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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 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 $y\in M$ ?

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  • $\begingroup$ $G$ is a comonad, I think you get a cosimplicial object $G : \Delta\to N$. $\endgroup$
    – fosco
    Commented Feb 25, 2019 at 10:45
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    $\begingroup$ @FoscoLorengian You get both from a comonad. The canonical resolution arising from a comonad gives a simplicial object, while the cobar construction gives a cosimplicial object. An example of a simplicial object coming from a comonad is the left adjoint to the coherent nerve restricted to nerves of categories. This produces a simplicial object from the free-forgetful comonad on Cat. See: ncatlab.org/nlab/show/canonical+resolution $\endgroup$
    – Harry Gindi
    Commented Feb 25, 2019 at 11:07
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    $\begingroup$ Fosco, $\Delta$ (the augmented simplex category) is initial among strict monoidal categories equipped with a monoid. So given a monad $T = UF$, i.e., a monoid in the endofunctor category $[M, M]$, we get an induced cosimplicial object $\Delta \to [M, M]$, and dually we get a simplicial object $\Delta^{op} \to [N, N]$ which upon evaluation at $x \in Ob(N)$ gives a simplicial object $\Delta^{op} \to N$. $\endgroup$
    – Todd Trimble
    Commented Feb 25, 2019 at 12:53

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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.

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