# Kleisli categories of monads on Kleisli categories

In general, the composition of two monadic adjunctions is not necessarily itself monadic. Two known counterexamples include $$\mathbf{TorsionFreeAb} \to \mathbf{Ab} \to \mathbf{Set}$$ and $$\mathbf{Cat} \to \mathbf{Quiver} \to \mathbf{Set} \times \mathbf{Set}$$ (the latter is because $$\mathbf{Cat}$$, the category of small categories, is not regular).

Now, let $$C$$ be a category with a monad $$T$$ and $$S$$ be a monad on the Kleisli category $$C_T$$. Then, for any two objects $$X$$ and $$Y$$ in $$C$$, morphisms from $$X$$ to $$Y$$ in the Kleisli category $$(C_T)_S$$ correspond to morphisms from $$X$$ to $$SY$$ in $$C_T$$, which in turn correspond to morphisms from $$X$$ to $$TSY$$ in $$C$$. This suggests that $$(C_T)_S$$ is also the Kleisli category of a monad $$\bar{S}$$ on $$C$$ for which $$\bar{S}X = TSX$$ on objects.

Question:

Does there always actually exist such a monad $$\bar{S}$$? In other words, is the composition of two Kleisli adjunctions always itself a Kleisli adjunction?

• Related (but doesn't answer) : you can find an answer, not for the Kleisli category but for the Eilenberg-Moore category, with the keywords distributive law - there is for instance an exercise sheet on Samuel Mimram's website about these : lix.polytechnique.fr/Labo/Samuel.Mimram/teaching/cat – Max Jul 12 at 17:31

Surprisingly to me, yes-there's no need for any distributive laws or anything. If $$U:D\leftrightarrows C:F$$ is an adjunction, then by Peter Lumsdaine's answer here, our adjunction is equivalent to the Kleisli adjunction of the monad $$UF$$ if and only if $$F$$ is essentially surjective, and isomorphic to it if and only if $$F$$ is bijective on objects. Both these properties are closed under composition of functors.