Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: category theory

Topic: Fib has conjoints, but not companions

Christian Williams (Dec 09 2022 at 00:43):

A fibration $\mathbb{A}\to \mathbb{I}$ is a "dependent category" : over each object $I$ there is a category $\mathbb{A}_I$, and over each morphism $f:I\to J$ there is a profunctor which is representable as a conjoint $\mathbb{A}_I(1,f^\ast):\mathbb{A}_I\to \mathbb{A}_J$
--- a representation is a choice of cartesian morphisms.
fib-cart.png

Christian Williams (Dec 09 2022 at 00:43):

So, a fibration is essentially a diagram of conjoints in Cat, whose collage gives back the total category $\mathbb{A}$.
fib-collage.png

Christian Williams (Dec 09 2022 at 00:46):

We can define a fibered profunctor the same way: over each heteromorphism, a conjoint.

Christian Williams (Dec 09 2022 at 00:50):

Let $\mathbb{A}, \mathbb{B}$ be fibrations, and $R_0:\mathbb{A}_0\to \mathbb{B}_0$ be a profunctor of base categories. Define a fibered profunctor over $R_0$ to be a profunctor $R:\mathbb{A}\to \mathbb{B}$ with a transformation $R\Rightarrow R_0$ so that the collage is a fibration.
preimg-trans.png fibprof-cart.png

Christian Williams (Dec 09 2022 at 00:52):

Similarly, a fibered transformation is a commuting square of transformations so that the collage is a fibered functor.

Christian Williams (Dec 09 2022 at 00:52):

This defines the double category $\mathrm{Fib}$ of fibered categories and functors, fibered profunctors and transformations.

Christian Williams (Dec 09 2022 at 00:53):

One might hope that this is a fibrant double category, a.k.a. an equipment; but no: there are conjoints, but not companions.

Christian Williams (Dec 09 2022 at 00:57):

The reason boils down to the fact that fibered categories and fibered profunctors are really "indexed conjoints". I'll draw a picture to show the problem.

Christian Williams (Dec 09 2022 at 01:02):

fib-no-comp.png

Christian Williams (Dec 09 2022 at 01:06):

Let $f:\mathbb{A}\to\mathbb{B}$ be a fibered functor, and consider the companions $\mathbb{B}_0(f_0,1):\mathbb{A}_0\to \mathbb{B}_0$ and $\mathbb{B}(f,1):\mathbb{A}\to \mathbb{B}$. Let $\phi: \mathbb{B}_0(f_0A_0,B_0)$ be an element of the base profunctor, and $B$ be in the fiber over $B_0$.

Christian Williams (Dec 09 2022 at 01:07):

Is there a "cartesian element" of $\mathbb{B}(f,1)$ over $\phi$?

Christian Williams (Dec 09 2022 at 01:08):

There is a perfectly good cartesian morphism $\phi^\ast B\to B$, but this is in $\mathbb{B}$, not the companion $\mathbb{B}(f,1)$.

Christian Williams (Dec 09 2022 at 01:09):

There is no reason to expect that the cartesian morphism is of the form $fA\to B$.

Christian Williams (Dec 09 2022 at 01:10):

This is a basic obstruction; extra properties of the functor or the fibrations do not help.

Christian Williams (Dec 09 2022 at 01:11):

So, $\mathrm{Fib}$ has conjoints but not companions. (And dually, $\mathrm{opFib}$ has companions but not conjoints.)

Christian Williams (Dec 09 2022 at 01:17):

(All thoughts welcome.)

Nathanael Arkor (Dec 09 2022 at 01:21):

If you want an equipment of fibred categories, I expect you need to restrict to categories fibred over a specific category, rather than allowing the base to vary (like how one usually considers $\textrm{Cat}(\mathcal E)$ for a specific $\mathcal E$). But I'm guessing there's a reason you've chosen not to do it that way?

Christian Williams (Dec 09 2022 at 01:23):

Hm okay, well why do you think fixing the base could help?

Christian Williams (Dec 09 2022 at 01:30):

The motivation: a fibrant double category is a category $\mathbb{C}_0$ with a fibration $\mathbb{C}\to \mathbb{C}_0^2$. I'm constructing the triple category $\mathrm{FDC}$, and it should be fibrant: a double functor should have both a companion and a conjoint.

Christian Williams (Dec 09 2022 at 01:33):

I don't think that fixing the base would help, but I need to draw it out.

Christian Williams (Dec 09 2022 at 01:38):

Yeah, I think it doesn't change the problem. In that picture "fib-no-comp" above, even if the bottom profunctor is the hom of a fixed base, there's still no reason to expect the cartesian morphism to be of the form $fA\to B$.

