For the following we will work in the smooth category. (But examples in the topological category is also welcome.)
The usual definition of a fibre bundle is
Def A fibre bundle is the quadruple $(E,B,\pi,F)$ where $E,B,F$ are differentiable manifolds, $\pi: E\to B$ is a surjection which is locally trivial. Locally trivial means that for every $e\in E$, there exists an open neighbourhood $B\supset U\ni \pi(e)$ such that $\pi^{-1}(U)$ is diffeomorphic to $U\times F$.
One also encounters the following definition of a fibred manifold in the literature
Def A fibred manifold is the triple $(E,B,\pi)$ where $E,B$ are differentiable manifolds and $\pi$ is a surjection from $E\to B$ with maximal rank such that the dimension of $E$ at $e\in E$ is greater than or equal to the dimension of $B$ at $\pi(e) \in B$.
One also would encounter the remark that a fibre bundle is in particular an example of a fibred manifold.
Questions
- I hope I am not mistaken in thinking that the definition of fibred manifold, with $\pi$ being surjective, maximal rank, and with the dimension condition, is equivalent to $\pi$ being a surjective submersion of $E\to B$. Is that correct or is there a counterexample (going either way)?
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The more important question is, what does locally trivial actually do for us?
a. There is a trivial example of fibred manifold that is not a fibre bundle if we allow our differential manifolds to be disconnected with connected components of different dimensions. Then the local fibres $\pi^{-1}(b_1)$ and $\pi^{-1}(b_2)$ for $b_1\neq b_2\in B$ may have different dimensions and cannot be the same manifold $F$.
b. We also have Ehresmann's fibration theorem, which states that $f:M\to N$ smooth, surjective, submersive, and proper implies that $(M,N,f)$ is a locally trivial fibration.
In particular, I am looking to gain some intuition on what fibred manifolds that aren't fibre bundles can look like, aside from case (a) above. It would also be nice to see a non-proper counterexample to Ehresmann's theorem.
Best Answer
For the first question, if the dimension of $E$ is greater than or equal to the dimension of $M$, then having maximal rank is the same as the differential map being surjective. The other direction is also obvious.
Since we are dealing with differentiable manifolds and we have that property on the differential map, the manifold $E$ has a local product structure, i.e. for any $e\in E$ there is a neighborhood $U$ of $e$ and a manifold $F_e$ such that $U$ is diffeomorphic to $\pi(U)\times F_e$ (this is a direct consequence of the implicit function theorem). However, this product structure is not restrictive enough and you might end up, for example, with fibres with different topologies or homotopy types, for example the fibred manifold $(\mathbb{R}^2-\{0\},pr,\mathbb{R})$ where $pr:\mathbb{R}^2-\{0\}\rightarrow\mathbb{R}$ is the projection to the first coordinate, illustrates that phenomenon (by the way, this is also an example of a non-proper map which is not a fiber bundle). So, with only that product structure is hard to relate the topology of the manifolds involved. However, once you impose the local trivialization condition, you get rid of those "pathologies" (for instance the fibres on each component of the base become diffeomorphic). Also, local trivializations give the fibre bundle the homotopy lifting property, so you can compute invariants of the manifolds via exact sequences of homotopy groups, or the Serre spectral sequence, etc.
I'm sure that local trivializations do much more for us, but that's what I know so far.
Another fibred manifold which is not a fiber bundle is the following (non-proper) map: define $\mathbb{R}^2/\mathbb{Z}^2\rightarrow\mathbb{R}$ by $(x,y)$ mod $\mathbb{Z}^2\mapsto y-\sqrt{2}x$. Then this is a surjective sumbersion onto its image which is not locally trivial.