The space \(P^1\) and the covering map \(p: S^1 \to P^1\) are familiar ones. What are they?
Solution
The real projective line \(P^1\) is the set of one-dimensional linear subspaces of \(\mathbb R^2\). A line through the origin is determined by either of the two antipodal unit vectors lying on it. Hence \[ P^1=S^1/(x\sim -x). \]
This quotient is itself homeomorphic to a circle. In complex notation, identify \(S^1\subset\mathbb C\). The map \[ q:S^1\to S^1,\qquad q(z)=z^2, \] has exactly the antipodal pairs \(\{z,-z\}\) as its fibers, so it factors through a continuous bijection \[ P^1\longrightarrow S^1. \] Since \(P^1\) is compact and \(S^1\) Hausdorff, this bijection is a homeomorphism.
Under this identification, the standard covering \[ p:S^1\to P^1, \qquad p(x)=\mathbb Rx, \] is exactly the familiar two-sheeted circle covering \(z\mapsto z^2\).