I was learning about bundles and my professor said that not every bundle with typical fiber is a fiber bundle. I asked him for an example, but he couldn't find one. Can anyone help me give that example or just prove the statement?
Just for clarification on the nomenclature used. A bundle $E\xrightarrow{\,\pi\,}B$ is just a surjective continuos function with this vanilla topological spaces E and B. A bundle with typical fiber F is a bundle that the fiber of every point $b \in B$ is F. A fiber bundle is a bundle that is locally isomorphic (as bundle) to a product bundle ($B \times F\xrightarrow{\,\pi\,}B$, such that $\pi(b,f)=b$).
So, I would like to see a bundle that has the same fiber for every point in the base space, but it isn't locally isomorphic to a product bundle.
Best Answer
Think of the universal cover of $S^1$ as a spiral going around a cylinder, where the covering map is the restriction of projecting the cylinder down to the unit circle in the plane.
Now, deform this spiral on the cylinder so that it intersects itself one time, and take the same projection map. This gives you a continuous map $\mathbb{R} \to S^1$ where every fiber is $\mathbb{Z}$, but if $x$ is below the intersection point on the spiral $x$ has no neighborhood $U$ whose preimage is $U \times \mathbb{Z}$.