Covering space(s) of $\mathbb{R}\text{P}^2$ minus one point 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.
 A: 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:
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
   \mathbb{R}^2\setminus\{0\}    & \ra{s}       &    \mathbb{R}P^2\setminus\{p\}     \\
  \da{h_1}     &              &  \da{h_2}               \\
   S^1       & \ras{\times 2} &    S^1                        \\
\end{array}
$$
