Let $p:E\to B$ be a locally trivial fibration of connected, non-compact smooth manifolds. Let $U\subset E$ be a connected open subset and $p|\_U:U\to p(U)$ has connected diffeomorphic fibers.


Can we conclude that $p|\_U$ is again a locally trivial fibration?


This question is related to my other question on fibering a certain hyperplane arrangement complement, I posted before. I find this general version interesting in its own right.



Restriction of a fibration to an open subset with connected diffeomorphic fibers
Here is a counterexample inspired by algebraic geometry:


**Example.** Let $E \to B$ be the first projection $\mathbf C^2 \to \mathbf C$ (or $\mathbf R^4 \to \mathbf R^2$, if you like), and let $U \subseteq \mathbf C^2$ be the complement of the divisor $\{(x,y) \in \mathbf C^2\ |\ xy = 1\}$ and the origin $\{(0,0)\}$. This is Zariski connected (as it is dense in an irreducible space) and hence also connected in the classical topology. Each fibre of $U \to \mathbf C$ is isomorphic to $\mathbf C^\times$: above nonzero points $x \in \mathbf C$ we removed the point $(x,1/x)$, and above $0$ we removed the point $(0,0)$.


But the restriction $f \colon U \to \mathbf C$ of $p$ to $U$ is not a fibre bundle. Indeed, for any open ball $V \subseteq \mathbf C$ around $0$, the fibre $f^{-1}(V)$ satisfies $H^3(f^{-1}(V),\mathbf Z) \neq 0$, so $f^{-1}(V)$ is never homeomorphic to $V \times \mathbf C^\times$.


