What is the homotopy colimit of the Cech nerve as a bi-simplical set?

Question

Let $f:Y_\bullet\to X_\bullet$ be an epimorphism of simplicial sets and define the bi-simplicial set $$ F_{\bullet\bullet}=\ldots Y\times_X Y\times_X Y\underset{\to}{\underset{\to}{\to}}Y\times_X Y \underset{\to}{\to}Y $$ as usual (link). Now $F_{\bullet\bullet}$ can be viewed as a diagram in simplicial sets and one can take its homotopy colimit $\operatorname{hocolim}(F_{\bullet\bullet})$ which is a simplicial set. This is weakly equivalent to the diagonal (or the realization) of $F_{\bullet\bullet}$.

Is $\operatorname{hocolim}(F_{\bullet\bullet})$ weakly equivalent to $X_\bullet$? What is a reference for this?

I believe this to be true since there is a similar statement called the nerve theorem for spaces and simplicial spaces instead of simplicial sets and bi-simplicial sets. Perhaps it is not sufficient for $f$ only being an epimorphism for the statement to hold. In this case, my question would be what the exact conditions on $f$ are.

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Comment on a comment to this question: If $X$ and $Y$ are discrete simplicial sets, the diagonal of $F_{\bullet \bullet}$ is the simplicial set $$ F_{\bullet}=\ldots Y\times_X Y\times_X Y\underset{\to}{\underset{\to}{\to}}Y\times_X Y \underset{\to}{\to}Y $$ where each $X$ and $Y$ is viewed as a set (and not as a simplicial set). This simplicial set $F_\bullet$ is weakly equivalent to the colimit $C$ (in the category of sets) of $$ Y\times_X Y \underset{\to}{\to}Y $$ and, if I understand it correctly, because an epimorphism $f$ is the same as an effective epimorphism of sets, $X$ is the colimit of this diagram and therefore $\operatorname{hocolim}(F_{\bullet\bullet})\cong diagonal(F_{\bullet\bullet})\cong F_{\bullet}\cong C \cong X$ in this discrete case.

On the other hand, in the example with $X$ being a point and $Y$ being two points, I think that $C$ is a space with two points which is mysterious. What am I doing wrong?

Answer 1

I am not sure what the precise conditions under which the weak equivalence is true, because the simplicial space you've written down isn't homotopy invariant unless $Y \to X$ is a fibration, or unless you take homotopy fiber products at every stage. But if you take homotopy fiber products at every stage, then the geometric realization is always $X$ (if $Y \to X$ is an epimorphism on $\pi_0$): this is a sort of homotopical descent property. However, I'm not sure if this is what you want.

One way to see this is that the simplicial object above is augmented: it lives naturally in the category of spaces over $X$. The (homotopy) pullback functor from spaces over $X$ to spaces over $Y$ preserves homotopy colimits (e.g., at the level of model categories, it is a left Quillen functor), and it's conservative (reflects homotopy equivalences) since $\pi_0 Y \to \pi_0 X $ is surjective, so it's sufficient to check that after you pull the simplicial object over $X$ back to $Y$, its geometric realization is equivalent to $Y$. But when you pull it back, it's augmented over $Y$ and has an extra degeneracy, so its realization is weakly equivalent to $Y$.

Answer 2

This is true.

Start with the map $Y_\bullet \to X_\bullet$ of simplicial sets and form the bisimplicial set $F_{\bullet \bullet}$ as you've already done. Fiber products in simplicial sets are calculated levelwise, and so $F_{p,q}$ is the ${p+1}$-fold fiber product of $Y_q$ over $X_q$, and there is a natural augmentation $F_{0,q} \to X$.

To check that you get an equivalence, it suffices to check that $|F_{\bullet \bullet}| \to |X_\bullet|$ is an equivalence. The former can be realized several ways: realize with respect to $p$ first, with respect to $q$ first, or take the diagonal and then realize. All of these are homeomorphic.

Let's do realization in the $p$ component first first. Then for any $q$, we get a map of spaces $|F_{\bullet,q}| \to X_q$, where the former space is built out of the iterated fiber products $(Y_q)^{p+1}$ over $X_q$. It turns out that this is a homotopy equivalence whenever $Y_q \to X_q$ is surjective.

(The shortest proof I know is that, given a section $s: X_q \to Y_q$, there is a simplicial contraction down to $s(x)$; given a simplex $(y_0,\ldots,y_p)$ over $x \in X_q$, we have a $(p+1)$-simplex $(s(x),y_0,\ldots,y_p)$ that contracts it down to $s(x)$, and this respects boundary identifications.)

Therefore, the map $|F_{\bullet,q}| \to X_q$, viewed as a map of simplicial spaces, is a levelwise equivalence, and so it becomes an equivalence on geometric realization $|F_{\bullet \bullet}| \to |X_\bullet|$. (Geometric realization takes weak equivalences to weak equivalences for "good" simplicial spaces, and bisimplicial sets always give "good" simplicial spaces when you realize one direction.)