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Let $F: \mathcal A^\to_\leftarrow \mathcal B: U$ be an adjunction, and suppose we want to know whether the comparision functor $\mathcal B \to Alg^{UF}$ is an equivalence, where $Alg^{UF}$ is the category of algebras for the monad $UF$. The Beck monadicity theorem gives a necessary and sufficient condition for this to be the case, which can be verified by looking just at $U$. That is, all we need to know about $F$ in order to apply Beck's theorem is that $F$ exists; the condition "$U$ creates coequalizers of $U$-split pairs" refers only to $U$ and not to $F$. I wonder if there is dually some necessary and sufficient criterion for monadicity which can be checked by looking at $F$ only (so that all we need to know about $U$ is that it exists)?

Question: Given a functor $F$ which is known to have a right adjoint $U$, is there some way to check whether $U$ is monadic by looking just at $F$ (and $\mathcal A, \mathcal B$), so that all we need to know about $U$ is that it exists?

For my purposes, I'm not at all averse to making strong assumptions about $\mathcal A, \mathcal B$, like (co)completeness assumptions, exactess assumptions, etc. Just so long as I don't have to explicitly consider $U$.

To put a finer point on it, if the hypotheses of the adjoint functor theorem hold, then Beck's theorem can be used to show that $U$ is monadic without referring to $F$ at all -- $F$ can be verified to exist by verifying that $U$ preserves limits and satisfies the solution set condition, and then the other condition for the monadicity theorem likewise refers only to $U$. So dually, I'm looking for a criterion which would give monadicity of the right adjoint of $F$ which, in the presence of the adjoint functor theorem, might never require me to explicitly refer to that right adjoint at all.

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    $\begingroup$ 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. $\endgroup$ – Paul Taylor Mar 2 at 14:50
  • $\begingroup$ @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... $\endgroup$ – Tim Campion Mar 2 at 14:53
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    $\begingroup$ @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 $U$. $\endgroup$ – Tim Campion Mar 2 at 15:08
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    $\begingroup$ 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. $\endgroup$ – varkor Mar 2 at 15:12
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    $\begingroup$ Up to equivalence, a left adjoint is Kleisli if it is essentially surjective. For instance, see this question. $\endgroup$ – varkor Mar 2 at 15:19
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Let $F: C \to D$ be a left adjoint functor. I hope I'm not saying anything stupid, but I think you can just rephrase the two conditions of Beck Monadicity theorem in terms of the left adjoint:

The condition that $U$ is conservative translate as:

  • The $Hom(F(x),\_)$ are jointly conservatives.

It can also be replaced by the apparently stronger condition but more familiar:

  • The essential image of $F$ is dense,

As this condition is known to holds for monadic functors and implies the previous one.

The other condition is a bit harder. But I think we can manage if we assume that the domain of $F$ is Cauchy complete using the following lemma.

Lemma: Let $C$ be a Cauchy complete category. Then a pair $X \rightrightarrows Y$ admit a split coequalizer if and and only if $Hom(A,X) \rightrightarrows Hom(A,Y)$ admits a split coequalizer for each object $A \in C$ functorially in $A$.

I suspect one can remove the "functorially in $A$", but that was anoying to check, if someone has the motivation to do it, let me know ! In any case, the functoriality makes the proof very simple: this provides a split hence absolute coequalizer in the category of presheaves on $C$, but as $C$ is Cauchy complete any absolute colimits of representables is representable, hence the split coequalizer is already in $C$.

So assuming $C$ is Cauchy complete the second condition can be rephrased as:

  • Every pair $X \rightrightarrows Y$ in $D$ such that $Hom(F(A),X) \rightrightarrows Hom(F(A),Y)$ admit (functorially in $A$) a split coequalizer for each $A \in C$, admits a coequalizer in $D$ that is preserved by $Hom(F(A),\_)$ for each $A \in C$.

Of course if you can remove the functoriality in the lemma, then you can remove it here as well.

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    $\begingroup$ I think that what you say is imprecise, but true. $U$ conservative only implies that the essential image of $F$ is a strong generator. Linton proved that in the case of a category of algebras, the free algebras are dense, but it does not trivially follow from the conservativity of $U$. $\endgroup$ – Ivan Di Liberti Mar 2 at 15:44
  • $\begingroup$ I think you are right, I missed that However the condition that the essential image of $F$ is dense implies what we want and we know it holds for monadic functor, so that is indeed correct. $\endgroup$ – Simon Henry Mar 2 at 15:49
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    $\begingroup$ I love "functrotially". :-) $\endgroup$ – Paul Taylor Mar 2 at 15:52
  • $\begingroup$ @Ivan what is the correct name for the condition "jointly conservative" I have written ? it is equivalent to strong generator ? or is it only under special additional conditions ? I never remember... $\endgroup$ – Simon Henry Mar 2 at 15:55
  • $\begingroup$ Using that $U$ must be also faithful, yes this is the same of strong generator, otherwise it would not be. Although, a conservative right adjoint between categories with equalizers is always faithful. $\endgroup$ – Ivan Di Liberti Mar 2 at 15:58

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