Timeline for Cohomology Ring of a small category $\mathsf{C}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2017 at 21:51 | history | edited | mayer_vietoris | CC BY-SA 3.0 |
added 12 characters in body
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| Jun 21, 2017 at 1:24 | comment | added | Benjamin Steinberg | The higher inverse limits are the Ext functors from the constant functor. Hom from the constant functor is the inverse limit. | |
| Jun 20, 2017 at 23:47 | comment | added | Benjamin Steinberg | If you used the constant functor then it doesn't matter if there are finitely many or infinitely many objects. The functor category has enough projectives and injectives to derive Ext. I personally don't think the notation R is good to denote the RC module corresponding to the constant functor in the case of more than one object. | |
| Jun 20, 2017 at 22:19 | answer | added | Nicholas Kuhn | timeline score: 7 | |
| Jun 20, 2017 at 21:51 | comment | added | mayer_vietoris | Thank you for your comment, you mean that the definition I've written in the finite case is incorrect? Regarding the first comment, by writing this definition, I mean that we have the constant sheaf indeed. | |
| Jun 20, 2017 at 21:39 | comment | added | Benjamin Steinberg | You can only use R in your definition of cohomology for a monoid. In general you have to take the direct sum of one copy of R for each object of C. | |
| Jun 20, 2017 at 21:38 | comment | added | Benjamin Steinberg | You can define the cohomology of $C$ using Ext from the constant functor sending each object to R and each morphism to the identity to your functor. | |
| Jun 20, 2017 at 21:35 | answer | added | Johannes Hahn | timeline score: 2 | |
| Jun 20, 2017 at 20:51 | history | asked | mayer_vietoris | CC BY-SA 3.0 |