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Stream: theory: category theory

Topic: Fibrations and pullback-preserving functors

James Deikun (Oct 17 2024 at 14:29):

Pullback-preserving functors allow one to push forward internal categories. But in some sense an internal category "is just" a small fibration. Is it possible somehow to push forward fibrations along a pullback-preserving functor?

John Baez (Oct 17 2024 at 14:39):

I'm ignorant of how an internal category is a small fibration - is that easy to explain?

James Deikun (Oct 17 2024 at 15:03):

I'd like to link you to the Stacks Project where I learned it but the page seems to be gone. Section B2.3 of the Elephant has a description, but in short:

Given an internal category $C$ in $\mathcal B$, for each object $I$ of $\mathcal{B}$, you can get a category in the external sense, by taking as objects the arrows from $I$ to $C_0$, and as arrows the arrows $I$ to $C_1$; composition with the structural arrows of $C$ gives the structure of the external category. By composing with arrows of $B$ on the other side you get pullback functors; this gives you a split indexed category $[C] : B^\mathrm{op} \to \mathbf{Cat}$. You take the Grothendieck construction to get a fibration.

James Deikun (Oct 17 2024 at 15:05):

(And from $C_0$ and $C_1$, you can construct the evidence that the fibration is locally small and has a generic object.)

James Deikun (Oct 17 2024 at 15:21):

The whole thing is a full embedding of 2-categories. The construction above, when you consider every object $I$, captures all the information about the internal category for Yoneda-like reasons. The interesting thing is that you can end up in some cases with isomorphisms of indexed categories/fibrations when the internal categories they came from were equivalent but not isomorphic! But most things work out very nicely.

John Baez (Oct 17 2024 at 15:44):

Wow, thanks - that's really cool. I never knew about this, even though I'm used to "externalizing" an internal thing by looking at homs from a "test object" $I$ to that internal thing, and how letting $I$ range over all objects gives us all possible information about our internal thing.

John Baez (Oct 17 2024 at 15:46):

Is there a reverse process, where given an indexed category $B^{\text{op}} \to \mathbf{Cat}$ you can get from it a category internal to $B$... when some conditions hold?

James Deikun (Oct 17 2024 at 15:48):

Yes! But the conditions for that one are pretty hard to describe, especially the "local smallness" condition--I can't just rattle them off.

John Baez (Oct 17 2024 at 15:52):

That's fine - even if you did I would probably not follow them or remember them! It's just nice to know people have worked that out.

James Deikun (Oct 17 2024 at 15:55):

Ah, if you're interested in the details [[locally internal category]] and [[small fibration]] have some of them, though neither has information on the "generic object" condition.

Fernando Yamauti (Oct 17 2024 at 15:57):

Any split cartesian fibration with small fibers gives a strict 2-sheaf (assuming your base is a topos with the canonical top). But any sheaf valued in models of an algebraic theory gives an internal model inside the topos. Isn’t it just that?

Fernando Yamauti (Oct 17 2024 at 15:59):

And any cartesian fibration is equivalent to a split one, so the unique problem is the size. But that can be solved by enlarging the base topos to a bigger universe (by taking sheaves valued in bigger sets)

Fernando Yamauti (Oct 17 2024 at 16:20):

Does the base change functor of cartesian fibrations always preserve limits? I think that's true for right fibrations, but I'm unsure about cartesian fibrations. If that's the case, you will likely get a left adjoint to the base-change.

Daniel Gratzer (Oct 17 2024 at 16:36):

Fernando Yamauti said:

Does the base change functor of cartesian fibrations always preserve limits? I think that's true for right fibrations, but I'm unsure about cartesian fibrations. If that's the case, you will likely get a left adjoint to the base-change.

Up to strictness issues (solved by say, by treating everything as (2,1)-categories), pullback of cartesian fibrations has both a left and right adjoint: by straightening one has $\mathrm{cart}(\mathcal{C}) \simeq \mathrm{Fun}(\mathcal{C}^{\mathrm{op}},\mathrm{Cat})$ and pullback along $f$ corresponds under this equivalence to precomposition, which has left and right adjoints given by Kan extension.

James Deikun (Oct 17 2024 at 16:54):

Hm. Does pushforward of an internal category along a pullback-preserving functor really correspond to left Kan extension of indexed categories? If so it's doing a lot of work, taking all these coends in $\mathrm{Cat}$! It also feels like a left adjoint to pullback/precomposition doesn't involve preservation of pullbacks at all? But when a functor doesn't preserve pullbacks, it doesn't seem like it does anything straightforward at all to internal categories.

Fernando Yamauti (Oct 17 2024 at 17:02):

Daniel Gratzer said:

Fernando Yamauti said:

Does the base change functor of cartesian fibrations always preserve limits? I think that's true for right fibrations, but I'm unsure about cartesian fibrations. If that's the case, you will likely get a left adjoint to the base-change.

