The relevant Wikipedia article about Fiber bundles defines them as structures $(E,B,\pi,F)$ with $\pi:E\to B$ a continuous surjection such that

  1. For every $x\in B$, there is a neighborhood $U\subseteq B$ with $x\in U$ such that there is a homeomorphism $$\varphi:\pi^{-1}(U)\to U\times F,$$
  2. The maps $\pi$ and $\varphi$ "agree" with the projection onto the first factor, meaning that $$\pi = \operatorname{proj}_1\circ\,\varphi.$$

I don't quite understand if and why this second condition is required. More precisely, it feels like we should not need to add it as a further requirement for the definition of fiber bundle.

I imagine $\varphi$ as a map that "locally straightens out" the total space. For example, for the Mobius strip, if $U$ is a neighborhood of a point in $S^1$, then $\pi^{-1}(U)$ is the set of points of $\mathbb R^3$ that gets projected to points in $U$, that is, a set of lines pointing in different directions but all intersecting $U$ at some point. The map $\varphi$ should then, I suppose, "straighten up" all these lines, thus in some sense recognizing that all the fibers are, in fact, lines (i.e. one-dimensional vector spaces).

It would seem obvious then that projecting on the first part of the result of the application of $\varphi$ would give back the original $x\in U$. Is this not the case? If not, what is an example in which not adding this as a further assumption gives an object which we would not like to call a "fiber bundle"?

  • $\begingroup$ I don't understand what you are doing with a Möebius strip (what does it have to do with $\mathbb R^3$?). However, think about this: if you don't require the commutativity of the diagram, $\varphi$ could take a fibre over the point $p$ to the fibre over the point $q$ where $p$ and $q$ are different. $\endgroup$ Jul 10, 2021 at 20:33
  • $\begingroup$ Also, you wrote the condition wrong. $\pi^{-1}$ as a function does not exists in most cases, since $\pi$ may not be injective. The condition you want is $\pi=\mathrm{proj}_1\circ\varphi$. $\endgroup$ Jul 10, 2021 at 20:35
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    $\begingroup$ "It would seem obvious then that projecting on the first part of the result of the application of $\varphi$ would give back the original $x \in U$". It may be obvious to you, but it's not true! Note that $\varphi$ is just a homeomorphism $\pi^{-1}(U) \to U\times F$, so $\operatorname{proj}_1\circ\varphi$ is a map $\pi^{-1}(U) \to U$, there's no reason it has to be the map $\pi$ and if you don't choose $\varphi$ appropriately, it won't be. $\endgroup$ Jul 10, 2021 at 20:35
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    $\begingroup$ @JackozeeHakkiuz I mentioned $\mathbb R^3$ for the Moebius strip because I was thinking it as embedded in $\mathbb R^3$ (i.e., with the total space being $\mathbb R^3$). Is that not a standard picture for it? About the condition, thanks, I fixed it. That aside, I think I understand now: we need the condition to actually be able to understand $\varphi$ as a map that only "straightens the fibers", without changing how the fibers are attached to be base points $\endgroup$
    – glS
    Jul 10, 2021 at 22:39
  • $\begingroup$ You can embed the Mobius strip in $\mathbb{R}^3$ if you want, but it's also commonly given as a quotient space. For example, take $\mathbb{R}^2$ modulo the group generated by the transformation that translates by $1$ in the $x$ direction while reflecting over the $y$ axis. The $x$ axis under this quotient is homeomorphic to $S^1$, and the projection onto the $x$ axis descends to the projection map from the Mobius strip to $S^1$. This forms the fiber bundle. $\endgroup$ Jul 10, 2021 at 22:56

1 Answer 1


Let us consider systems $\mathfrak B = (E,B,\pi,F)$ consisting of a continuous surjection $\pi : E \to B$ and a space $F$. Let us say that $\mathfrak B$ has fibers of type $F$ if all fibers $\pi^{-1}(b)$ over the points $b \in B$ are copies of $F$, i.e. $\pi^{-1}(b)$ and $F$ are homeomorphic. This is certainly a minimal requirement that should be satisfied by a fiber bundle with fiber $F$.

You ask why we do not define a fiber bundle with fiber $F$ to be any system $\mathfrak B$ satisfying condition 1. Unfortunately such a thing does in general not have fibers of type $F$ and is therefore totally unsuitable to be used as a concept of fiber bundle.

Here is an example.

