Fundamental group of the complement of $3$ disjoint hypersurfaces in $\mathbb{C}P^2$

Question

Let $X$ be the union of $3$ hypersurfaces in $\mathbb{C}P^2$, then how to compute the $\pi_1(\mathbb{C}P^2\setminus X)$?

What I know is the complement of a hypersurface in $\mathbb{C}P^2$ is $\mathbb{C}^2$. I can't go further.

Answer 1

In the lines of @MoisheKohan after a change of co-ordinates you can assume one of your hypersurfaces is $H_3=\{[z_1:z_2:z_3]\in \mathbb C\mathbb P^2 | \ z_3= 0 \}$

$ \mathbb C \mathbb P^2-H_3\cong\mathbb C^2$

Thus you are reduced to computing $\pi_1(\mathbb C^2-l_1\cup l_2)$ where $l_1, \ l_2$ are two complex straight lines. Again after a change of co-ordinates you can assume $l_1=\{(z_1,z_2)\in \mathbb C^2 :z_1=0\}$ and $l_2=\{(z_1,z_2)\in \mathbb C^2 :z_2=0\}$

Then you have $\mathbb C^2 -l_1\cup l_2=\mathbb C^*\times \mathbb C^*$

So you get $$\pi_1(\mathbb C^2-l_1\cup l_2)=\pi_1(\mathbb C^*)\times \pi_1(\mathbb C^*)=\mathbb Z^2$$

Thus $$\pi_1(\mathbb C\mathbb P^2-H_1\cup H_2 \cup H_3)=\mathbb Z^2$$