I have a problem in showing that the free-forgetful adjunction $F\dashv U: \mathbf{Mon} \to \mathbf {Set}$ (call $\eta$ the unity and $\varepsilon$ the counity) is monadic, and it seems that the problem is in the definition of the comparison functor $K$.
The adjunction above gives rise to a monad $(T:=UF, \ \eta,\ \mu :=U*\varepsilon*F)$, where $*$ is the horizontal composition. The associated adjunction $F^T\dashv U^T: \mathbf{Set}^T \to \mathbf {Set}$ consists of the forgetful functor $U^T$, whose definiton is obvious, and the functor $F^T$ that on the objects (sets) is defined $X\mapsto (\mu _X:T^2X\to TX)$ and on the arrows acts like $T$.
The comparison functor is defined by $K=F^TU$ so, for a monoid $M$, $KM$ should be $F^TUM=\mu_{UM}:T^2UM\to TUM$, where $\mu_{UM}$ is the component in $UM$ of the $\mu$ defined above via horizontal composition. However I read that the action of $K$ should be $M\mapsto (\varepsilon_M:TM\to M)$, that actually makes clear why $K$ is essentially surjective (since $\varepsilon_M$ is the realization of the product we obtain again the initial monoid $M$). How do this two definition conciliate? Or am I misunderstanding something? (The definitions that I wrote come from my lessons). Thanks in advance
Best Answer
You have the wrong comparison functor. The functor $K$ must satisfy $U^TK=U$, so for a monoid $M$, the underlying set of the T-algebra $K(M)$ should be the underlying set of $M$, thus $K(M)=(U(M),h)$ for some function $h\colon TU(M)\to U(M)$ that satisfy the axioms of a $T$-algebra. In fact, in order for $K$ to also satisfy $KF=F^T$, this $h$ must be given by $U(\epsilon_M)\colon UFU(M)\to U(M)$.