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Suppose we define two types of bundles,

(1) a triple $(E,\pi,B)$ of two topological spaces $E$ and $B$, and a continuous surjective map $\pi : E \to B$, such that all fibres $F_b = \pi^{-1}(\{b\})$ (for $b \in B$) are pairwise homeomorphic, and

(2) a triple $(E,\pi,B)$ of two smooth manifolds $E$ and $B$, and a smooth surjective map $\pi : E \to B$, such that all fibres $F_b = \pi^{-1}(\{b\})$ (for $b \in B$) are pairwise diffeomorphic.

We say that a bundle allows a local trivialization if each $b \in B$ has a neighbourhood $O \subseteq B$ and an isomorphism (homeomorphism, diffeomorphism) $\phi : O \times F_b \to \pi^{-1}(O)$ such that $\pi \circ \phi (p,f) = p$ for any $p \in O$ and any $f \in F_b$.

The question is: Are there examples of bundles of type (1) or type (2) which don't allow local trivializations? In other words, what could go wrong?

Ivica Smolić
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1 Answers1

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Restrict $\exp(it)$ (map to the unit circle) to the interval $(0,\infty)$. More interesting examples exist even among proper maps where the projection is the quotient map given by a group action. Such examples exist even it you assume that the action is free. See my answer here.

I am sure this was asked many times on MSE.

Moishe Kohan
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  • Hm, let me see if I understood well that example: you have $E = \left< 0, \infty \right>$, $B$ is a circle (both with standard topologies), $\pi(t) = \exp(it)$, and each fibre is a discrete set of points, right? Where does it fail to be locally trivializable? – Ivica Smolić Jul 27 '17 at 14:31
  • @IvicaSmolić Near the point 1 on the circle. This is one of standard examples of local diffeomorphisms which are not covering maps, for instance because the path lifting property fails. – Moishe Kohan Jul 27 '17 at 15:18
  • Aha, ok, now I see it, nice example! Thank you! – Ivica Smolić Jul 27 '17 at 16:11