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Let $\mathbb{P}^n$ denote the real projective n-space. How can I show that the quotient map $\pi: \mathbb{R}^{n+1}\setminus\{0\} \rightarrow \mathbb{P}^n$ defined by $(x_1, ..., x_{n+1}) \mapsto [x_1: \dots : x_{n+1}]$ is in fact a smooth map?

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    $\begingroup$ Ohh I just saw that you are using the quotient from $\mathbb{R}^{n+1}$ and not from $\mathbb{S}^{n}$. I'll adjust my answer very soon. $\endgroup$
    – Laz
    Oct 7, 2018 at 2:03

1 Answer 1

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Your quotient map $\pi:\mathbb{R}^{n+1}\setminus\{0\}\rightarrow \mathbb{RP}^n$ is just the composition $\pi\circ p$, where $p: \mathbb{R}^{n+1}\setminus\{0\}\rightarrow \mathbb{S}^n$ is the map $p(x)=\frac{x}{|x|}$ (smooth) and $\pi:\mathbb{S}^n\rightarrow \mathbb{RP}^n$ is the restriction of the original $\pi$ to $\mathbb{S}^n$ (that's why I keep calling it $\pi$). Then it's enough to prove that the restriction of $\pi$ to $\mathbb{S}^n$ is smooth.
To do so, observe that since the new $\pi$ is a local homeomorphism, you can take any chart in $\mathbb{S}^n$, $(x,U)$ such that $\pi|_U: U\rightarrow \pi(U)\subset \mathbb{RP}^n$ is a homeomorphism. Now with this you can use the smooth atlas in $\mathbb{S}^n$, $\{(x,U)\}$ to construct a smooth atlas $\{(\mathbb{x}, \pi(U))\}$ such that each $\mathbb{x}|_{\pi(U)}$ is just $x\circ (\pi|_U)^{-1}|_{\pi(U)}$ (take care with the domains).
Prove that in these two charts, the restricted $\pi$ looks locally like the identity of $x(U)\subset \mathbb{R}^n$. Then it has to be smooth, actually, a local diffeomorphism.
Hence, your old $\pi$ is smooth.

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