Extending a quotient map to a covering map on $\mathbb{RP}^2$

Question

Why can we not extend the quotient map $q:[0,1]\times[0,1] \to \mathbb{RP}^2$ to a covering map, $\mathbb{R}^2 \to \mathbb{RP}^2$?

Answer 1

Recall that the map $f:S^2\to \mathbb{RP}^2$ which identifies antipodal points is a covering, and that $S^2$ is simply-connected, hence $S^2$ is the universal cover of $\mathbb{RP}^2$. If we had a covering $p:\mathbb R^2\to \mathbb{RP}^2$ then since $\mathbb R^2$ is also simply connected it would also be a universal cover of $\mathbb{RP}^2$. Since universal covers are unique up to homeomorphism, this implies that $S^2$ is homeomorphic to $\mathbb R^2$, which is false (for example, because $S^2$ is compact while $\mathbb R^2$ is not). Thus no such covering $p:\mathbb R^2\to \mathbb{RP}^2$ can exist.

Answer 2

I assume that $q$ is the map that identifies $(0,y)$ with $(1,1-y)$ and $(x,0)$ with $(1-x,1)$. Tiling the plane with squares and extending the quotient map using reflections does lead to a map $\mathbb{R}^2\rightarrow\mathbb{RP}^2$, but it is not a covering map: notice that a small disc around a point on one of the square's edges gets mapped to a half disc, and around one of the vertices to a quarter-disc; thus, the map is not locally a homeomorphism.

EDIT: A subtler idea (one that, once I thought of it, was in the back of my head bothering me a little till I got to sit down and think about it tonight) would be to extend $q$ using glide reflections. This is closer to a covering map but it's still branched at the corners of the square. E.g. $(0,y)$ has the same image as $(1,1+y)$, which has the same image as $(0,2-y)$, so by considering what happens as $y\to 1$ we see that this way of extending the map is 2-to-1 on the $y$-axis in a neighborhood of $(0,1)$.

EDIT: Olivier Begassat's answer to this question clarifies the situation to me. In this latter construction, $\mathbb{RP}^2$ is realized as the quotient of $\mathbb{R}^2$ by the group of isometries generated by unit glide reflections in the lines $y=1/2, x=1/2$; call them $\nu,\rho$ respectively. This group contains rotations about the corners of the squares, i.e. the integer lattice points of $\mathbb{R}^2$; for example, $\rho^{-1}\nu$ is the 180 degree rotation about the origin. Thus the quotient map identifies points that are symmetric about the integer lattice points. In particular, it is 2-to-1 in small neighborhoods of these points.