Timeline for Are compact objects in presheaf categories finite colimits of representables?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 1:22 | answer | added | Todd Trimble | timeline score: 6 | |
| May 27, 2020 at 22:25 | answer | added | Tim Campion | timeline score: 8 | |
| May 25, 2020 at 14:32 | history | edited | David White | CC BY-SA 4.0 |
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| May 25, 2020 at 13:13 | history | became hot network question | |||
| May 25, 2020 at 12:31 | comment | added | Dylan Wilson | FWIW this is no longer true in the $\infty$-categorical setting, where "finite" has a slightly different meaning, and taking retracts is no longer a finite colimit. (It fails already when $C$ is a point, by the Wall finiteness obstruction.) | |
| May 25, 2020 at 11:36 | comment | added | Ivan Di Liberti | Maybe a relevant motivational analogy is that compact objects in the poset $2^X$ are precisely finite subsets of $X$. | |
| May 25, 2020 at 6:28 | answer | added | Aurélien Djament | timeline score: 17 | |
| May 25, 2020 at 6:28 | comment | added | William Balderrama | @DmitriPavlov Ah, so they are. For some reason I had in mind the sequential colimit along the given idempotent morphism as the way of splitting it. | |
| May 25, 2020 at 6:24 | comment | added | Dmitri Pavlov | @WilliamBalderrama: Retracts are colimits over a category with a single object and a single nonidentity morphism, which is idempotent. | |
| May 25, 2020 at 6:17 | comment | added | William Balderrama | If you don't want to assume that X is idempotent-complete, then I think you can get a counterexample by looking at retracts of representables that don't exist in your base category, since these will be compact and I don't think you can write them as finite colimits of representables in general. But a particular example with a proof that it isn't a finite colimit of representables isn't coming to me right now. | |
| May 25, 2020 at 5:10 | history | asked | John Baez | CC BY-SA 4.0 |