Up to strictness issues (solved by say, by treating everything as (2,1)-categories), pullback of cartesian fibrations has both a left and right adjoint: by straightening one has $\mathrm{cart}(\mathcal{C}) \simeq \mathrm{Fun}(\mathcal{C}^{\mathrm{op}},\mathrm{Cat})$ and pullback along $f$ corresponds under this equivalence to precomposition, which has left and right adjoints given by Kan extension.

Hmm...Maybe I'm confusing it with the (2,2)- case. But I recall there was a problem with exponentiability (the right adjoint).

Daniel Gratzer (Oct 17 2024 at 17:08):

James Deikun said:

Hm. Does pushforward of an internal category along a pullback-preserving functor really correspond to left Kan extension of indexed categories? If so it's doing a lot of work, taking all these coends in $\mathrm{Cat}$! It also feels like a left adjoint to pullback/precomposition doesn't involve preservation of pullbacks at all? But when a functor doesn't preserve pullbacks, it doesn't seem like it does anything straightforward at all to internal categories.

Sorry I should say, I'm not sure how this relates to the pushforward of internal categories: this is just a description of the left and right adjoints to the pullback of cartesian fibrations. It could be that when one restricts to the subcategories of cartesian fibrations corresponding to internal categories, things are different.

Daniel Gratzer (Oct 17 2024 at 17:11):

Fernando Yamauti said:

Daniel Gratzer said:

Fernando Yamauti said:

Does the base change functor of cartesian fibrations always preserve limits? I think that's true for right fibrations, but I'm unsure about cartesian fibrations. If that's the case, you will likely get a left adjoint to the base-change.

Up to strictness issues (solved by say, by treating everything as (2,1)-categories), pullback of cartesian fibrations has both a left and right adjoint: by straightening one has $\mathrm{cart}(\mathcal{C}) \simeq \mathrm{Fun}(\mathcal{C}^{\mathrm{op}},\mathrm{Cat})$ and pullback along $f$ corresponds under this equivalence to precomposition, which has left and right adjoints given by Kan extension.

Hmm...Maybe I'm confusing it with the (2,2)- case. But I recall there was a problem with exponentiability (the right adjoint).

I think you're correctly recalling that $\mathrm{Cat}$ doesn't have right adjoints to pullback in general, but this is a weaker observation: we're only describing how to pushforward cartesian fibrations and this is only possible because we can straighten those.

Fernando Yamauti (Oct 17 2024 at 17:17):

Daniel Gratzer said:

Fernando Yamauti said:

Daniel Gratzer said:

Fernando Yamauti said:

Does the base change functor of cartesian fibrations always preserve limits? I think that's true for right fibrations, but I'm unsure about cartesian fibrations. If that's the case, you will likely get a left adjoint to the base-change.

Up to strictness issues (solved by say, by treating everything as (2,1)-categories), pullback of cartesian fibrations has both a left and right adjoint: by straightening one has $\mathrm{cart}(\mathcal{C}) \simeq \mathrm{Fun}(\mathcal{C}^{\mathrm{op}},\mathrm{Cat})$ and pullback along $f$ corresponds under this equivalence to precomposition, which has left and right adjoints given by Kan extension.

Hmm...Maybe I'm confusing it with the (2,2)- case. But I recall there was a problem with exponentiability (the right adjoint).

I think you're correctly recalling that $\mathrm{Cat}$ doesn't have right adjoints to pullback in general, but this is a weaker observation: we're only describing how to pushforward cartesian fibrations and this is only possible because we can straighten those.

Ah. Ok. Thanks. Yes. I've just checked. Fibrations are exponentiable (in any (2,2)-topos).

Fernando Yamauti (Oct 17 2024 at 17:35):

James Deikun said:

Hm. Does pushforward of an internal category along a pullback-preserving functor really correspond to left Kan extension of indexed categories? If so it's doing a lot of work, taking all these coends in $\mathrm{Cat}$! It also feels like a left adjoint to pullback/precomposition doesn't involve preservation of pullbacks at all? But when a functor doesn't preserve pullbacks, it doesn't seem like it does anything straightforward at all to internal categories.

If you're willing to internalise inside presheaves instead of a small cats, the pullback of internal categories should be the pullback of fibrations. I'm just worried whether the pullback of a split fibration will still be split... up to equivalence we are fine I guess... but up to isos...

James Deikun (Oct 18 2024 at 08:43):

I still feel like this is an open question. When you have a functor $F : \mathcal{B \to B'}$ between finitely complete categories preserving pullbacks, you get a small fibration over $\mathcal{B'}$ from one over $\mathcal{B}$. Nothing needs to be a topos, and there's no relation that's been explicated between the original fibration over $\mathcal{B}$ and the one you would get by pulling back the one over $\mathcal{B}'$ along $F$. I'm interested in specifically extending this way of getting a fibration, not any unrelated way to get a fibration on $\mathcal{B}'$. I'm afraid if there's an answer to this in the above it completely flew over my head.

