I want to show that $S^n/C_2$ is homeomorphic to $\mathbb{R}P^n$, where $C_2$ is the group with 2 elements and acts on $S^2$ by $x\mapsto -x$. I think I've found the map $F:\mathbb{R}P^n\to S^n/C_2$ by $[x_0:...:x_n]\mapsto [\frac{x_0}{\sqrt{\sum_0^nx_k^2}},...,\frac{x_n}{\sqrt{\sum_0^nx_k^2}}]$, where in $\mathbb{R}P^2$, bracket notation is used for $x\sim \lambda x$ ($\lambda\neq 0)$ and in $S^n/C_2$, it is used for $x\sim -x$. Firstly, I want to make sure this is indeed a homeomorphism between the two desired spaces, and secondly (most importantly), I don't know how to find the inverse map.

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    $\begingroup$ The inverse map is easy to define : if you have a point in $S^n$ this define a line. This define a map $S^n \to RP^n$. Since it is invariant by the $C_2$ action this define a map : $S^n/C_2 \to RP^n$ which is clearly bijective. $\endgroup$ – user171326 Oct 3 '16 at 17:31
  • $\begingroup$ Related: math.stackexchange.com/questions/1032094 $\endgroup$ – Watson Oct 10 '16 at 6:37

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