8
$\begingroup$

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.


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?

$\endgroup$
4
  • 1
    $\begingroup$ The diagonal of a bisimplicial set is weakly equivalent to its homotopy colimit: see here. $\endgroup$ – Zhen Lin Apr 24 '13 at 0:12
  • $\begingroup$ What happens if $X$ is a point? Furthermore what if $Y$ is two points? $\endgroup$ – Baby Dragon Apr 24 '13 at 1:40
  • $\begingroup$ Dear @BabyDragon, I've made an edit to the question concerning your example. $\endgroup$ – Ronald Bernard Apr 24 '13 at 9:12
  • $\begingroup$ I am not sure that you are doing anything wrong. Thank you for your reply. I thought that contemplating of the simplest possible nontrivial case would hopefully be enlightening. $\endgroup$ – Baby Dragon Apr 24 '13 at 14:02
8
$\begingroup$

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 $Y$.

$\endgroup$
4
  • $\begingroup$ Dear @Akhil Mathew, thank you for the great answer. I am trying to grasp the things you said in the terms of model categories or $\infty$-topoi. Let us assume we know that $Sets^{\Delta^{op}}$ is such. Let $f:Y\to X$ be the fibration. The first property ($f^*$ preserves homotopy colimits) seems to be true since homotopy colimits are universal (Higher topos theory, definition 6.1.1.2) but I can't find the second property ($f^*$ is ''conservative''). Do you know how this property is called in Lurie's? Is it one of the $\infty$-topos properties? It seems to be more crucial than the first one. $\endgroup$ – Ronald Bernard Apr 25 '13 at 13:03
  • 1
    $\begingroup$ @Ronald: The conservativity property follows from the long exact sequence of a fibration. One way to see it, though, is that you can think of fibrations over $X$ as being functors from $X$ (considered as an $\infty$-category) into spaces, and the pull-back of a fibration from $Y$ to $X$ corresponds to restricting the functor from $X$ to $Y$. Since every object in $X$ is isomorphic to something in the image of $Y$, the pull-back on functors is conservative. $\endgroup$ – Akhil Mathew Apr 25 '13 at 13:18
  • 2
    $\begingroup$ Akhil gave excellent replies. As to where to find this in Lurie's book: around 7.2.1.14, 7.2.1.15. A digest is here: ncatlab.org/nlab/show/… $\endgroup$ – Urs Schreiber Apr 25 '13 at 14:15
  • $\begingroup$ Dear AkhilMathew, thank you. Dear @Urs Schreiber, thank you for the references. $\endgroup$ – Ronald Bernard Apr 25 '13 at 14:42
3
$\begingroup$

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.)

$\endgroup$

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.