TITLE: Monad, algebras and reflexive coequalizer
QUESTION [5 upvotes]: Suppose we have an adjunction of categories $F:M\leftrightarrows N:U$. We define the associated (co)monad $G=F\circ U$. For any object $x\in N$ we define the simplicial resolution of $x$ given by 
$$
G_{\bullet}(x)=\dots  G^{2}(x)\rightrightarrows G(x)
$$
I was wondering if $\operatorname{colim}_{n} G_{n}(x)=x$ ? 
I'm under the impression that in general it is false (in the case when the adjunction is not monadic); maybe it is true if we suppose that $x=F(y)$ for some $y\in M$ ?

REPLY [4 votes]: Let $\eta: 1_M \to UF$ denote the unit and $\varepsilon: FU \to 1_N$ the counit of the adjunction. For $x = Fy$, we certainly get a split coequalizer 
$$FUFUFy \rightrightarrows FUFy \to Fy$$ 
in $N$, where the splitting is given by the pair of arrows $F\eta y: Fy \to FUFy$, $FUF\eta y: FUFy \to FUFUFy$, since $\varepsilon Fy \circ F\eta y = 1_{Fy}$ and $FU \varepsilon FY \circ FUF\eta y = 1_{FUFy}$ by triangular equations, and $\varepsilon FUFy \circ FUF\eta y = F\eta y \circ \varepsilon Fy$ by naturality of $\varepsilon$. 
I may add more later when I get a spare moment.