# Why must we require the local trivialization of fiber bundles, $\varphi:\pi^{-1}(U)\to U\times F$, to satisfy $\pi={\rm proj}_1\circ\,\varphi$?

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"?

• 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. Jul 10, 2021 at 20:33
• 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 $\pi=\mathrm{proj}_1\circ\varphi$. Jul 10, 2021 at 20:35
• "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. Jul 10, 2021 at 20:35
• @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
– glS
Jul 10, 2021 at 22:39
• 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. Jul 10, 2021 at 22:56

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 $$\pi^{-1}(U) \approx U \times F$$.

Update:

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.

• 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?
– glS
Feb 5, 2022 at 16:41
• 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
– glS
Feb 5, 2022 at 16:44