Nathanael Arkor (Dec 09 2022 at 01:39):

Christian Williams said:

Hm okay, well why do you think fixing the base could help?

It's known that $\mathcal S$-indexed categories form a Yoneda structure: see, for instance, §6 of Conspectus of variable categories, hence an equipment.

Christian Williams (Dec 09 2022 at 01:45):

Oh great thanks, I hadn't found any literature like this. So what is the notion of fibered profunctor" here...

Christian Williams (Dec 09 2022 at 01:49):

Oh, this is different. This is two-sided fibrations, i.e. pseudofunctors $A^{op}\times B\to \mathrm{Cat}$.

Christian Williams (Dec 09 2022 at 01:50):

These are bimodules, loose morphisms in a fibrant double category. I'm talking about a double category in which fibered categories are objects.

Christian Williams (Dec 09 2022 at 01:58):

Anyway, this thread is for any thoughts about: the concept of profunctor between fibered categories, and their formal theory.

Kenji Maillard (Dec 09 2022 at 09:17):

If fibred categories have conjoints and opfibered categories have companions, do bifibred categories have both (so form a fibrant double category) ?

Nathanael Arkor (Dec 09 2022 at 12:35):

Christian Williams said:

These are bimodules, loose morphisms in a fibrant double category. I'm talking about a double category in which fibered categories are objects.

The objects of the Yoneda structure in that paper are indexed categories, i.e. (up to equivalence) fibred categories. I haven't looked at the paper for a while, but I would have expected the proarrows to be indexed profunctors.

Christian Williams (Dec 09 2022 at 16:52):

That's what I hoped! But then I realized no, bifibrations don't change the problem --- there's still no reason for the cartesian morphism $\phi^\ast(B)\to B$ to be of the form $fA\to B$.

Christian Williams (Dec 09 2022 at 16:53):

Ah okay, I need to look at that paper more closely.

Christian Williams (Dec 09 2022 at 18:00):

Wow, I forgot --- of course fibrations are monads; they're categories in $[C^{op},\mathrm{Set}]$. So yes, $\mathrm{Fib}(C)$ is an equipment, and I'll be working to understand those bimodules.

Christian Williams (Dec 09 2022 at 18:00):

So yes, I see what you were saying about a fixed base. But I think the difficulty is in changing base along a profunctor.

Christian Williams (Dec 09 2022 at 21:26):

So for Street, a profunctor in $\mathrm{Fib}(\mathbb{I})$ $R:\mathbb{A}\to \mathbb{B}$ of fibrations over $\mathbb{I}$ gives for each object $I:\mathbb{I}$ a profunctor between fibers $R_I:\mathbb{A}_I\to \mathbb{B}_I$, and for each morphism $f:I\to J$ a reindexing transformation $R_J\Rightarrow R_I(\mathbb{A}_f,\mathbb{B}_f)$.

Christian Williams (Dec 09 2022 at 21:28):

So for each fixed object, it gives an arbitrary profunctor between fibers, "functorially" with respect to base morphisms.

Christian Williams (Dec 09 2022 at 21:31):

Whereas the definition I propose above, there is a profunctor between base categories $Q_0:\mathbb{X}\to \mathbb{Y}$, and over each heteromorphism $q:Q_0(X,Y)$ a conjoint $Q(q)\simeq \mathbb{A}_X(1,q^\ast):\mathbb{A}_X\to \mathbb{B}_Y$.

Christian Williams (Dec 09 2022 at 23:37):

Okay, I think I'm seeing the bigger picture. Street's equipments Fib(I) are slices over a fixed base Fib/I, but Fib itself is a collage over all of Cat, over both functors and profunctors. Being fibered over an identity profunctor is very special (hence forming an equipment), but the more general notion of fibered profunctor must be defined purely in terms of heteromorphisms.

That leads to this simple characterization via collage of a transformation. Yet it also leads to this obstacle with companions, above.

So I think the double category of all fibrations has only conjoints, because ultimately, both fibered categories and profunctors are just (matrices of families of) conjoints.

Christian Williams (Dec 10 2022 at 00:12):

What if we just collage opFib, BiFib, and Fib all together?

Christian Williams (Dec 10 2022 at 00:12):

08ff21b6-644a-4095-96fd-97e424c08722.jpg

Christian Williams (Dec 10 2022 at 00:12):

(deleted)

Christian Williams (Dec 10 2022 at 00:16):

Companions on the left, conjoints on the right: so this double category is not fibrant but rather a "two-sided fibration" (2-profunctor) from left to right.

Christian Williams (Dec 10 2022 at 00:25):

I think there is a double profunctor from opFib to Fib, which gives categories of bifibered profunctors and transformations.

and I am thinking of its collage.

