Konrad Waldorf shows in his paper one may realize a Grothendieck topology on the category of diffeological spaces. Is there any work exploring stacks over the category of diffeologies?

2$\begingroup$ What do you mean by the category of diffeologies? If you mean the category of diffeological spaces, this is not much different from stacks on the site of Euclidean spaces, if I'm not mistaken. If you mean something else, then please clarify. $\endgroup$ – David Roberts Nov 9 '13 at 9:46

$\begingroup$ Yes, I mean the category of diffeological spaces; I'm not quite sure I've seen this similarity to stacks on Euclidean spaces... does this follow from some more general relationship between the sheaves on a site and stacks defined over the category of such sheaves? $\endgroup$ – Seth Wolbert Nov 9 '13 at 14:43

$\begingroup$ @Seth: Almost, but since Konrad defined an explicit Grothendieck pretopology using subductions, one has to check that this generates the topology induced from the ordinary one on manifolds. $\endgroup$ – David Carchedi Nov 9 '13 at 14:51
I will show that stacks over diffeological spaces are "the same" (in the sense of equivalence of 2categories) as ordinary stacks on manifolds.
The Grothendieck pretopology in question is the Grothendieck topology of "subductions". A map $f:X \to Y$ between diffeological spaces is a subduction if for every map $g:M \to Y$ with $M$ a manifold, the pullback $$X \times_Y M \to M$$ admits local sections. The pretopology consists of singleton subductions as covers.
Notice that a map between smooth manifolds is a subduction if and only if it admits local sections. Given an open cover $U_\alpha$ of a manifold $M,$ $$\coprod\limits_\alpha U_\alpha \to M$$ is a subduction, and any subduction between manifolds can be refined by such a map. It follows that the restriction of the subduction pretopology to manifolds generates the same Grothendieck topology (i.e. the same covering sieves) as the standard open cover pretopology.
Denote, the full and faithful inclusion of manifolds into diffeological spaces by $$i:Mfd \hookrightarrow DiffSp,$$ where I allow arbitrary disjoint unions of manifolds to be considered manifolds. Consider a diffeological space $X.$ Let $$X_n=\coprod\limits_{f \in Hom\left(\mathbb{R}^n,X\right)} \mathbb{R}^n.$$ $X_n$ is a manifold and has a canonical map $$X_n \to X.$$ Putting all these together we get a map $$\pi_X:\tilde X=\coprod\limits_n X_n \to X,$$ and $\tilde X$ is a manifold. Since any point of a manifold has a neighborhood diffeomorphic to $\mathbb{R}^n,$ it follows that $\pi_X$ is a subduction. By the comparison Lemma (see SGA 4, III), it follows that $i$ induces an equivalence $$Sh\left(Mfd\right) \simeq Sh\left(DiffSp\right)$$ between the topoi of sheaves on manifolds and diffeological spaces respectively, where on the left we have open covers, and on the right we have subductions. Moreover, by applying the comparison lemma again, we can replace the category $Mfd$ with the subcategory $Man$ consisting of only embedded submanifolds of Euclidean spaces. A standard argument implies that in fact this equivalence extends to an equivalence $$St\left(Man\right) \simeq St\left(DiffSp\right)$$ between their 2categories of stacks. (E.g. you could model these as 1truncated hypersheaves using the Jardine model strtucture, and the result follows from the ordinary comparison lemma).
In summary:
Stacks over diffeological spaces are "the same" as stacks over manifolds.

$\begingroup$ I should also mention that the equivalence can be realized by sending a stack $\mathscr{X}$ on $DiffSp$ to its restriction to $Man.$ $\endgroup$ – David Carchedi Nov 9 '13 at 15:06

$\begingroup$ Excellent, thank you. Is there any work done actually exploiting this relationship? I can think of situations where working over a complete/cocomplete site may have advantages... $\endgroup$ – Seth Wolbert Nov 9 '13 at 15:44

1$\begingroup$ I don't know of any work doing this. However, if all you want is to work over a complete and cocomplete site, why not work over the entire category of sheaves, rather than just concrete sheaves? You can use the Grothendieck topology generated by jointly surjective epimorphisms this is basically the "subduction topology" but without mentioning that your sheaves are concrete. $\endgroup$ – David Carchedi Nov 9 '13 at 15:47