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Apr 20, 2012 at 13:28 comment added Jesse Wolfson Yes, but we can conclude a lot of things in mathematics if we just adjust the notion of equivalence :)
Apr 20, 2012 at 6:50 comment added David Roberts Regarding: "the nerve of a good cover is not equivalent in this setting to its levelwise $\pi_0$" - one just needs the right sort of equivalence :) See ncatlab.org/nlab/show/nerve+theorem
Apr 20, 2012 at 6:44 comment added David Roberts Hmm, yes, there are some subtleties when the coefficient sheaf isn't 'discrete'. For a discrete abelian group I think it should be ok. I only partially understand the $(\infty,1)$-topos stuff that goes on at the nLab, so I hope someone with a better grasp tunes in.
Apr 20, 2012 at 3:03 comment added Jesse Wolfson in getting an answer to this question, I'll try again with the Cech cohomology question and see if that is more successful. They're equivalent at the end of the day in any case, and I thought this was the more basic/concrete formulation to ask.
Apr 20, 2012 at 3:02 comment added Jesse Wolfson Hi David, that's actually the interesting thing about smooth manifolds: the nerve of a good cover is not equivalent in this setting to its levelwise $\pi_0$. For instance, consider the sheaf of smooth functions into $U(1)$. $H^0(-,U(1))$ is just the ring of smooth $U(1)$-valued functions, which is quite different when evaluated on $\mathbb{R}^n$ or the point. The stuff on n-lab doesn't answer this question, at least not so far as I can see. Thanks for taking the time to write such a lengthy comment by the way. I'm glad you find both questions interesting, and if I don't succeed . . .
Apr 20, 2012 at 1:49 comment added David Roberts One thing to think about might be the sequence of maps of sites: $GoodCovers/M \hookrightarrow Submersion/M \hookrightarrow Diff/M \to Diff$ which is the composite of inclusion and the forgetful functor. The two subcategories on the left are full. One can pull sheaves back to $GoodCovers/M$ this way, so if one had an example of a sheaf with non-vanishing cohomology on $\mathbb{R}^n$ on $Diff$, one could see what it would look like over the small site $GoodCovers/\mathbb{R}^n$.
Apr 19, 2012 at 23:56 comment added David Roberts object. This is a bit of a vague answer, hence it is a comment. If you had asked the question you are really interested in, it would have been answered quickly. However, the question you did ask is very interesting in its own right!
Apr 19, 2012 at 23:54 comment added David Roberts finite-dimensional paracompact second-countable manifolds one only needs ordinary covers, and the simplicial manifold that arises from taking the nerve of the Cech groupoid of the cover is levelwise a coproduct of representables when considered as a sheaf on $Cart$. Since each representable is smoothly contractible, I believe this means you find that the manifold is equivalent (in the sense of homotopy of simplicial sheaves) to the (Kan) simplicial set which is levelwise $\pi_0$ of this simplicial manifold. Then the Cech cohomology is given by maps from this simplicial set to the coefficient
Apr 19, 2012 at 23:50 comment added David Roberts The category of smooth manifolds is a subcategory of the category of sheaves on the site with objects the $\mathbb{R}^n$s for $n\geq 0$ with the good open cover coverage (note that this coverage is not a pretopology - see nLab for terms). Urs Schreiber calls this site $Cart$ (=cartesian spaces). I'm sure there is stuff on the nLab that will answer your question, I can't think exactly of the precise result. I'm sure it is true, because the whole point about Cech cohomology is that it works if you have fine enough open covers. A priori one should use hypercovers to calculate it, but working with
Apr 18, 2012 at 15:35 comment added Jesse Wolfson ps. I'm sorry if that comes off as pedantic. It's certainly not intended as such, I just want to be clear about what I'm trying to ask.
Apr 18, 2012 at 15:24 comment added Jesse Wolfson @Misha: no push forwards. Pre-sheaves are contravariant functors from the category of smooth manifolds to sets (or groups, or abelian groups). So we can pull back along any map but nothing else. Sheaves are those presheaves which satisfy descent with respect to surjective submersions (or open covers if you prefer). For example, for any k, differential k-forms are a sheaf on the smooth site.
Apr 18, 2012 at 15:07 comment added Misha @Jesse: I am still confused. Namely, do you allow push-forward of sheaves via surjective submersions of smooth manifolds? Do you allow pull-back of sheaves via open immersions (or open embeddings)? (I understand that non-open embeddings are not allowed as in Donu's example.)
Apr 18, 2012 at 14:16 comment added Jesse Wolfson Dona, the point about considering the smooth site is that you don't have direct images. These depend on a map of sites or topoi (in your example, the direct image is a map from the category of sheaves on $S^1$ to the category of sheaves on $\mathbb{R}^2$ induced by the embedding of the circle). I only want to consider sheaves defined naturally for all smooth manifolds.
Apr 18, 2012 at 14:11 history edited Jesse Wolfson CC BY-SA 3.0
clarified question
Apr 18, 2012 at 13:58 comment added Donu Arapura Jesse: OK, I think, I understand your question more clearly now. You should still have direct images in this world, right? So take the constant sheaf $\mathbb{Z}$ on $S^1$ and form the direct image $i_*\mathbb{Z}$ under $i:S^1\to \mathbb{R}^2$. This seems like a candidate, but I admit my intuition here may be off here. (David, yes, point taken.)
Apr 18, 2012 at 3:15 comment added Jesse Wolfson Smooth manifolds has two natural pretopologies, where the covers are either the surjective submersions or the open covers. These generate the same topology (since open covers are smooth submersions, and by the implicit function theorem, every surjective submersion can be refined by an open cover).
Apr 18, 2012 at 3:13 answer added John Hubbard timeline score: 3
Apr 18, 2012 at 2:05 comment added Joël I like the question, but I am not sure I really understand it. What is exactly the "site of smooth manifold". I mean what topology of you put on it ?
Apr 18, 2012 at 0:48 comment added David Roberts Not even the category theorists' generalised 'the' can excuse talking about the constant sheaf on a space. :-)
Apr 17, 2012 at 13:00 comment added Jesse Wolfson Though John's answer may be such an example . . .
Apr 17, 2012 at 12:58 history edited Jesse Wolfson CC BY-SA 3.0
clarified question
Apr 17, 2012 at 12:57 comment added Jesse Wolfson Sorry, I should have clarified: I'm looking for sheaves on the site of smooth manifolds which have nontrivial cohomology on $\mathbb{R}^n$. Donu, your point about support suggests that there be also be a counterexample of this type, but I haven't put my hands on it yet.
Apr 17, 2012 at 11:46 answer added John Hubbard timeline score: 13
Apr 17, 2012 at 9:45 comment added Donu Arapura David, it's as Dustin says. My point really was that a sheaf on $\mathbb{R"^n$ is at least as complicated as its support.
Apr 17, 2012 at 7:11 comment added Dustin Cartwright @David, for any non-zero group $G$, the constant sheaf with values in $G$ on $S^1 \subset \RR^2$ has first cohomology equal to $G$.
Apr 17, 2012 at 4:27 comment added David Roberts Donu - constant sheaf with values in what?
Apr 17, 2012 at 0:00 comment added Donu Arapura Sure, take a circle in $S^1\subset \mathbb{R}^2$, and extend the constant sheaf on it by zero. Its first cohomology is the same as for the circle, which is nonzero.
Apr 16, 2012 at 23:48 history asked Jesse Wolfson CC BY-SA 3.0