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Aug 10 at 8:29 comment added Francesco Polizzi @HenrikRüping: also in my example both fibre bundles are trivial.
Aug 10 at 7:19 comment added HenrikRüping In many of these examples we can express a space as a fiber bundle in two ways, and one of the fiber bundles is not trivial. But even if both fiber bundles are trivial and they have the same fiber, there are still ways to obtain different base spaces. See for example mathoverflow.net/a/26404/3969.
Aug 9 at 8:26 answer added Francesco Polizzi timeline score: 7
Aug 9 at 8:06 answer added HenrikRüping timeline score: 3
Aug 8 at 20:26 history became hot network question
Aug 8 at 20:13 answer added Robert Bryant timeline score: 17
Aug 8 at 12:56 comment added Michael Albanese Here's a silly example that you probably want to rule out: $\mathbb{Z}\times\{0,1\}$ is the total space of a fiber bundle over $\{0, 1\}$ (projection onto second factor), with fiber $\mathbb{Z}$, but also the total space of a fiber bundle over a point (constant map) with fiber $\mathbb{Z}\times\{0, 1\}$. The fibers are diffeomorphic, but the bases are not.
Aug 8 at 12:16 history asked Nicolas Medina Sanchez CC BY-SA 4.0