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Stream: learning: questions

Topic: computing 2-colimits

sarahzrf (Nov 12 2020 at 20:20):

so i asked about computing a 2-limit, now i want to know about computing 2-colimits. for mostly unrelated reasons, even!

sarahzrf (Nov 12 2020 at 20:22):

i've found on the nlab that you can compute 2-colimits in Cat that are indexed by 1-categories by taking a grothendieck construction of the diagram and then localizing at the opcartesian morphisms

sarahzrf (Nov 12 2020 at 20:23):

but i don't know all that much about localizations—i know the idea with reflective localizations and ive read a tiny bit about calculi of fractions, but i dont think either of those is gonna tend to be helpful here

sarahzrf (Nov 12 2020 at 20:23):

are there specific useful presentations of the localizations that arise in this situation?

sarahzrf (Nov 12 2020 at 20:25):

for the record: the particular 2-colimit i'm interested in computing at the moment is—let Fin be a skeleton of FinSet (im not super invested in it being a skeleton but that seems convenient); then i want to know the 2-colimit of the diagram

$$\mathrm{Fin} \xrightarrow{{\bullet} + \mathbf{1}} \mathrm{Fin} \xrightarrow{{\bullet} + \mathbf{1}} \mathrm{Fin} \xrightarrow{{\bullet} + \mathbf{1}} \cdots$$

Reid Barton (Nov 12 2020 at 20:27):

since it's a filtered colimit, life is a lot simpler

sarahzrf (Nov 12 2020 at 20:27):

(wondering if it might be some sort of interesting categorification of Z—alternatively, the permutation groupoid might be appropriate to try instead of Fin)

sarahzrf (Nov 12 2020 at 20:29):

Reid Barton said:

since it's a filtered colimit, life is a lot simpler

what do filtered 2-colimits look like?

Reid Barton (Nov 12 2020 at 20:29):

and especially since you've set it up as a colimit along functors which are injective on objects

Reid Barton (Nov 12 2020 at 20:29):

let me remember what the right thing to say here is

sarahzrf (Nov 12 2020 at 20:31):

...hmmm, for the permutation groupoid, i wonder if it would result in, like... Z-many copies of an object w/ an automorphism group that's some sort of finite-support thing

Reid Barton (Nov 12 2020 at 20:32):

So one way to say it is that mapping from the categories $[0] = \{*\}$, $[1] = \{* \to *\}$, ... will commute with this colimit because they're finite and the colimit is filtered.

Reid Barton (Nov 12 2020 at 20:32):

and so the objects are just the colimit of the objects

sarahzrf (Nov 12 2020 at 20:33):

ooh

sarahzrf (Nov 12 2020 at 20:33):

filtered 2-colimits commute w/ finite 2-limits?

Reid Barton (Nov 12 2020 at 20:33):

the maps between two objects, well those objects both appeared at some stage and then the maps between their images in every subsequent stage form a new directed colimit with transition maps given by the action of the functors on maps, and the maps in the colimit is the colimit of that diagram

sarahzrf (Nov 12 2020 at 20:33):

oh wait no sorry misspoke

sarahzrf (Nov 12 2020 at 20:34):

meant: finite categories are "2-compact"?

Reid Barton (Nov 12 2020 at 20:34):

Since your functors are injective on objects (= cofibrations in the canonical model structure on Cat) and the diagram is a sequential composition, the 2-colimit is also computed by the colimit in the 1-category Cat

sarahzrf (Nov 12 2020 at 20:34):

actually maybe i should just try thinking this through a bit lol

Reid Barton (Nov 12 2020 at 20:34):

that's what I'm using to avoid thinking about that question

sarahzrf (Nov 12 2020 at 20:35):

Reid Barton said:

Since your functors are injective on objects (= cofibrations in the canonical model structure on Cat) and the diagram is a sequential composition, the 2-colimit is also computed by the colimit in the 1-category Cat

god dammit i should probably learn some homotopy theory shouldn't i :weary:

