# Prove the quotient map of real projective n space is smooth

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?

• 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.
– Laz
Oct 7, 2018 at 2:03

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.