Skip to main content
14 events
when toggle format what by license comment
Feb 8, 2022 at 11:53 vote accept Valery Isaev
Feb 8, 2022 at 10:50 comment added Valery Isaev @TimCampion The less weak equivalences you have, the more objects you'll get in the homotopy category. So, with the "maximal" model structure, you get the largest homotopy category. I just think it's more convenient to think about it in this way.
Feb 8, 2022 at 2:12 comment added Tim Campion I'm confused by the terminology "maximal model structure". I suppose it has the maximal number of fibrations for given cofibrations, but I think I've more often heard it called "minimal" since it has the fewest weak equivalences. Rosicky and Tholen call it the "left-determined" model structure for a class of cofibrations.
Aug 2, 2019 at 0:36 answer added Matt Feller timeline score: 7
Aug 1, 2017 at 7:03 comment added Andrea Gagna And there is even another issue. On his paper Catégories dérivables he defines a variation of the notion of model structure, which has all the right to be called a Cisinski model structure (although it is way less popular then model structures on presheaves categories).
Jul 31, 2017 at 19:39 comment added David White My point was that the term should be defined in the question. It's certainly not well-known terminology. Cisinski's papers on this are only about 10 years old!
Jul 31, 2017 at 17:56 comment added Valery Isaev @DavidWhite ncatlab.org/nlab/show/Cisinski+model+structure
Jul 31, 2017 at 17:54 comment added David White What is a Cisinski model structure?
Jul 29, 2017 at 13:52 answer added Andrea Gagna timeline score: 8
Jul 29, 2017 at 10:08 comment added Valery Isaev @DavidRoberts Yes, this is the condition and it implies that the trivial fibrations are the same as in the Quillen (or Joyal) model structure.
Jul 29, 2017 at 10:06 comment added David Roberts Ah, I was under the impression that the condition was that the cofibrations were the monos.
Jul 29, 2017 at 9:47 comment added Valery Isaev @DavidRoberts No since trivial fibrations must be weak equivalences. If $X$ is a nontrivial contractible Kan complex, then $X \to 1$ is a trivial fibration, but not an iso.
Jul 29, 2017 at 9:41 comment added David Roberts So taking the class of weak equivalences to be the isos doesn't work?
Jul 29, 2017 at 7:13 history asked Valery Isaev CC BY-SA 3.0