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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, this might be relevant. In this case $E$ is $B\times F$, but you need to know that there is a global trivialization.2012-09-14
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    $F$ 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.2012-09-14
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    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...2012-09-14
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    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.2012-09-14
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    @Arthur The fibres of a fibre bundle do not need to be vector spaces. A vector bundle is a more specialized concept.2017-06-26

1 Answers 1

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You seem to have gotten as much as possible from the given analogy. Observe that if we were to take the definition of a fiber bundle as our designing principle we wouldn't be able to obtain a hairbrush:

  1. There should be a model fiber $F$ to which each fiber is homeomorphic, whence we would have only a (hollow) handle, or a hairy thing that has no handle, if we insist that $\pi:E\to B$ be surjective (which we need not to do, see e.g. May's Concise, p.51). Let's choose the latter for the sake of argument.
  2. Also observe that we are not interested in putting a fiber "on top of" a point in the inner region of the middle cylinder, thus the bundle really is over the surface of a cylinder.
  3. We should see a copy of $F$ "on top of" each point of the middle cylinder, which would prohibit having distinct bristles. Instead all bristles would be glued together, and we would end up having a solid cylinder that has its center punched out, i.e. our bundle would be $(S^1\times I)\times I\stackrel{\operatorname{proj}_1}{\to} S^1\times I$, where $I$ is the unit interval and $S^1$ is the circle.

A priori, a fiber bundle $F\to E\stackrel{\pi}{\to} B$ need not include its base, since its definition is categorical, i.e., up to isomorphism. Even in the case of a trivial bundle, we don't have containment, but an injection, by way of considering $\operatorname{id}_B:B\to B$ as a fiber bundle and mapping it into $ B\times F\stackrel{\operatorname{proj}_1}{\to} B$. Note that since we need to respect the base $B$ this will guarantee that any such mapping of bundles, called a section of the bundle $E\stackrel{\pi}{\to} B$, will be injective. Existence of these (defined globally) are not guaranteed in general.

Let's consider the first example of a nontrivial fiber bundle given on Wikipedia, namely, let $M$ be the Möbius strip obtained from the unit square $I\times I$ by identifying $(0,y)\sim(1,1-y)$. Define $p:M\to S^1\times\{1/2\} \to S^1$ by $[x,y]\mapsto (x,1/2)\mapsto x$. Then we can consider $p$ to be an $I$-bundle over $S^1$. Observe that we could also have used the notation $[x,1/2]$ to mean that we first collapse the strip onto its central circle, but really here we are making a choice, as indicated above. (As an exercise it might be instructive to come up with other sections.)

Finally let me note that we are required that $\pi^{-1}(U)\cong U\times F$, not $\pi^{-1}(U)\cong B\times F$ (unless of course $U=B$), in case there was no typo in your comment.