Timeline for Does $\mathsf{Ind}_\lambda^\mu(\mathsf{Ind}_\kappa^\lambda(\mathcal C)) = \mathsf{Ind}_\kappa^\mu(\mathcal C)$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Jan 29, 2021 at 21:06 | history | bounty ended | Tim Campion | ||
| S Jan 29, 2021 at 21:06 | history | notice removed | Tim Campion | ||
| S Jan 28, 2021 at 19:17 | history | bounty started | Tim Campion | ||
| S Jan 28, 2021 at 19:17 | history | notice added | Tim Campion | Reward existing answer | |
| Jan 27, 2021 at 2:07 | history | edited | Tim Campion | CC BY-SA 4.0 |
Thanks, LSpice! Since you've made up for my laziness, I now have the freedom to set things to my preferred font for general constructions like this.
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| Jan 27, 2021 at 2:04 | history | edited | LSpice | CC BY-SA 4.0 |
\mathit
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| Jan 27, 2021 at 1:21 | vote | accept | Tim Campion | ||
| Jan 27, 2021 at 0:43 | answer | added | Zhen Lin | timeline score: 5 | |
| Jan 26, 2021 at 23:34 | comment | added | Tim Campion | @ZhenLin By your Prop 3.5, if $\kappa \triangleleft \lambda$, then the $\lambda$-presentable objects of $Ind_\kappa(C)$ form (in your terminology) a $(\kappa,\lambda)$-accessibly-generated category. Therefore (in my terminology, and assuming that $C$ has split idempotents) $I_\kappa^\lambda(C) = Ind_\kappa^\lambda(C)$. Thus by your work, if $C$ is essentially small with split idempotents and $\kappa \triangleleft \lambda$, then $(\kappa,\lambda,Ord)$ answers to Question 1 and (hence) to Question 2. If you write this, or even just "answered in the comments" as an answer, I'll gladly accept! | |
| Jan 26, 2021 at 23:20 | comment | added | Tim Campion | This reduces the most important case of Question 2 to Question 1, which is fantastic! | |
| Jan 26, 2021 at 23:19 | comment | added | Tim Campion | @ZhenLin Thanks, this is great! Let me see if I've got this straight. If we assume (in my terminology) that $C$ is idempotent-complete and $I_\kappa^\lambda(C) = Ind_\kappa^\lambda(C)$ (i.e. $(\kappa,\lambda)$ answers to Question 1), then the inclusion $C \to Ind_\kappa^\lambda(C)$ is clearly (in your terminology) a $(\kappa,\kappa,\lambda)$-accessibly-generated extension. In this case, your Prop 3.10(iv) says (in my terminology) that if $\kappa \triangleleft \lambda$, then $Ind_\kappa^{Ord}(C) = Ind_\lambda^{Ord}(Ind_\kappa^\lambda(C))$, so that $(\kappa,\lambda,Ord)$ answers to Question 2. | |
| Jan 26, 2021 at 22:39 | comment | added | Zhen Lin | Regarding iterated ind-completion, see proposition 3.10(iv) here. | |
| Jan 26, 2021 at 15:37 | history | asked | Tim Campion | CC BY-SA 4.0 |