 Solution

A monad on $\mathcal{C}$ is an endofunctor $T$ with natural transformationssatisfying $\mu T\mu =\mu \mu T$ and $\mu T\eta =1_T=\mu \eta T$. Its Eilenberg-Moore category ${\mathcal{C}}^T$ has algebras for a monad $(A,a:TA\to A)$ satisfyingand morphisms of algebras for a monad $f:(A,a)\to (B,b)$ satisfying $fa=bT(f)$.

The Kleisli category ${\mathcal{C}}_T$ has the objects of $\mathcal{C}$ andIts identity is ${\eta }_A$, while the composite of $f:A\to TB$ and $g:B\to TC$ is ${\mu }_{C}T(g)f$. The free functor into a Kleisli category $F_T:\mathcal{C}\to {\mathcal{C}}_T$ is the identity on objects and sends $f:A\to B$ to ${\eta }_{B}f$.

On the functor category $[\mathcal{D},\mathcal{C}]$, postcomposition gives the pointwise monad on a functor categoryThe monad laws hold componentwise. A $T_{\ast }$-algebra is a functor $F:\mathcal{D}\to \mathcal{C}$ with a natural transformation $a:TF\to F$ whose components are $T$-algebras. Naturality says precisely that every $F(u)$ is an algebra morphism. Hence sending $(F,a)$ to the lifted functor $\mathcal{D}\to {\mathcal{C}}^T$ gives an isomorphism, and in particular an equivalence,

On $[\mathcal{C},\mathcal{D}]$, precomposition gives the precomposition monad on a functor categoryA $T^{\ast }$-algebra is a natural transformation $a:GT\to G$ satisfying $aG\eta =1_G$ and $a(aT)=aG\mu$. From it define $\bar{G}:{\mathcal{C}}_T\to \mathcal{D}$ byThe two algebra laws say exactly that $\bar{G}$ preserves identities and Kleisli composition, and $\bar{G}{F}_T=G$. Conversely, a factorization $G=\bar{G}{F}_T$ giveswhere $1_{TA}$ represents a Kleisli arrow $TA\to A$. These constructions are inverse on objects and morphisms, so