The quotient map $q: \mathbb R^{n+1} \setminus \{0\} \to \mathbb P^n$ is open.

Question

I'd like to show that the quotient map $q: \mathbb R^{n+1} \setminus \{0\} \to \mathbb P^n$ is open, where I'm considering $\mathbb P^n$ as the quotient space of $\mathbb R^{n+1} \setminus \{0\}$ under the equivalence relation $(x_0, \dots, x_n)\sim (y_0,\dots,y_n)$ if there exists a non-zero real number $\lambda$ such that $(y_0,\dots,y_n) = \lambda(x_0,\dots, x_n).$

I believe I have a proof, but I'd like to know if there is a cleaner way to do it. I know that this can be done using group actions, but I want to avoid that.

My proof:

To prove that the quotient map is open, we take an open set in $\mathbb R^{n+1} \setminus \{0\}$ and show that it maps to an open set in $\mathbb P^n$. For an open set $U$ in $\mathbb R^{n+1}\setminus \{0\}$, to show that $q(U)$ is open, we must show that $q^{-1}(q(U))$ is open in $\mathbb R^{n+1}\setminus \{0\}$.

Let $x \in q^{-1}(q(U))$. We show that there exists an open set containing $x$ that is also contained in $q^{-1}(q(U))$. Since $x \in q^{-1}(q(U))$, there exists a $\lambda \in \mathbb R\setminus \{0\}$ such that $\lambda x \in U$. Since $U$ is open, there exists an $\varepsilon > 0$ such that $B(\varepsilon,\lambda x) \subseteq U$. We claim that $B(\frac{\varepsilon}{|\lambda|},x)$ is an open set containing $x$ that is also contained in $q^{-1}(q(U))$. To show this, we choose an arbitrary point $y \in B(\frac{\varepsilon}{|\lambda|},x)$, and show that $y \in q^{-1}(q(U))$. Since $y \in B(\frac{\varepsilon}{|\lambda|},x)$, we know that $\Vert{y - x}\Vert < \frac{\varepsilon}{|\lambda|}$, and moreover, $\Vert{\lambda y - \lambda x}\Vert < \varepsilon$ so that $\lambda y \in B(\varepsilon,\lambda x)\subseteq U$. Thus, $q(y) \in q(U)$ and $y \in q^{-1}(q(U))$ so that $x \in B(\frac{\varepsilon}{|\lambda|},x)\subseteq q^{-1}(q(U))$. This means that $q^{-1}(q(U))$ is open, and that $q$ is an open map.

Answer 1

There is a simpler proof.

It's enough to prove it on a basis of $\mathbb R^{n+1}\setminus\{0\}$, which can be taken as the set of balls whose radius is smaller than the norm of their center.

Let $B$ be such a ball. To show $q(B)$ is open, we must show that $\bigcup q(B)$ is open in $\mathbb R^{n+1}\setminus\{0\}$. But $$\bigcup q(B)=\{\lambda x\mid x\in B,\lambda\in\mathbb R\setminus\{0\}\}=\bigcup_{\lambda\in\mathbb R\setminus\{0\}}\lambda B,$$ which is just a union of open balls, so it's open.

Answer 2

This proof looks pretty good, there's not a lot of room to make it 'cleaner'.

But one modification you might like (or might not):

Suppose $x \in q^{-1}(q(U))$. Then there exists $\lambda \in \mathbb{R}\backslash \{0\}$ such that $\lambda x \in U$. Let $m:\mathbb{R}^{n} \backslash \{0\} \rightarrow \mathbb{R}^{n} \backslash \{0\}$ denote multiplication by $\lambda$. Then $m$ is continuous, and $q \circ m = q$, so we get $x \in m^{-1}(U) \subset q^{-1}(q(U))$, and $m^{-1}(U)$ is open.

Answer 3

Quite similar to @Alex Zorn: proof, this fact is rather standard

If the topological group $G$ acts on $X$, then the map $X \to X/G$ is open. The formal proof: $q^{-1}(q(U)) = \cup_{g\in G}\,\, g\cdot U$ is open.

To see the fact, the proof of @user2345215: is useful.