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I want to show that there is no holomorphic surjection from $\mathbb{C}^n$ to $\mathbb{CP}^n$.

But Consider a holomorphic surjection $f: \mathbb{C}^n \to \mathbb{CP}^n$. I find I really can say little thing about this map, is this statement true? If it's true, how to prove it?

Thank you for your answer！

taiat
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    Not sure what happens for $n \geq 2$ – I’d like to say it’s impossible, but I’m less sure than ever – anyway, these MSE counterexamples for $n=1$ are interesting: https://math.stackexchange.com/questions/1070860/does-there-exist-a-regular-map-mathbba1-to-mathbbp1-which-is-surjective#1071198 . For $n \geq 2$, I’m reasonably confident that such a map lifts into $\mathbb{C}^{n+1} \backslash {0}$ and we’re reduced to finding (or disproving the existence of) holomorphic surjections $\mathbb{C}^m \rightarrow \mathbb{C}^m \backslash {0}$ where $m=n+1 \geq 3$. – Aphelli May 30 '22 at 14:58
  • @Aphelli why can we lift the map to $\mathbb{C}^{n+1}\blackslash {0}$? – taiat May 31 '22 at 02:51
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    Because $H^1(\mathbb{C}^n,\mathcal{O}^{\times})$ (which parametrizes holomorphic line bundles) vanishes, due to the exponential LES and (eg) “Cartier B”. – Aphelli May 31 '22 at 06:47
  • @Aphelli Thanks！ – taiat May 31 '22 at 09:11
  • @Aphelli But the lift $\mathbb C^n \to \mathbb C^{n+1} \setminus {0}$ need not be surjective, for $f$ to be surjective? – red_trumpet May 31 '22 at 11:12
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    @red_trumpet: it doesn’t need to (otherwise the problem would be trivial), but the map $h(z,w)=e^wf(z)$ must be (from $\mathbb{C}^{n+1}$ to $\mathbb{C}^{n+1} \backslash {0}$). – Aphelli May 31 '22 at 11:16
  • @Aphelli I see, thanks for clarifying this. – red_trumpet May 31 '22 at 11:37

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