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Suppose we have the following equivalence relation on $\mathbb{R}^{n+1}\setminus \{0\}$: $$x \sim y :\Leftrightarrow \exists \lambda \in \mathbb{R}^\times ,x = \lambda y$$ How can I show that the canonical projection $\pi: \mathbb{R}^{n+1}\setminus \{0\} \to \left(\mathbb{R}^{n+1}\setminus \{0\}\right)/\sim$ is an open quotient map? Somehow the term open confuses me a little bit.

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    $\begingroup$ A continuous map $f:X\to Y$ is called open if $U\subseteq X$ open implies $f(U)\subseteq Y$ open. $\endgroup$ – Arthur Oct 8 '16 at 21:52
  • $\begingroup$ @Arthur Yes. So I take open balls in $\mathbb{R}^{n+1}\setminus \{0\}$ and look at the images? $\endgroup$ – TheGeekGreek Oct 8 '16 at 21:56
  • $\begingroup$ Technically you need to do any kind of open sets, not just balls. But if you can handle balls (or any other basis for the standard topology of $\Bbb R^{n+1}\setminus\{0\}$) then the rest should follow as a consequence. $\endgroup$ – Arthur Oct 8 '16 at 22:02
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For $\lambda \in \mathbb{R}^\times$ define a mapping $$\psi_\lambda: \begin{cases} \left(\mathbb{R}^{n+1}\setminus\{0\}\right) \to\left(\mathbb{R}^{n+1}\setminus\{0\}\right)\\ x \to \lambda x \end{cases}$$

Then $\psi_\lambda^{-1} = \psi_{\lambda^{-1}}$, so $\psi_\lambda$ is a bijection. Furthermore, since $\psi_\lambda \in C^\infty$ we have that $\psi_\lambda$ is a homeomorphism. If $U$ is open in $\mathbb{R}^{n+1}\setminus\{0\}$ then $\psi_\lambda(U)$ is open in $\mathbb{R}^{n+1}\setminus\{0\}$. Hence by

$$\pi^{-1}\left( [U]_\sim \right) = \bigcup_{\lambda \in \mathbb{R}^\times} \psi_\lambda(U)$$

we have that $[U]_\sim$ is open in $\mathbb{RP}^n$.

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