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Stream: theory: category theory

Topic: slice of abelian category

Matteo Capucci (he/him) (Mar 11 2024 at 13:20):

Is the slice ${\cal A}/X$ of an abelian category over any of its objects still Abelian? it seems no to me but i'd like to be reassured

Matteo Capucci (he/him) (Mar 11 2024 at 13:20):

the problem seems to be already at the level of 0-objects: $X=X$ is terminal and not initail in ${\cal A}/X$, so not a zero

Matteo Capucci (he/him) (Mar 11 2024 at 13:21):

but maybe i'm taking the slice in $\bf Cat$ and not in the right 2-category

Tim Hosgood (Mar 11 2024 at 14:02):

yeah, in this setting the initial object is $0\to X$, so unless you slice over the zero object then you'll already fail that axiom

Tim Hosgood (Mar 11 2024 at 14:02):

Matteo Capucci (he/him) said:

but maybe i'm taking the slice in $\bf Cat$ and not in the right 2-category

but this question is interesting

Tim Hosgood (Mar 11 2024 at 14:10):

what's true is that for any nice enough category $\mathcal{C}$ and any object $X\in\mathcal{C}$, the category $\mathrm{Ab}(\mathcal{C}/X)$ of abelian group objects in $\mathcal{C}/X$ is an abelian category

Tim Hosgood (Mar 11 2024 at 14:11):

so a sort of related question is maybe "are there any abelian categories $\mathcal{A}$ such that $\mathrm{Ab}(\mathcal{A}/X)$ is a strict subcategory of $\mathcal{A}/X$?"

Tim Hosgood (Mar 11 2024 at 14:11):

Tim Hosgood said:

what's true is that for any nice enough category $\mathcal{C}$ and any object $X\in\mathcal{C}$, the category $\mathrm{Ab}(\mathcal{C}/X)$ of abelian group objects in $\mathcal{C}/X$ is an abelian category

(the nlab says that "nice enough" = "effective regular", in [[Beck module]])

Matteo Capucci (he/him) (Mar 12 2024 at 13:56):

I had a more mundane thought, that the comma objects in the 2-category of abelian categories might not be the same as in Cat

Matteo Capucci (he/him) (Mar 12 2024 at 13:56):

(provided they exist)

Tim Hosgood (Mar 12 2024 at 15:20):

i don't know much about comma objects, are they easy to compute? can we write down what e.g. $\mathsf{Ab}/X$ is?

Kevin Carlson (aka Arlin) (Mar 12 2024 at 16:38):

I'm pretty sure abelian categories (with two-sided exact functors, I guess) are monadic over categories, so the form of the slice won't be novel since a comma is a limit. The problem is the functor $1\to \mathcal A$ picking out $X$ is not a morphism of abelian categories! I guess you'd need to replace it with a functor from the free abelian category $F(1)$ on a single (nonzero) object.

Kevin Carlson (aka Arlin) (Mar 12 2024 at 16:38):

I don't know an exact description of $F(1)$; it surely contains the free finite product category $\mathrm{FinSet}^{\mathrm{op}}$ but you need much more...maybe it's just the category of finitely presented abelian groups?

Kevin Carlson (aka Arlin) (Mar 12 2024 at 16:39):

In any case, the comma object would then involve objects over $X^n$ for every $n$, and perhaps also over cotensors of $X$ with more interesting fp abelian groups than $n,$ i.e. than $\mathbb Z^n.$ So it's certainly true that you get a more complicated slice construction. Interesting idea, I'm not sure what else to say about it right now.

Matteo Capucci (he/him) (Mar 13 2024 at 07:49):

It seems free abelian categories are hell of a construction :exploding_head: https://arxiv.org/pdf/2103.08379.pdf

John Baez (Mar 13 2024 at 17:19):

I like links to abstracts better than PDFs, so: https://arxiv.org/abs/2103.08379

Kevin Carlson (aka Arlin) (Mar 13 2024 at 17:23):

I wish that article would compute an example of an Adelman category! I can't be bothered to look at other references right now, but it's pretty wild that apparently the free abelian category on an additive category $A$ is $A-\mathrm{mod}-\mathrm{mod},$ where $\mathrm{mod}$ constructs finitely presented $A$-modules. I have no idea what that second $\mathrm{mod}$ is doing there.

