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Let $ X$ be a space with a (free and properly discontinuous) $\mathbb{Z}/2$-action and $$p: X\to X/(\mathbb{Z}/2) $$ be a $2$-sheeted covering map. Then we have an associated vector bundle $$ \xi: \mathbb{R}^2\to X\times_{\mathbb{Z}/2}\mathbb{R}^2\to X/(\mathbb{Z}/2) $$ where $\mathbb{Z}/2$ acts on $\mathbb{R}^2$ by reversing the order of coordinates $(x,y)\mapsto(y,x)$.

Question: if $\xi$ is a trivial bundle, can we conclude that the covering map $p$ is a trivial $2$-sheeted covering map? i.e. $X=(X/(\mathbb{Z}/2))\times \mathbb{Z}/2$?

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    $\begingroup$ Note that, if you allow non-free actions, the answer is negative. The quotient $Y$ of $X = \mathbb{R}^2$ by the action $(x, \, y) \to (y, x)$ is a half-plane, hence contractible. Therefore every vector bundle on $Y$ is trivial, but clearly $X \to Y$ is not the trivial covering. $\endgroup$
    – Francesco Polizzi
    Oct 26, 2015 at 15:58
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    $\begingroup$ @FrancescoPolizzi Thanks! but this is not a covering space. $\endgroup$
    – QSR
    Oct 26, 2015 at 16:24
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    $\begingroup$ Well, it depends on your definition :-) For me, this is a branched covering space (the branching locus is the line $y=x$) $\endgroup$
    – Francesco Polizzi
    Oct 26, 2015 at 16:24

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Yes, you can conclude that, because the construction $$(\wedge^2 \xi \setminus \text{zero section})/\mathbb{R}_{>0} \to X/(\mathbb{Z}/2)$$ recovers the original double cover.

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