Timeline for $\operatorname{Ind}(C^I) = \operatorname{Ind}(C)^I$?
Current License: CC BY-SA 4.0
9 events
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| Mar 8, 2023 at 17:29 | comment | added | Simon Henry | @BenWieland : yes, but the idea is to use Makkai's result to Ind(C), which is always accessible, and deduce the result in my question for C when C is Cauchy complete. But I'm now convinced that Makkai's theorem is false thanks to your exemples (well, small modification so that they are Cauchy complete). I'll add some details in the next few days. | |
| Mar 7, 2023 at 20:12 | comment | added | Ben Wieland | Makkai requires $C$ to be accessible, not just Cauchy complete. | |
| Mar 6, 2023 at 18:30 | comment | added | Simon Henry | @BenWieland yes but only in the sense that Ind(C) have them. Being $\omega$-accessible is equivalent to be of the form Ind(C) for C any category and the $\omega$-presentable objects of Ind(C) are the retracts of objects of $C$. So If Makkai's theorem is correct (which I'm seriously starting to doubt due to your remarks) this implies the results in the question - at least in the case where $C$ is Cauchy complete ( = all idempotent splits) I need to think more to figure out if your example does really contradicts this, but it definitely feels like it does... | |
| Mar 6, 2023 at 1:10 | comment | added | Ben Wieland | Doesn't accessibility require all countable filtering colimits? | |
| Mar 5, 2023 at 23:49 | comment | added | Simon Henry | So... at this point, I neither fully understand Makkai's proof nor the counter-example proposed by Ben Wieland, so I'm really not sure of anything ^^ | |
| Mar 4, 2023 at 15:05 | comment | added | Simon Henry | I guess it only shows it for Cauchy complete categories, but I'm happy with this. Thanks! | |
| Mar 4, 2023 at 15:05 | vote | accept | Simon Henry | ||
| Mar 16, 2023 at 1:50 | |||||
| Mar 4, 2023 at 13:57 | history | edited | Peter LeFanu Lumsdaine | CC BY-SA 4.0 |
added link for convenience
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| Mar 4, 2023 at 11:32 | history | answered | Ivan Di Liberti | CC BY-SA 4.0 |