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Stream: learning: questions

Topic: Are sets accessible categories?

Josselin Poiret (Dec 01 2022 at 14:18):

Hi, I'm trying to learn some things about accessible categories, and it seems to me that sets, seen as categories with only identity morphisms, are $\lambda$-accessible for any regular cardinal $\lambda$, no matter the set's cardinality. All objects seem to be $\lambda$-presentable because $\lambda$-directed diagrams in a set have to be constant. Is that really the case?

Morgan Rogers (he/him) (Dec 01 2022 at 14:34):

Yes!

Reid Barton (Dec 01 2022 at 16:23):

It's also important that the set is, well, a set, and not a proper class.

Jan Pax (Jan 04 2024 at 21:19):

@Reid Barton Could you explain for a beginner what would go wrong if the set weren't set but a proper class, what would fail ? If the cardinality is greater than $\lambda$ then it can be arbitrarily large and even a proper class, why "not" ? Maybe the reason is, that it wouldn't have a set of representable objects ?

Matteo Capucci (he/him) (Jan 04 2024 at 21:30):

Because the definition of [[accessible category]] requires a small set of $\lambda$-presentable objects. So even if the category is large, you know you can use just a small worth of objects to present all the other ones. The reason every set is accessible is that every object present itself, so this cuts it only if the objects form a small set, else you'd need too many objects to present the set!

Mike Shulman (Jan 04 2024 at 21:40):

More generally, any small category is accessible as long as it has split idempotents. In particular, any (small) poset is accessible.

David Michael Roberts (Jan 05 2024 at 02:11):

@Mike Shulman huh, what goes wrong in a small category if idempotents don't split?

Mike Shulman (Jan 05 2024 at 02:15):

A $\lambda$-accessible category is required to have $\lambda$-filtered colimits. Splitting of idempotents is a $\lambda$-filtered colimit for any $\lambda$.

David Michael Roberts (Jan 05 2024 at 02:15):

I guess it's the minimum necessary to generate all the needed colimits?

Is the index of accessiblity the cardinality of the set of morphisms? Or is it something like a really small regular cardinal, eg 2?

David Michael Roberts (Jan 05 2024 at 02:16):

Oh, heh, snap

David Michael Roberts (Jan 05 2024 at 02:17):

So I guess we can take $\lambda = 2$?

David Michael Roberts (Jan 05 2024 at 02:18):

No, wait. A split idempotent is a limit of a diagram with more than one arrow.

Mike Shulman (Jan 05 2024 at 02:18):

I don't think the concept of accessibility works very well for finite regular cardinals.

David Michael Roberts (Jan 05 2024 at 02:19):

Hmm. I can imagine.

Mike Shulman (Jan 05 2024 at 02:20):

As proven in Adamek-Rosicky, the result is that if a small category has fewer than $\lambda$ morphisms, then it is $\lambda^+$-accessible.

Mike Shulman (Jan 05 2024 at 02:21):

They also say this is the best you can do, exhibiting a countable category with split idempotents that is not $\aleph_1$-accessible.