Subset $V$ of projective space is open iff $q^{-1}(V)$ is open?

Question

Let $\mathbb{P}^n$ denote the set of all lines through the origin in the coordinate space $\mathbb{R}^{n+1}$. Define a function$$q: \mathbb{R}^{n+1} - \{0\} \to \mathbb{P}^n$$by $q(x) = \mathbb{R}x =$ line through $x$. How do I see that a subset $V \subset \mathbb{P}^n$ is open if and only if $q^{-1}(V)$ is open?

Answer 1

This is because the topology used for $\mathbb P^n$ is the quotient topology.

And the quotient topology is exactly the one for which $V \subset \mathbb P^n$ is open if and only if $q^{-1}(V) \subset \mathbb R^{n+1} \setminus \{0\}$ is open.

Answer 2

If $V \subset \mathbb{P}^n$ is open, certainly $q^{-1}(V)$ is open by continuity. Conversely, if $q^{-1}(V)$ is open, for any point, say with $x_{n+1} \neq 0$ so we can use that coordinate patch, with coordinates$$p = [x_1, \dots,\, x_n,\,1],$$the preimage point $(x_1, \dots, x_n, 1)$ has an open neighborhood in $q^{-1}(V)$, where $x_{n+1}$ is still nonzero. This neighborhood, as an open set of $\mathbb{R}^{n+1}$, contains an open neighborhood in $q^{-1}(V)$, where $x_{n+1}$ is still nonzero. This neighborhood, as an open set of $\mathbb{R}^{n+1}$, contains an open neighborhood of the intersection with the hyperplane $x_{n+1} = 1$. This open neighborhood of the point in the hyperplane maps homeomorphically under $q$ to a neighborhood of $p$, since we can factor the coordinate patch map defined here through the map; it is $q(x_1, x_2, \dots, x_n, 1)$. So every point of $V$ has an open neighborhood.