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I am trying to understand the hairbrush example of a fiber bundle from the Wikipedia article on fiber bundles.

If I am understanding this, in the hairbrush example E is the hairbrush, ie. all the bristles with the cylinder they are attached to, B is the cylinder attaching the bristles, F are the bristles, and $\pi$ maps a bristle to the point on the cylinder it is attached to. So is $E$ in this case equal to $B\times F$? Now pick a bristle $x$, and a small neighbourhood $U$ of $\pi(x)$. Then $\pi^{-1}(U)$ are all the bristles attached to the cylinder somewhere in $U$, and does this include $U$ itself?

I think the idea is that the hairbrush looks like a cylinder itself, but I am a bit confused.

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Since this can be thought of as the normal bundle, <a href="…; might be relevant. In this case $E$ is $B\times F$, but you need to know that there is a global trivialization. – Matt Sep 14 '12 at 18:00
$F$ is a single bristle, otherwise I see nothing wrong with your argument. Of course, in a topological setting, there would be an infinite number of infinitely thin bristles, one atached to every point of the cylinger. In addition, any single bristle is a vector space. – Arthur Sep 14 '12 at 18:01
So how is $\pi^{-1}(U)\cong B\times F$. $B\times F$ is a cylinder isn't it? Whereas $ $\pi^{-1}(U)$ is just a bunch of bristles... – JLA Sep 14 '12 at 18:05
Because the base space is the surface of the cylinder the bristles are atatched to, and thus infinitely thin. The cylinder together with the bristles makes a thick cylinder, which is the whole space. – Arthur Sep 14 '12 at 18:09

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