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. – Tom Goodwillie Nov 11 '12 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. – Fabio Nov 11 '12 at 20:28
• Small remark: if you wanted the fiber bundle to be smooth in question 1, then this fails even for $n=0$. – Kevin Ventullo Nov 12 '12 at 6:27
• I added question 3 later. – Fabio Nov 12 '12 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 $g(x,y)=\arctan(y-f(x))$. 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. – Tom Goodwillie Nov 12 '12 at 15:27