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## Cech nerve as homotopy colimit?

Given a category $\mathcal{C}$ with a notion of covering $\{ U_{i} \rightarrow X \}$ for an object $X$ (say $\mathcal{C}$ is a Grothendieck site), we can form the Cech nerve

$$\cdots \coprod_{i}{U_{ijk}} \overrightarrow{\overrightarrow{\rightarrow}}\coprod_{i}{U_{ij}} \overrightarrow{\rightarrow} \coprod_{i} U_{i}$$

(In the notation, I've suppressed degeneracy maps going from right to left.)

This can be viewed in two ways. 1. As a simplicial object in simplicial presheaves, by considering each $U_{i}$ as a simplicial presheaf constant in the simplicial direction. I'll denote that by $U_{\bullet}$. 2. As a simplicial object in presheaves and hence a simplicial presheaf. I'll denote that by $\check{U}_{\bullet}$.

The latter $\check{U}_{\bullet}$ can be shown to be level-wise weakly equivalent to $colim(\coprod_{i}{U_{ij}} \overrightarrow{\rightarrow} \coprod_{i} U_{i})$, where we consider this as a simplicial presheaf constant in the simplicial direction.

On the other hand, we could compute $hocolim(U_{\bullet})$, and from things I read, this is supposed to be identified with/weakly equivalent to $\check{U}_{\bullet}$. One reason I am having difficulty seeing this is that I don't really understand $hocolim(U_{\bullet})$. Since each object in $U_{\bullet}$ is cofibrant, I would guess I could just take the usual colimit, but this seems to produce something constant in the simplicial direction, which is clearly wrong.

So the question is:

How to see if there is a weak equivalence $hocolim(U_{\bullet}) \rightarrow \check{U}_{\bullet}$?

Probably if I read through the many pages of material suggested in the comments to this question, I'd be able to figure this out. But a more direct answer would make that reading more fruitful for me, I think. At least, pointing out what things I need to know to figure this out would help.

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 I don't understand the question. For instance, you don't specify in what category you're taking homotopy colimits. And I don't understand when you say "...can be shown to be levelwise equivalent to...". That looks like wrong. – Fernando Muro Feb 7 2012 at 17:26 Sorry. I think I meant object-wise. I'll think about this and edit. – dhagbert Feb 7 2012 at 18:30

More generally, let $X_\cdot$ be a simplicial presheaf. As such, we can consider it as a simplicial object in presheaves, which in particular may be thought of as a simplicial object in simplicial presheaves $X'_\cdot.$ So we have:
$$X_\cdot:\Delta^{op} \to Set^{C^{op}}$$ and $$X'_\cdot =\left( \mspace{3mu} \cdot \mspace{3mu}\right)^{(id)} \circ X_\cdot:\Delta^{op} \to Set_{\Delta}^{C^{op}}$$ where $$\left( \mspace{3mu} \cdot \mspace{3mu}\right)^{(id)}:Set^{C^{op}} \to Set_{\Delta}^{C^{op}}$$ is the evident inclusion of presheaves into simplicial presheaves.
The homotopy colimit of $X'_\cdot$ in simplicial presheaves is computed "object-wise". Hence, for all $c \in C$ we have $$hocolim\left(X'\right)(C)=hocolim \left( X'\left(C\right)\right).$$ The right-hand side is the homotopy colimit of a simplicial object in simplicial sets, and can be computed by taking the diagonal of the corresponding bisimplicial set. But the diagonal of $X'\left(C\right)_\cdot$ is simply $X\left(C\right)_\cdot$ since $X'_\cdot$ in constant in one simplicial direction. Hence $$hocolim\left(X'\right)=X.$$