$\mathbb{P}^2(\mathbb{R})$ can be seen as a 2-d unit disk $D^2\subseteq \mathbb{R}^2$ with the antipodal points on $\partial D^2$ identified. Suppose $q: D^2 \rightarrow \mathbb{P}^2(\mathbb{R})$ is the quotient map that takes $(x,y) \in D^2$ to the point $(x,y,\sqrt{1-x^2-y^2}) \in \mathbb{S}^2$ and then to its equivalence class in $\mathbb{P}^2(\mathbb{R})$.
Define $H_z:=\{[x,y,z] \in \mathbb{P}^2(\mathbb{R}), z=0\}$. Then $q^{-1}(H_z)=\partial D^2$.
Is it true that $\mathbb{P}^2(\mathbb{R})\setminus H_z$ is homeomorphic to $D^2\setminus\partial D^2$?
Clearly $q:D^2\setminus\partial D^2 \rightarrow \mathbb{P}^2(\mathbb{R})\setminus H_z$ is bijective and continuous, however I would need to show it is either open or closed to conclude it is a homeomorphism. I think it is closed but I've been unable to prove it, do you have any hint?
Best Answer
The restricted map $q : D^2 \setminus \partial D^2 \to \mathbb{P}^2(\mathbb{R}) \setminus H_z$ is indeed an open map, for the following reasons.
The restricted domain $D^2 \setminus \partial D^2$ is a saturated open set with respect to the original unrestricted map $q : D^2 \to \mathbb P^2(\mathbb R)$, meaning that for each $z \in D^2 \setminus \partial D^2$ we have $q^{-1}(q(z)) \subset D^2 \setminus \partial D^2$.
Also, the restricted map is one-to-one, and it follows that every open subset of the restricted domain is also a saturated open set of the original map.
By definition of a quotient map, the image of every saturated open subset is an open set.
Therefore, the image under $q$ of every open subset of the restricted domain $D^2 \setminus \partial D^2$ is an open subset of $\mathbb P^2(\mathbb R)$ disjoint from $H_z$, and is therefore an open subset of $\mathbb{P}^2(\mathbb{R}) \setminus H_z$.