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