bio  website  math.mit.edu/~hoyois 

location  Cambridge, MA, USA  
age  28  
visits  member for  3 years, 8 months 
seen  1 hour ago  
stats  profile views  1,304 
I'm a postdoc at MIT.
1h

comment 
About the relation between the categories $\text{Sch}$, $\text{LRS}$ and $\text{RS}$
@user40276 I should have said that LRS ⊂ RS creates colimits. This is not the statement of Prop 1.6 but that's how it's proved. For limits it's clear for instance that $Spec(k)\times Spec(k')=\emptyset$ in LRS if $k$ and $k'$ are fields with different characteristics. I did mean section 5.1 for the statement that Sch ⊂ LRS preserves finite limits, which is also proved in the paper you linked to, but I don't think it's as formal as you suggest. It's definitely not true that any map from an affine scheme factors through an affine open... 
18h

comment 
About the relation between the categories $\text{Sch}$, $\text{LRS}$ and $\text{RS}$
In your last example, a pullback of fields will be a field... In fact LRS ⊂ RS preserves colimits, though of course not limits. This is Prop 1.6 in "Groupes algébriques" by Demazure and Gabriel. On the other hand, Sch ⊂ LRS preserves finite limits (section 5.1 in loc. cit.). I think it also preserves cofiltered limits with affine transition morphisms. 
Aug
28 
comment 
A “universally non Hypercomplete” $\infty$topos?
As you say, any $1$excisive functor $\mathcal{S}^\mathrm{fin}_* \to \mathcal{S}$ belongs to this localization, since $(*,S^0)$ is ∞connected, but I don't know if there's an unpointed analog of this (what's $\mathrm{Exc}^1(\mathcal{S}^\mathrm{fin},\mathcal{S})$ anyway? torsors under bundles of spectra?). In any case, I don't see why the reverse inclusion should hold, but I'm no expert in calculus either... 
Aug
28 
comment 
A “universally non Hypercomplete” $\infty$topos?
Right, the classifying ∞topos for (pointed) $n$connected objects is the left exact localization of $\mathrm{Fun}(\mathcal{S}^\mathrm{fin}_{(*)}, \mathcal{S})$ generated by the map $\tau_{\leq n} \to *$, and for $n=\infty$ it's their intersection. 
Aug
28 
comment 
A “universally non Hypercomplete” $\infty$topos?
So the ∞topos that classifies $(1)$connected objects is the subtopos of $\mathrm{Fun}(\mathcal{S}^\mathrm{fin},\mathcal{S})$ consisting of functors that are RKE of their restriction to nonempty spaces. Does a $1$excisive functor have this property? 
Aug
28 
comment 
A “universally non Hypercomplete” $\infty$topos?
Being a localization of presheaves on $\mathcal{S}^\mathrm{fin,op}_*$, the ∞topos of parametrized spectra should classify pointed objects with some property. So it can't be exactly the classifying topos you're looking for. 
Aug
17 
reviewed  Approve Sheaves and Differential Equations 
Aug
17 
reviewed  Approve Does Riemann map depend continuously on the domain? 
Aug
17 
reviewed  Approve Exceptional points for generalized northeastern knight walks in a quarter plane 
Aug
17 
awarded  Necromancer 
Aug
17 
awarded  Revival 
Aug
17 
answered  Why does this setting imply that a category is Grothendieck? 
Aug
17 
comment 
Triangulated structure on $\mathbf{SH}(S)$: $\mathbb{P}^1$suspension versus classical suspension
This is not true: suspension is always $\wedge S^1$. I think that's also what Riou means (he would say ``$T$suspension'' otherwise). 
Aug
15 
awarded  Enlightened 
Aug
15 
awarded  Nice Answer 
Jul
17 
comment 
Are there connections between Homotopy type theory and Grothendieck's theory of motives?
Well, motivic homotopy theory is a presentable locally cartesian closed ∞category, and therefore it admits a presentation by a typetheoretic model category. Of course, univalence does not hold... 
Jul
16 
comment 
What is the (co)homology of $K(\mathbb{R}_\delta,n)$?
$K(\mathbb R_\delta,n)$ cannot be a retract of a manifolds since a manifold is a countable CW complex and hence has countable homotopy groups. 
Jul
5 
answered  Does the forgetful functor from presentable $\infty$categories to $\infty$categories preserve filtered colimits? 
Jul
4 
comment 
What is the cokernel of a map of presentable stable $\infty$categories?
Yes, exactly: you can factor the functor $D\to C$ as a colocalization $D\to C'$ followed by a conservative functor $C'\to C$, and you get the same pushout if you replace $C$ by $C'$. 
Jul
4 
answered  What is the cokernel of a map of presentable stable $\infty$categories? 