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I apologize for the vagueness of the following.

Informally, in the site of commutative rings, one roughly get the notion of a derived stack by swapping out the commmutative rings with its subcategory of simplicial objects. This is part of the story told by Toen and Vezzosi in their many expository accounts on derived geometry.

Now, there appears to be a fairly well-studied story for stacks defined in the site of topological spaces (topological stacks), notably in a series of papers by Noohi. Then it's natural to try to obtain a notion of a derived topological stack by replacing the source category by simplicial topological spaces. Has this been done by anyone? Is the story there trivial or uninteresting for some reason?

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    $\begingroup$ I don't think simplicial commutative rings form a subcategory of commutative rings in a natural way. $\endgroup$ – S. Carnahan Jan 20 '18 at 8:04
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    $\begingroup$ Simplicial rings correspond to cosimplicial affine schemes, so the source category for derived topological stacks would have to be cosimplicial topological spaces. And you would have to decide a sensible notion of weak equivalence for them. $\endgroup$ – Jon Pridham Jan 20 '18 at 9:58
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At Andre's suggestion, I'll turn my comments into an answer. If we correct the OP by asking about cosimplicial topological spaces, then we have to decide on a notion of equivalence for them, and the obvious one (tying in with DAG and derived differential geometry) is to ask for weak equivalence on the associated simplicial ring of $\mathbb{R}$-valued functions. But then it turns out that the derived structure is meaningless, as everything is equivalent to something with constant cosimplicial structure:

The reason for this is that the ring $A$ of $\mathbb{R}$-valued functions on a topological space has the structure of a $C^0$-ring, meaning that for every continuous function $f : \mathbb{R}^n \to \mathbb{R}$, we have a systematic way of evaluating $f(a_1, \ldots,a_n) \in A$ for all $a_1, \ldots, a_n \in A$. But any simplicial $C^0$-ring is discrete in the sense that $\pi_0A\simeq A$.

This can be proved by looking at the simplicial $C^0$-rings $P^n$ representing $\pi_n$ in the homotopy category. We find that the functor $\pi_n$ is identically $0$ because $\pi_n(P^n)=0$. For the standard cofibrant representative (given by $P^n_i=\mathbb{R}$ for $i<n$, $P^n_n=C^0(\mathbb{R})$, $P^n_{n+1}=C^0(\mathbb{R}^{n+1})$ etc.), this amounts to saying that any function $f$ defined on the line $\langle (1,1, \ldots,1)\rangle \subset \mathbb{R}^{n+1}$ which vanishes at the origin can be extended to a function on $\mathbb{R}^{n+1}$ vanishing on the co-ordinate axes. For continuous functions this can be done, whereas for smooth functions the derivative at $0$ is overdetermined.

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  • $\begingroup$ Have a few questions about this because I actually needed this result. 1) Why does it suffice to check that the homotopy groups of P^n vanish? 2) Why are those the standard cofibrant representatives here? 3) Why is their vanishing equivalent to the condition you stated about extending continuous functions? $\endgroup$ – John Rached Aug 30 at 16:40
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    $\begingroup$ 1) Let $u \in \pi_n(P_n)$ be the element corresponding to the identity on $P_n$. By functoriality, if $a \in \pi_n(A)$ is represented by $f : P_n \to A$, then $a=f_*u$. but $u=0$. $\endgroup$ – Jon Pridham Aug 30 at 19:25
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    $\begingroup$ 2) On pointed simplicial sets, $\pi_n$ is represented by $\Delta^n/\partial\Delta^n$. You then obtain $P_n$ by applying the relevant free functor, which sends a pointed set $(S,s)$ to $C^0(\mathbb{R}^{S\setminus \{s\}})$. $\endgroup$ – Jon Pridham Aug 30 at 19:31
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    $\begingroup$ 3) Homotopy groups are homology groups of the associated normalised chain complex. In this case, $N_nP^n$ consists of continuous functions on $\mathbb{R}$ vanishing at the origin, while $N_{n+1}P^n$ consists of functions on $\mathbb{R}^{n+1}$ vanishing on the co-ordinate axes. The chain differential is given by identifying $\mathbb{R}$ with the diagonal line in $\mathbb{R}^{n+1}$. $\endgroup$ – Jon Pridham Aug 30 at 19:36
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    $\begingroup$ Lest anyone try to apply my formula in (2) too widely, I should mention it only applies to finite pointed sets. For infinite sets, you have to take filtered colimits of the finite formula. $\endgroup$ – Jon Pridham Aug 30 at 19:41
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I am not aware of any theory of "derived topological stack", and I believe that the answer to your question is: No, this has not been considered by anyone yet.

I have no reason to believe that this yet to be considered theory would be trivial or uninteresting. But I am also not aware of any mathematical situation/problem that would naturally require one to develop such a theory.

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    $\begingroup$ Perhaps topological manifolds that are not smoothable would benefit from having access to a topological version of derived geometry, even if not arbitrary topological spaces. $\endgroup$ – David Roberts Jan 20 '18 at 23:05
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    $\begingroup$ Also a derived $C^\infty$ stack is an interesting and well studied notion - i.e. a stack in the setting where you replace smooth manifolds by cosimplicial smooth manifolds. In general derived geometry comes up when you have a good reason that you're unhappy with the given notions of intersection or fiber products, just as stacks come up when you're unhappy with your notion of quotients. The question is thus whether you have a need to improve fiber products of topological spaces. $\endgroup$ – David Ben-Zvi Jan 21 '18 at 1:25
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    $\begingroup$ There is an almost-example in the literature. A Kuranishi structure is presentation of a quasi-smooth $C^\infty$ derived stack. Pardon defines topological Kuranishi structures and builds vfc for them. However, he doesn't define a category of derived stacks, just goes directly from chart to vfc. Moreover, in his applications, the objects should be $C^\infty$, it's just difficult to prove that. $\endgroup$ – Ben Wieland Jan 22 '18 at 18:40
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    $\begingroup$ I'll just mention here that because continuous functions don't satisfy Hadamard's lemma, simplicial $C^0$-rings don't behave very well at all. They might even all be discrete (i.e. $A \simeq \pi_0A$). $\endgroup$ – Jon Pridham Jan 23 '18 at 21:56
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    $\begingroup$ @Jon Pridham: I think that your comments might be worthy of typing them as a response to the OP's question. It's a pretty convincing argument that the theory would indeed be quite dull. $\endgroup$ – André Henriques Jan 24 '18 at 21:17

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