 Solution

For the monad $\mathbb{T}=(T,\eta ,\mu )$ define the free algebra functor byThe unit and associativity identities of the monad make $(TX,{\mu }_X)$ an algebra for a monad, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $G^{\mathbb{T}}$ be the forgetful functor. For $(A,a)$ in the Eilenberg-Moore category, setThe inverse candidate is a monad algebra morphism, sinceThe two composites are identities:These use respectively the unit law for a monad algebra, the algebra-morphism equation, and a unit identity of the monad. The formulas commute with precomposition in $X$ and postcomposition by monad algebra morphisms, so they form a natural bijection. Therefore the free-algebra functor is left adjoint to forgetting:Its adjunction unit is ${\eta }_X$, and its adjunction counit at $(A,a)$ is $a:(TA,{\mu }_A)\to (A,a)$.