I have read in the paper of Meigniez "Submersions, fibrations and bundles" that a smooth surjective submersion $f: E \rightarrow B$ whose fibers are all diffeomorphic to $\mathbb{R}^{n}$ is locally trivial, i.e. a fiber bundle (corollay 31).

1) Is there a counterexample for $f$ not a submersion, which is not a fiber bundle even topologically?

2) Is there a smooth family of vector spaces all isomorphic to $\mathbb{R}^{n}$ which is not a vector bundle, even topologically?

3) Is there a surjective smooth map $f: E \rightarrow B$, with $E$ and $B$ compact, such that all the fibers are pairwise diffeomorphic, but which is not a fiber bundle, even topologically? (Here of course I do not require any more that the fibers are diffeomorphic to $\mathbb{R}^{n}$)

• I don't understand question 2. What does "smooth family of vector spaces" mean? I sort of understand question 1, but let me ask: You want an example a map $E\to B$ that is: smooth? continuous with $E$ and $B$ topological manifolds, and such that near each point in $E$ it looks like projection of a product of disks on one factor? continuous with $E$ and $B$ topological manifolds? continuous? and with each fiber homeomorphic to $\mathbb R^n$ but not locally trivial. Nov 11, 2012 at 19:14
• For question 1, I would like to see an example of a smooth function between smooth manifolds $\pi: E \rightarrow B$ which is surjective and such that $\pi^{-1}(x)$ is diffeomorphic to $\mathbb{R}^{n}$ for every $x \in B$, but which is not a fiber bundle. Anyway, even a continuous example is ok. For 2, I mean a map $\pi: E \rightarrow B$ as in question 1, with a vector space structure on each fiber such that the sum is smooth as a function $E \times_{B} E \rightarrow E$ and the exterior product is smooth as a function $\mathbb{R} \times E \rightarrow E$, but which is not locally trivial. Nov 11, 2012 at 20:28
• Small remark: if you wanted the fiber bundle to be smooth in question 1, then this fails even for $n=0$. Nov 12, 2012 at 6:27
• I added question 3 later. Nov 12, 2012 at 13:14

Here is a counter-example to Q1. Consider a Reeb-like foliation on the annulus $$E=S^1\times\mathbb R$$. Namely fix a point $$p\in S^1$$, let $$I=S^1\setminus \{p\}$$ and fix a smooth function $$h:I\to\mathbb R$$ which tends to $$+\infty$$ at both ends of the interval $$I$$. One leaf of the foliation is $$\{p\}\times\mathbb R$$, and the remaining strip $$I\times\mathbb R$$ is foliated by graphs of the form $$y=h(x)+const$$. Note that all leaves are diffeomorphic to $$\mathbb R$$.
There is a map $$f:E\to S^1$$ whose fibers are leaves of this foliation. First define $$g:I\times\mathbb R\to(-\pi/2,\pi/2)$$ by $$g(x,y)=\arctan(y-h(x))$$. Extend $$g$$ to the entire annulus $$S^1\times\mathbb R$$ by setting $$g(p,y)=-\pi/2$$. Now we have a continuous map $$g$$ from the annulus onto $$[-\pi/2,\pi/2)$$ whose fibers are leaves of our foliation. It remains to compose it with a continuous bijection from $$[-\pi/2,\pi/2)$$ to $$S^1$$, e.g. $$t\mapsto(\cos 2t,\sin 2t)$$.
The resulting map $$f:E \to S^1 =: B$$ is not a fiber bundle as a small neighbohood of a point on $$\{p\} \times \mathbb R$$ cannot have a connected inersection with a nearby leaf.
Replacing $$\arctan$$ by a function which converges to its asymptotic values sufficiently fast, one can make the map $$f$$ smooth (with zero derivatives at $$\{p\}\times\mathbb R$$).
• Note that in this example the continuous surjection $f:E\to B$ is not open, not even a quotient map, which is another reason why it's not a fiber bundle. You could ask whether there are examples in which $f$ is a quotient map. Nov 12, 2012 at 15:27