Timeline for Example of a saturated class of morphisms which is not _obviously_ saturated?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2015 at 22:09 | vote | accept | Tim Campion | ||
| Jan 2, 2015 at 18:01 | comment | added | Mike Shulman | Well, I think it's not quite obvious, it relies on the existence of an interval; but you're right that that's a fairly easy argument. It doesn't work for higher categories, though, because in that case in order to define an "equivalence of categories" by the existence of a quasi-inverse, you need to allow weak functors; but at least with algebraic models for higher categories, the weak functors don't generally form a well-behaved 1-category that you can put a model structure on. | |
| Jan 2, 2015 at 7:54 | comment | added | Tim Campion | In the case $n=1$, it's obvious that when you mod out by the congruence identifying isomorphic functors, you invert exactly the equivalences of categories. I guess I assumed that a similar observation should work for every value of $n$. In some sense, this good control over equivalences is really the point of the homotopy hypothesis, isn't it? | |
| Jan 1, 2015 at 21:36 | comment | added | Mike Shulman | Why would "naturalness" imply that saturation should be obvious? Even in the case n=1, I can't think of any obvious functor giving the equivalences of categories as the preimage of the isomorphisms. | |
| Dec 31, 2014 at 18:16 | comment | added | Tim Campion | Interesting. I thought the idea of a "canonical model structure" on $n$-categories was that the weak equivalences should be "the evident generalization of equivalences of categories", and the naturalness of this choice should suggest that saturation should be obvious. As I think about it, maybe "the evident generalization" isn't so obvious when $n = \omega$. | |
| Dec 31, 2014 at 4:54 | history | answered | Mike Shulman | CC BY-SA 3.0 |