Souriau's definition of diffeology may be phrased as defining a concrete sheaf on the category $\mathsf{Open}$ of open subsets of Euclidean/coordinate spaces. It seems to me, unless I am missing something, that any such sheaf extends to a sheaf on the site $\mathsf{Man}$ of all smooth manifolds. Is there a proof written down somewhere?

More generally it seems to me that the inclusion $\mathsf{Open} \hookrightarrow \mathsf{Man}$ should induce an equivalence of 2-categories between the 2-category of stacks on $\mathsf{Man}$ and the stacks on $\mathsf{Open}$. Is this true? If so, is there a reference? (Metzler seems to mentions something like this in passing in here: arXiv:math/0306176 [math.DG] .)


2 Answers 2


This would be the "comparison lemma" from SGA 4-1, Éxposé III, Thm. 4.1: if $C$ is a full subcategory of a site $D$, equipped with the induced topology, and if every object of $D$ is covered by objects of $C$, then the restriction functor $Shv(D)\to Shv(C)$ is an equivalence of categories.

More generally, for any finite $n$, the $n$-topoi of $n$-stacks on $C$ and $D$ are equivalent iff the $1$-topoi of $1$-stacks (=sheaves) are, so the answer to your second question is also yes. This is because the $n$-topos of $n$-stacks on a $1$-site is $1$-localic, which means that it belongs to the image of the fully faithful embedding of $1$-topoi into $n$-topoi. This is disussed in Higher Topos Theory, 6.4.5.

  • $\begingroup$ Thanks. Would you know of a more "elementary" reference for part 2 of my question? $\endgroup$ Jul 26, 2013 at 19:41
  • $\begingroup$ I don't know any other reference, sorry. $\endgroup$ Jul 27, 2013 at 8:30
  • $\begingroup$ Actually Metzler references J. Giraud, Cohomologie non abélienne, p. 91, where the special case $n=2$ is proved (even for stacks of categories). $\endgroup$ Jul 27, 2013 at 10:22
  • $\begingroup$ You can also piggyback from the $n=1$ case, by taking the Jardine model structure on simplicial sheaves (for hypersheaves), and left Bousfield localizing to recover the 1-truncated objects (the stacks of groupoids). Since the two categories of simplicial sheaves would become equivalent (as 1-categories), in particular you get a Quillen equivalence, and hence an equivalence between their 2-categories of stacks. $\endgroup$ Oct 4, 2013 at 21:57

Actually, one doesn't need the comparison lemma in this case. As it turns out, $\mathbf{Man}$ is the Karoubi envelope of $\mathbf{Open},$ (see the Examples section of http://ncatlab.org/nlab/show/Karoubi+envelope), which implies that if $$i:\mathbf{Open} \hookrightarrow \mathbf{Man}$$ is the canonical inclusion, the induced restriction functor $$i^*:Psh_n\left(\mathbf{Man}\right) \to Psh_n\left(\mathbf{Open} \right)$$ between their categories of presheaves of $n$-groupoids for any $n$ is already an equivalence.

Edit: In this question (Proof that the category of presheaves on a category $C$ is equivalent to the category of presheaves on its Karoubi envelope) it discusses that presheaves of sets on the Karoubi envelope of a small category is equivalent to presheaves on the original category, and gives a reference.

Another Edit: First I gave the following argument, but it seems to have a gap (skip over this to get to the direct answer):

Now, consider the functor $i_!:Psh_\infty\left(\mathbf{Open} \right) \to Psh_\infty\left(\mathbf{Man}\right)$ which is left adjoint to $i^*.$ Consider the composite $i^*i_!$ which is colimit preserving. It also restricts to an equivalence on $0$-truncated objects, by the above (wait: Why shoudl $i_!$ send $0$-truncated objects to $0$-truncated objects?). If $F$ is an arbitrary presheaf on $\mathbf{Open}$, then $F$ can be represented as a simplicial presheaf, hence there exists a simplicial diagram $c_F:\Delta^{op} \to Psh_\infty\left(\mathbf{Open} \right)$ for which each ${c_F}_n$ is $0$-truncated and such that the colimit of $c_F$ is $F.$ Since $i^*i_!$ is colimit preserving, it must send $F$ to itself. A similary argument works using the composite $i_!i^*$, and one concludes that $i_!$ and $i^*$ form an equivalence of $\infty$-categories. In particular, they restrict to an equivalence between $n$-truncated objects for any $n$, hence the induced map $$i^*:Psh_n\left(\mathbf{Man}\right) \to Psh_n\left(\mathbf{Open} \right)$$ is an equivalence for all $n$.

