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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?

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  • $\begingroup$ 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 $\endgroup$ – Max Jul 12 at 17:31
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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.

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