# Prove that there is no holomorphic map $s: \mathbb{P}^n \to \mathbb{C}^{n+1} \setminus \{0\}$ with $p \circ s = \operatorname{id}$.

Let $$p:\mathbb{C}^{n+1} \setminus \{0\} \to \mathbb{P}^n$$ be the natural projection. Prove that there is no holomorphic map $$s: \mathbb{P}^n \to \mathbb{C}^{n+1} \setminus \{0\}$$ with $$p \circ s = \operatorname{id}$$.

I'm seeking for intuition for the problem. I started to think about what the map $$s$$ would have to be in order for this to hold and for example if $$s([z_0:\dots:z_n]) = (z_0,\dots,z_n)$$, the the map would give the identity, but this $$s$$ isn't even well-defined let alone holomorphic since $$[z_0:\dots:z_n]=[\lambda z_0:\dots:\lambda z_n]$$ for any non-zero $$\lambda$$, but clearly $$(z_0,\dots,z_n) \ne (\lambda z_0,\dots, \lambda z_n)$$ unless $$\lambda = 1$$.

Now my understanding of the projective space is very limited so I don't have any intuition as to why such map wouldn't exist and I would appreciate if anyone could shed some light in to why this shouldn't be true?

• Do you have knowledge of algebraic topology/homological algebra? If so, you could show that the existence of such a map (which would be an isomorphism between $\mathbb{C}\backslash\{0\}$ and $\mathbb{P}^n$ would have to induce an isomorphism on homology groups (or fundamental groups, based on your knowledge)), from which you could derive a contradiction. It is a quick way of proving this, though admittedly not the most elementary one
– Azur
Commented Sep 25, 2023 at 11:32
• I have enough to see that this would be quite a nice solution, but I would be happy to get more intuition for the projective space rather than use the machinery from algebraic topology. Thanks for the nice idea though! @Azur Commented Sep 25, 2023 at 11:36
• Recall that the only holomorphic maps $\Bbb P^n\to \Bbb C$ are constants. Commented Sep 26, 2023 at 23:29
• @TedShifrin So $p \circ z_j : \Bbb P^n\to \Bbb C$ is constant for each of the coordinate functions $z_j$. Will this automatically imply that $s$ is itself constant also? Commented Sep 27, 2023 at 6:43
• You mean $z_j\circ s$? Commented Sep 27, 2023 at 14:42

Since $$\mathbb{P}^n$$ is compact, $$s(\mathbb{P}^n)$$ is compact and thus closed and bounded. But that means there is a $$p \in \mathbb{P}^n$$ with $$|s(p)|$$ maximized. Pick an unitary matrix $$M$$ that takes $$s(p)$$ to $$(a, 0, \ldots 0)$$ where $$a = |s(p)|$$. Then for $$g = Ms$$, $$|g|$$ is also maximized at $$p$$, but now $$g(p) = (a, 0, .., 0)$$. This is clearly not maximal if $$g_1$$ is non-constant since $$g_1$$ would then be an open mapping. But if $$g_1$$ is constant, then this $$|g(p)|$$ can only be maximal if all the $$g_i$$ are constant, but then $$s$$ would be constant, contradicting $$p \circ s = \mbox{id}$$.