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Stream: learning: questions

Topic: fibrations of EI categories

Joe Moeller (Nov 03 2021 at 00:35):

A category where every endomorphism is an isomorphism, or equivalently where every map between isomorphic objects is invertible, is called an [[EI category]]. Fix an EI category X. I want to know exactly which indexed categories $F \colon X^{op} \to \mathsf{Cat}$ correspond to fibrations over X where the total category is also EI.

An endomorphism $(f,k) \colon (x,a) \to (x,a)$ in the Grothendieck construction of F has components $f \colon x \to x$ in X, and $k \colon a \to Ff(a)$ in Fx. Since f is an endo in X, it is invertible. So (f,k) is invertible if and only if k is invertible. This will be true if Fx is an EI category and Ff(a) happens to be isomorphic to a in Fx.

So here's my question: given a fibration $P \colon A \to X$ where both categories are EI, an endo $f \colon x \to x$ in X, and an object a in the fiber over x, how do you show a is isomorphic to $f^*a$? Or maybe I mixed something up and this isn't what I want? Note that you can show the fibers are EI quickly.

Reid Barton (Nov 03 2021 at 01:50):

Consider the functor $F : B\mathbb{Z} \to \mathrm{Cat}$ given by the (strict) action of the group $\mathbb{Z}$ by translation on the poset $\mathbb{Z}$ (with the usual order). The single fiber $F(*)$ is EI (it's the poset $\mathbb{Z}$) but I think the total category is not EI (it's equivalent to $B\mathbb{N}$, if I'm not mistaken).

Joe Moeller (Nov 03 2021 at 12:51):

Thanks for the example. That seems not have the property that Ffa is isomorphic to a.

Joe Moeller (Nov 03 2021 at 15:44):

So my question still stands. The cartesian lift of f is a map with the right co/domain, but it's obviously not in the fiber.

Reid Barton (Nov 03 2021 at 15:49):

I think I don't understand the question, or maybe the terminology.

Reid Barton (Nov 03 2021 at 15:50):

Joe Moeller said:

I want to know exactly which indexed categories $F \colon X^{op} \to \mathsf{Cat}$ correspond to fibrations over X where the total category is also EI.

This part I understand. Do you have a guess for what the answer should be?

Joe Moeller (Nov 03 2021 at 15:54):

Yeah, my guess is it should be an indexed category on X where the fibers are EI and Ff(a) is isomorphic to a when f is an endomorphism. In fact, this seems somewhat obviously true in the indexed category perspective. But I'm having trouble seeing the equivalent thing on the fibration side.

Joe Moeller (Nov 03 2021 at 15:56):

I may be making some fatal slip that's making it so I can't see why what I'm saying doesn't make sense.

Reid Barton (Nov 03 2021 at 15:58):

The condition that Ff(a) should be isomorphic to a without having any prior map between these objects seems strange to me.

Reid Barton (Nov 03 2021 at 15:58):

What about the following example: Like above, except $F(*)$ is just the set $\mathbb{Z}$ (with no non-identity morphisms).

Reid Barton (Nov 03 2021 at 16:00):

Then everything in sight is a groupoid, so definitely EI.

John Baez (Nov 03 2021 at 16:05):

The condition that Ff(a) should be isomorphic to a without having any prior map between these objects seems strange to me.

Yes, my categorical "spidey sense" says the condition should instead be that some morphism we already have is an isomorphism.

Joe Moeller (Nov 03 2021 at 16:16):

Yes, that's essentially what I'm asking for. What map should it be though?

John Baez (Nov 03 2021 at 16:28):

I don't know what's going on - I'm not really thinking about this stuff, just following gut instinct. But I think Reid meant that this sounds funny:

So (f,k) is invertible if and only if k is invertible. This will be true if Fx is an EI category and Ff(a) happens to be isomorphic to a in Fx.

I don't see how Ff(a) being isomorphic to a can help you show k is invertible, unless you mean isomorphic via some specific isomorphism.

So, how are you showing

Fx is an EI category and Ff(a) $\cong$ a $\qquad \implies \qquad$ k is invertible?

Mike Shulman (Nov 03 2021 at 16:50):

It made my spidey sense tingle too, but then I realized that it makes sense. The point is that in an EI category, you can show that a morphism is invertible by showing that it is an endomorphism... or by showing that its domain and codomain are isomorphic! Because if $k:u\to v$ is a morphism and there is any isomorphism $\ell : u\cong v$, then $\ell^{-1} k$ is an endomorphism, hence invertible, and thus so is $k$.

