An intuitive vision of fiber bundles

In my mind it is clear the formal definition of a fiber bundle but I can not have a geometric image of it. Roughly speaking, given three topological spaces $X, B, F$ with a continuous surjection $\pi: X\rightarrow B$, we "attach" to every point $b$ of $B$ a closed set $\pi^{-1}(b)$ such that it is homeomorphic to $F$ and so $X$ results a disjoint union of closed sets and each of them is homeomorphic to $F$. We also ask that this collection of closed subset of $X$ varies with continuity depending on $b\in B$, but I don't understand why this request is formalized using the conditions of local triviality.

Solution 1:

Maybe it is helpful to take a map $f\colon\mathbb R\to\mathbb R$ and consider its graph $\Gamma(f):=\{(x,f(x)) |x\in \mathbb R\}$. We get a continuous map $\pi\colon \Gamma(f)\to \mathbb R$, $(x,f(x)) \mapsto x$. For every $p\in \mathbb R$, the preimage $\pi^{-1}(p)$ is again a single point, so this has a chance to be a fiber bundle. However, it is a fiber bundle, if and only if $f$ is continuous:

If $\pi$ is a fiber bundle near $x$, then there is a local trivialization, i.e. a homeomorphism $\pi^{-1}(U)\to U$ over $U$ for $U\subset \mathbb R$ a neighborhood of $x$. This implies that this homeomorphism is precisely given by $\pi$ and so the assignment $x\mapsto (x,f(x))\mapsto f(x)$ (which is the inverse of the homeomorphism composed with the projection) is continuous near $x$ and hence $f$ is continuous near $x$. On the other hand, if $f$ is continuous, the map $\pi$ itself is already a homeomorphism (an inverse is given by $x\mapsto (x,f(x))$) and hence a fiber bundle.

So, local triviality in this case means that the map $f$ is locally continuous and the fiber which is essentially $f(x)$, varies continuously with the base point. Also, it is not hard to see that for the sign-function (as an example of a non continuous function), the non continuity at $0$ ruins the local triviality.

I hope this is a little helpful.