Christian Williams (Dec 10 2022 at 00:39):

Ah, but a bifibered profunctor composed with a fibered one is only fibered - so the left and right act on the middle, not vice versa. So it's not a profunctor from opFib to Fib.

And yet I think the "collage" is a perfectly nice double category! Things just "flow outward" horizontally - a bifibration in the middle will go either to the "left double cat" opFib or the right Fib as soon as you compose on the left or right.

Christian Williams (Dec 10 2022 at 00:45):

And we do have companions on the left and conjoints on the right, so it's a "two-side fibered" double category.

Christian Williams (Dec 10 2022 at 00:48):

But what should it be called? "BiFib" should probably mean the nicer sub-double category of bifibered categories and their functors, profunctors and transformations.

Christian Williams (Dec 10 2022 at 00:50):

I guess you just need to pick a word for "either an opfibration or fibration". Maybe "family category", "family" for short.

Christian Williams (Dec 10 2022 at 00:58):

I think it's the pushout of BiFib including into opFib and Fib.

Christian Williams (Dec 10 2022 at 01:43):

"FamCat" is the union of opFib and Fib along their intersection, BiFib. A square is either one in opFib ($\alpha$) one in Fib ($\omega$) or in both $(\gamma)$.
Fam.png

Christian Williams (Dec 10 2022 at 01:52):

Fib is fibered over Cat: the pullback of a fibration is a fibration. Same for opFib. Moreover, they're bifibered: both have "pushforward" / direct image, given by left Kan extension.

Christian Williams (Dec 10 2022 at 01:54):

Ironically, BiFib is not a bifibration. The pushforward of a bifibration is not a bifibration.

Christian Williams (Dec 10 2022 at 01:55):

But hopefully there is a theorem that the pushforward is always either a fibration or opfibration.

Christian Williams (Dec 10 2022 at 01:56):

These pushforwards are needed to define matrix composition $\sum b. R(a,b)\times S(b,c)$, thinking of a fibration $R\to A\times B$ as an $AB$-matrix of categories; see Framed Bicategories and Monoidal Fibrations.

Christian Williams (Dec 10 2022 at 02:18):

Ah, it depends how you define bifibration. If it is only "both an opfibration and a fibration", then I think bifibrations are clearly stable under pushforwards, because each property is. But if you also include the Chevalley condition, then it's more complex.

Christian Williams (Dec 10 2022 at 02:20):

I think the former is all that's needed here. So, "FamCat" is bifibered over Cat. Great!

Christian Williams (Dec 10 2022 at 17:56):

Okay, so I wasn't thinking all the way through: we can't glue opFib and Fib together along BiFib, because a composite opfib;bifib;fib is not any kind of op/fibered profunctor! It's just an arbitrary transformation. So I think we have to accept that profunctors of the form $\mathbb{C}(f,1)\circ \mathbb{C}(1,g)\simeq \mathbb{C}(f,g)$ are neither opfibered nor fibered, and include them in a larger universe.

Hence we need to "glue along their common ground", arbitrary functors. Let $\iota_0: \mathrm{opFib}\to \mathrm{Cat}^{\to}$ and $\iota_1: \mathrm{Fib}\to \mathrm{Cat}^{\to}$ be the inclusion of each into the arrow double category of $\mathrm{Cat}$, and consider the collage of $\mathrm{Cat}^{\to}(\iota_0,\iota_1)$.

Christian Williams (Dec 10 2022 at 18:05):

Preimage along an opfibration is a diagram of companions in Cat; preimage along a functor is a (lax) diagram of arbitrary profunctors; and preimage along a fibration is a diagram of conjoints. famcat.png

Christian Williams (Dec 10 2022 at 18:10):

In this picture, X is an opfibration and Y is a fibration. Q is a profunctor of their base categories, a profunctor of their total categories, and any transformation filling the square. (Hence can be drawn as a diagram of arbitrary profunctors, one over each heteromorphism.) Similarly for R.

Christian Williams (Dec 10 2022 at 18:14):

It might feel weird to just "glue opFib and Fib along the unstructured common ground Arrow(Cat)", but this whole diagrams-in-Cat idea is very intuitive to me. Arbitrary profunctors in the middle seem to complete the whole picture.

Christian Williams (Dec 10 2022 at 18:24):

Ah wait! Opfibrations and fibrations have more common ground than being arbitrary functors - they're both exponentiable. Then the middle is pseudo-diagrams of profunctors, rather than lax.

Christian Williams (Dec 10 2022 at 18:36):

So take the inclusions of opFib and Fib into Exp, then define FamCat to be the collage of Exp(inc_0, inc_1).

I think this is a good union of the two ways to index categories.

Christian Williams (Dec 10 2022 at 18:36):

All thoughts welcome.