My question is :

Is there a classifying $\infty$-topos for $\infty$-connected objects ? Does this $\infty$-topos has a nice description (as an $\infty$-category ) ?

What I mean by $\infty$-connected object is an object $X$ such that all the $\pi_n(X)$ (seen as truncaturated objects of the $\infty$-topos) are isomorphic to the terminal object.

And by classifying $\infty$-topos I mean that for any other $\infty$-topos $\mathcal{E}$ there is a (natural) equivalence of $\infty$-category between $\infty$-connected object in $\mathcal{E}$ and geometric morphisms from $\mathcal{E}$ to our topos $\mathcal{T}$, this (inverse of this) equivalence being induced by $f \mapsto f^* (X)$ where $X$ is a given ("universal") $\infty$-connected object of $\mathcal{T}$.

I am relatively convinced that such an $\infty$-topos exists, so I'm more interested in knowing if it has an interesting description (something simpler than than a localization of the category of simplicial presheaves over the category of finite simplicial set would be great ! ).

This topos should have a unique point which should be its hypercompletion, I know one $\infty$-topos with this property (mentioned for example in this answer ). I guess it would be way too beautiful if this was the answer, but there is at least be a geometric morphism between them...

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    $\begingroup$ Tangentially, what does the $\infty$-topos of local systems of spectra classify? $\endgroup$ – Zhen Lin Aug 27 '15 at 10:05
  • $\begingroup$ 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. $\endgroup$ – Marc Hoyois Aug 28 '15 at 12:59
  • $\begingroup$ Well I wouldn't mind if one get a description of the classifying topos for pointed $\infty$-connected spaces instead. but even like that I don't see reason for point parametrized spectra to be the solution...... $\endgroup$ – Simon Henry Aug 28 '15 at 13:08
  • $\begingroup$ 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? $\endgroup$ – Marc Hoyois Aug 28 '15 at 13:20
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    $\begingroup$ The localization can be described more concretely: it's sheaves with respect to the Grothendieck topology where every morphism generates a covering. Equivalently, it's functors F: {finite spaces} -> {spaces} (or you can do a pointed version if you like) with the property that for any X, F(X) is the totalization of the cosimplicial space given by applying F to the "Cech nerve" of the map X->* (in the opposite of finite spaces). This is much larger than the class of 1-excisive functors: it contains all n-excisive functors for any n, and more (such as products of n-excisive functors as n varies). $\endgroup$ – Jacob Lurie Aug 30 '15 at 15:08

As discussed in the comments, I'm writing here the proof of the following fact:

Let $\mathscr{X}_∞$ be the ∞-topos of sheaves on $\mathrm{FinTop}^{op}$ under the atomic topology (the topology where a sieve is covering iff it is nonempty). Then for any other ∞-topos $\mathscr{X}$, there is a natural equivalence between the ∞-category of geometric morphisms $\mathscr{X}\to \mathscr{X}_∞$ and the ∞-category of ∞-connective objects in $\mathscr{X}$.

To do so, we will need a lemma.

Lemma: Let $\mathscr{X}$ be an ∞-topos and let $X\in\mathscr{X}$ be an object. Then all the following statements are equivalent

  1. The object $X$ is ∞-connected. That is $\pi_nX\to X$ is an equivalence for each $n\ge 0$ and $X\to *$ is an effective epimorphism.
  2. For every $n\ge -1$ the diagonal map $X\to X^{S^n}$ is an effective epimorphism (where $S^{-1}=\varnothing$)
  3. For every finite space $T$ the diagonal map $X\to X^T$ is an effective epimorphism.
  4. For every nonempty collection of maps of finite spaces $\{T\to T'_i\}_{i\in I}$ the map $\coprod_{i\in I} X^{T_i'}\to X^T$ is an effective epimorphism.

The lemma immediately implies the fact we were looking for thanks to proposition in Higher Topos Theory. In fact our condition 4 is exactly the condition that the left exact functor $\mathrm{FinTop}^{op}\to \mathscr{X}$ $$T\mapsto X^T$$ sends covering families to effective epimorphisms.

