Timeline for derived functors and triangulated categories
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2011 at 0:07 | comment | added | Sándor Kovács | Dear unknown, I think I understand what you are saying, but the point I am trying to make is that $RF$ is relative to $F$. It is not saying that an abstract functor is a right derived functor but that this particular one is the right derived functor of $F$ and that does make sense. You are right that this distinction would not make sense for an arbitrary functor on a triangulated category. | |
| Nov 3, 2011 at 23:39 | vote | accept | Nicolás | ||
| Nov 3, 2011 at 23:39 | comment | added | Nicolás | ... composition, irrespective if the functors appearing are left exact or right exact or one and one: you just use that they are triangulated. From this perspective using the letters $R$ and $L$ seem to only serve the purpose of remembering that the negative cohomologies are zero (in the case of $R$) or its positive cohomologies are zero (in the case of $L$). Rereading what I wrote, I'm not sure anyone will understand anything... | |
| Nov 3, 2011 at 23:37 | comment | added | Nicolás | Dear Sandor, thanks for your answer. My question had to do with the fact that if one stays in the derived/triangulated world there is no difference between left and right exact functors: they are both triangulated. For example, a left exact functor will give you a long exact sequence that goes to the left in cohomology. A right exact functor will give you a long exact sequence going to the right in cohomology. But in the derived category they both 'just' preserve triangles. There is also a spectral sequence that relates the cohomology of any two derived functors to the cohomology of their... | |
| Nov 3, 2011 at 20:24 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
added 9 characters in body
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| Nov 3, 2011 at 20:04 | history | answered | Sándor Kovács | CC BY-SA 3.0 |