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Stream: theory: category theory

Topic: Slice categories of slice categories

Jonathan Beardsley (Jul 29 2024 at 22:38):

Let f:X→Y be a morphism in some sufficiently nice category-like thing C (I'm thinking about quasicategories actually, but a sufficiently general proof should suffice for whatever context). I am trying to show that the slice category C\f is equivalent to (C\X)\f (where in the latter I am thinking about f as an object of the slice category C\X). Both of these things can be produced as 2-pullbacks, but I can't quite figure out how to produce even a functor between them for "universal reasons," much less an equivalence. What is the 2-categorical/purely diagrammatic proof that these two things are equivalent?

Jonathan Beardsley (Jul 29 2024 at 22:38):

I mean I guess just take C to be an object in an ∞-cosmos. The proof should be the same.

Kevin Carlson (Jul 29 2024 at 22:41):

I guess it's $C/X/f = C/\mathrm{dom}(f)$ where $\mathrm{cod}(f)=X,$ to be sure?

Jonathan Beardsley (Jul 29 2024 at 22:42):

Just checking, you're doing the opposite case?

Jonathan Beardsley (Jul 29 2024 at 22:42):

I'm trying to show $(C^{\backslash X})^{\backslash f}\simeq C^{\backslash f}$

Kevin Carlson (Jul 29 2024 at 22:43):

Uh hmm, where I'm from the "slice" has objects $A\to X$, I don't love that Lurie notation

Kevin Carlson (Jul 29 2024 at 22:43):

You want objects $X\to A$? Not that it'll be a huge deal

Jonathan Beardsley (Jul 29 2024 at 22:43):

Oh maybe I meant coslice then.

Jonathan Beardsley (Jul 29 2024 at 22:43):

I can't keep it straight.

Kevin Carlson (Jul 29 2024 at 22:43):

sure sure

Kevin Carlson (Jul 29 2024 at 22:44):

But then also what exactly is $C^{\backslash f}$?

Jonathan Beardsley (Jul 29 2024 at 22:44):

Roughly, cocones on $f$

Jonathan Beardsley (Jul 29 2024 at 22:45):

I think I'm probably even screwing up the Lurie notation. I just mean "things under $f$."

Joe Moeller (Jul 29 2024 at 22:45):

specifically the comma on $(\cdot\to\cdot) \xrightarrow f C$ and $id_C$.

Jonathan Beardsley (Jul 29 2024 at 22:45):

Or the comma category

Jonathan Beardsley (Jul 29 2024 at 22:45):

Yeah.

Kevin Carlson (Jul 29 2024 at 22:46):

Right right, and aren't these things both probably the same as $C^{\backslash Y}$?

Jonathan Beardsley (Jul 29 2024 at 22:46):

So you could produce it as the 2-pullback $\{f\}\to C^{[1]}\leftarrow C$

Jonathan Beardsley (Jul 29 2024 at 22:46):

Kevin Carlson said:

Right right, and aren't these things both probably the same as $C^{\backslash Y}$?

Ah okay... right. Yes. So $C^{\backslash f}$ is definitely equivalent to $C^{\backslash Y}$.

Jonathan Beardsley (Jul 29 2024 at 22:47):

This notation that I somehow have ended up using is awful because you have to constantly type \backslash

Kevin Carlson (Jul 29 2024 at 22:47):

I personally adore $X\downarrow C$

Jonathan Beardsley (Jul 29 2024 at 22:47):

It's so wide though! =P

Kevin Carlson (Jul 29 2024 at 22:48):

eh, screens have scrollbars :)

Jonathan Beardsley (Jul 29 2024 at 22:48):

Anyway, yeah, so I know how to reduce this to showing then that $C^{\backslash Y}$ is equivalent to $(C^{\backslash X})^{\backslash f}$

Jonathan Beardsley (Jul 29 2024 at 22:49):

But I don't immediately see how to do that either.

Jonathan Beardsley (Jul 29 2024 at 22:49):

In both cases it's pretty much "obvious."

Jonathan Beardsley (Jul 29 2024 at 22:51):

I mean I guess it's a little bit ridiculous that I'm even trying to write out a proof of this. I just hate when people aren't careful, and it seems like the ∞-cosmos framework should basically reduce this to a "standard" 2-category argument.

Jonathan Beardsley (Jul 29 2024 at 22:52):

I've been trying to write these both down as (the same) full subcategories of $C^{\Delta^2}$, without much success.

Kevin Carlson (Jul 29 2024 at 23:01):

I'm having some success defining the equivalences using the universal property of the comma in the homotopy 2-category

Kevin Carlson (Jul 29 2024 at 23:02):

Haven't got so far as checking the equivalences yet but would it be useful to send some Quivers of the definitions I'm writing down?

Jonathan Beardsley (Jul 29 2024 at 23:02):

Sure, I'd love to see it!

Jonathan Beardsley (Jul 29 2024 at 23:03):

That's exactly the sort of thing I couldn't get. There clearly should be some "universal maps" between these commas.

