Timeline for Injective resolution for right derived functor
Current License: CC BY-SA 3.0
9 events
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| Jul 30, 2022 at 2:26 | comment | added | PrimeRibeyeDeal | You may be interested in this paper, which investigates the consequences of treating tensor/hom as a left/right derived functor, anti-respectively. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 9, 2014 at 16:26 | comment | added | Pablo Zadunaisky | Well, yes, in this way you the right derived functors of tensor products. But since Tensor products are right exact, its $i$-th right derived functor is identically zero, save for $i = 0$. | |
| Mar 7, 2014 at 23:11 | comment | added | Li Yutong | I see where I was wrong. But why derived tensor has to be a left derived functor? It could well be a right derived functor as I defined -- this make sense, but maybe meaningless -- I don't have any idea about this... | |
| Mar 7, 2014 at 23:01 | vote | accept | Li Yutong | ||
| Mar 7, 2014 at 19:52 | answer | added | Denis Nardin | timeline score: 3 | |
| Mar 7, 2014 at 14:18 | comment | added | user36931 | Ok, I still don't understand, but my advice is really to revisit some basic points. Why do you think this is a RIGHT derived functor????? derived tensor product is a LEFT derived functor. | |
| Mar 7, 2014 at 14:17 | answer | added | answer_bot | timeline score: 3 | |
| Mar 7, 2014 at 13:56 | history | asked | Li Yutong | CC BY-SA 3.0 |