, we factor it as f=mef=m e where mm is ff and ee is eso; but then mm is also cofaithful, hence an equivalence, and so ff, like ee, 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:ABm:A\to B be ff, and consider the 2-congruence on B+BB+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 BB) 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 2Y\hookrightarrow B ^{\mathbf{2}} is defined to be the ff image of the “composition” morphism (1 B/m/1 B)B 2(1_B/m/1_B) \to B ^{\mathbf{2}}; in other words it is “the object of arrows in BB which factor through some element of AA.” The composition is easy to define making this into a 2-congruence, and if BB is posetal, then it is a (1,2)-congruence.

Let [p,q]:B+BC[p,q]:B+B\to C be the quotient of this congruence. Analogously to the proof of Theorem 1, the fork defining this quotient gives a 2-cell ϕ:pq\phi:p\to q such that ϕm\phi m is an isomorphism. Thus, to show that mm is the inverter of ϕ\phi, it suffices to show that for any x:XBx:X\to B with ϕx\phi x invertible, xx factors through mm. Now ϕx\phi x is induced by the fork defining CC together with the composite XBB 2X\to B \to B ^{\mathbf{2}}. If ϕx\phi x is invertible, then its inverse is given by some map y:XYy:X\to Y, which must lie over the diagonal XBB×BX\to B \to B\times B.

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

Corollary

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.

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, 1. every liberal rff is an equivalence, and 1. 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 mm is retract-closed rather than the assumption that BB 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{\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 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{\mathbf{2}}. And we have also seen that since every liberal is cso, it is cofaithful.

However, even CatCat fails the final condition, as shown in (CJSV, Prop. 3.4). This can be remedied by passing to the 2-category Cat ccCat_{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)