Timeline for If a colimit of distinguished triangles exists, is it also a distinguished triangle?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 17, 2010 at 11:05 | answer | added | Matthias Künzer | timeline score: 1 | |
| Aug 8, 2010 at 20:51 | answer | added | Mikhail Bondarko | timeline score: 3 | |
| Jul 27, 2010 at 19:24 | comment | added | Andreas Holmstrom | Tom, I asked two new questions based on your reply, in the hope that more people will see the questions. I would still be very interested in what exactly was wrong and what was right in your original answer though. | |
| Jul 27, 2010 at 12:19 | comment | added | Tom Goodwillie | My answer was so very very wrong that I have deleted it. | |
| Jul 27, 2010 at 0:33 | comment | added | Andreas Holmstrom | Aha, I guess I never had to think of uncountable homotopy colimits so far ;-) and thanks, I will check the reference to May. | |
| Jul 26, 2010 at 21:30 | comment | added | Greg Stevenson | Yes, provided one has countable coproducts one can always define linear countable homotopy colimits as you say. This does not give the correct definition for larger cardinalities of indexing set though even in the linear case. If one has a countable sequence of triangles I believe it is true that the homotopy colimits of their terms in this sense do form a triangle. One should be able to obtain this fairly directly from the 3x3 Lemma (as in May's The additivity of traces in triangulated categories). In particular it doesn't use anything more than the octahedral axiom. | |
| Jul 26, 2010 at 21:01 | comment | added | Andreas Holmstrom | Actually I was not completely sure myself what I meant. I am trying to understand this for a specific application and wasn't sure which case I need. When writing the question I was thinking of a situation where the colimit happens to exist, but I also had in mind a homotopy colimit. Now I learnt from Tom that they coincide, which was a big surprise. About your last question: As far as I understand one can define sequential hocolim in any triangulated category, without assuming existence of a model, as a cone of the shift map on the direct sum of all terms, or something like that. | |
| Jul 26, 2010 at 11:26 | comment | added | Greg Stevenson | I am not completely sure what you mean by colimit here. Are you assuming that your triangulated category has a model so that you can take a homotopy colimit? | |
| Jul 26, 2010 at 5:53 | answer | added | Sean Tilson | timeline score: 1 | |
| Jul 26, 2010 at 3:17 | comment | added | thel | A colimit in your category is a limit in the opposite category, so perhaps we can find a counterexample in the opposite category of complexes of abelian groups. | |
| Jul 26, 2010 at 0:49 | history | asked | Andreas Holmstrom | CC BY-SA 2.5 |