Problem
Consider the domain $$\Omega=\mathbb{C}^4\setminus\{z_0(z_1^2+z_2^2+z_3^2)=0\}$$ and the map $$F:\Omega\to\mathbb{CP}^1\qquad F(z_0,z_1,z_2,z_3)=[z_0^2:z_1^2+z_2^2+z_3^2]\;.$$ Define $U=\mathbb{CP}^1\setminus\{[1:0],[0:1],[1:a_1],\ldots, [1:a_k]\}$ with $a_1,\ldots, a_k\in\mathbb{R}_{>0}$.
I would like to compute $\pi_1(X)$ with $X=F^{-1}(U))$.
A fibration
- $F$ is a fibration (it is a smooth submersion)
- $\pi_1(U)=F_{k+1}$ (i.e. the free group on $k+1$ elements)
- the fibers of $F$ have fundamental group $\mathbb{Z}$
Proof they are are double covers of the graph of $$f:\mathbb{C}^3\setminus\{z_1^2+z_2^2+z_3^2=0\}\to\mathbb{C}\qquad f(z_1,z_2,z_3)=z_1^2+z_2^2+z_3^2$$ and this graph has the fundamental group of $\mathbb{C}^3\setminus\{z_1^2+z_2^2+z_3^2=0\}$; this fundamental group can be computed by noticing that $f$ is a submersion on that set and the fibers are diffeomorphic to the total bundle of $\mathbb{S}^2$, so are simply connected, then from the long sequence for the fibration, we obtain $\pi_1(\mathbb{C}^3\setminus\{z_1^2+z_2^2+z_3^2=0\})=\pi_1(\mathbb{C}^*)=\mathbb{Z}$.
A double cover
Define $S=\{(0,z_1,z_2,z_3)\in\mathbb{C}^4\ :\ z_1^2+z_2^2+z_3^2=1\}$ and $$p: S\times\mathbb C^2\to\mathbb C^4\qquad p(s,(w_0,w_1))=w_0^2e_0+w_1^2s$$ where $e_0=(1,0,0,0)$. Then
- $p:p^{-1}(X)\to X$ is a $2$-to-$1$ map.
- if $\rho:\mathbb C^2\setminus\{(0,0)\}\to \mathbb{CP}^1$ is the usual projection, then $\rho(p^{-1}(X))$ is the complement of $2k+2$ points
- the map $G:p^{-1}(X)\to\mathbb{CP}^1$ given by $G(s,(w_0,w_1))=\rho(w_0,w_1)$ is a submersion, hence a fibration (in fact, it's the Hopf fibration)
- the fundamental group of $p^{-1}(X)$ is $\mathbb Z\times F_{2k+1}$ (it is hinted on some MO comment that I can't find now and in this paper in example 4.5)
Short exact sequences
The previous observations imply that $$0\to\mathbb{Z}\to\pi_1(X)\to F_{k+1}\to 0$$ $$0\to\mathbb{Z}\times F_{2k+1}\to\pi_1(X)\to\mathbb{Z}/2\mathbb{Z}\to 0$$ And here I am stuck! I can write down the obvious semidirect product from the firts one, given that on the right I have a free group, however, I cannot find how to conclude. My claim is that the semidirect product is indeed direct, hence $$\textbf{Claim: }\pi_1(X)=Z\times F_{k+1}$$
Any ideas? Thanks!