I am trying to understand the proof of Theorem 4.19 in this paper, describing when does the comparison functor $\textbf{D}\to\textbf{C}^T$ gives an isomorphism. However, I am not quite sure what the author really trying to prove though. The argument he gives seems like a copy of Mac Lane's Theorem 1 in Section VI.7, from the "Categories for the Working Mathematicians".
The most puzzling part for me is $(i)\implies (ii)$. We need to show that whenever we have a pair of parallel morphism $f,g:C\to D$ whose image under $G$ has a absolute coequalizer, then $G$ (strictly) creates coequalizers for $f,g$. So we need to start with an absolute coequalizer $q:G(D)\to X$ of $Gf,Gg$ in $\textbf{C}$ and end up with a unique morphism $p:D\to Y$ in $\textbf{D}$ such that $Gp=q$ and $p$ gives a coequalizer of $f,g$. However, when I look at the proof (either the one in the link or Mac Lane's), it looks like that's not what it is doing. For example, the proof in the link starts with an absolute coequalizer in $\textbf{D}$, the thing that is supposed to be obtained at the end of the proof.
Can someone tell me what is the logic of the proof?
I think the proof is written in a slightly ambiguous way. The important point is that when the comparison functor $\textbf{D}\to\textbf{C}^T$ is an isomorphism, it is enough to prove that the forgetful functor $\textbf{C}^T\to\textbf{C}$ creates the good coequalizers. In other words, you can act like $\textbf{D}=\textbf{C}^T$.
The problem in that linked paper is that the identification of $\textbf{D}$ and $\textbf{C}^T$ is not done very explicitly, so there is some confusion about whether $f,g,q$ are in $\textbf{D}$ or $\textbf{C}$.
What happens is that the maps $f,g$ are in $\textbf{D}=\textbf{C}^T$, so they are really maps in $\textbf{C}$ which are compatible with the algebra structures of their domain and codomain; we assume that the corresponding maps in $\textbf{C}$ have an absolute coequalizer $q:D\to Q$ in $\textbf{C}$, and the goal is to prove that there is an algebra structure on $Q$ that makes $q$ the coequalizer of $f$ and $g$ in $\mathbf{C}^T$.