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To avoid problems with the notoriously inconsistent notation/formalisms of geometry, I will define bundles and fiber bundles (as I call them) below:

  • We define a bundle $E \xrightarrow{\pi} B \ $ as a triplet $(E, B, \pi)$ with $E, B$ topological spaces and $\pi: E \to B$ a continuous surjection. We further define the fiber at $x \in B$ as the set $\pi^{-1}(\{x\})$.
  • A Fiber bundle is a 4-tuple $(E, B, F, \pi)$ with $E, B, F$ topological spaces, $\pi: E \to B$ a continuous surjection, together with a local trivialization $\{(U_i, \varphi_i)\}$, where $\{U_i\}$ is an open cover of $B$ and $\varphi_i: \pi^{-1}(U_i) \to U_i \times F$ is a homeomorphism such that $\text{proj}_1 \circ \varphi_i \equiv \pi \ $ on $\ \pi^{-1}(U_i)$.

Now, in a lecture series on differential geometry that I am watching, bundles were defined in much the same way except that topological spaces were replaced by topological manifolds. No complaints here. However, a fiber bundle was subsequently defined as a bundle $E \xrightarrow{\pi} B \ $ such that all of the fibers $\pi^{-1}(\{x\})$ were homeomorphic to some common topological manifold $F$—with no mention of the local triviality condition.

My question is: In the case of topological manifolds, does this condition naturally induce the locally trivial condition that appeared before? Does it for general topological spaces?

My intuition is that it does not for general topological spaces: Clearly, there exists a family of homeomorphisms $\varphi_x: \pi^{-1}(\{x\}) \to \{x\} \times F$, and for any $U \subset B$, we can glue these maps together to obtain $\varphi_U: \pi^{-1}(U) \to U \times F$ defined point-wise by $\varphi_U(y) := \varphi_{\pi(y)}(y)$. It's easy to see that this map is bijective independent of $U$, but I see no reason to suspect bi-continuity even with "nice" choices in $U$ as—roughly speaking—we don't have any information about how this map behaves when moving transverse to the fibers. Perhaps this construction is too artificial, but it seemed compelling enough to me.

In the case where the base space, total space, and common fiber are topological manifolds, I wouldn't be surprised if the chart maps could be used to construct a local trivialization but I'm not sure how to explicitly construct it at the moment.

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This is indeed not true. Counterexample:

Let $E = \mathbb Z \times \mathbb C$ and $B=\mathbb C$ with $\pi (p, z) = z^2$. Then all the fibers are homeomorphic to $\mathbb Z$ with the discrete topology.

Remark One can have a connected example: let $f : T \to \mathbb S^2$ be a branched covering (such an example is found here. Indeed there is a lot of such examples). Let $g:\mathbb R^2 \to T$ be the universal covering map. Then

$$ \pi= f\circ g:\mathbb R^2 \to \mathbb S^2$$

is a bundle so that $\pi^{-1} (p)$ is homoeomorphic to $\mathbb Z$ with the discrete topology.

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    $\begingroup$ It would be a trivial fiber bundle if $\pi (p, z)=z$. Note that a bundle if a triplet $(E, B, \pi)$. @infinitylord $\endgroup$
    – Arctic Char
    Dec 15, 2019 at 4:24
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    $\begingroup$ Just like $(\mathbb S^1, \mathbb S^1, \pi_n)$, where $\pi_n (e^{i\theta}) = e^{in \theta}$ are all different fiber bundles. $\endgroup$
    – Arctic Char
    Dec 15, 2019 at 4:28
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    $\begingroup$ Apologies, I'm still pretty new to bundles. It's clear that my example is not compatible with the projection map in hindsight (and that helps motivate the compatibility condition in the first place!). I suppose the only natural choice for global trivialization is $\varphi(p, z) = (p, z^2)$, which fails because $z^2$ is not uniquely invertible. $\endgroup$
    – infinitylord
    Dec 15, 2019 at 4:37
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    $\begingroup$ $\mathbb Z\times \mathbb C$ is a topological manifold, it is just a countable disjoint union of $\mathbb C$. @infinitylord . It is not connected though so the example is a bit artificial. $\endgroup$
    – Arctic Char
    Dec 15, 2019 at 5:00
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    $\begingroup$ Good point. I also just found that there is a classification of $0$-dimemsional manifolds as well, of which discrete spaces are inherently. A rather strange idea, and indeed begs for some more "intuitive" examples, but I'm sufficiently satisfied to keep moving in the lectures. Thanks again. $\endgroup$
    – infinitylord
    Dec 15, 2019 at 5:13

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