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. $X=(X/(\mathbb{Z}/2))\times \mathbb{Z}/2$?



triviality of a $2$-sheeted covering map and the triviality of the associated vector bundle
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.