Kevin Carlson (aka Arlin) (Mar 13 2024 at 17:57):

I guess since we're looking at covariant $A$-modules, doing modules twice is a sort of double dual construction, but doing a double dual using the contravariant Yoneda embedding doesn't ring a lot of other bells for me.

John Baez (Mar 13 2024 at 20:36):

If $\mathsf{Ab}$ is the category of abelian groups, I tend to think of an $\mathsf{Ab}$-enriched category as a 'ringoid', since a one-object one is just a ring. I define a module of an $\mathsf{Ab}$-enriched category $A$ as an $\mathsf{Ab}$-enriched functor $F: A \to \mathsf{Ab}$.

I define the category $A \mathsf{Mod}$ to be the category of modules and natural transformations between them. In the case where $A$ is a ring, this is the usual category of modules of that ring.

Since $A \mathsf{Mod}$ is is naturally $\mathsf{Ab}$-enriched, we can then go on and define $A \mathsf{Mod} \mathsf{Mod}$... at least if we're not worrying about size issues, or we choose a way to deal with them.

Kevin Carlson (aka Arlin) (Mar 13 2024 at 20:37):

Yeah, and this is finitely presentable modules so there's no size issues. It just seems kind of crazy to see a ModMod having a nice universal property.

John Baez (Mar 13 2024 at 20:43):

If we let $\mathsf{fgProj}(\mathsf{A})$ be the category of finitely generated presentable modules of a ring or ringoid $\mathsf{A}$, I believe $\mathsf{fgProj}(\mathsf{A})$ is the [[Cauchy completion]] of that ringoid in the 2-category of $\mathsf{Ab}$-enriched categories. So my guess is that $\mathsf{fgProj}$ has fairly nice properties when you apply it twice - like a [[lax idempotent 2-monad]].

John Baez (Mar 13 2024 at 20:44):

I may be omitting some necessary "ops" here, though!

Matteo Capucci (he/him) (Mar 14 2024 at 08:47):

So it seems $F1 = 1 \sf ModMod = AbMod = [Ab,Ab]$...??

Kevin Carlson (aka Arlin) (Mar 14 2024 at 16:16):

But add "finitely presented" everywhere. It's finitely presented functors from finitely presented abelian groups to abelian groups.

Morgan Rogers (he/him) (Mar 14 2024 at 16:23):

What does it mean for a functor to be finitely presented in this context?

Kevin Carlson (aka Arlin) (Mar 14 2024 at 16:27):

Maps out commute with filtered colimits, or equivalently it's the cokernel of a map between finite sums of representables.

Morgan Rogers (he/him) (Mar 14 2024 at 16:50):

Okay, it's not as huge as Matteo's description suggested then :grinning_face_with_smiling_eyes:

Matteo Capucci (he/him) (Mar 14 2024 at 17:24):

Ah, I missed that caveat Kevin

Matteo Capucci (he/him) (Mar 14 2024 at 17:25):

That object still baffles me :flushed:

Kevin Carlson (aka Arlin) (Mar 14 2024 at 17:26):

Agreed! Exact functors out of fp abelian groups, I can handle, or even just right exact ones, but additive ones are loco.

Kevin Carlson (aka Arlin) (Mar 14 2024 at 17:26):

I think.

Reid Barton (Mar 18 2024 at 16:49):

The category $1$ is not the free (Grothendieck) topos on an object in much the same way that it is not the free abelian category on an object.

Reid Barton (Mar 18 2024 at 16:50):

Maybe a better abelian category analogue for a slice category of a topos would be something like the category of comodules for a coalgebra?

Reid Barton (Mar 19 2024 at 17:16):

(of course, this assumes the abelian category is, at a minimum, monoidal)