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I would like to see why here the name "fibration" has been chosen. What precisely are the fibres (threads) in this definition? The name is standard, but I have never seen its origin.

user122424
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1 Answers1

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A fibration is a generalization of a fiber bundle (as it says at the link in your post), where the intuition is clearer. There is even a helpful picture of a hairbrush on that wikipedia page, to aid one's intuition.

Lee Mosher
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  • To which parts in the hairbrush correspond the names $X$, $E$ and $B$ form this definition ? And the maps $h,p,\tilde{h}_0$ ? – user122424 Feb 23 '23 at 11:42
  • The caption on that picture answers your questions regarding $X$, $E$ and $B$ precisely. – Lee Mosher Feb 23 '23 at 11:47
  • Regarding $h$ and $p$ and $\tilde h_0$, that seems to be a question for another post, namely the question of why every fiber bundle is a fibration, i.e. why a fibration is a generalization of a fiber bundle. – Lee Mosher Feb 23 '23 at 11:50
  • OK.I just do not follow what is the problem with taking $E=B\times F$ and $\pi$ the projection for the hairbrush. So why this is not globally possible. – user122424 Feb 23 '23 at 11:51
  • It sounds like you have bunches of further questions about the mathematical meanings of fibrations and fiber bundles. That's great, however, in that case the proper thing to do is to post new questions, rather than asking further questions in the comments of the answer to this question (where no-one will see them but the poster and the answerer, i.e. you and me, and anyone else who perchance wanders by). – Lee Mosher Feb 23 '23 at 11:56
  • I've done it, please see: here. – user122424 Feb 23 '23 at 12:09
  • Also, in the Wikipedia link you shared: Properties The fibers $p^{-1}(b)$over $b\in B$ are homotopy equivalent for each path component of $B$. – Bob Dobbs Feb 26 '23 at 08:18