$S^2$ not a fiber bundle?

Question

I am new to the fiber bundle concept. As I understand, fiber bundles are very special manifolds. That is, not every manifold is a fiber bundle. In particular, by intuition, the common $S^2$ is not a fiber bundle.

Is there any proof of this intuitive feeling?

Answer 1

A fiber bundle is more than a space, it is a total base $E$, a base space $B$, and a map $\pi: E \to B$ (satisfying various properties). If the base is connected we can speak of 'the' fiber, which is the preimage of a point of $B$.

The 2-sphere can appear as the total space, base space, or the fiber of various fiber bundles:

1. The map $S^2 \to \mathbb{R}P^2$ identifying antipodal points is a fiber bundle.
2. The tangent bundle $TS^2 \to S^2$ is a fiber bundle, the fibers are $\mathbb{R}^2$.
3. Inside the the total space of the tangent bundle $TS^3 \to S^3$ we can take the unit sphere in each fiber to get a bundle $T^1S^3\to S^3$ which is a bundle where the fiber is $S^2$. You can do this construction for any $3$-manifold. This particular bundle is trivial (since the three sphere has trivial tangent bundle) but it is not trivial in general.