Michael Shulman
comprehensive factorization

In Cat there is a factorization system (E,M)(E,M) where EE is the class of initial functors and MM is the class of discrete opfibrations; see

  • Street, Walters, The comprehensive factorization of a functor. Bull. Amer. Math. Soc. 79, 1973.

We can construct an analogous factorization system in any 1-exact and countably-coherent 2-category.

Definition

A morphism f:ABf:A\to B in a 2-category KK is initial if it is left orthogonal to discrete opfibrations. That is, whenever g:CDg:C\to D is a discrete opfibration, the following square is a pullback in CatCat:

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

Every morphism in a 1-exact countably-coherent 2-category (in particular, in an nn-pretopos for n1n\ge 1) factors, up to isomorphism, as an initial morphism followed by a discrete opfibration.

Proof

By the 2-categorical analogue of a standard theorem about factorization systems in 1-categories, it suffices to show that

  1. discrete opfibrations are closed under composition, and
  2. for every XX, the discrete opfibrational slice DOpf(X)DOpf(X) is a full reflective subcategory of the slice 2-category K/XK/X.

The first point is true in any 2-category. For the second, we factor the inclusion as

DOpf(X)Opf(X)K/X DOpf(X) \to Opf(X) \to K/X

and observe that both forgetful functors have left adjoints. The left adjoint to DOpf(X)Opf(X)DOpf(X) \to Opf(X) is discretization in the 1-exact and countably-coherent 2-category Opf(X)Opf(X). The left adjoint to Opf(X)K/X Opf(X) \to K/X constructs the “free opfibration” on a functor f:AXf:A\to X, which is given by A× XX 2A\times_X X^{\mathbf{2}}.

Corollary

In any 1-exact countably-coherent 2-category, (initial, discrete opfibration) is a (2-categorical) factorization system.

Corollary

In any 1-exact countably-coherent 2-category, each pullback functor f *:DOpf(Y)DOpf(X)f^*:DOpf(Y)\to DOpf(X) has a left adjoint Lan fLan_f.

Proof

Lan fLan_f is given by composing with ff, then factoring into an initial morphism followed by a discrete opfibration.

Examples

  • In a 1-category, every morphism is a discrete opfibration, so the only initial morphisms are isomorphisms. The factorization system (isomorphisms,all morphisms) is well-known.

  • In a (2,1)-category, the discrete opfibrations are precisely the faithful morphisms. Thus, in this case the comprehensive factorization system coincides with the (eso+full, faithful) factorization system.

Beck-Chevalley conditions

The appropriate Beck-Chevalley condition for the adjunctions Lan ff *Lan_f\dashv f^* refers not to pullback squares, but to comma squares. The easiest proof of this fact uses the internal logic. We start with an internal characterization of initial morphisms, analogous to the classical characterization in CatCat as the functors f:ABf:A\to B for which each comma category (f/b)(f/b), for bBb\in B, is nonempty and connected.

For the rest of this page we make the standing assumption that KK is a 1-exact and countably-coherent 2-category.

Lemma

A morphism f:ABf:A\to B in KK is initial iff the following are internally valid:

b:B|(a:A)(α:hom B(f(a),b)) b:B,a 1:A,a 2:A,α 1:hom B(f(a 1),b),α 2:hom B(f(a 2),b)| iφ i \begin{gathered} b:B | \top \vdash (\exists a:A)(\exists \alpha:hom_B(f(a),b))\\ b:B, a_1:A, a_2:A, \alpha_1:hom_B(f(a_1),b), \alpha_2:hom_B(f(a_2),b) | \top \vdash \bigvee_{i\in \mathbb{N}} \varphi_i \end{gathered}

where for each ii, φ i\varphi_i expresses “there is a zigzag of length ii connecting α 1\alpha_1 and α 2\alpha_2 over bb.”

Thus, for instance, we have

φ 1 (γ:hom A(a 1,a 2))(α 2f(γ)=α 1) φ 2 (c:A)(β:hom B(f(c),b))(γ 1:hom A(a 1,c))(γ 2:hom A(a 2,c))(α 1=βf(γ 1)α 2=βf(γ 2)) \begin{aligned} \varphi_1 \equiv& (\exists \gamma:hom_A(a_1,a_2)) (\alpha_2 \circ f(\gamma) = \alpha_1)\\ \varphi_2 \equiv& (\exists c:A)(\exists \beta:hom_B(f(c),b))(\exists \gamma_1:hom_A(a_1,c))(\exists \gamma_2:hom_A(a_2,c)) (\alpha_1 = \beta \circ f(\gamma_1) \wedge \alpha_2 = \beta \circ f(\gamma_2)) \end{aligned}

and so on.

