This question is inspired by [Characterization of functors whose right adjoint is monadic?](https://mathoverflow.net/questions/385363/characterization-of-functors-whose-right-adjoint-is-monadic).


Let $F : \mathbf C \rightleftarrows \mathbf D : U$ be an adjunction, and suppose that we want to establish when the canonical comparison functor $\mathbf{Kl}(UF) \to \mathbf D$ is an equivalence. A necessary and sufficient condition is that [$F$ be essentially surjective on objects](https://mathoverflow.net/questions/26075/characterization-of-kleisli-adjunctions).


Is it possible to characterise this condition in terms only of the right adjoint $U : \mathbf D \to \mathbf C$?



Characterisation of functors whose left adjoint is Kleisli
Try this. I confess I haven't written out the proofs yet, but will do so if you (and the community) think this is an appropriate answer. Please excuse my renaming your categories according to my personal convention.


First, $U:{\mathcal A}\to{\mathcal S}$ must be faithful and reflect invertibility, cf Beck's theorem.


We rewrite essential surjectivity of $F:{\mathcal S}\to{\mathcal A}$ in terms of univeral properties:


For every "algebra" $A$, ie object of $\mathcal A$,


there are a "set" $X$ and a "function" $e:X\to U A$ in $\mathcal S$ that is universal in the sense that


for every other "algebra" $B$ and "function" $f:X\to U B$ in $\mathcal S$ there is a unique "homomorphism" $h:A\to B$ in $\mathcal A$ that extends $f$ in the sense that $f=e;U h$.


How do we find the object $X$? This is going to invoke $F$ (if that's allowed by the question!)


It need not be unique (up to iso), but the construction of the replacement for $\mathcal S$ is essentially the one in [this paper of mine](http://www.paultaylor.eu/ASD/sobsc) and the ones by Hayo Thielecke and Peter Selinger that it cites.


If appropriate equalisers exist in $\mathcal S$ then the canonical $e:X\to U A$ (in fact the terminal one) is given by the equaliser of $U A\rightrightarrows U F U A$.


In terms of $X$, the two maps are $\eta U F X=\eta U A$ and $U F\eta X$. We can derive the second of these from the same problem starting with $F U A$ in place of $A$, in which case we get a split equaliser.


