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Let $\pi: G\to X$ be a continuous open (!) surjection of locally compact Hausdorff spaces. Assume that each fiber $G_x=\pi^{-1}(x)$, $x\in X$ carries a group structure making it a locally compact group. Suppose the unit section $\varepsilon:X\to G$ is continuous as well as the multiplication $M:G^{(2)}\to G$, where $G^{(2)}\subset G\times G$ is the union of all $G_x\times G_x$, $x\in X$. Finally I want the inversion $i:G\to G$ to be continuous as well.

My question is, whether there are enough local sections, i.e., whether $G$ is the union of the images of continuous local sections?

More explicitly, this means that for every $g\in G$ there exists an open set $U\subset X$ and a continuous map $s:U\to G$ with $\pi\circ s=Id_U$ such that $g$ lies in the image of $s$?

Natural counterexamples to the existence of local sections, like the square map ${\mathbb C}\to \mathbb C$, $z\mapsto z^2$ seem to be ruled out by the existence of the unit section and the continuity of the group structure.

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  • $\begingroup$ I know only one result in this spirit: if $X$ is a locally paracompact topological space and $G$ is a semi-continuous bundle of Banach space (so one forget compactness, keep the "open" map and in addition to your axioms, each fiber is a real banach space, the action of $\mathbb{R}$ is globally continuous and the norm function is upper semi-continuous) then it has enough local sections. But the proof uses partition of unity and hence the multiplication by $\mathbb{R}$ a lot, so it is not clear that the idea of this results can be adapted here. $\endgroup$
    – Simon Henry
    Commented Mar 15, 2016 at 11:58

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