Reid Barton (Nov 12 2020 at 20:35):

for a general directed diagram, I think it's easier to say when a cocone that you've guessed is a (2-)colimit

sarahzrf (Nov 12 2020 at 20:35):

ouch

Reid Barton (Nov 12 2020 at 20:37):

namely, $(C_i \to C)$ is a colimit (over a directed poset $I$) when:

• every object in $C$ is isomorphic to the image of some object in some $C_i$
• for any $x_i$, $y_i \in C_i$, let's write $x_j$, $y_j$ for their images in any $C_j$ for $i \le j$, and $x$, $y$ for their images in $C$; then $\mathrm{Hom}_C(x, y)$ should be the colimit of the $\mathrm{Hom}_{C_j}(x_j, y_j)$.

sarahzrf (Nov 12 2020 at 20:39):

i wanna make some kind of joke about guess-and-check and integration but i guess integral signs are for oplax colimits not pseudo colimits

Reid Barton (Nov 12 2020 at 20:39):

but this latter colimit is also filtered, so it just means that

• every map in $\mathrm{Hom}_C(x, y)$ is the image of a map in some $\mathrm{Hom}_{C_j}(x_j, y_j)$,
• whenever two maps in $\mathrm{Hom}_{C_i}(x_i, y_i)$ become equal in $\mathrm{Hom}_C(x, y)$, they already became equal in some $\mathrm{Hom}_{C_j}(x_j, y_j)$. "every equation appears at some finite stage"

Reid Barton (Nov 12 2020 at 20:40):

I think you could probably read this as a formula as well, I just find this description more useful sometimes.

Reid Barton (Nov 12 2020 at 20:41):

I like this question by the way, and I think I know what the answer is in the (FinSet, iso) case but I won't spoil it.

sarahzrf (Nov 12 2020 at 20:45):

well, i have a phone call in 15 minutes, so we'll see if i figure it out before getting sidetracked enough to forget about this for like 3 days

Reid Barton (Nov 12 2020 at 20:45):

The non-answer summary of the above is that you could take the objects to be the disjoint union of all the object sets, and then you get a map from $x \in C_i$ to $y \in C_j$ for every map between their images in $C_k$ for $k \ge \max\{i,j\}$, but identify two of these whenever they're carried to the same map in yet another $C_l$

sarahzrf (Nov 12 2020 at 20:46):

hmmm

sarahzrf (Nov 12 2020 at 20:47):

that's like uhhh

sarahzrf (Nov 12 2020 at 20:47):

when you take a filtered colimit of setoids

sarahzrf (Nov 12 2020 at 20:47):

or sth

sarahzrf (Nov 12 2020 at 20:48):

i'm bullshitting, i dunno what kind of colimit of setoids that would be, it might be weak somehow—really i mean it's notionally what happens when you take a filtered colimit of sets

Reid Barton (Nov 12 2020 at 20:50):

right, that's because any kind of finitary algebraic structure will pass through a filtered colimit of sets

sarahzrf (Nov 12 2020 at 20:50):

as far as im concerned every filtered colimit is a stalk

sarahzrf (Nov 12 2020 at 20:51):

well, i mean, you can probably make that literally true in a boring way by using the ind-completion of the diagram as a site or something...

sarahzrf (Nov 12 2020 at 20:51):

whatever

Reid Barton (Nov 12 2020 at 20:58):

The point about the filtered colimit is that normally when you compute, say, a pushout, it could happen that a morphism in one of the categories becomes composable with a morphism in the other category; and then in the pushout their composition is a morphism that didn't exist before

Reid Barton (Nov 12 2020 at 20:58):

and the same thing can happen with an equation in one category and a morphism in the other

Reid Barton (Nov 12 2020 at 20:59):

but when the diagram is filtered, you know stuff like that already got sorted out at some finite stage

Reid Barton (Nov 12 2020 at 21:00):

Anyways, if I'm remembering correctly, the answer for (FinSet, iso) is something interesting that probably wouldn't be one of your first few guesses.