Unfortunately (see the bold wait) I don't see how to show that $i_!$ preserves $0$-truncated objects (i.e. agrees with the 1-categorical left Kan extension when restricted to presheaves of sets) until I show its an equivalence, so here's another way to finish the argument:

(Start reading again here if you skipped): Consider the restriction functor $$i^*:Psh\left(\mathbf{Man}\right) \to Psh\left(\mathbf{Open} \right)$$ of presheaves of sets, which has a right adjoint $i_*$. (Explicitly, by the Yoneda lemma, one has that $i_*(F)(M)\cong Hom(i^*y(M),F)$ where $y$ is the Yoneda embedding.). Since $i^*$ also has a left adjoint $i_!,$ $i^*$ is left exact, and since $i$ is full and faithful, so is $i_*.$ So we have that the adjunction $(i^*,i_*)$ exhibits $Psh\left(\mathbf{Open} \right)$ as a left exact localization of $Psh\left(\mathbf{Man}\right)$. Hence, there is a unique Grothendieck topology $J$ on $\mathbf{Man}$ for which $Sh_J\left(\mathbf{Man}\right)\cong Psh\left(\mathbf{Open}\right)$ (more precisely, such that the localization induced by sheafification agrees with the one above). However, since we know that $i^*$ is an equivalence, it implies that this Grothendieck topology must be the trivial one. Notice that we also have an induced adjunction $(i^*,i_*)$ between $\infty$-presheaves. (Here there is no danger of the abuse of notation, since both functors are left exact, so preserves $n$-truncated objects for all $n$, so their restriction to presheaves of sets agree with the ones above). This adjunction is still a left exact localization, and since $Psh_\infty\left(\mathbf{Man}\right)$ is $1$-localic, it must again correspond to a unique Grothendieck topology. This left exact localization factors uniquely as a topological localization (one coming from a Grothendieck topology), followed by a cotopological one (one for which the only monos sent to equivalences are equivalences). The covering sieves of the topology corresponding to the topological part of the localization correspond exactly to those monos $f:S \to y(M)$ such that $i^*(f)$ is an equivalence. However, subobjects of representable objects in $Psh_\infty\left(\mathbf{Man}\right)$ are the same as subobjects in $Psh\left(\mathbf{Man}\right),$ so one sees the resulting class of covering sieves must be the same as for the $1$-categorical case, which we have argued only gave the maximal sieves (the trivial Grothendieck topology). It follows that the localization must be cotopological. Since the $\infty$-category of $\infty$-groupoids is hypercomplete and colimits are computed pointwise in presheaves, $Psh_\infty\left(\mathbf{Man}\right)$ is hypercomplete, and hence the localization must be trivial. Hence $$i^*:Psh_\infty\left(\mathbf{Man}\right) \to Psh_\infty\left(\mathbf{Open} \right)$$ is an equivalence, and by restriction to $n$-truncated objects, $$i^*:Psh_n\left(\mathbf{Man}\right) \to Psh_n\left(\mathbf{Open} \right)$$ is as well.

  • $\begingroup$ David: could you please elaborate on the "which implies that" part of your answer. $\endgroup$ Aug 5, 2014 at 20:11
  • $\begingroup$ Hi Andre, I was just signing on because of that reason. I thought this was true in general, but perhaps that's not right. I'll leave this up for now, so that someone can verify or deny this claim. $\endgroup$ Aug 5, 2014 at 21:12
  • $\begingroup$ The wiki-article claims that this is true for presheaves of sets (though I'd like a better reference). To go from here to presheaves of $n$-groupoids, one can use the model structure on simplicial presheaves, and Bousfield localize to get a presentation for presheaves of $n$-groupoids. $\endgroup$ Aug 5, 2014 at 23:05
  • $\begingroup$ @AndréHenriques: I have edited my answer to address this. $\endgroup$ Aug 6, 2014 at 7:33
  • $\begingroup$ Could whoever downvoted, please explain to me why? Because if there is math error, I would like to know. $\endgroup$ Aug 6, 2014 at 22:28

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