Re: cogenerator in omegaCat ?

[Michael Barr <barr@triples.math.mcgill.ca>]

I imagine that omega-categories, however defined, will be a locally
c-presentable category (c=cardinal of the continuum) in the sense of
Gabriel-Ulmer, equivalently complete and c-accessible in the sense of
Makkai-Pare and hence cocomplete.  In other words, the colimit will grow
but not by much.  Actually, aleph_1 is all you are really going to need.


On Wed, 21 Oct 1998, Philippe Gaucher wrote:

> 
> 
> > 	No, it seems not since a co-generator for omega cat
> > would surely give rise to one for cat in particular, but such
> > does not exist. This contrasts with the situation for the
> > "larger" universe of simplicial sets.  A category of "small"
> > sets is a kind of approximation to a co-generator, but each
> > enlargement of the meaning of "small" creates new categories
> > which are not co-generated.
> 
> 
> The argument sounds reasonable. Before this question, I was 
> convinced of the existence of this cogenerator. I have to find
> something else for the lemma I would like to prove...
> 
> Since it does not exist, I have another questions (I suppose well-
> known) and any reference abou the subject would be welcome : 
> 
> How does one prove the cocompleteness of omegaCat (small & strict) ?
> The only idea of proof I had in mind until this question was : omegaCat
> is obviously complete (and the forgetful functor towards the category of Sets 
> preserves projective limits), and well-powered and a cogenerator 
> => the cocompleteness (Borceux I, prop 3.3.8 p 112).
> 
> Without cogenerator, how can one prove the cocompleteness ? The explicit 
> construction of the colimit seems to be very hard : the forgetful
> functor towards Set does not preserve colimits because the 
> underlying set of the colimit might be bigger than the colimit of the
> underlying sets. Every time two n-morphisms are identified in the 
> colimit of the underlying sets, p-morphisms (with p>n) might be "created"
> by the colimit.
> 
> Thanks in advance for any answer. pg.
> 
>