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This is probably not a research level question, I'm sorry if it is inappropriate. I'm reasking here this question on

Suppose that $\xi: E \to B$ is a bundle (by which I mean simply a continuous mapping), and there exists a manifold $F$ such that $\Gamma(\xi) \cong C(B, F)$ as sheaves over $B$ (here by $\Gamma$ I mean the sheaf of sections, and by $C$ I mean the sheaf of continuous functions from $B$ to $F$). Does this imply that $\xi$ is locally trivial? If the proof is simple, I would like a hint on where to start, otherwise I would like a reference that studies this and related topics in more detail, please.

UPD: I guess it boils down to whether a bundle $\xi: E \to B$ such that $\Gamma(\xi) \cong C(B, \mathbb{R}^n)$ is (locally) trivial, then one can take preimages of charts to obtain the cover of $B$ that I hope can be used in the proof, but not having really worked with sheaves before I'm not sure.

UPD: Please comment if you think this quesion should be retagged, rephrased, or something like that. Having received no comments whatsoever on both and MO, I don't know what to think :(

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I suspect that you haven't gotten comments, because people are unsure about meaning of the question. For example, could you please clarify what you mean by "bundle". Obviously it is something more general than a locally trivial fibre bundle. Is $C(B,F)$ the sheaf of continuous $F$-valued functions on $B$? Are you expecting $F$ to be the fibre? – Donu Arapura Sep 17 '11 at 16:41
Depending on the precise definitions, the following might be a counterexample: take sufficiently nontrivial $B$ and $F$ and let $E$ be the etale space of $C(B,F)$. – user2035 Sep 17 '11 at 16:58
@Donu: Thank you, I clarified the question. @a-fortiori: Thank you, I'll look into it. – Alexei Averchenko Sep 18 '11 at 1:03

1 Answer 1

up vote 1 down vote accepted

Let $E$ be the union of the coordinate axes in $\mathbb R^2$, and let $\xi:E\to\mathbb R$ be the projection on the first coordinate. Then $\Gamma(\xi)$ is a trivial sheaf, so it is isomorphic to $C(\mathbb R,F)$ with $F$ any one-point space. Yet $\xi$ is not locally trivial. (This can be modified so that the fiber at $0$ of $\xi$ is pretty much anything...)

If this is too trivial an example, let $F$ be any manifold, let $E$ be as before, and let $\xi=E\times F\to\mathbb R$ such that $\xi((x,y),f)=x$. Then $\Gamma(\xi)\cong C(\mathbb R,F)$, just as before.

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You'll complain, probably, that the total space of $\xi$ in these examples is not a manifold... – Mariano Suárez-Alvarez Sep 18 '11 at 1:20
Yes, this is a great counter example! – Alexei Averchenko Sep 18 '11 at 1:30
I don't seem to be able to produce an example in which everything in sight is a manifold. – Mariano Suárez-Alvarez Sep 18 '11 at 2:47

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