 Solution

Write $F:\mathcal{X}\to \mathcal{A}$ and $G:\mathcal{A}\to \mathcal{X}$, with adjunction counit $\epsilon$. The induced monad has $T=GF$ and $\mu =G\epsilon F$. Define the Eilenberg-Moore comparison functor byThe action $a_A=G{\epsilon }_A:TGA\to GA$ satisfies $a_A{\eta }_{GA}=1_{GA}$ by a triangular equation. Naturality of $\epsilon$ at ${\epsilon }_A:FGA\to A$ givesApplying $G$ yields $a_{A}T(a_A)=a_A{\mu }_{GA}$, the other algebra identity. For $f:A\to A^{\prime }$, naturality givesso $Gf$ is a monad algebra morphism. Identities and compositions are respected because $G$ is a functor. Finally,and the same equalities hold on morphisms. Thus both comparison identities hold, strictly when $\mathbb{T}$ is the specified induced monad: