A small question about fiber bundle

Question

I'm recently studying fiber bundles,and I find many examples of them are motivated from maps like $p:E\to B$ with homeomorphic fiber $F=p^{-1}(b)$.Therefore I'm wonder if the converse is true,that is if we are given a map $p:E\to B$ with homeomorphic fiber $F=p^{-1}(b)$,would it be a fiber bundle?

Answer 1

No, take $B$ to be any set with cardinality greater than one and give it the indiscrete topology. Then take $E$ to be $B$ but with the discrete topology. Then take $p$ to be identity.

Edit: If you require $E$ and $B$ to be manifolds take $E$ to be the disjoint union of countably many unit circles and $B$ to be the closed unit interval. and let $p$ be projection to the $x$-coordinate. Then show that an endpoint of the interval has no neighbourhood whose preimage is a product in the way required of a fibre bundle.

Edit 2: If you require $E$ and $B$ to be manifolds without boundary, a similar idea works. Let $B$ be the real line and let $E$ be the union of all circles in the plane with positive integer radii and centered at the origin, and let $p$ be projection to the $x$-coordinate.