Rémy Tuyéras (Oct 18 2024 at 11:38):

It looks like you are dealing with a factorization problem that attempts to capture "representable categories" simultaneously. I don't think I have a solution to your problem, but I wanted to throw out some ideas that might lead to a solution, depending on some reformulation of my statements below.

Please let me know if I am missing something here. As far as I understand, you have a functor $F:\mathcal{B} \to \mathcal{B}'$ and a fibration $G:\mathcal{E} \to \mathcal{B}$. This fibration seems to have the following property:

$G^{-1}(I) \cong (\mathcal{B}(I,C_1)\rightrightarrows \mathcal{B}(I,C_0))$

where $C:T_{\mathsf{cat}} \to \mathcal{B}$ is your internal category. When you compose your fibration $G$ with the functor $F$, you seem to want a factorization:

$\mathcal{E} \mathop{\to}\limits^{F'} \mathcal{E}' \mathop{\to}\limits^{G'} \mathcal{B}'$

of $F \circ G$ with the following property:

${G'}^{-1}(J) \cong (\mathcal{B}'(J,F(C_1))\rightrightarrows \mathcal{B}'(J,F(C_0)))$.

However, what you start with is an isomorphism as follows for every $I \in F^{-1}(J)$:

${G}^{-1}(I) \cong (\mathcal{B}(I,C_1)\rightrightarrows \mathcal{B}(I,C_0))$.

So maybe you could first construct (note that this might not be well-defined):

$H_J := \int^I\coprod_{I \in F^{-1}(J)} {G}^{-1}(I)$.

This might not give you a functor $H:{\mathcal{B}'}^{\mathsf{op}} \to \mathbf{Cat}$ (this is the point of the factorization I mentioned above), but you could investigate ways to turn these colimits into such a functor $G':{\mathcal{B}'}^{\mathsf{op}}\to \mathbf{Cat}$, which would then give you a fibration $G':\mathcal{E}' \to \mathcal{B}'$. While doing this, you might need to control how the colimit:

$\int^I\coprod_{I \in F^{-1}(J)} (\mathcal{B}(I,C_1)\rightrightarrows \mathcal{B}(I,C_0))$

"converges" toward a category of the following form:

$\mathcal{B}'(J,F(C_1))\rightrightarrows \mathcal{B}'(J,F(C_0))$.

One variant I think is worth considering is when you define:

$H_J := \int^I\coprod_{F^{-1}(J)\downarrow I} {G}^{-1}(I)$.

Then I believe you have a functor $H:{\mathcal{B}'}^{\mathsf{op}} \to \mathbf{Cat}$, and the representability problem mentioned above becomes:

$\int^I\coprod_{F^{-1}(J)\downarrow I} \mathcal{B}(I,C_t)\cong \mathcal{B}'(J,F(C_t))$.

Fernando Yamauti (Oct 18 2024 at 11:39):

(deleted)

Fernando Yamauti (Oct 18 2024 at 11:47):

Ah! My bad. Now I've seen you want to push foward a fibration instead of pulling back an internal cat along a lex functor

Fernando Yamauti (Oct 18 2024 at 12:08):

Sorry for not reading properly. As long everything is split, you can work with presheaves. Then applying your $F$ to an internal category is exactly applying $F_!$ to an internal cat such that objects and arrows are representable presheaves. $F^*$ applied to internal cats will correspond to the pullback of fibrations. I don't know how to do that with non-split fibrations though.

Kevin Carlson (Oct 18 2024 at 13:04):

I think James was complaining that it seems intuitively surprising that something as simple as composing with a left exact functor, for internal categories in the base category, turns into something as relatively complicated as left Kan extension when written in terms of presheaves, and that it's not obvious how left exactness interacts with this result; can you comment on either of those questions?

Fernando Yamauti (Oct 18 2024 at 18:19):

Kevin Carlson said:

I think James was complaining that it seems intuitively surprising that something as simple as composing with a left exact functor, for internal categories in the base category, turns into something as relatively complicated as left Kan extension when written in terms of presheaves, and that it's not obvious how left exactness interacts with this result; can you comment on either of those questions?

I think it's because $F_! (h_X) = h_{F(X)}$.

Kevin Carlson (Oct 18 2024 at 18:21):

Right, and then something about left Kan extension along left exact functors preserving left exact functors?

Fernando Yamauti (Oct 18 2024 at 18:25):

Well. The lex is only required because $F$ applied to a cat need not to be a cat in general. Everything seems to work well if we substitute the theory of cats by any essentially algebraic theory.

Fernando Yamauti (Oct 18 2024 at 18:28):

Btw, sorry everybody for the sloppy writing. My communication skills are not that good, but I was also typing from my phone all this time.