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Feb 15, 2021 at 1:55 vote accept Tim Campion
Feb 13, 2021 at 20:29 history became hot network question
Feb 13, 2021 at 20:23 comment added Tim Campion Quick, someone throw Asaf a set theory question, the isolation is getting to him!
Feb 13, 2021 at 20:16 comment added Asaf Karagila @IvanDiLiberti: What are whiteboards? I forgot.
Feb 13, 2021 at 14:28 answer added Kevin Carlson timeline score: 4
Feb 13, 2021 at 13:08 comment added Ivan Di Liberti On a conceptual level, I do see a pairing $<-,->: \mathsf{RSub} \times \mathsf{RQuo} \to 2$, then of course whether this is the correct pov, it's hard to tell.
Feb 13, 2021 at 13:01 comment added Ivan Di Liberti That was my idea. It's hard to say without spending time on a whiteboard.
Feb 13, 2021 at 13:00 comment added Tim Campion @IvanDiLiberti That sounds promising. The usual ways I know of relating subobjects and quotients -- taking kernels / cokernels, taking kernel pairs / quotienting by a congruence etc. seem to require stronger exactness properties than I'm assuming here. It would be interesting if there were something like this which didn't require any sort of exactness.
Feb 13, 2021 at 12:59 comment added Ivan Di Liberti One way to think about this could be to study the presheaf $\mathsf{RSub}$ of regular subobject and the copresheaf $\mathsf{RQuo}$ of regular quotient. Now one could try to say that there must be an injective map $\mathsf{RQuo} \to 2^\mathsf{RSub}$ (notice that the variance now match). Similar tricks are used by Freyd in Sec 5 of "On the concreteness of certain categories", for very different purposes.
Feb 13, 2021 at 12:51 comment added Tim Campion @AsafKaragila Exactly. Unfortuately, "co-well-powered" means that it's also connected to a continuous energy drain. So the above theorem that $(1) \Leftrightarrow (2)$ is really a weak form of the 0th law of thermodynamics.
Feb 13, 2021 at 12:45 comment added Asaf Karagila By well-powered do you mean "connected to a continuous energy source"?
Feb 13, 2021 at 12:28 history asked Tim Campion CC BY-SA 4.0