Reid Barton (Nov 12 2020 at 21:00):

So if you get stuck, ask me about it again some time.

Reid Barton (Nov 12 2020 at 21:08):

Oh, I think I was thinking of something slightly different

John Baez (Nov 12 2020 at 21:25):

sarahzrf said:

for the record: the particular 2-colimit i'm interested in computing at the moment is—let Fin be a skeleton of FinSet (im not super invested in it being a skeleton but that seems convenient); then i want to know the 2-colimit of the diagram

$$\mathrm{Fin} \xrightarrow{{\bullet} + \mathbf{1}} \mathrm{Fin} \xrightarrow{{\bullet} + \mathbf{1}} \mathrm{Fin} \xrightarrow{{\bullet} + \mathbf{1}} \cdots$$

What do you guess the answer is? This is a nice question so it should have a nice answer, and it should be possible to guess it... and it's easier to compute things when you already have a guess as to the answer.

John Baez (Nov 12 2020 at 21:27):

So far Reid seems to have said it doesn't matter whether you're doing a 2-colimit or a 1-colimit.

sarahzrf (Nov 12 2020 at 21:27):

cmon i only just got back from my call

sarahzrf (Nov 12 2020 at 21:27):

give me a minute to work on this

John Baez (Nov 12 2020 at 21:40):

I meant: instead of computing it, guessing the answer. John Wheeler said "it's never good to do a computation unless you already know the answer" - and while he was exaggerating, I would never do a computation without first guessing the answer. I might have more than one guess. Then as I do the calculation I can see which guess is winning.

sarahzrf (Nov 12 2020 at 21:41):

i mean sure but in this case for once im actually not sure what im gonna get beyond "the objects are prolly gonna be Z" and a vague sense of flavor

John Baez (Nov 12 2020 at 21:42):

My first guess was "you get a category where the one objects is $\mathbb{N}$ and the morphisms are functions that equal the identity except on some finite initial segment". But then I decided that was screwed up.

sarahzrf (Nov 12 2020 at 21:43):

no spoilers!

sarahzrf (Nov 12 2020 at 21:43):

im still chewing on the more general info about filtered 2-colimits

Reid Barton (Nov 12 2020 at 21:43):

It's not a spoiler if it's a guess :upside_down:

John Baez (Nov 12 2020 at 21:43):

If "spoiler" means "crappy guess that spoils your ability to make a better guess", then maybe I just gave you a spoiler.

Reid Barton (Nov 12 2020 at 21:45):

I can't even figure out what question the thing I thought was going to be the answer is actually the answer to.

John Baez (Nov 12 2020 at 21:45):

Whew, that's a deep level of confusion. :upside_down:

sarahzrf (Nov 12 2020 at 21:46):

lmao

John Baez (Nov 12 2020 at 21:46):

Did you say what that thing is, or have you kept it secret to avoid spoiling Sarah's day?

John Baez (Nov 12 2020 at 21:46):

(I didn't read all the messages carefully.)

sarahzrf (Nov 12 2020 at 21:47):

he's been very courteous

sarahzrf (Nov 12 2020 at 21:49):

sarahzrf said:

meant: finite categories are "2-compact"?

wait if that is true then now i want to know the codensity monad (codensity 2-monad???) of FinCat's inclusion, or maybe just of the simplex category's (maybe theyre the same?)

Reid Barton (Nov 12 2020 at 21:49):

I didn't say what it was, but there's a certain theorem proven by guys with names like Barratt and Priddy and Quillen which I thought was going to be relevant.

Reid Barton (Nov 12 2020 at 21:50):

But now I'm not sure whether it is.

John Baez (Nov 12 2020 at 21:51):

I'm not even sure precisely what theorem you mean, but I was thinking about stuff vaguely like that.

