[proofplan] We must show first that the displayed map $G\varepsilon_d: T(Gd) \to Gd$ makes $Gd$ into a $T$-algebra. The unit axiom follows from the triangle identity for the adjunction, while the associativity axiom follows by applying $G$ to the naturality square of the counit at the morphism $\varepsilon_d: FGd \to d$. We then check that each $Gh$ is a morphism of $T$-algebras by applying $G$ to naturality of the counit at $h$, and ordinary functoriality follows from the functoriality of $G$. [/proofplan] [step:Verify the unit axiom for the algebra on $Gd$] Fix an object $d \in \mathcal D$. Define \begin{align*} a_d: T(Gd) = GFGd \to Gd \end{align*} by \begin{align*} a_d := G\varepsilon_d. \end{align*} The unit axiom for a $T$-algebra on $Gd$ requires \begin{align*} a_d \circ \eta_{Gd} = \operatorname{id}_{Gd}. \end{align*} Substituting $a_d = G\varepsilon_d$, this becomes \begin{align*} G\varepsilon_d \circ \eta_{Gd} = \operatorname{id}_{Gd}, \end{align*} which is exactly one of the triangle identities for the adjunction $F \dashv G$. Hence $a_d$ satisfies the unit axiom. [/step] [step:Verify the associativity axiom using naturality of the counit] The associativity axiom for the $T$-algebra structure $a_d: T(Gd) \to Gd$ requires \begin{align*} a_d \circ T(a_d) = a_d \circ \mu_{Gd}. \end{align*} By definition of $T = GF$, of $a_d = G\varepsilon_d$, and of $\mu = G\varepsilon F$, the two sides are \begin{align*} a_d \circ T(a_d) &= G\varepsilon_d \circ GF(G\varepsilon_d), \\ a_d \circ \mu_{Gd} &= G\varepsilon_d \circ G\varepsilon_{FGd}. \end{align*} Since $G$ is a functor, it is enough to prove in $\mathcal D$ that \begin{align*} \varepsilon_d \circ FG(\varepsilon_d) = \varepsilon_d \circ \varepsilon_{FGd}. \end{align*} This is precisely the naturality of the counit $\varepsilon: FG \Rightarrow \operatorname{id}_{\mathcal D}$ applied to the morphism \begin{align*} \varepsilon_d: FGd \to d. \end{align*} Indeed, naturality gives \begin{align*} \varepsilon_d \circ FG(\varepsilon_d) = \varepsilon_d \circ \varepsilon_{FGd}. \end{align*} Applying $G$ yields \begin{align*} G\varepsilon_d \circ GF(G\varepsilon_d) = G\varepsilon_d \circ G\varepsilon_{FGd}. \end{align*} Therefore \begin{align*} a_d \circ T(a_d) = a_d \circ \mu_{Gd}, \end{align*} so $K(d) = (Gd,a_d)$ is a $T$-algebra. [/step] [step:Check that $Gh$ is a morphism of $T$-algebras] Let $h: d \to d'$ be a morphism in $\mathcal D$. We must show that \begin{align*} Gh: Gd \to Gd' \end{align*} is a morphism from the $T$-algebra $(Gd, G\varepsilon_d)$ to the $T$-algebra $(Gd', G\varepsilon_{d'})$. This means verifying \begin{align*} Gh \circ G\varepsilon_d = G\varepsilon_{d'} \circ T(Gh). \end{align*} Since $T = GF$, the right-hand side is \begin{align*} G\varepsilon_{d'} \circ GF(Gh). \end{align*} Naturality of the counit $\varepsilon: FG \Rightarrow \operatorname{id}_{\mathcal D}$ applied to $h: d \to d'$ gives \begin{align*} h \circ \varepsilon_d = \varepsilon_{d'} \circ FG(h). \end{align*} Applying the functor $G$ gives \begin{align*} Gh \circ G\varepsilon_d = G\varepsilon_{d'} \circ GFG(h). \end{align*} Since $GFG(h) = GF(Gh) = T(Gh)$, this is exactly the algebra morphism condition. Hence $K(h)=Gh$ is a morphism in $\mathcal C^{\mathsf{T}}$. [/step] [step:Use functoriality of $G$ to obtain functoriality of $K$] For every object $d \in \mathcal D$, \begin{align*} K(\operatorname{id}_d) = G(\operatorname{id}_d) = \operatorname{id}_{Gd} = \operatorname{id}_{K(d)} \end{align*} as a morphism in $\mathcal C^{\mathsf{T}}$. If $h: d \to d'$ and $k: d' \to d''$ are morphisms in $\mathcal D$, then \begin{align*} K(k \circ h) = G(k \circ h) = Gk \circ Gh = K(k) \circ K(h). \end{align*} The preceding step shows that these underlying morphisms in $\mathcal C$ are morphisms of $T$-algebras, so the equalities hold in $\mathcal C^{\mathsf{T}}$. Therefore $K: \mathcal D \to \mathcal C^{\mathsf{T}}$ is a well-defined functor with the stated object and morphism assignments. [/step]