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?



Derived topological stacks?
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.


