In my mind it is clear the formal definition of a fiber bundle but I can not have a geometric image of it. Roughly speaking, given three topological spaces $X, B, F$ with a continuous surjection $\pi: X\rightarrow B$, we "attach" to every point $b$ of $B$ a closed set $\pi^{-1}(b)$ such that it is homeomorphic to $F$ and so $X$ results a disjoint union of closed sets and each of them is homeomorphic to $F$. We also ask that this collection of closed subset of $X$ varies with continuity depending on $b\in B$, but I don't understand why this request is formalized using the conditions of local triviality.

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    $\begingroup$ A fiber bundle looks "locally" like a product. In some ways, it is to topological spaces what the "semi-direct product" is to groups. $\endgroup$ Oct 22, 2012 at 16:01
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    $\begingroup$ In particular, it is more than just $\pi^{-1}(b)$ homeomorphic to $F$. For example, the disjoint union of copies of $F$, one for each $b\in B$, is not a fiber bundle, even though there is a function that satisfies $\pi^{-1}(b)\cong F$ which is continuous. $\endgroup$ Oct 22, 2012 at 16:05
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    $\begingroup$ The (open) Möbius strip is a nice fiber bundle to use as an example. It is a nontrivial bundle whose base space is the circle and whose fiber is the real line. You make it from a strip of paper (think of it as $[0,1]\times\mathbb{R}$), which is a trivial bundle. Now identify $(0,x)$ with $(1,-x)$, and you have the Möbius strip. Understanding all the concepts for this simple example is a good start. $\endgroup$ Oct 22, 2012 at 16:15
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    $\begingroup$ If I understand OP correctly, he is saying that he understands the definition but doesn't find it motivated by the geometry. His intuition is that the right geometric definition should be "all the fibers look alike, and they vary continuously," and so the issue with making the correct definition is figuring out how to formalize "they vary continuously." He doesn't see how that "the fibers vary continuously" is made formal by "looks locally like a product." He wants to be geometrically convinced that these are the same. $\endgroup$
    – user29743
    Oct 22, 2012 at 17:05
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    $\begingroup$ (continued) I tried to type something up, but don't really know what his picture is. Certainly it's clear that things which "look locally like products" have "continuously varying fibers." For the other direction, "pick a small enough neighborhood that the fibers are so close that you can bend them to look like a product." $\endgroup$
    – user29743
    Oct 22, 2012 at 17:06

2 Answers 2


Maybe it is helpful to take a map $f\colon\mathbb R\to\mathbb R$ and consider its graph $\Gamma(f):=\{(x,f(x)) |x\in \mathbb R\}$. We get a continuous map $\pi\colon \Gamma(f)\to \mathbb R$, $(x,f(x)) \mapsto x$. For every $p\in \mathbb R$, the preimage $\pi^{-1}(p)$ is again a single point, so this has a chance to be a fiber bundle. However, it is a fiber bundle, if and only if $f$ is continuous:

If $\pi$ is a fiber bundle near $x$, then there is a local trivialization, i.e. a homeomorphism $\pi^{-1}(U)\to U$ over $U$ for $U\subset \mathbb R$ a neighborhood of $x$. This implies that this homeomorphism is precisely given by $\pi$ and so the assignment $x\mapsto (x,f(x))\mapsto f(x)$ (which is the inverse of the homeomorphism composed with the projection) is continuous near $x$ and hence $f$ is continuous near $x$. On the other hand, if $f$ is continuous, the map $\pi$ itself is already a homeomorphism (an inverse is given by $x\mapsto (x,f(x))$) and hence a fiber bundle.

So, local triviality in this case means that the map $f$ is locally continuous and the fiber which is essentially $f(x)$, varies continuously with the base point. Also, it is not hard to see that for the sign-function (as an example of a non continuous function), the non continuity at $0$ ruins the local triviality.

I hope this is a little helpful.

  • $\begingroup$ Doesn't locally continuous mean continuous? $\endgroup$
    – Nancium
    Oct 9, 2022 at 21:55

One example: A branched cover is a fiber bundle, where the fiber is a set of points.

See the section "Representing Covering Spaces by Permutations", p. 68. You can build a branched cover by covering your space with open sets $X \subset \bigcup U_i$ and your bundle with be covered by $U_i \times \{ 1, 2, \dots, n\}$ Then you need to consider what happens on $U_i \cap U_j $. The transition function will be a bijective map $$\{ 1, 2, \dots, n \} \to \{ 1, 2, \dots , n \} $$ which is a permutation.

Therefore, branched covers can be thought of as fiber bundles over spaces where the fibers are finite sets.

By covering your space with open sets $X \subset \bigcup U_i$, taking direct products with your fiber $U_i \times V$ and considering what happens over interections $(U_i \cap U_j) \times V$ you can "patchwork" a fiber bundle together.

This construction is very general. If your fiber $V$ is a vector space, your transitions maps are invertible linear maps $V \to V$ which take values in the general linear group $GL(V)$. So the set of vector bundles can be thought of as the space representations of the fundamental group $\pi_1(X) \to GL(V)$.

Then you can ask your transition functions be holomorphic and then it's a holomorphic vector bundle. Or you can ask your transition functions be continuous or have 5 derivatives or "reasonable" restriction (i.e. consistent with vector bundle axioms).

Alternatively, you can study a vector bundle by looking at its section, which are maps from the base space into the vector space. In high school and college, we deal mostly with the trivial bundle $\mathbb{R}^2$ and where the base space is $\mathbb{R}$ and the fibers are $\mathbb{R}$. Then we look at sections $$\{ (x, f(x)\} \in \mathbb{R}^2$$ In this way we can consider trivial bundles over the circle $S^1 \times \mathbb{R}$ and consider only those sections which are square-integrable i.e. $$ \int_{S^1} |f(x)|^2 dx < \infty$$ Morally, this vector bundle is still the cylinder, but we are ruling out certain sections.

Question: What is the analogue of Fourier analysis for the Möbius band in this picture?

The sphere can be thought of as a fiber bundle over the line. Indeed the fiber $$ \{ x^2 + y^2 + z^2 = 1 \} \cap \{ z = k\} = \{ (x,y,k): x^2 +y^2 = 1- k^2 \}$$ is a circle except at the endpoints $k = \pm 1$, where the fiber degenerate to points. Also the torus is a fiber bundle $S^1 \times S^1$.

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    $\begingroup$ Branched cover is not a fiber bundle. $\endgroup$
    – Grigory M
    Nov 5, 2012 at 17:06
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    $\begingroup$ Branched covering is a fiber bundle only when it is, well, not branched. Of course, (ordinary) covering is a fiber bundle (and Hatcher writes about coverings, not branched coverings). $\endgroup$
    – Grigory M
    Nov 6, 2012 at 16:01
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    $\begingroup$ Oh, and the sphere is not a fiber bundle over a line! Fiber bundle can't have different fibers over different points -- that's the main point of the definition, actually. $\endgroup$
    – Grigory M
    Nov 6, 2012 at 16:06
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    $\begingroup$ A more focused answer would be more helpful, I think. Bringing in branched coverings at the point the OP is is also not a great idea! $\endgroup$ Nov 6, 2012 at 16:20
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    $\begingroup$ @cactus314: Bundles that have singular fibers are useful! But they aren't fiber bundles. $\endgroup$
    – user14972
    Sep 11, 2017 at 11:52

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