Fix a base scheme $S$, and let $Sm_S$ be the (ordinary) category of smooth schemes over $S$.

  • Denote by $Psh(Sm_S)$ the $\infty$-category of $\infty$-presheaves on $Sm_S$.
  • Let $Sh_{Nis}(Sm_S)\subseteq Psh(Sm_S)$ be the full subcategory of presheaves which are sheaves for the Nisnevich topology. Denote $a_{Nis}: Psh(Sm_S)^\to_\leftarrow Sh_{Nis}(Sm_S): i_{Nis}$ the inclusion / sheafification adjunction. Note that $a_{Nis}$ and $i_{Nis}$ are left exact and accessible.
  • Let $Sh_{\mathbb A^1}(Sm_S) \subseteq Psh(Sm_S)$ be the full subcategory of presheaves which are $\mathbb A^1$-local. Denote by $a_{\mathbb A^1}: Psh(Sm_S)^\to_\leftarrow Sh_{\mathbb A^1}(Sm_S): i_{\mathbb A^1}$ the inclusion / localization adjunction. Again, $a_{\mathbb A^1}$ and $i_{\mathbb A^1}a_{\mathbb A^1}$, are left exact (EDIT: This is the problem -- $a_{\mathbb A^!}$ is not left exact!) and accessible.
  • Let $Spaces_S \subseteq Psh(Sm_S)$ be the intersection $Sh_{Nis}(Sm_S) \cap Sh_{\mathbb A^1}(Sm_S)$. I'll call this the unstable motivic category.

It's a fact that $Spaces_S$ is not an $\infty$-topos (cf. section 4.3 here). Nevertheless, here is a proof that it is an $\infty$-topos:

We exhibit $Spaces_S$ as a left exact localization of $Psh(Sm_S)$. Let $F: Psh(Sm_S) \to Psh(Sm_S)$ be the composite $i_{\mathbb A^1}a_{\mathbb A^1}i_{Nis} a_{Nis}$. Then $F$ is left exact and accessible. Then for any $Y \in Spaces_S$, we have that $Y$ is local with respect to the map $X \to F(X)$ for any $X \in Psh(Sm_S)$. By induction, $Y$ is also local with respect to $X \to F^\kappa(X)$ for any ordinal $\kappa$. Since $F$ is accessible, the chain $F^\bullet(X)$ eventually stabilizes at some $\kappa$; at this point we have an object of $Spaces_S$, so that the localization functor $L_{Mot}: Psh(Sm_S) \to Spaces_S$ is given by $X \mapsto F^{\infty}(X)$, where $\infty$ is the size of the universe.

Now, $L_{Mot} = F^\infty$ is a filtered colimit of left exact functors on a presheaf category, and therefore left exact. So $Spaces_S$ is an $\infty$-topos.

Question: What went wrong? Is the argument flawed, or one of the premises? I've cut a few corners, but not in ways that I think are inherently dodgy. I'd be happy to expand on such points.

  • $\begingroup$ I'm pretty confident that the construction of $L_{Mot}$ is correct -- at least that's what it sounds like when motivic people describe it. Formally, I think the argument shows that for sufficiently large $\kappa$, (1) $X \to F^\kappa(X)$ is a map with respect to which motivic spaces are local and (2) $F^\kappa(X)$ is a motivic space. This just means that $F^\kappa(X)$ is a reflection of $X$ in $Spaces_S$. $\endgroup$ – Tim Campion Jul 25 '18 at 3:21
  • $\begingroup$ Why does $F$ so-defined preserve colimits? PS I deleted my last comment since I realized that L should be a left rather than right adjoint. $\endgroup$ – Harry Gindi Jul 25 '18 at 3:24
  • $\begingroup$ $F$ doesn't preserve colimits. $F^\infty$ does (after restricting its codomain to $Spaces_S$), but that's only established indirectly. This is analogous tot he Grothendieck plus construction in ordinary topos theory or the $T_n$ functor in Goodwillie calculus -- although the iterates of the functor converge to something that preserves colimits (after restricting the codomain), the functor itself does not. $\endgroup$ – Tim Campion Jul 25 '18 at 3:29
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    $\begingroup$ That's the one thing I don't know how to prove, although I believe I've been told before that it's true. Perhaps I'm misremembering. And perhaps the subtlety is this: from one localization you can always universally generate a stronger localization which is left exact, so there is "a" topos that one might call $Sh_{\mathbb A^1}(Sm_S)$ which does have a left exact localization functor. But perhaps it's just not the same as $\mathbb A^1$ localization. If this is the case, then the reason for the failure to be a topos is in some sense much less subtle than I'd been led to believe. $\endgroup$ – Tim Campion Jul 25 '18 at 4:05
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    $\begingroup$ The homotopy localization functor preserves finite products but not all finite limits. $\endgroup$ – Dylan Wilson Jul 25 '18 at 7:52

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