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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 impression that in general it is false (in the case if it is not monadic), may be it is true if we suppose that $x=F(y)$ for some $y\in M$ ?

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 $y\in M$ ?

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 impression that in general it is false (in the case if it is not monadic), may be it is true if we impose that $x=F(y)$ for some $y\in M$ ?

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 impression that in general it is false (in the case if it is not monadic), may be it is true if we suppose that $x=F(y)$ for some $y\in M$ ?

Suppose we have an adjunction of categories $F:M\leftrightarrow N:U$. 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 $colim_{n} G_{n}(x)=x$ ?

I'm under impression that in general it is false (in the case if it is not monadic), may be it is true if we impose that $x=F(y)$ for some $y\in M$ ?

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 impression that in general it is false (in the case if it is not monadic), may be it is true if we impose that $x=F(y)$ for some $y\in M$ ?