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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.

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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 $V \times \mathbf C^\times$.

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  • $\begingroup$ Thanks a lot. Kindly share a reference for the cohomology non-vanishing result. Also I am curious to know if $0$ is the only troubling point. I mean if we did not add the point $(0,0)$, is the restriction a fibre bundle? $\endgroup$
    – RKS
    Commented Jul 2, 2023 at 5:30
  • $\begingroup$ The non-vanishing is just because the intersection of $f^{-1}(V)$ with a small ball around $(0,0)$ in $\mathbf C^2$ is diffeomorphic to $\mathbf R^4 \setminus \{0\}$. Away from $0$, the map $f^{-1}(\mathbf C^\times) \to \mathbf C^\times$ is in fact a trivial fibre bundle: the map $f^{-1}(\mathbf C^\times) \to \mathbf C^\times \times \mathbf C^\times$ given by $(x,y) \mapsto (x,xy-1)$ is a biholomorphism with inverse $(x,z) \mapsto (x,(z+1)/x)$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jul 2, 2023 at 11:33
  • $\begingroup$ Oh! It is that simple. I was thinking in the real plane sense and thought that the line $xy=1$ will delete some part of $f^{-1}(V)$, away from $(0,0)$. Thank you! $\endgroup$
    – RKS
    Commented Jul 2, 2023 at 11:53
  • $\begingroup$ You can almost do the same example over the reals as well. The curve $xy=1$ is the standard parabola $y = 1/x$. Away from $0$, this is a trivial fibre bundle $\mathbf R^\times \times \mathbf R^\times \to \mathbf R^\times$. Moreover, $H^1(f^{-1}(V),\mathbf Z) \neq 0$ for any small ball $V \subseteq \mathbf R$ around $0$. The reason it's not a counterexample is that $U$ is not connected in this case. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jul 2, 2023 at 11:57
  • $\begingroup$ Yes, I see it now. $\endgroup$
    – RKS
    Commented Jul 2, 2023 at 12:05

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