Timeline for homotopy limits of dg categories
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2011 at 21:58 | comment | added | Fernando Muro | You're right :-( | |
| Apr 28, 2011 at 20:41 | comment | added | D.-C. Cisinski | @Shengao. There is absolutely no chance that the formula $H^0 holim = holim H^0$ holds: to get a counter-example, you may restrict to dg categories with one objects (aka dg algebras); note that, for a diagram of dg algebras $A_i$, $holim A_i$ is computed as the homotopy limit of the underlying complexes. Therefore, such a formula would imply that any tower of algebras $A_{n+1}\to A_n$, $n\geq 0$, would satisfy the Mittag-Leffler condition... @Fernando. Hence $H^0$ cannot be a right adjoint. | |
| Apr 28, 2011 at 17:02 | comment | added | shenghao | It is possible that, as suggested to me by Y. Laszlo, the commutativity $H^0\ holim=holim\ H^0$ still holds, not as a consequence of some formal argument, but of the so-called strictification theorem. I don't know this stuff, and hope some experts can explain. BTW, when the index category $I$ is infinite, how to show the existence of the holim? \\ @Denis-Charles: It's great to see you on MO! | |
| Apr 28, 2011 at 13:48 | history | edited | Fernando Muro | CC BY-SA 3.0 |
added 19 characters in body; added 4 characters in body
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| Apr 28, 2011 at 13:40 | comment | added | Fernando Muro | @Denis-Charles: oh, you're right! I forgot about cofibrations... Could it still be possible that $H^0\colon\operatorname{Ho}(\mathscr{M})\rightarrow \operatorname{Ho}(\mathscr{N})$ were right adjoint to $\operatorname{Ho}(\mathscr{N})\subset \operatorname{Ho}(\mathscr{M})$? | |
| Apr 28, 2011 at 1:32 | comment | added | D.-C. Cisinski | Unfortunately, the inclusion of linear categories into dg categories is not a left Quillen functor at all! | |
| Apr 28, 2011 at 0:34 | history | answered | Fernando Muro | CC BY-SA 3.0 |