# Is it possible to build a fiber bundle whose fibers are different? (Or we should not call it a fiber bundle?)

Suppose there is a fiber bundle $E$. The base space is $M$ so that $\pi:E\rightarrow M$ is the projection. By the definition, the bundle has a typical fiber $F$ such that the local trivialization over one open subset $U_i$ of $M$ is defined in this way $\phi_i:U_i\times F\rightarrow \pi^{-1}(U_i)$. So we can find out that no matter which open set $U_i$ we choose, we always choose the same typical fiber $F$. I guess this is because we want to have a smooth structure. If at different points of $M$, we choose different fibers, for example, at point $p\in M$ we attach a 3 dimensional vector space, while at point $q\in M$ we attach a 2 dimensional vector space and so on, then we are not going to have a smooth structure.

So my question is: can we attach different fibers to different points of a base manifold? If we insist to do so, is this structure still smooth? Is it still called a fiber bundle?

• Sure. The base could be disconnected. (If the base is connected, then no: it's a nice exercise to show that a fiber bundle over a connected base has homeomorphic fibers, and similarly a smooth fiber bundle over a connected base has diffeomorphic fibers.) Nov 27 '14 at 21:39

The projection $\mathbb R^3\setminus \{0\}\to \mathbb R: (x,y,z)\mapsto x$ is a smooth submersion between connected smooth manifolds and all its fibers are connected.
However the fibers are not diffeomorphic: the fiber of $0\in \mathbb R$ is difffeomorphic to $\mathbb R^2\setminus \{0\}$ whereas all other fibers are diffeomorphic to $\mathbb R^2$.