5
$\begingroup$

Let $L_{Nis}(sPre(Sm_S))$ be the Nisnevich localization of the category of simplicial presheaves, how to see that whether $\mathbb{A}^1$-projections $\mathbb{A}^1\times_S X\to X$ are closed under homotopy pullback in $L_{Nis}(sPre(Sm_S))$? How to compute the homotopy pullback of maps $\mathbb{A}^1\times_S X\to X$ along itself in $L_{Nis}(sPre(Sm_S))$?

$\endgroup$
1
$\begingroup$

First, in any model category, weak equivalences are closed under homotopy pullback. That's the whole point of the "homotopy" in "homotopy pullback."

In your question, you are missing a step. Generally, $Sm_S$ is the category of smooth and finite type schemes over a scheme S of finite dimension. This category doesn't have good categorical behavior (it's not bicomplete, for example), so Voevodsky embedded it into simplicial presheaves $SPre(Sm_S)$, i.e. functors into simplicial sets. This has a model structure and you can do the localization $L_{Nis} SPre(Sm_S)$ that you wanted. To get the statement about $\mathbb{A}^1$-weak equivalences, you do a further localization to make $\mathbb{A}^1$ contractible, and it's this model structure that tells you homotopy pullbacks preserve weak equivalences. See Morel-Voevodsky.

The resulting model category is proper, meaning that if you look at a diagram $A \to B \gets C$ where one of those maps is a fibration, then you can compute homotopy pullback as just the regular old pullback, which is easy in presheaf categories. For a proof, see Jardine. If neither of those maps is a fibration, you can do fibrant replacement to replace one by a fibration.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The $A_1$-localization is for the class $S$ of all maps $A_1\times X\to X$. What will happen if localizing $L_{Nis}sPre$ at the homotopy pullback closure of $S$? $\endgroup$ – L. Xie Aug 20 '19 at 10:40
  • $\begingroup$ You would get the same result. In general, if you localize a model category with respect to a class of maps contained in the weak equivalences, nothing changes. So, if you first localize with respect to S, then with respect to the homotopy pullback closure of S (or just do both together, with this one localization), you get the same as if you just did it with respect to S. $\endgroup$ – David White Aug 20 '19 at 13:20
  • $\begingroup$ If just localizing with respect to $\bar{S}$ once, I expect them not to be the same. $L_{\bar{S}}$ is left exact. But it's a fact that motivic homotopy theory is not a model topos. What's the problem here? $\endgroup$ – L. Xie Aug 20 '19 at 14:44
  • $\begingroup$ I think we have different definitions of what it means to be "closed under homotopy pullbacks" - in my answer, it means that if you have a map of diagrams that is an objectwise weak equivalence then it induces a weak equivalence on homotopy limits. Googling your comment about "model topos" I find this paper by Raptis and Strunk, where it closure seems to mean something else arxiv.org/pdf/1704.08467.pdf $\endgroup$ – David White Aug 20 '19 at 15:28
  • 2
    $\begingroup$ @Alexis The homotopy pullback of $A^1\times X\to X$ along any map $F\to X$ is $A^1\times F\to F$, which is still an $A^1$-weak equivalence, so indeed the localization does not change. On the other hand, the class of all $A^1$-weak equivalences in $L_{Nis}sPre$ is not closed under homotopy pullbacks, so if you localize at the closure you will get a different localization (which still may or may not be an ∞-topos!). To actually get a left exact localization, you have to localize at the simultaneous closure of $A^1$-weak equivalences under homotopy colimits and homotopy pullbacks. $\endgroup$ – Marc Hoyois Aug 20 '19 at 16:18

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.