Covering space(s) of $\mathbb{R}\text{P}^2$ minus one point

Question

I know that the covering space of $\mathbb{R}P^2$ is $S^2$, and it is unique unless than isomorphism of covering spaces. Now, $S^2$ minus one point is homeomorphic to $\mathbb{R}^2$ (by stereographic projection), and so $S^2$ minus two points is homeomorphic to $\mathbb{R}^2$ minus one point, which is homotopically equivalent to $S^1$ (more specifically, $S^1$ is a deformation retract of $\mathbb{R}^2$ minus one point). Then, we have a surjective map from $\mathbb{R}^2$ minus one point to $\mathbb{R}\text{P}^2$ minus one point. Is it a covering map? And if yes, how can I see this? Any help would be greatly appreciated.

Answer 1

Are you ok with knowing how to describe the map up to homotopy? If so, $\mathbb{R}^2\setminus\{0\}$ is homotopy equivalent to a circle and $\mathbb{R}P^2\setminus\{p\}$ is also homotopy equivalent to a circle. If you factor the map $\mathbb{R}^2\setminus\{0\}\to\mathbb{R}P^2\setminus\{p\}$ through these homotopy equivalences, we get the doubling map $S^1\to S^1$ which is a covering space.

That is, the deformation retractions $h_1$ and $h_2$,the surjection $s$ onto the punctured projective plane, and the doubling map on the circle fit into a commutative (up to homotopy) diagram: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \mathbb{R}^2\setminus\{0\} & \ra{s} & \mathbb{R}P^2\setminus\{p\} \\ \da{h_1} & & \da{h_2} \\ S^1 & \ras{\times 2} & S^1 \\ \end{array} $$