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In the category of smooth manifolds (without corners), what are some examples of Hurewicz fibrations which are not fiber bundles?

The minimal topological example I know is to project the standard 2-simplex onto the $x$-axis. The right-most fiber degenerates into a point while the others are homeomorphic to a closed interval.

I don't understand how to produce such a "dimensional degeneration" phenomenon in the smooth world. In fact this seems sort of impossible to me: a Hurewicz fibration is a submersion and the vertical bundle of a submersion has locally constant rank, so the fibers are homotopy equivalent equidimensional embedded submanifolds which foliate the source (assume the base is connected).

I don't see what other kind of degeneration (other than dimensional) might preclude a fibration from being a fiber bundle.

Out of helplessness, since fibrations are submersions, I was tempted to examine submersions which are not fiber bundles. The classical example $\mathbb R^2\to \mathbb R,\; (x,y)\mapsto(x^2-1)e^y$ is not a Hurewicz fibration because fibers of negative numbers are connected while those of non-negative numbers are disconnected. I don't know any other examples.

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  • $\begingroup$ It seems to me the fibers would have to be submanifolds, all of the same dimension, and homotopy equivalent. So I would look for an example where you get fibers that are homotopy equivalent but not homeomorphic, say, using lens spaces. I can't quite see how to do that, though. $\endgroup$
    – Steve Costenoble
    Commented May 10, 2019 at 22:36
  • $\begingroup$ On the other hand: Ehresmann's lemma implies that, if the fibers are compact, then your map is a fiber bundle. So any counterexample will have to have noncompact fibers. $\endgroup$
    – Steve Costenoble
    Commented May 10, 2019 at 23:28
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    $\begingroup$ When you say "fibrations are submersions" I get the idea that what you mean by fibration here is a lifting property with respect to smooth maps only. Is that right? I mean, you can have a smooth map that is a homeomorphism but not a submersion. $\endgroup$
    – Tom Goodwillie
    Commented May 10, 2019 at 23:32
  • $\begingroup$ @Tom, yes. Sorry for being unclear. $\endgroup$
    – Arrow
    Commented May 11, 2019 at 0:10
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    $\begingroup$ @Steve to get properness for Ehresmann I think you need the fibers to be both compact and connected. $\endgroup$
    – Arrow
    Commented May 11, 2019 at 0:12

1 Answer 1

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Some poking around led to an example in G. Meigniez, Submersions, fibrations, and bundles, Trans. Amer. Math. Soc. 354 (2002), 3771-3787. It's Example 21 in that paper and described briefly as follows: Let $W\subset\mathbb R^3$ be the Whitehead manifold, an open, contractible subset not homeomorphic to $\mathbb R^3$. Let $E$ be the set of $(x,y,z,t)\in\mathbb R^4$ with $(x,y,z)\in W$ or $t \neq 0$. Then the projection $\pi\colon E\to \mathbb R$, $\pi(x,y,z,t) = t$, is a smooth submersion, a fibration (with contractible fibers), but not locally trivial because the fiber over $0$ is not homeomorphic to the others.

That paper also refers to other counterexamples given in S. Ferry, Alexander duality and Hurewicz fibrations, Trans. Amer. Math Soc. 327 (1991), 201-219.

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  • $\begingroup$ Well that's a bummer. I can't visualize the Whitehead manifold example at all, so I don't really get the picture. The validity of the examples in both papers relies on seemingly intricate theorems revolving around new concepts, which is also a bummer. $\endgroup$
    – Arrow
    Commented May 11, 2019 at 0:36
  • $\begingroup$ Yeah, that's why I quoted and didn't try to explain. Not sure how well I understand them, either. But I think it's clear that any example would need to be somewhat subtle. $\endgroup$
    – Steve Costenoble
    Commented May 11, 2019 at 0:39
  • $\begingroup$ Dear @Steve, how about $(x,y)\to (xy)^2$ from the punctured plane to the non-negative real line? Since the answer is subtle I'm guessing no, but I lack the intuition beyond path lifting. $\endgroup$
    – Arrow
    Commented May 12, 2019 at 11:15
  • $\begingroup$ That's still not a fibration. The same (now deleted) example I gave before still works: If you take the line from (0,1) to (1,0) and, in the base, contract its image to 0, the homotopy doesn't lift. The problem is that each component of a fiber close to 0 is trying to approach two components of the fiber over 0. $\endgroup$
    – Steve Costenoble
    Commented May 12, 2019 at 19:30
  • $\begingroup$ Completely forgot about your crystal clear example from before. I am still struggling with intuition for what goes wrong with a Hurewicz connection. Where must there be a discontinuity? I thought to fix a loop at zero downstairs and lift it, and perhaps the dependence of the lifts upon the initial condition will be discontinuous. Can't quite see it though. $\endgroup$
    – Arrow
    Commented May 12, 2019 at 20:40

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