$\require{AMScd}$We have a neat way to lift a monad along a *monadic* right adjoint, through a distributive law: in a setting like
$$
\begin{CD}
X @. X \\ 
@VUVV @VVUV\\
C @>>T> C
\end{CD}$$
if $U$ is monadic there is a monad on $X$ making the square (pseudo)commute if and only if there is a distributive law between $T$ and the monad $S$ such that $X=D^S$.


Now, consider the same monad $S$, but *another* adjunction $F\dashv G$ inducing it, with right adjoint $G : Y \to C$, not monadic.


I'd like to lift $T$ to $Y$, so that I have a pseudo-commutative square
$$
\begin{CD}
Y @>>> Y \\ 
@VGVV @VVGV\\
C @>>T> C
\end{CD}$$
can this be done, or monadicity is impossible to drop in these lifting theorems?


(I attempted to think about it, and I *think* one can at least lift $T$ to an *endofunctor* of $Y$ via a distributive law: but maybe monadicity is strictly necessary for this lifting to be a monad?)



Lift a monad along a generic right adjoint
When $C$ is complete, and $U$ is a fibration with complete fibers, we do find results of this kind. A concrete example of this idea is given by topological functors. I reccomend the introduction of the paper below.


**Semi-topological functors III: Lifting of monads and adjoint functors**. *Street, Tholen, Wischenewsk, Harvey*. JPAA 16 (1980) 299-314.