Jonathan Beardsley (Jul 29 2024 at 23:04):

I was also trying to do it by using the join/slice adjunction....

Jonathan Beardsley (Jul 29 2024 at 23:04):

That's the closest I got to actually having a proof, since that's the closest to having actual "diagrams" that I can visualize, as simplices in some simplicial set. But that feels ugly, haha.

Kevin Carlson (Jul 29 2024 at 23:08):

Screenshot-2024-07-29-at-4.08.01PM.png
So here's the first map

Kevin Carlson (Jul 29 2024 at 23:08):

The first diagram is the definition of $f\downarrow X\downarrow C.$

Kevin Carlson (Jul 29 2024 at 23:08):

The second diagram is the upper-left square of the first, whiskered with $\pi_2.$

Kevin Carlson (Jul 29 2024 at 23:08):

This is the data you need for a map into $Y\downarrow C$!

Kevin Carlson (Jul 29 2024 at 23:09):

And we can hopefully see it's the right map.

Kevin Carlson (Jul 29 2024 at 23:10):

For the map in the other direction there are more steps. You need a map $\beta:Y\downarrow C\to X\downarrow C$ first, which you get by gluing the 2-cell for $f$ onto the square defining $Y\downarrow C.$ Then you need to define a map into $\beta$ from the constant map $Y\downarrow C \to X\downarrow C$ at $[f].$

Kevin Carlson (Jul 29 2024 at 23:13):

But we have a universal property for 2-morphisms into $X\downarrow C$ too; the two given morphisms $Y\downarrow C\rightrightarrows X\downarrow C$ are defined by these diagrams:
Screenshot-2024-07-29-at-4.13.48PM.png

Kevin Carlson (Jul 29 2024 at 23:14):

To get a map from the map induced by the left to the map induced by the right diagram, you just need 2-morphisms $Y\Rightarrow \pi_2$ and $(Y\downarrow C\to \bullet)\to (Y\downarrow C\to \bullet)$ making a certain square commute in $C...$

Kevin Carlson (Jul 29 2024 at 23:15):

And the only data that needs is the 2-morphism $\pi_3: Y\Rightarrow \pi_2$ that's part of the definition of $Y\downarrow C$!

Jonathan Beardsley (Jul 29 2024 at 23:15):

Haha, okay, this looks great....now I will begin trying to decode/digest/rewrite.

Jonathan Beardsley (Jul 29 2024 at 23:20):

It does feel like this should be doable purely by gluing the pullbacks themselves.
Screenshot-2024-07-29-at-4.20.40PM.png.

Jonathan Beardsley (Jul 29 2024 at 23:21):

Which is how I was trying to do it....

Kevin Carlson (Jul 29 2024 at 23:23):

"The pullbacks themselves" as in, the strict pullbacks in the simplicially enriched guy?

Kevin Carlson (Jul 29 2024 at 23:23):

I dunno, I suppose one could think of the strict guys as ontologically prior... :innocent:

Kevin Carlson (Jul 29 2024 at 23:24):

I'm dubious about literally gluing but I do think you should be able to define maps similarly via the strict universal properties, if that's the kind of thing you're into. Avoids some business about smothering anyway.

Jonathan Beardsley (Jul 29 2024 at 23:24):

Oh god. Smothering.

Jonathan Beardsley (Jul 29 2024 at 23:24):

:evil_eye:

Kevin Carlson (Jul 29 2024 at 23:24):

it's not that baaaaaad

Jonathan Beardsley (Jul 29 2024 at 23:32):

But, uh, to be clear....does your argument use "smothering" in some place that you just didn't mention explicitly?

Jonathan Beardsley (Jul 29 2024 at 23:32):

Because I honestly don't even know what smothering is.

Jonathan Beardsley (Jul 29 2024 at 23:32):

Except a section in things written by Emily that I run away from.

Kevin Carlson (Jul 30 2024 at 00:02):

No, so the deal is that these comma objects are only limits in the sense that the map $(A\to X\downarrow C) \to [(A\to C)\downarrow (A\to \bullet)],$ which would be an isomorphism for a legitimate comma object, is instead only surjective on objects, full, and conservative. So I haven't said anything dishonest so far, because I've only used surjectivity (real actual surjectivity, not essential) and then fullness.

Kevin Carlson (Jul 30 2024 at 00:04):

Now, once you compose the two maps we defined earlier you should get endomorphisms of $f\downarrow X\downarrow C$ and of $Y\downarrow C$ such that the cones to which they correspond have all components identities. That's where smotheringness will become slightly annoying: a map into a smothering-limit induced by a cone whose components are identities doesn't quite have to be an identity!

Kevin Carlson (Jul 30 2024 at 00:04):

But it's easy to use fullness and surjectivity to show that such a map is isomorphic to the identity, and so smothering-limits are still unique up to equivalence, which is usually good enough.