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We have a natural quotient map $$\phi\colon S^{2n+1}\rightarrow (\mathbb{C}^{n+1}\setminus{\{0\}})/\mathbb{C}^{*}=\mathbb{P}^n\mathbb{C}$$ and I want to see, that it's open. Denote by $$\iota\colon S^{2n+1}\hookrightarrow\mathbb{C}^{n+1}\setminus{\{0\}}$$ the inclusion and by $$\pi\colon \mathbb{C}^{n+1}\setminus{\{0\}}\rightarrow(\mathbb{C}^{n+1}\setminus{\{0\}})/\mathbb{C}^{*}$$ the quotient map.

It holds $\phi=\pi\circ\iota$. Let's take an open $U\subset S^{2n+1}$, so we find a open $V\subset (\mathbb{C}^{n+1}\setminus{\{0\}})/\mathbb{C}^{*}$, such that $U=V\cap S^{2n+1}$.

$\phi(U)$ is open iff $\pi^{-1}(\pi(V\cap S^{2n+1}))=\bigcup_{\lambda\in\mathbb{C}^{*}}\lambda(V\cap S^{2n+1})$ is open in $\mathbb{C}^{n+1}\setminus{\{0\}}$, but I don't see, why this holds.

Thanks in advance.

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ι:S 2n+1 ↪C n ∖{0} What's this inclusion map? –  lee Jan 12 '13 at 15:10
My mistake. I corrected n to n+1 –  max Jan 12 '13 at 15:12
Then I know. Do you know what is rank theorem? –  lee Jan 12 '13 at 15:18
The one of linear maps which occurs in linear algebra? –  max Jan 12 '13 at 15:23
Not the same, but they are similar. The map you have here is differentiable, and you can check that actually when you give S^(2n+1) and CP^n a local coordinate, the Jacobians are always of rank 2n. The rank theorem then states that it must be of the form(when you give a good local coordinate),f(x1,x2,...,x2n+1)=(x1,x2,...,x2n). You can easily check this map must be locally open. Thus the global map is open. –  lee Jan 13 '13 at 1:42
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