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There are fibrations $E \rightarrow B$ which are not fiber bundles. Example: $E = [0,1]^2 / \text{middle vertical line segment}$ and $B=[0,1]$.

In this example, $E$ has the homotopy type of a total space in a fiber bundle over $B$: namely $E'=[0,1]^2$. Is this always true?

If you don't know, then still I would appreciate any new examples of fibrations that aren't fiber bundles.

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    $\begingroup$ Some time ago, I asked a topologist the same question.. The answer was "Fibration is a categorical notion, so it depends on the category of spaces you consider. For a nice category like CW-complexes it's probably true, but for too general spaces/maps it's probably false", but he had no counterexample either. I'd really like to see one too. $\endgroup$
    – Ben
    Aug 17, 2015 at 16:03
  • $\begingroup$ @Ben, i want to see the proof in $CW$-case more than counterexample in general. (I had some ideas to prove, but they failed) $\endgroup$ Aug 17, 2015 at 17:09
  • $\begingroup$ also i had an answer that "Fibration arise in homotopic problems, and fiber bundles in geometric ones; so these are two different instruments to work with, and connection between them is useless and non-interesting," -- here i'm not agree with the last statement. $\endgroup$ Aug 17, 2015 at 17:16
  • $\begingroup$ Dear @AndreyRyabichev, I don't know how it works for sure either, but maybe it's time to reconsider the question. $\endgroup$
    – Ben
    Aug 17, 2015 at 17:22

2 Answers 2

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If the fibration is smooth and the spaces involved are compact, then it is a fiber bundle. Just to make clear what I mean by smooth fibration:

Definition. A smooth map $p\colon E \to B$ is said to satisfy the homotopy lifting property in the smooth category if given the following commutative diagram where all maps are smooth:

there exists an smooth map $\widetilde{F}$ making the following diagram smooth:

Definition. A smooth map is said to be a smooth (Hurewicz) fibration if it satisfies the homotopy lifting property in the smooth category for all manifolds $Y$.

Definition. A smooth map is said to be a smooth Serre fibration if it satisfies the homotopy lifting property in the smooth category for all discs $I^n$, $n\ge 0$.

Now the idea of the proof:

1) A (Serre) fibration $p\colon E \to B$ where $B$ is path-connected and $E\neq \emptyset$ is surjective.

2) A smooth (Serre) fibration is a submersion. I asked the question and it was answered here.

3) Compactness combined with the above guarantees we can apply Ehresmann and we are done.

Remark. This works for weak or Serre fibrations. You don't need to assume you are working with Hurewicz fibrations. So the answer is a bit more general than the question you asked.

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Good news as long as we only care for CW complexes. In D. Barnes, The simplicial bundle of a CW Fibration (jstor link), to every fibration with base and fibres being CW, there is associated a fibre-homotopic (simplicial) fibre bundle.

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