Timeline for A category with weak equivalences that is not a model category
Current License: CC BY-SA 3.0
14 events
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| Jan 17 at 1:52 | comment | added | David White | A published reference for the non-existence of this model structure is page 2 (just after Theorem 1.2) of the paper Model categories with simple homotopy categories by Jean-Marie Droz and Inna Zakharevich. Probably there are other references, but I didn't spot any in the published works of Thibault. | |
| Jun 8, 2013 at 17:09 | comment | added | Fernando | Nice! Thank you! Although my question wasn't formulated precisely enough, professor Peter May answered exactly what I was looking for! Thank you very much! | |
| Jun 8, 2013 at 17:05 | vote | accept | Fernando | ||
| Jun 8, 2013 at 15:42 | comment | added | Peter May | Alice in Wonderland here: I edited my answer to complete the proof 20 hours ago, before the last two comments. | |
| Jun 8, 2013 at 10:09 | comment | added | Karol Szumiło | @Tom: you are absolutely right, for some reason I missed your comment. In that case perhaps the quickest way to finish off the argument is to say that if every map is a cofibration, then every pushout is a homotopy pushout which is absurd. | |
| Jun 7, 2013 at 20:55 | comment | added | Tom Goodwillie | @Karol: See my comment above. Any such model structure would be such that every trivial fibration is an isomorphism, therefore every map is a cofibration. | |
| Jun 7, 2013 at 19:15 | history | edited | Peter May | CC BY-SA 3.0 |
Proved claim
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| Jun 7, 2013 at 17:29 | comment | added | Karol Szumiło | Originally I thought that these weak equivalences coincide with the classical ones. But it turns out that the image of any finite dimensional semisimplicial set under the right adjoint is empty. So weak equivalences created by this functor are indeed very different. Can you give an argument for why there is no model structure with classical weak equivalences? I can see that there is no such model structure if the one-point semisimplicial set is cofibrant. | |
| Jun 7, 2013 at 17:07 | comment | added | Peter May | The weak equivalences in that reference are a priori different from those I prescribed (which are the most natural ones). It uses a right adjoint rather than the left adjoint of the forgetful map to create the weak equivalences. | |
| Jun 7, 2013 at 16:47 | comment | added | Karol Szumiło | This note uf-ias-2012.wikispaces.com/file/view/semisimplicialsets.pdf/… claims that there is such a model structure. What's certainly true is that there is no model structure where all monomorphisms are cofibrations or even where all objects are cofibrant. | |
| Jun 7, 2013 at 14:34 | comment | added | Tom Goodwillie | If the class of weak equivalences is as Peter says, or smaller, then every trivial fibration $X\to Y$ is an isomorphism. Proof by induction that $X_n\to Y_n$ is bijective: Given $y:\Delta^n\to Y$, pull back the map to get a trivial fibration $y^\ast X\to \Delta^n$. An object mapped to $\Delta^n$ is at most $n$-dimensional (this is really the crux), and if the map to $\Delta^n$ is an isomorphism on the $(n-1)$-skeleton then it cannot be a weak equivalence without being an isomorphism. | |
| Jun 7, 2013 at 13:19 | comment | added | Dylan Wilson | Can you give a little indication of the argument for why this does not have a model structure? (This is a neat example!) | |
| Jun 7, 2013 at 12:51 | comment | added | Omar Antolín-Camarena | That's a very surprising example. | |
| Jun 7, 2013 at 12:50 | history | answered | Peter May | CC BY-SA 3.0 |