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.

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 15:08leftadjoint $F$. This doesn't seem like a coincidence. $\endgroup$ – varkor Mar 2 at 15:122more comments