Timeline for Characterization of functors whose right adjoint is monadic?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2021 at 13:18 | vote | accept | Tim Campion | ||
| Mar 2, 2021 at 22:44 | history | became hot network question | |||
| Mar 2, 2021 at 15:30 | answer | added | Simon Henry | timeline score: 11 | |
| Mar 2, 2021 at 15:19 | comment | added | varkor | Up to equivalence, a left adjoint is Kleisli if it is essentially surjective. For instance, see this question. | |
| Mar 2, 2021 at 15:16 | comment | added | Tim Campion | @varkor That's interesting -- I don't think I'm familiar with this characterization, what is it? I tried thinking about a Kleisli object as an EM object in the dual 2-category, but that didn't seem to immediately give the sort of thing you're talking about. | |
| Mar 2, 2021 at 15:12 | comment | added | varkor | Perhaps it's worth noting that to characterise when an adjunction is Kleisli involves only the left adjoint $F$. This doesn't seem like a coincidence. | |
| Mar 2, 2021 at 15:09 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Mar 2, 2021 at 15:08 | comment | added | Tim Campion | @PaulTaylor Alright, in that case, let's change the rules. Then name of the game is we can't refer to $U$, but we can talk about $\mathcal A, F, \mathcal B$ all we want. I agree the terminology "monadic adjunction" is more natural here than "monadic functor". I don't expect the condition will exactly mirror the Beck condition on $U$. | |
| Mar 2, 2021 at 15:01 | comment | added | Paul Taylor | If you're phrasing the condition in terms of the category $\mathcal B$ of would-be algebras, you're breaking the rules that you set. Besides, reflexive coequalisers are not the precise Beck condition. That would seem unavoidably to involve the forgetful functor (right adjoint). Shame, because my own usage is to call the adjunction monadic, rather than just the right adjoint. | |
| Mar 2, 2021 at 14:53 | comment | added | Tim Campion | @PaulTaylor I agree, it seems tricky. For instance, this putative condition would have to recognize when $\mathcal B$ is the EM category as opposed to the Kleisli category for the monad. Perhaps some notion of $\mathcal B$ having all reflexive coequalizers and having them be "free" with respect to $F$ would do it... | |
| Mar 2, 2021 at 14:50 | comment | added | Paul Taylor | Nice question, but my hunch is not. I think of monadicity as a kind of "completeness" of the adjunction. Beck's condition on the right adjoint says that there are enough coequalisers in what should be the category of algebras. The left adjoint only knows about the carriers, for each of which it gives a free algebra: it doesn't know about the other algebras. | |
| Mar 2, 2021 at 14:50 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Mar 2, 2021 at 14:41 | history | asked | Tim Campion | CC BY-SA 4.0 |