Fiber Bundle: Hairbrush 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.
 A: 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:


*

*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.

*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.

*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.