Mike Shulman (Nov 03 2021 at 16:51):

Put differently, the property of "being an endomorphism" is evil, and its isomorphism-invariant version is "having an isomorphic domain and codomain (by, of course, a specified isomorphism)".

Joe Moeller (Nov 03 2021 at 16:55):

I said at the top that being EI is equivalent to having the property that maps between isomorphic objects are invertible.

Reid Barton (Nov 03 2021 at 16:55):

I wondered if it was something like this, but the example I gave shows that the guess is wrong anyways, right?

Joe Moeller (Nov 03 2021 at 16:56):

I don't think your example has the property I guessed.

Mike Shulman (Nov 03 2021 at 16:57):

Joe Moeller said:

I said at the top that being EI is equivalent to having the property that maps between isomorphic objects are invertible.

That's right, you did!

Reid Barton (Nov 03 2021 at 16:57):

Right, but the total category is EI nevertheless

Joe Moeller (Nov 03 2021 at 16:57):

Oh, you said the total category was not EI.

Reid Barton (Nov 03 2021 at 16:57):

I mean the second example

Joe Moeller (Nov 03 2021 at 16:58):

Oh whoops, I missed the second example.

Mike Shulman (Nov 03 2021 at 16:58):

So it looks like your condition is sufficient but not necessary for the total category to be EI.

Joe Moeller (Nov 03 2021 at 16:59):

So it's still BZ acting on Z by translation? That still has $Ff(a) \not \cong a$, right?

Joe Moeller (Nov 03 2021 at 17:00):

Oh yeah, I see that's the point.

Mike Shulman (Nov 03 2021 at 17:01):

I wonder if conservativity of the fibration, as a functor, could be relevant.

Reid Barton (Nov 03 2021 at 17:04):

I usually think of EI categories instead as categories that admit a conservative functor to a poset. But there is a subtlety here because the poset is not necessarily uniquely determined by the category, in cases like this one where the category has multiple connected components. And if two objects of the category are isomorphic then they must live above the same element of the poset, but the converse isn't necessarily true. This might also be related to Mike's most recent comment.

Reid Barton (Nov 03 2021 at 17:11):

There's always a "finest" poset to which an EI category $C$ admits a conservative functor, in which two objects of $C$ become equal if and only if they were isomorphic in $C$. Your condition "$Ff(a) \cong a$" is saying that the action of isomorphisms in the indexing category needs to leave objects of $C = F(*)$ lying over the same element of this poset. This isn't true in the second example I gave but it would become true if instead you regarded $C$ as equipped with a conservative functor to the poset $P = *$.

Mike Shulman (Nov 03 2021 at 17:36):

Ok, any category that admits a conservative functor to an EI category is EI. So conservativity of the fibration is also a sufficient condition. But it's not necessary either, since for example the functor from any poset to 1 is not conservative but its domain is EI.

Joe Moeller (Nov 03 2021 at 18:03):

So I think there's some sort of quantification problem with my original guess. Let F be an indexed category with $\int F$ an EI category. Let $f \colon x\to x$ be an endo in X. It's an iso. Do we get $Ff(a) \cong a$ for all a? An endo looks like $(f,k) \cong (x,a) \to (x,a)$ with $k \colon a\to Ff(a)$. Since $\int F$ is EI, then (f,k) is invertible, so k is invertible. But such a k doesn't always exist. There may be no maps $a \to Ff(a)$ in Fx. But if there is, it must be an isomorphism :thinking:

Joe Moeller (Nov 03 2021 at 18:15):

I'd like to think that a common refinement of my guess and the conservative condition would be a good step forward, but I don't know what that would be. Some inkling is telling me these are already very closely related.

Joe Moeller (Nov 03 2021 at 19:18):

Oh, I feel like I should my new official guess. Let X be an EI category, $F\colon X^{op} \to \mathsf{Cat}$ an indexed category. Then $\int F$ is EI if and only if all Fx are EI and for an endo $f\colon x \to x$ and object a in the fiber over x, every map $a\to Ff(a)$ in Fx is an iso.

Mike Shulman (Nov 03 2021 at 20:02):

That new guess sounds plausible to me.

Joe Moeller (Nov 03 2021 at 20:44):

Yeah, works perfect.