John Baez (Nov 12 2020 at 21:52):

I used to know some "Barratt-Priddy theorem" or something. I probably still know what it says, just not which one is the Barratt-Priddy theorem.

John Baez (Nov 12 2020 at 21:55):

Okay, yeah - I looked it up, I know that thing.

John Baez (Nov 12 2020 at 21:55):

That's connected to the groupoid of finite sets.

sarahzrf (Nov 12 2020 at 21:57):

okay, i got it i think

Reid Barton (Nov 12 2020 at 21:57):

Right, that was a second version of the question which was also discussed.

sarahzrf (Nov 12 2020 at 21:57):

objects are Z

sarahzrf (Nov 12 2020 at 21:57):

a morphism n → m is...

sarahzrf (Nov 12 2020 at 21:58):

oh wait sorry the case i was tihnking about doesnt generalize obviously enough to constitute a full answer, one minute

sarahzrf (Nov 12 2020 at 21:58):

almost there though... let's see

sarahzrf (Nov 12 2020 at 21:59):

:sweat_smile:

sarahzrf (Nov 12 2020 at 22:05):

okay, i think you can present it as... a map N → N which is defined by x ↦ x + (m - n) at all but finitely many points

sarahzrf (Nov 12 2020 at 22:06):

but i wouldnt be surprised if that can be rephrased more nicely

Matteo Capucci (he/him) (Nov 12 2020 at 22:08):

If we go one level down and replace Fin with $\Bbb N$, then we'd get
$\Bbb N \times \Bbb N / \sim$, where $(h,k) \sim (n,m)$ iff $k = n + m - k$, i.e. $h+k=m+n$.

sarahzrf (Nov 12 2020 at 22:08):

yup, that's what i wastrying to categorify :>

Matteo Capucci (he/him) (Nov 12 2020 at 22:09):

Does this give you the integers? :thinking:

sarahzrf (Nov 12 2020 at 22:09):

yeah!

sarahzrf (Nov 12 2020 at 22:09):

famously so

Matteo Capucci (he/him) (Nov 12 2020 at 22:09):

shouldn't it be $h+m = n+k$?

sarahzrf (Nov 12 2020 at 22:10):

for the other, groupoidy version of the question, i think you should indeed get a groupoid with Z-many objects and where each one's automorphism group is the subgroup of Aut(N) on the permutations that fix cofinitely many elements

sarahzrf (Nov 12 2020 at 22:10):

oh wait i wasnt reading carefully lmao i just thought it looked about right

sarahzrf (Nov 12 2020 at 22:10):

i mean, this colimit does give you the integers

Reid Barton (Nov 12 2020 at 22:11):

Another way you could think of the maps is as maps $\{\ldots, n-1, n\} \to \{\ldots, m-1, m\}$ which fix all but finitely many values

sarahzrf (Nov 12 2020 at 22:11):

sarahzrf said:

for the other, groupoidy version of the question, i think you should indeed get a groupoid with Z-many objects and where each one's automorphism group is the subgroup of Aut(N) on the permutations that fix cofinitely many elements

which is approximately what i was trying to get at with
sarahzrf said:

...hmmm, for the permutation groupoid, i wonder if it would result in, like... Z-many copies of an object w/ an automorphism group that's some sort of finite-support thing

sarahzrf (Nov 12 2020 at 22:11):

:triumph:

sarahzrf (Nov 12 2020 at 22:12):

Reid Barton said:

Another way you could think of the maps is as maps $\{\ldots, n-1, n\} \to \{\ldots, m-1, m\}$ which fix all but finitely many values

right, i figured there was probably a way to translate things and make it nicer

Reid Barton (Nov 12 2020 at 22:23):

So I'm still a bit confused about what I was thinking about earlier, but it should have something to do with the fact that, though our original category FinSet (or (FinSet, iso)) was monoidal, this new category is not

