# Is the site of (smooth) manifolds hypercomplete?

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.

• "However, the proof cannot be generalized since there are non-contractible manifolds": you don't need non-contractible manifolds. Wether you include them or not, you get the same theory of sheaves. – André Henriques May 17 '13 at 21:06
• It suffices to show that the contractible opens generate the infinity topos under colimits in a canonical way, i.e. that it is a dense sub -infinity category. – David Carchedi May 17 '13 at 21:07
• @Andre: You get the same theory of sheaves, but maybe not 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
• @Andre: But we are saved by the existence of good covers :) – David Carchedi May 17 '13 at 21:57

I think your idea to reduce the question to small slice topoi works perfectly. I will use it to show that every sheaf on $Man$ (either the continuous or the smooth version) is the limit of its Postnikov tower. This implies that the topos is hypercomplete since truncated objects are hypercomplete.

Let $F$ be a sheaf. Since limits are computed objectwise, we must show that for every $M\in Man$, $F(M)$ is the limit of $\{(a\tau_{\leq n} F)(M)\}$, where $a$ means sheafification and $\tau_{\leq n}$ objectwise truncation. Given $M\in Man$, let $Open(M)$ denote the small site of $M$, and let $i: Open(M) \to Man$ be the forgetful functor. The key is that the restriction functor $i^\ast$ (on presheaves) preserves sheaves and commutes with sheafification; I guess this is intuitively obvious, but as David points out in the comments to the OP we must be careful with intuition here, so I will explain in more details below. Assuming this, we get

$$(a\tau_{\leq n}F)(M)=(a\tau_{\leq n} i^\ast F)(M).$$

We know that $Sh(M)$ is locally of homotopy dimension $\leq dim(M)$ since objects of $Open(M)$ have covering dimension $\leq dim(M)$, so that $i^\ast (F)$ is the limit of its Postnikov tower in $Sh(M)$. Evaluating at $M$ gives what we wanted.

This argument even shows that "Postnikov towers are convergent" in the topos of sheaves on $Man$, in the sense of HTT (which is stronger than just saying that sheaves are limits of their Postnikov towers, and much stronger than hypercompleteness).

Back to the claim that $i^\ast$ preserves sheaves and commutes with sheafification. Descent is equivalent to Cech descent for open covers so it's clear that $i^\ast$ preserves sheaves. To prove that it commutes with sheafification it suffices to show that $i^\ast(S)\subset T$ for some classes of maps $S$ and $T$ such that sheaves on $Man$ (resp. on $Open(M)$) are the $S$-local (resp. $T$-local) presheaves. We can use for $S$ and $T$ the classes of maps of the form $F\to \tilde F$ where $F$ is a simplicial presheaf and $\tilde F$ is its degreewise sheafification (see Dugger-Hollander-Isaksen, Theorem A.6). That $i^*(S)\subset T$ now follows from the fact that $i^\ast$ commutes with sheafification for sheaves of sets.

• Very nice Marc! I'm glad this works for the continuous case too, where one does not have good covers. – David Carchedi May 18 '13 at 14:12
• Thank you Marc. I chose your answer because it applies to a more general situation. Do you have a reference or an easy proof that the Postnikov tower of a sheaf can be described in this way? – Alexander Körschgen Jun 7 '13 at 1:18
• @Alexander: The sheafification functor $a$ commutes with truncation by HTT 5.5.6.28. So to truncate in the category of sheaves you first truncate in the category of presheaves and then sheafify. – Marc Hoyois Jun 8 '13 at 14:13
• Does this argument also demonstrate the hypercompleteness of the site of topological manifolds? – Dmitri Pavlov Nov 6 '14 at 17:02
• @DmitriPavlov I believe so, provided your topological manifolds are paracompact (so that homotopy dimension = covering dimension, HTT §7.3). – Marc Hoyois Nov 14 '14 at 0:36

I think you can argue like this:

Let $l:Cart \to Mfd$ be the inclusion of the full subcategory of smooth manifolds consisting of the Euclidean ones ($\mathbb{R}^n$'s). It induces an essential geometric morphism of infinity topoi:

you have $l^*:Sh_\infty(Mfd) \to Sh_\infty(Cart),$ with a left adjoint $l_!,$ where $$l_!=Lan_{y_{Cart}} \left(y_{Mfd} \circ l\right).$$

Let's replace $M$ by a Cech nerve of a good cover, $C_\bullet,$ so that $C_n$ is (isomorphic to) a coproduct of Eucldiean manifolds. So we have $M=hocolim C_\bullet.$ Then, since $l^\star$ has a right adjoint, $l_*,$ $l_! \circ l^\star$ preserves colimits. So far each $C_n,$ $l_! l^\star C_n\simeq C_n,$ and hence $$l_!l^\star M\simeq hocolim l_!l^\star C_\bullet \simeq hocolim C_\bullet \simeq M.$$

Now, since every infinity sheaf $F$ can be written as a colimit of representables, it follows that the co-unit of $l_! \dashv l^\star$ is always an equivalence. This in turn implies that $l^\star$ is fully faitful and that $Cart$ is a dense infinity subsite, and hence $$Sh_\infty(Mfd) \simeq Sh_\infty(Cart).$$ Since $Sh_\infty(Cart)$ is hypercomplete, we are done.

• The nLab states (ncatlab.org/nlab/show/CartSp) that the topological version of $Cart$ is also a dense subsite of the topological version of $Mfd$ (without giving a reference). So is there a distinction between dense subsite and dense infinity subsite? Does this boil down to considering (iterated) intersections of covers? – Alexander Körschgen Jun 7 '13 at 1:39
• In general being a dense $\infty$-subsite implies being a dense subsite, but not conversely. See nforum.mathforge.org/discussion/4457/…. – Marc Hoyois Jun 8 '13 at 14:28