 Solution

Define the free algebra functor $F^T$ by $F^{T}X=(TX,{\mu }_X)$ and $F^{T}f=Tf$. The monad laws make ${\mu }_X$ an algebra action, and naturality of $\mu$ makes $Tf$ a monad algebra morphism. For an algebra for a monad $(A,a)$, definewith proposed inverse $g\mapsto aTg$. This inverse is a monad algebra morphism, sinceOne composite is $aTg{\eta }_X=a{\eta }_{A}g=g$. For a monad algebra morphism $h$, the other is $aT(h{\eta }_X)=aThT{\eta }_X=h{\mu }_{X}T{\eta }_X=h$. Both formulas commute with composition in a category in $X$ and with monad algebra morphisms in $(A,a)$, so the bijection is natural. Thus