Find the fundamental group of $\Bbb C^2 \setminus \{(x,y):xy=0 \}$.

Question

What is the fundamental group of $\Bbb C^2 \setminus \{(x,y):xy=0 \}$?

How do I proceed? Please help me in this regard.

Thank you very much.

Answer 1

Sketch, but almost a complete solution: First observe that topologically, the space is $ \mathbb{R}^4 \setminus (H_1 \cup H_2) $ where $$ H_1 = \{(x_1, x_2, x_3, x_4) | x_1=x_2=0 \} , H_2 = \{(x_1, x_2, x_3, x_4) | x_3=x_4=0 \} $$ are two dimensional subspaces. This space clearly deformation retracts to $ S^3 $ minus two copies of $ S^1 $, which are given by the equations $ C_1:x_3^2 + x_4^2 = 1 $ and $ C_2:x_1^2 + x_2^2 = 1 $. Now use stereographic projection from the north pole $ (0,0,0,1) $ to see that $ C_2 $ goes to the same circle in $ \mathbb{R}^3 $ while $ C_1 $ maps to the $ z$-axis of $ \mathbb{R}^3 $. So finally, you need to compute the fundamental group of $ \mathbb{R}^3 \setminus (\{ x^2 + y^2 = 1 \} \cup \{ (0,0,z) | z \in \mathbb{R} \}) $. This space deformation retracts to a torus, whose fundamental group is $ \mathbb{Z} \oplus \mathbb{Z} $.

Answer 2

Your space $X$ is exactly equal to $\mathbb C^\ast \times \mathbb C ^\ast$ so that $$\pi_1(X)= \pi_1(\mathbb C^\ast \times \mathbb C ^\ast)=\pi_1(\mathbb C^\ast )\times \pi_1(\mathbb C^\ast )=\mathbb Z\times \mathbb Z$$