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I am learning about fiber bundles and I understand both the formal definition and the intuitive picture of the space being "twisted". However one point I am having trouble with is why we cannot express a fiber bundles in terms of a product.

More specifically, let $\pi: E \rightarrow M$ be a fiber bundle with fiber $F$. What is wrong with viewing $E$ as $M \times F$, and expressing every $e \in E$ as $(m, f)$ for some $m \in M$ and $f \in F$?

For example take the Mobius strip. Even though it is "twisted", we can describe any point on it by specifying a point in the base manifold $x \in S^1 = M$ and where we are in the fiber at that point by specifying $y \in (-1, 1) = F$. What is wrong with this view?

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    $\begingroup$ I suppose you do realize that you're asking "why isn't every fiber product trivial?" :) $\endgroup$ Jul 21, 2023 at 21:21
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    $\begingroup$ Already in the case of the Mobius strip, your attempt runs into the problem that the assignment of the coordinate in $(-1,1)$ is either not well-defined or is not continuous as a function on the thing. One way or another, that's the non-triviality of this fiber-bundle. $\endgroup$ Jul 21, 2023 at 21:26
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    $\begingroup$ Ok, well, perhaps by definition, fiber products might be locally trivial (certainly vector bundles are required to be so...) But/and we simply cannot assign (continuous...) coordinates in base x fiber, in general. Maybe locally, but not globally. So it's not so much an "advantage" of fiber products, but a fact about them. :) $\endgroup$ Jul 21, 2023 at 21:41
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    $\begingroup$ You can of course write all fiber bundles as a cartesian product. The point, as you've seen in the Moebius band example, is that it won't be a continuous description of the total space. There will be discontinuities, and the structure of those discontinuities will depend on the nature of your bundle. $\endgroup$ Jul 21, 2023 at 23:05
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    $\begingroup$ To summarize all the comments and answers, The fiber bundle is a product as a set. But it does not always have the product topology. $\endgroup$ Jul 22, 2023 at 14:32

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Suppose that the Möbius strip can be expressed as $E=S^1\times\mathbb{R}$ and consider a "section" of this fibre bundle, i.e. a map $s:S^1\to E$ such that $\pi\circ s=\text{id}$, the identity map. For instance, we could take the section $\theta\mapsto (\theta,1)$. According to the product topology, this is a continuous map which is nowhere vanishing. However, consider the Möbius strip as $I\times\mathbb{R}/\sim$, where $(0,x)\sim(1,-x)$. Then you can check that continuous sections of $E$ are precisely the same as functions $f:I\to\mathbb{R}$ such that $f(0)=-f(1)$. By the intermediate value theorem, every such function must vanish somewhere on $I$. Hence, the section $s$ from above cannot exist, and so $E\neq S^1\times\mathbb{R}$.

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The definition of fiber bundles only gives us local triviality, so for each point $m_0\in M$, there is a neighborhood $U$ of $m_0$ in $M$ such that we have a homeomorphism $\pi^{-1}(U)\to U\times F$ which respects the fibers. But from this information alone, there is no way to “glue” together these various local trivializations to provide a continuous global trivialization. I can definitely do it discontinuously: for example, for each point $m_0$, the fiber $E_{m_0}$ is homeomorphic to $F$, meaning there is a homeomorphism $\phi_{m_0}:F\to E_{m_0}$. There are infinitely many such homeomorphisms, so just fix one. So, we now have a collection of homeomorphisms $\{\phi_m:F\to E_m\}_{m\in M}$. This allows us to define a map $\Phi:M\times F\to E$, $\Phi(m,f):=\phi_m(f)$. The map $\Phi$ is a bijection which is fiber-preserving and a fiberwise homeomorphism, but it is not going to be continuous in general, let alone have a continuous inverse.

You may object that I’ve been way too crude here; but even if I use a local trivialization (which locally guarantees I have continuity in the $m$-variable as well), there’s no guarantee that I can glue the various local trivializations to form a global (continuous/smooth/analytic…) trivialization. The famous example is of course the Mobius strip, where if you try the naive procedure, you’ll get a discontinuity at the place of the “join” (i.e for each $p\in S^1$, if you let $U_p=S^1\setminus\{p\}$ be the complement, then $\pi^{-1}(U_p)$ is trivializable, $U_p\times (-1,1)$; it’s this one lonely little point which causes all the troubles). These remarks don’t prove that the Mobius strip is not trivializable, they just indicate the difficulty. The other answer provides a proof for non-triviality.

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It's possible to use the Cartesian product as a coordinate system for the fiber bundle as you described, but the resulting function $M\times F\to E$ won't generally be a homeomorphism. In your Mobius strip example, you'll have a discontinuity where the $y=1$ edge meets the $y=-1$ edge.

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