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?
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 $p \circ s = \mbox{id}$.