Timeline for Obtaining derived functors from derived functors of similar complexes or "bluntly truncated" unbounded complexes (without adding 0's to the left)
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2012 at 17:56 | history | edited | Louis A | CC BY-SA 3.0 |
added 182 characters in body; edited title
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| Sep 15, 2012 at 3:01 | answer | added | Sasha | timeline score: 1 | |
| Sep 14, 2012 at 22:17 | answer | added | Mariano Suárez-Álvarez | timeline score: 1 | |
| Sep 14, 2012 at 21:56 | comment | added | Louis A | This is the first time I've heard the term perturbation in this context, so I don't know what it means for a perturbartion to travel $d$ places, apply the functor to what? To the perturbation? To see what at once? Sorry if I sound clueless but I'm totally unfamiliar with what you're saying | |
| Sep 14, 2012 at 20:44 | comment | added | Mariano Suárez-Álvarez | By perturbation I simply mean a change one of the terms in the complex (for example, replacing it with zero) | |
| Sep 14, 2012 at 20:40 | comment | added | Louis A | Sorry I'm a bit lost here, what's a perturbation in this context? Thanks for answering | |
| Sep 14, 2012 at 20:18 | comment | added | Mariano Suárez-Álvarez | If the domain category has finite global dimension $d$, for example, perturbations in a complex only travel $d$ (or maybe $d+1$?) places. If you go through the steps of constructing a projective resolution of your initial complex and then applying the functor to that, you can see this at once. | |
| Sep 14, 2012 at 20:16 | comment | added | Mariano Suárez-Álvarez | As an extreme case, imagine the initial complex has exactly one term! :-) | |
| Sep 14, 2012 at 20:06 | history | asked | Louis A | CC BY-SA 3.0 |