Let $\mathcal{D} \underset{G}{\overset{F}{\leftrightarrows}}\mathcal{C}$ with $F\dashv G$ and $G$ fully faithful. Write $T=GF$. $\epsilon$ denotes the counit of the adjunction.
The comparison functor $K:\mathcal{D}\rightarrow TAlg$ for this adjunction is given by $K(D)=(GD,G\epsilon_D)$ and $K(f:A\rightarrow B) = G(f):G(A)\rightarrow G(B)$.
I want to prove that $K$ is an equivalence.
My attempt:
If I prove that $K$ is fully faithful and essentially surjective on objects, I'm done.
It was easy to see that $K$ inherits the fullness property of $G$. Similarly, using faithfulness of $G$ I was able to prove straight away that $K$ is faithful.
To prove that $K$ is essentially surjective: Let $(X,h)$ be a T-algebra. We want to show that there is $D\in\mathcal{D}$ such that $K(D)\overset{\phi}{\cong} (X,h)$.
Well, we can write $h:T(X)\rightarrow X$, so setting $D=T(X)$ and $\phi=h$ seems like a good bet. But I don't see why $h$ should be an isomorphism. I was thinking on trying to find $h^{-1}$ using the fullness of $G$, but failed.
Any hints?