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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 27th 2011

added a little bit of content to (0,1)-category. Just a little bit. (0,1)-content, so to say.

• CommentRowNumber2.
• CommentAuthorColin Tan
• CommentTimeJun 29th 2014

Added to (0,1)-category the observation that a (0,1)-category with the property of whose each morphism is iso (that is, a $(0,1)$-groupoid) is a setoid. Are there constructive aspects which I should take notice of?

Also updated this description to equivalence relation in the section on setoids.

• CommentRowNumber3.
• CommentAuthorColin Tan
• CommentTimeJun 29th 2014
• (edited Jun 29th 2014)
On further inspection, I find that in the Lab "n-groupoid" is a concept but not "(n,r)-groupoid".

Should I write the full "(0,1)-category of whose each morphism is iso" instead of my shorthand "(0,1)-groupoid"? (For now, I write the long form in all my edits.)
• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJun 29th 2014

A $(0,1)$-category in which every $1$-morphism is invertible is precisely a $(0,0)$-category, which is the same as a $0$-category (because the first and second number are the same) and the same as a $0$-groupoid (because the second number is zero).

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeJun 29th 2014

More generally, we don’t talk about (n,r)-groupoids because they are the same as n-groupoids, the only difference between n and r being invertibility.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeJun 29th 2014
• (edited Jun 29th 2014)

Is an $(n, r)$-groupoid any different from an $n$-groupoid? It seems that the extra condition that every $j$-cell is invertible when $j \gt r$ isn’t an extra condition at all, if we already are in an $\infty$-groupoid or $n$-groupoid where every arrow is invertible to begin with.

(I had the tab open without refreshing for some time, which explains why I didn’t see Mike’s reply before posting. The soccer game’s on.)

• CommentRowNumber7.
• CommentAuthorColin Tan
• CommentTimeJun 30th 2014
Thank you Toby, Mike and Todd for explaining to me why groupoids are indexed by only one number instead of two!
• CommentRowNumber8.
• CommentAuthorColin Tan
• CommentTimeJun 30th 2014

A further comment, at 0-category, a 0-category is defined as a set. I mentioned that this is the case only up to equivalence and changed the actual definition to setoid, to be consistent with (0,1)-category.

• CommentRowNumber9.
• CommentAuthorColin Tan
• CommentTimeJul 1st 2014
There seems to be a certain conflict of principles here. On one hand, we are saying that a 0-category is a setoid. However, another principle is that each hom object in a n- category is a (n-1)-category. Typically, we consider each hom object in a 1-category to be a set, rather than a setoid.
• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeJul 1st 2014

You’re the one who said that a 0-category is a setoid. I don’t think that’s quite accurate, even if at the level of literal data it’s technically true (although even that depends on a particular definition of 0-category) — generally one treats setoids differently than 0-categories. When 0-categories are treated categorically, they really are no different than sets.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeJul 1st 2014

Colin, I’m not clear on what the real difficulty is supposed to be. Let me try to rephrase what I think Mike might be saying:

generally one treats setoids differently than 0-categories

I think this means there is a category of setoids and setoid maps (those which preserve the equivalence relation). This is quite different from $Set$ (it’s not a topos for instance). But from the POV of higher category theory, it’s what we might call a naive truncation of the $(\infty, 1)$-category of 0-categories, where we simply discard $n$-cells for $n \gt 1$.

When 0-categories are treated categorically, they really are no different than sets.

If we more reasonably form the “homotopy category” consisting of setoids/0-categories and functors between them modulo natural isomorphism, then we get a category equivalent to $Set$. (For the moment “equivalence” means weak equivalence: the evident functor $Set \to Ho(0\text{-}Cat)$ is fully faithful and essentially surjective.) A better way of saying it might be that if we regard the 1-category $Set$ as an $(\infty, 1)$-category, then we have an $(\infty, 1)$-equivalence $Set \to 0\text{-}Cat$.

