Universal Cover of $\mathbb{R}P^{2}$ minus a point

Question

I've already calculated that the fundamental group of $\mathbb{R}P^{2}$ minus a point as $\mathbb{Z}$ since we can think of real projected space as an oriented unit square, and puncturing it we can show it is homotopy equivalent to the boundary which is homotopy equivalent to the unit circle.

I am having some trouble finding the universal cover of punctured real projective space, however. I'm not sure how to construct one. I know that punctured real projective space is homeomorphic to the mobius band, but I'm not sure if that helps me here. Thank you for any help.

Answer 1

So it's a Moebius band, as you say. The Moebius band is a quotient of a cylinder, which is a quotient of the real plane.

Answer 2

The space $E=\mathbb P^2(\mathbb R)\setminus\{0\}$ obtained by deleting $0=[0:0:1]$ from the $[x:y:z]$ plane $\mathbb P^2$ can be sen as the tautological line bundle on the line at infinity $z=0$ of the plane and since that line is homeomorphic to $\mathbb P^1(\mathbb R)\cong S^1$ we obtain the non trivial (=Möbius) bundle $p:E\to S^1.$

Pulling back that bundle to the universal covering $u:\mathbb R\to S^1$ we obtain the cartesian diagram $$\begin{array} uu^*(E) & \stackrel{p^*(u)}{\longrightarrow} & E=\mathbb P^2(\mathbb R)\setminus\{0\} \\ \downarrow{u^*(p)} & & \downarrow{p} \\ \mathbb R & \stackrel{u}{\longrightarrow} & S^1 \end{array} $$ The pulled back line bundle $u^*E \stackrel{u^*(p)}{\longrightarrow} \mathbb R$ is trivial, like all line bundles on $\mathbb R$, and thus $u^*E=\mathbb R\times \mathbb R$.
But then we see that by pulling back the universal covering $u:\mathbb R\to S^1$ by $p$ to $E$ we get the pulled back covering space $u^*E=\mathbb R\times \mathbb R \stackrel{p^*(u)}{\longrightarrow} E=\mathbb P^2(\mathbb R)\setminus\{0\}$, which must be the required universal covering since it is simply connected (and even contractible).

Summing up: The universal covering of $\mathbb P^2(\mathbb R)\setminus\{0\}$ is $\mathbb R\times \mathbb R$.

Answer 3

Letting $M$ denote the Mobius band, there is a universal covering map $\mathbb{R}^2 \to M$ with infinite cyclic deck transformation group $\langle r \rangle$ generated by the "glide reflection" $$r : (x,y) \to (x+1,-y) $$ As with any universal covering map, $M$ is the orbit space of this action. One can verify this by noticing that the vertical strip $[0,1] \times \mathbb{R}$ is a fundamental domain for $r$, its left side $0 \times \mathbb{R}$ being glued to its right side $1 \times \mathbb{R}$ by identifying $(0,y)$ to $(1,-y)$. This is exactly the gluing pattern for constructing the Mobius band that we all learned on our mother's knee.