Reid Barton (Nov 12 2020 at 22:23):

which is not surprising, since we inverted a functor $- \amalg 1$ which was not monoidal

sarahzrf (Nov 12 2020 at 22:25):

interesting

sarahzrf (Nov 12 2020 at 22:26):

sarahzrf said:

Reid Barton said:

Another way you could think of the maps is as maps $\{\ldots, n-1, n\} \to \{\ldots, m-1, m\}$ which fix all but finitely many values

right, i figured there was probably a way to translate things and make it nicer

ohhh right this is more evident if you put it as 1 + - instead of - + 1

sarahzrf (Nov 12 2020 at 22:26):

lol

Reid Barton (Nov 12 2020 at 22:28):

I actually thought about $\{-n, -n + 1, \ldots\}$ first and decided that was a bit silly

sarahzrf (Nov 12 2020 at 22:28):

yeah hah

John Baez (Nov 13 2020 at 00:04):

sarahzrf said:

Reid Barton said:

Another way you could think of the maps is as maps $\{\ldots, n-1, n\} \to \{\ldots, m-1, m\}$ which fix all but finitely many values

right, i figured there was probably a way to translate things and make it nicer

Is this the same or different from my guess: we get a category with one object $\mathbb{Z}$, and morphisms are functions that are the identity for all sufficiently large $n$?

John Baez (Nov 13 2020 at 00:06):

Actually my guess was a bit different:

John Baez said:

My first guess was "you get a category where the one objects is $\mathbb{N}$ and the morphisms are functions that equal the identity except on some finite initial segment".

John Baez (Nov 13 2020 at 00:06):

But I'm mainly not quite sure what category Reid is proposing.

sarahzrf (Nov 13 2020 at 00:13):

objects are sets (-∞, n] for n ∈ Z; maps as described

sarahzrf (Nov 13 2020 at 00:15):

if $n \neq m$ then $(-\infty, n] \not\cong (-\infty, m]$ i think

sarahzrf (Nov 13 2020 at 00:16):

so it's gonna be different from yours

John Baez (Nov 13 2020 at 00:18):

Oh, okay - lots of nonisomorphic objects.

John Baez (Nov 13 2020 at 00:18):

Let me think about that...

John Baez (Nov 13 2020 at 00:22):

So each functor in the diagram you're taking a 2-colimit of, which I'll call

$- + 1 : \mathsf{FinSet} \to \mathsf{FinSet}$

or

$- + 1 : \mathsf{Fin} \to \mathsf{Fin}$

is an equivalence onto its essential image. In other words, it's full and faithful. So there's no reason to think nonisomorphic objects are gonna become isomorphic in the 2-colimit.

I'll make a guess: given a filtered diagram of categories where all the maps are full and faithful, they all become full subcategories of the 2-colimit.

John Baez (Nov 13 2020 at 00:27):

I.e. they don't get "squashed down", with nonisomorphic objects becoming isomorphic in the 2-colimit, or unequal parallel morphisms becoming equal.

sarahzrf (Nov 13 2020 at 00:30):

plausible!

sarahzrf (Nov 13 2020 at 00:31):

wait, no

sarahzrf (Nov 13 2020 at 00:31):

faithful and/or conservative, maybe—but i certainly don't think we get full

sarahzrf (Nov 13 2020 at 00:33):

actually wait, im reasoning from a wrong example—i dont think that functor is fully faithful at all

sarahzrf (Nov 13 2020 at 00:33):

consider: there's exactly one map 0 → 1, but there are 2 maps 0 + 1 → 1 + 1

John Baez (Nov 13 2020 at 05:16):

Right, that functor is not full, just faithful. So whenever I said "full and faithful", maybe I should have said "full and conservative".

sarahzrf (Nov 13 2020 at 06:30):

...did you typo again, or did you just mean to make a statement that didn't apply to this situation?

John Baez (Nov 13 2020 at 07:25):

Typo: I meant conservative and faithful. Which are sort of like "0-injective" and "1-injective".