• CommentRowNumber12.
• CommentAuthorTobyBartels
• CommentTimeJul 2nd 2014

I would rather say that a $0$-category is a setoid up to isomorphism and a set up to equivalence. And of course, up to equivalence is better.

Actually, even speaking of what things are up to isomorphism is a bit arbitrary, since what you get depends on where you start, but this way of saying it reflects what is probably a common starting point for people first coming to higher category theory: ordinary category theory.

To be honest, it is a bit dicey even to say that a $0$-category is a setoid up to isomorphism, but the issue there has to do with the term ‘setoid’. Although it was me who started saying things like ‘A setoid is a set equipped with an equivalence relation.’, the people who use setoids are people who start with an impoverished notion of set (specifically a notion that cannot handle quotient sets) and introduce setoids to fix this. And the category that they use (possibly implicitly) is not the naïve category of setoids that Todd refers to but the homotopy category that he refers to next.

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeJul 2nd 2014

I essentially “rolled back” the first sentence of the definition section at 0-category so as not to mention setoid, and then tried to summarize some of the discussion above into remarks, specifically the second remark, where setoids are mentioned. Perhaps those who were in this discussion could read this over (it feels heavy to me, but it may be useful for those who feel some puzzlement as they hop from one nLab page to another).

• CommentRowNumber14.
• CommentAuthorTobyBartels
• CommentTimeJul 2nd 2014

I moved the remark about setoids to equivalence relation, where it is really needed.

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeJul 2nd 2014

If we more reasonably form the “homotopy category” consisting of setoids/0-categories and functors between them modulo natural isomorphism, then we get a category equivalent to $Set$. (For the moment “equivalence” means weak equivalence: the evident functor $Set \to Ho(0\text{-}Cat)$ is fully faithful and essentially surjective.)

Your parenthetical suggests you are not assuming AC, but I think this is only true if you do assume AC. E.g. if $p:A\to B$ is a surjection without a section, then $ker(p) \to B$ is a map of setoids inducing an isomorphism in $Set$, but it’s not an isomorphism in that homotopy category of setoids (and more generally, I don’t think $ker(p)$ will be isomorphic to any set in that homotopy category). To make it an equivalence without AC I think you have to include “anafunctors” between 0-groupoids, which setoid-theorists rarely do.

It’s also not clear to me that people who use setoids are actually working in this homotopy category. Sometimes it seems like they are, but I think often they actually do use given representatives of the morphisms.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeJul 2nd 2014
• (edited Jul 2nd 2014)

Thanks for the note, Mike. (Noted, but I don’t think anything I wrote into the entry needs to be changed.)

Edit: actually, yes, I was assuming AC without saying so. Sorry if the phrasing was confusing.

• CommentRowNumber17.
• CommentAuthorTobyBartels
• CommentTimeJul 4th 2014

To make it an equivalence without AC I think you have to include “anafunctors” between 0-groupoids, which setoid-theorists rarely do.

It's not exactly clear what the users of setoids (I wouldn't want to say ‘setoid-theorists’) are really doing, since they're usually using a notion of ‘set’ that doesn't really match ours. (That means, of course, that they're not going to agree that the category of setoids is equivalent to the category of sets no matter how we set things up … that's why they use setoids, because the category of their sets is not enough to support mathematics, while the category of setoids is.)

But often setoid-users assume the axiom of choice in the sense that all of the sets that they accept as ‘sets’ are projective (being ‘completely presented’) and therefore the homotopy category that Todd describes does work. And on the other hand, some people (well, at least one person1, since FOLDS is the only foundation that I can think of now that does this, although it doesn't use this terminology) would use functional, entire, equivalence-respecting relations as morphisms between setoids, where ‘functional’ is defined using the given equivalence rather than equality, and these (which are in fact saturated anafunctors) also give the correct category.

1. It is probably no coincidence that this is also the person who invented anafunctors!

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeJul 4th 2014

I guess that’s right.