Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 23 / 8 / a / Solution

With $T=GF$, unit $\eta$ and multiplication $G\epsilon F$, the Eilenberg-Moore comparison functor isHere $G{\epsilon }_d:T(Gd)=GFGd\to Gd$ is the monad algebra action, and the underlying functor satisfies $UK=G$. The triangle identities and counit naturality provide the monad algebra laws and the compatibility of $Gh$, as required for this definition.