Let $B = [0,2], E = [0,2] \times \{0\} \cup [0,1] \times \{1\} \cup \{0\} \times [0,1] \subset \mathbb R^2 , F = [0,1]$ and $\pi : E \to B, \pi(x,y) = x$. The fibers of $\pi$ are $\pi^{-1}(0) = \{0\} \times [0,1], \pi^{-1}(t) = \{t\} \times \{0,1\}$ for $t \in [0,1]$ and $\pi^{-1}(t) = \{(t,0)\}$ for $t \in (1,2]$. Thus we have three non-homeomorphic types of fibers. In other words, the above system does not have fibers of any type $F$ and does certainly not deserve to be regarded as a fiber bundle. But we have $\pi^{-1}(B) = E \approx [0,2] \approx B \times \{0\}$, thus 1. is satisfied with $F = \{0\}$.

It is condition 2 which assures that a system $\mathfrak B$ satisfying condition 1 has fibers of type $F$.

Also note that a system $(E,B,\pi,F)$ having fibers of type $F$ may look fairly weird. As an example take $B = \mathbb R, E = \mathbb Q \times [0,1] \cup (\mathbb R \setminus \mathbb Q) \times[-1,0], F = [0,1]$ and $\pi : E \to B, \pi(x,y) = x$. All fibers have type $F$, but there is no open $U \subset B$ such that $\pi^{-1}(U) \approx U \times F$.


Call $U \subset B$ a trivializing subset if there exists a homeomorphism $φ:π^{−1}(U)→U×F$. Condition 1 requires the existence of a trivializing open neighborhood $U$ for each $b∈B$. Of course we need the same fiber $F$ for all such neigborhoods $U$, but it is not required that we can take arbitrary $U$. In the above example we can take $U=B$ for all $b∈B$ and this gives us a local trivialization with a one-point fiber. However, we cannot shrink the "universal neigborhood" $U=B$ to arbitrary small open neigborhoods $V$ of $b$; this prevents to find local trivializations over $V$ with a common $F$. But remember that condition 1 only requires the existence of some neighborhood which is true in the example. In presence of condition 2 you can of course shrink $U$ to arbitarily small $V⊂U$ without destroying the local trivialization property for $V$. This shows again the weakness of condition 1 if you take it alone.

Here is one more example.

Let $B = [0,1], E = \{(x,y) \in \mathbb R^2 \mid 0 \le x \le y \le 1 \}, F = [0,1]$ and $\pi(x,y) = x$. The space $E$ is a plane triangle with vertices $(0,0), (1,0), (1,1)$. The fibers are $\pi^{-1}(0) = \{(0,0)\}$ and $\pi^{-1}(t) = \{0\} \times [0,t]$ for $t > 0$. Thus we have two non-homeomorphic types of fibers. Nevertheless we can show that for each open $U \subset B$ we have $\pi^{-1}(U) \approx U \times F$. In fact, we can trivialize componentwise, i.e. take an individual trivialization for each connected component $C$ of $U$. These are intervals of the form $C = [0,1]$, $C = [0,b)$ with $0 < b \le 1$, $C =(a,b)$ with $0\le a < b \le 1$ or $C = (a,1]$ with $0 \le a < 1$. If $0 \notin C$, we may take a fiber-preserving trivialization. If $0 \in C$, it is easy to see that $\pi^{-1}(C) \approx C \times [0,1]$. Such a homeomorphism does of course not preserve fibers.

  • $\begingroup$ thanks for the answer. To clarify: how exactly are you understanding "condition 1"? Because I took it to mean that each local homeomorphism $\varphi:\pi^{-1}(U)\to U\times F$ is using the same $F$. In your example, I would thought this condition to not be satisfied: as you observe, you have $F=\{0,1\}$ for opens with $U\ni t$ and $t\in (0,1)$, but $F=\{0\}$ for $t\in (1,2)$. So, yours is a counterexample of a case in which the base space is "locally trivial", but with fibers not homeomorphic to each others, yes? $\endgroup$
    – glS
    Feb 5, 2022 at 16:41
  • $\begingroup$ From the other comments, I thought the core of the matter was more about ensuring each local homeomorphism to maintain information about the base point, so that we don't have things such as $\varphi(\pi^{-1}(0))=\{100\}\times [0,1]$, where projecting on the first element we don't agree with $\pi$, using again your first example. This aside, I appreciate the (counter)examples, thanks $\endgroup$
    – glS
    Feb 5, 2022 at 16:44

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