If that's the definition, then how do we know double covers of $SO(n)$ exist?
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This is a consequence of covering theory. Its central theorem says that for a reasonable topological space (like the manifold $\operatorname{SO}(n)$) path-connected coverings correspond in a certain way to subgroups of the fundamental group, and the degree of such a covering corresponds to the index of the subgroup.
The fundamental group of $\operatorname{SO}(n)$ is $\mathbb Z/2\mathbb Z$ hence the trivial group is an index 2 subgroup. The mentioned theorem now implies the existence of a (essentially unique) path-connected double covering.
tth2507
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1Thanks. Is there a reference that explains this in more detail? – Alex Feb 21 '24 at 12:21
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1@Alex Covering theory is discussed in Chapter 1.3 of Hatchers “Algebraic topology”. A very detailed exposition is given in Part VIII of these lecture notes. – tth2507 Feb 21 '24 at 12:33