holomorphic disk and crosscap as quotients of $\mathbb{CP}^1$ by antiholomorphic involutions

Question

Consider $\mathbb{CP}^1 \ni (u:v)$ and the maps $$ \sigma_{\pm}: \quad (u:v) \mapsto (\overline{v}:\pm\overline{u})$$ How do we show that the quotient $\mathbb{CP}^1/\sigma_+$ gives a disk, and $\mathbb{CP}^1/\sigma_-$ gives a crosscap?

Answer 1

If $(x: y)$ is fixed by $\sigma_+$ then there exists $\lambda$ with $x = \lambda \overline{y}$ and $y = \lambda\overline{x}$. We get that $x = \lambda\overline{\lambda}x$. We can't have $x = 0$ (as then $y = \lambda x = 0$) so $\lambda$ lies on the unit circle. These points are exactly $[1: \lambda]$ for $|\lambda| = 1$, i.e. identify with the unit circle under the standard identification of $\mathbb{CP}^1$ with the sphere.

Now it is clear that the quotient is a disk topologically (bounded by the circle of fixed points), since every point in the north hemisphere is glued to a distinct point in the south hemisphere (the inclusion map of the disk into this surface is bijective and continuous, and the disk is compact, so it's a homeomorphism).

On the other hand, if $(x: y)$ is fixed by $\sigma_-$, then there exists $\lambda$ with $x = \lambda\overline{y}$ and $y = -\lambda\overline{x}$. It follows that $x = -\lambda\overline{\lambda}x$ (which can't happen since $x$ can't be zero as above and $\lambda$ must have positive norm). This is thus a free action of $\frac{\mathbb{Z}}{2}$ on the sphere, so the quotient is a manifold and the quotient map is a covering map of degree $2$; it follows that the quotient is $\mathbb{RP}^2$.