Proof

f:ABf:A\to B will be initial iff the discrete-opfibration part of its factorization is an equivalence. By construction, this factorization is the discrete reflection of X=A× BB 2BX = A\times_B B ^{\mathbf{2}} \to B in Opf(B)Opf(B), which is constructed as the quotient of the equivalence relation X 1X_1 generated by X 2X ^{\mathbf{2}} (the power taken in Opf(B)Opf(B)). Therefore, this discrete reflection will be the terminal object 1 B1_B iff

  1. XBX\to B is eso and
  2. the kernel of XBX\to B in Opf(B)Opf(B) is X 1X_1.

It is fairly evident that XBX\to B is eso iff the first displayed sequent holds in KK. For the second, note that the kernel of XBX\to B in Opf(B)Opf(B) (or equivalently, in K/BK/B) is just the pullback X× BXX\times_B X, since BB is discrete as an object of K/BK/B. By definition of XX, this pullback can be computed as the lax limit

Lax pullback X × B X X\times_B X A A A A B B f f f f

and therefore it is precisely the context of the second displayed sequent. Since X 1X_1 is the equivalence relation generated by X 2X ^{\mathbf{2}} , it is necessarily contained in this kernel, so it suffices to show the converse implication. But X 1X_1 is defined as the symmetric transitive closure of the image of X 2X ^{\mathbf{2}} , which makes it essentially a countable union of “zigzags” from X 2X ^{\mathbf{2}} , and this translates directly into the conclusion of the second sequent.

Lemma

Initial morphisms in KK are stable under transfer across comma squares, in the sense that if

Comma Square ( f / g ) (f/g) A A B B C C f f g g q q p p

is a comma square with ff initial, then qq is also initial.

Proof

Given the characterization of initial morphisms in Lemma 1, we can simply observe that the usual proof of this fact in CatCat takes place entirely in countably-coherent logic.

Theorem

The adjunctions Lan ff *Lan_f\dashv f^* for discrete opfibrations in KK satisfy the Beck-Chevalley condition for comma squares. That is, if

Comma Square ( f / g ) (f/g) A A B B C C f f g g q q p p

is a comma square, then the canonical transformation Lan qp *g *Lan fLan_q p^* \to g^* Lan_f between functors DOpf(A)DOpf(B)DOpf(A)\to DOpf(B) is an equivalence.

Note that unlike the case for a pullback square, there is no possible “handedness” ambiguity in saying that a comma square satisfies the Beck-Chevalley condition; there is no transformation Lan pq *f *Lan gLan_p q^* \to f^* Lan_g at all.

The existence of the transformation in the correct direction also depends on the fact that opfibrational slices are functorial at the level of 2-cells. In particular, for a comma square there is no transformation Σ qp *g *Σ f\Sigma_q p^* \to g^*\Sigma_f between functors K/AK/BK/A\to K/B.

Proof

Given a discrete opfibration XAX\to A, let XLan fXCX\to Lan_f X \to C be the (initial,discrete-opfibration) factorization of the composite XAfCX\to A \overset{f}{\to} C. Now by properties of comma squares and pullbacks, the pasting composite

Composite Square 1 p * X p^* X X X ( f / g ) (f/g) A A B B C C f f g g q q p p

is also a comma square, and thus so is

Composite Square 2 p * X p^* X X X g * Lan f X g^* Lan_f X Lan f X Lan_f X B B C C g g

where the upper 2-cell is opcartesian for the opfibration Lan fXCLan_f X \to C, and the map p *Xg *Lan fXp^* X \to g^* Lan_f X is obtained from the universal property of the pullback square on the bottom. Again, it follows from general properties of pullbacks and comma squares that the top square in this latter diagram is also a comma square. Thus, by Lemma 2, its left-hand morphism is initial. But then the left-hand composite p *Xg *Lan fXBp^*X \to g^* Lan_f X \to B is an (initial, discrete-opfibration) factorization of p *X(f/g)Bp^*X \to (f/g) \to B, and hence it exhibits the desired equivalence Lan qp *Xg *Lan fXLan_q p^*X \simeq g^* Lan_f X.

Passing to 2-cell duals, we obtain:

Corollary

For a comma square as above, the canonical transformation Lan pq *f *Lan gLan_p q^* \to f^* Lan_g between functors DFib(B)DFib(A)DFib(B)\to DFib(A) is an equivalence.

Revised on June 12, 2012 11:10:00 by Andrew Stacey? (129.241.15.200)