The free–forgetful Eilenberg–Moore adjunction induces the given monad

Define $F^T(A)=(TA,{\mu }_A)$ and $F^T(f)=T(f)$. Naturality of $\mu$ gives $T(f)\circ {\mu }_A={\mu }_B\circ T^2(f)$, so $T(f)$ is an algebra homomorphism; the functor laws follow from those of $T$.

At an algebra $(A,a)$ define the counit component ${ϵ}_{(A,a)}:=a:(TA,{\mu }_A)\to (A,a)$; the algebra associativity law makes it an algebra homomorphism and the homomorphism equation makes these components natural. The equations $a\circ {\eta }_A=1_A$ and ${\mu }_A\circ T({\eta }_A)=1_{TA}$ are the two triangle identities, so $F^T⊣U^T$.

The composite $U^T{F}^T$ equals $T$, the adjunction unit is $\eta$, and $U^TϵF^T$ has component ${ϵ}_{(TA,{\mu }_A)}={\mu }_A$; hence the induced multiplication is $\mu$ and the induced monad is the given one on the nose.