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

$$ \cdots \coprod_{i}{U_{ijk}} \substack{\rightarrow \\ \rightarrow \\ \rightarrow}\coprod_{i}{U_{ij}} \rightrightarrows \coprod_{i} U_{i}$$

(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 $\operatorname{colim}(\coprod_{i}{U_{ij}} \rightrightarrows \coprod_{i} U_{i})$, where we consider this as a simplicial presheaf constant in the simplicial direction.

On the other hand, we could compute $\operatorname{hocolim}(U_{\bullet})$, 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 $\operatorname{hocolim}(U_{\bullet})$. 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 $\operatorname{hocolim}(U_{\bullet}) \rightarrow \check{U}_{\bullet}$?

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.

  • $\begingroup$ 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. $\endgroup$ Feb 7, 2012 at 17:26
  • $\begingroup$ Sorry. I think I meant object-wise. I'll think about this and edit. $\endgroup$
    – dhagbert
    Feb 7, 2012 at 18:30

1 Answer 1


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 $$hocolim\left(X'\right)=X.$$

  • $\begingroup$ Thanks. I think that answers the question. My basic misunderstanding must be that I didn't realize that homotopy colimits of simplicial presheaves could be computed object-wise. Of course the unhomotopy colimits are. (I suppose you mean hocolim instead of holim, in the above.) I'll think about this more and will probably accept the answer later. $\endgroup$
    – dhagbert
    Feb 7, 2012 at 18:34
  • $\begingroup$ HTT 4.2.4 states that colimits in an infinity category associated to a simplicial model category can be computed as homotopy colimits in the model category. But, by essentially the same proof as in the 1-categorical case, colimits in infinity presheaves can be computed object-wise. $\endgroup$ Feb 7, 2012 at 21:45
  • $\begingroup$ (I meant to say hocolim, yes) $\endgroup$ Feb 7, 2012 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.