Proof: $4 \Rightarrow 3\Rightarrow 2$ is obvious. Let us prove $1\Leftrightarrow 2$. We know that $\pi_nX$ is the 0-truncation of $X^{S^n}\to X$ given by the evaluation at the basepoint. So $\pi_nX\to X$ is an equivalence iff $X^{S^n}\to X$ is a 0-connected. But by HTT., this is true iff the diagonal map $X\to X^{S^n}$ (which is a section of the evaluation) is -1-connected, i.e an effective epimorphism.

In the following we will often use HTT., that is if $gf$ is an effective epimorphism, so is $g$.

$2\Rightarrow 3$ This follows from an induction on the number of cells of $T$. Let us assume that it is true for $T$ and we will show it is true for $T'=T\amalg_{S^{n-1}}D^n$. There is a cofiber sequence $$T\to T'\to T'/T=S^n$$ and so we obtain a pullback square

$$\require{AMScd} \begin{CD} X^{S^n} @>>> X^{T'}\\ @VVV @VVV\\ X @>>> X^T \end{CD} $$ Since effective epimorphisms are stable under pullbacks (HTT., it follows that $X^{S^n}\to X^{T'}$ is an effective epimorphism. Finally, since $X\to X^{S^n}$ is an effective epimorphism by hypothesis and by HTT. effective epimorphisms are stable under composition, we are concluded.

$3 \Rightarrow 4$ This follows from HTT. applied to the compositions $$ X\to X^{T_j'}\to \amalg_{i\in I}X^{T_i'}\to X^T$$ for some $j\in I$. $\square$

  • $\begingroup$ For those scared by technicalities, the equivalence $1\Leftrightarrow 2$ is really the only important part of the proof, the rest is checking that the definitions work as intended. $\endgroup$ – Denis Nardin May 20 '18 at 21:45
  • $\begingroup$ Interesting. Apparently you need to wait 24 hours after setting a bounty before awarding it. $\endgroup$ – Tim Campion May 20 '18 at 22:03

To summarize the discussion in the comments:

  • $Fun(\mathsf{FinTop},\mathsf{Top})$ classifies objects

    (Proof: since a left adjoint functor $Fun(\mathsf{FinTop},\mathsf{Top}) \to \mathcal E$ corresponds to an arbitrary functor $\mathsf{FinTop}^\mathrm{op} \to \mathcal E$, and if that functor is left exact, it just corresponds to a functor from the terminal category to $\mathcal E$.)

  • $\infty$-connected objects are objects with a certain property, so they are classified by a localization $\mathcal T$ of $Fun(\mathsf{FinTop},\mathsf{Top})$, which is precisely the full subcategory of $\infty$-connected objects [$Fun(\mathsf{FinTop},\mathsf{Top})$ is a presheaf topos, hence hypercomplete. So it's a bit more subtle.].

  • Jacob Lurie's comment describes this localization as being at the Grothendieck topology generated by all maps ( but I don't know why this is so This is explained in Denis Nardin's answer).

  • This localization can be computed using the fact that every object $X$ has a finest cover in this topology, the map in $\mathsf{FinTop}^\mathrm{op}$ corresponding to the map $X \to \ast$ in $\mathsf{FinTop}$. We take a limit over the Cech nerve of this singleton cover in $\mathsf{FinTop}^\mathrm{op}$, whose $n$th level consists of an "$n$-pointed cone on $X$".

  • These cones also show up in Goodwillie calculus, so we see that an $n$-excisive functor $F$ lies in $\mathcal T$ because the descent object stabilizes at the $n$th stage. Moreover, the localization is closed under limits, so it contains all limits of $n$-excisive functors for varying $n$.

  • This leaves the obvious question: does $\mathcal T$ consist just of the limits of polynomial functors?

  • $\begingroup$ If you want I took time to figure out the description of the topos in terms of the atomic topology (the one generated by all nonempty sieves), and I could send you, or write in another answer, the details. The key point is proving that $X$ is ∞-connected iff the diagonal map $X\to X^ {S^n}$ is an effective epimorphism for all $n\ge -1$. $\endgroup$ – Denis Nardin May 20 '18 at 21:03
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    $\begingroup$ @DenisNardin That would be great, thanks! Having come this far, I'm interested in seeing how this bounty mechanism works, so if it's all the same to you, putting it in another answer would be nice. $\endgroup$ – Tim Campion May 20 '18 at 21:06
  • $\begingroup$ Done .(plus characters) $\endgroup$ – Denis Nardin May 20 '18 at 21:47

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