Characterization of functors whose right adjoint is monadic?

Question

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.

Answer 1

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.