By site of manifolds Man, I mean the category of manifolds (maybe submanifolds to obtain a small category) with continuous maps between them. A Grothendieck topology is given by open covers. Actually, I am more interested in the corresponding smooth site but the question may be posed for both.

Daniel Dugger states this (implicitely) in "Sheaves and Homotopy Theory" (http://math.mit.edu/~dspivak/files/cech.pdf) by saying that Cech weak equivalences of presheaves on manifolds can be detected stalkwise. However, this paper is unfinished and the proof is missing. In the paper "Universal Homotopy Theories" he uses the hypercompletion of presheaves on manifolds instead.

In "Differential Cohomology in a cohesive (infinitiy)-topos" (http://ncatlab.org/schreiber/files/cohesivedocumentv032.pdf) Urs Schreiber proves that the subsite consisting of the manifolds R^n is cohesive which implies that it is hypercomplete. However, the proof cannot be generalized since there are non-contractible manifolds.

Since hypercompleteness is a local criterion it suffices to check that the subsites Man|X (overcategory) are hypercomplete. In an attempt to prove this, I found a criterion in HTT saying that an (infinity)-topos which is locally of homotopy dimension <=n is hypercomplete (7.2.1.12) and I hoped to show this for simplicial presheaves on a subsite Man|X (every manifold in such a subsite has the same dimension) using that the (representables of the) contractible open sets generate this (infinity)-topos under colimits. But I failed to identify the corresponding overcategory since this should be the (infinity)-overcategory.

in a canonical way,i.e. that it is a dense sub -infinity category. – David Carchedi May 17 '13 at 21:07maybenot of infinity-sheaves. You can have a subsite whose topos of sheaves is equivalent to the whole topos of sheaves, but whose topos of infinity-sheaves is not equivalent to the whole infinity topos of infinity sheaves. This can't happen if everything is hypercomplete, however, this is what he is trying to prove, so we go in a circle. – David Carchedi May 17 '13 at 21:09