Timeline for When is every element of a coend of abelian groups contained in one of the summands?
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| Apr 28, 2021 at 17:00 | comment | added | Martin Brandenburg | @TimCampion Good point. I need to think about the context again, but there was probably a reason that the coends where normal ($\mathbf{Set}$-enriched) coends. They appeared in the context of certain categories of algebras, which are not $\mathbf{Ab}$-enriched, for instance. | |
| Apr 28, 2021 at 16:38 | comment | added | Tim Campion | When doing coends of abelian groups, I'm usually thinking of $Ab$-enriched coends, which I think are relevantly different from the $Set$-enriched coends you seem to be using here. | |
| Apr 28, 2021 at 9:54 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Feb 15, 2020 at 16:49 | comment | added | Omar Antolín-Camarena | @MartinBrandeburg Oh, right! I was a little careless, I did think it meant the analogue of every tensor being pure. The correct thing has a much better chance of happening. :) | |
| Feb 15, 2020 at 7:08 | comment | added | Martin Brandenburg | @OmarAntolín-Camarena Actually, for $I=1$ the sufficient condition I mentioned is satisfied. My question is not when every tensor is pure (I corrected that in the post), but when every element belongs to a single tensor product then, or in the general case, to a single $D(i,i)$. I have clarified that in the post as well. | |
| Feb 15, 2020 at 7:07 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Feb 14, 2020 at 3:38 | comment | added | Omar Antolín-Camarena | Doesn't the special case of the tensor product (even for $I=1$) show the answer is "almost never"? I mean, it might still be interesting to work out when this happens, but I wouldn't expect to it too in useful cases. | |
| Feb 13, 2020 at 18:04 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Feb 13, 2020 at 8:02 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Feb 13, 2020 at 7:56 | history | asked | Martin Brandenburg | CC BY-SA 4.0 |