Why can't a fiber bundle be written as a cartesian product?

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?

• I suppose you do realize that you're asking "why isn't every fiber product trivial?" :) Jul 21, 2023 at 21:21
• 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. Jul 21, 2023 at 21:26
• 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. :) Jul 21, 2023 at 21:41
• 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. Jul 21, 2023 at 23:05
• 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. Jul 22, 2023 at 14:32

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 $$E\neq S^1\times\mathbb{R}$$.
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.
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.