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Stream: theory: category theory

Topic: exponentiable double categories

Christian Williams (Dec 29 2023 at 17:03):

A double category $\mathbb{D}$ for which $\mathbb{D}_0\leftarrow \mathbb{D}_1\to \mathbb{D}_0$ is a two-sided fibration is a double category with companions, and one whose opposite span is a two-sided fibration is a double category with conjoints. (or dually, depending on your convention)

It would be very useful to unify these two, by defining an "exponentiable double category" to be one for which $\mathbb{D}_1\to \mathbb{D}_0\times \mathbb{D}_0$ is exponentiable. This means every morphism in $\mathbb{D}_1$ over a composite in $\mathbb{D}_0$ has a factorization lying over each factor, unique up to reassociating.

I thought this would be fairly straightforward, but to define composition of "exponentiable spans" one needs to define pushforward of an exponentiable category along a functor; and so far it looks like this may not work as I hoped.

To pushforward an opfibration $A\to I$ along $f:I\to J$, we can compose the two-sided fibration $1\leftarrow A\to I$ with the companion coslice $I\leftarrow f/J\to J$. Dually for pushing fibrations, we can precompose with the conjoint.

But for exponentiable categories, either one of those operations seems like a way to pushforward, and since they're dual then neither would be the answer. Actually, those compositions would not even be defined in the same way, because composing two-sided fibrations is given by codescent (like a coend, but adjoining an associator isomorphism rather than quotienting), while composing "exp-spans" is give only by pullback --- such spans are no longer a kind of bimodule, so we lose co/Yoneda-reasoning for constructing a left adjoint to substitution.

While slice and coslice give notions of "free op/fibration" of a functor, it's not clear whether there is a similar operation of a "free exponentiable category".

I couldn't find anything online about Exp over Cat. If anyone has thoughts about how one might pushforward exponentiable categories, it would be greatly appreciated.

James Deikun (Dec 29 2023 at 17:33):

Could you give some more details? I'm kind of lost on why you need this pushforward operation in the first place.

Christian Williams (Dec 29 2023 at 17:37):

Oh, yes. I want to construct ExpSpan over Cat x Cat by pulling back Exp > Cat along the product of Cat (a categorification of Shulman's "frame construction" in Section 14 of framed bicategories).

Then composition of an exp-span $R\to A\times B$ and $S\to B\times C$ is given by $R\times S\to A\times B\times B\times C$, pulling back along $\Delta:B\to B\times B$, and then pushing forward along $!: B\to 1$. Similarly, the units involve a pullback and pushforward.

Christian Williams (Dec 29 2023 at 17:47):

It's possible that the simplest solution is to construct the "virtual triple category" of all double categories, and then define ExpDblCat as a substructure thereof -- which would then be representable rather than virtual, because composition by an exponentiable span preserves colimits.

But I think this question about constructing ExpSpan might be interesting.

James Deikun (Dec 29 2023 at 17:48):

In the span-like examples, though, pushing forward is just composition. Exponentiable morphisms are closed under composition, so don't you get something that way as soon as the diagonals and projections are exponentiable? And I think they are exponentiable; they're even bifibrations iirc...

Christian Williams (Dec 29 2023 at 17:49):

Oh, good point! I skipped to the general case too quick. Maybe that's all we need.

Mike Shulman (Dec 29 2023 at 17:51):

Product projections are bifibrations, but diagonals are not exponentiable. The morphism $\Delta(f) = (f,f) : (x,x) \to (y,y)$ factors as $(f,1) \circ (1,f)$, but this factorization doesn't lift along $\Delta$.

James Deikun (Dec 29 2023 at 17:54):

OK, so units are going to be a problem, at the very least.

Mike Shulman (Dec 29 2023 at 17:56):

Pushforwards of (op)fibrations correspond to Kan extensions of pseudofunctors into Cat, whereas an exponentiable functor $E\to B$ is equivalent to a pseudofunctor $B \to \mathsf{Prof}$. I wouldn't expect all Kan extensions of pseudofunctors into Prof to exist, since Prof doesn't have all bicategorical limits and colimits. But perhaps you could come up with some Kan-extension-like thing. In any case, you might get some mileage out of deciding first what you want the putative operation to mean in terms of such pseudofunctors.

Christian Williams (Dec 29 2023 at 17:58):

Even if ExpSpan doesn't have units, the (pseudo)monad construction freely creates units, so ExpDblCat will have them.

James Deikun (Dec 29 2023 at 18:05):

Yeah that's a good point; OTOH I don't know if the construction with Fr and composition will match the kind of thing you get from two-sided fibrations, which is really not an instance of the Fr construction as such and corresponds to the tensor product of bimodules. Exponentiable spans really ought to be looked at as "disoriented bimodules".

James Deikun (Dec 29 2023 at 18:49):

I think you can use the same raw objects and arrows as in the case of two-sided fibrations but with a finer quotienting that only looks at isos; this should be okay since uniqueness in the definition of Conduche' fibrations is allowed to be up to zigzags.

Christian Williams (Dec 29 2023 at 18:59):

Okay so after a brief conversation with Nathanael, I realized I was making a mistake.

If $A\leftarrow R\to B$ is a two-sided fibration, that does not imply that $R\to A\times B$ is exponentiable; it only implies that each leg is exponentiable. See the diagram below.
comp-not-exp.png

Christian Williams (Dec 29 2023 at 19:02):

We can't factor the square over each pair of tight morphisms, because we can only "bend" one side of each pair. So we get a three-fold factorization, which doesn't fit the definition.

Christian Williams (Dec 29 2023 at 19:03):

So the whole idea of these "exponentiable spans" was miguided.

Christian Williams (Dec 29 2023 at 20:01):

Christian Williams said:

It's possible that the simplest solution is to construct the "virtual triple category" of all double categories, and then define ExpDblCat as a substructure thereof -- which would then be representable rather than virtual, because composition by an exponentiable span preserves colimits.

So I think this is probably the way to go; we can define an "exponentiable double category" $\mathbb{D}$ to be one for which each leg of $\mathbb{D}_0\leftarrow \mathbb{D}_1\to \mathbb{D}_0$ is exponentiable.

Mike Shulman (Dec 29 2023 at 20:59):

Christian Williams said:

We can't factor the square over each pair of tight morphisms, because we can only "bend" one side of each pair. So we get a three-fold factorization, which doesn't fit the definition.

Ah, indeed: this is one of the examples of a [[ternary factorization system]].