Consider two lines $A$ and $B$ in $\Bbb C^2$ intersecting in one point.
Let's denote C and D respectively the lines $X=0$ and $Y=0$ in $\Bbb C^2$.
Let $X = \Bbb C^2 \setminus (A \cup B)$.
prove that X is homeomorphic to $\{(x,y) \in \Bbb C^2 : xy\ne 0\}$.
I tried to do the following:
let $z\in X$. I try to build an homeomorphism from $X$ to $\Bbb C^2\setminus (C \cup D)$.
Translation is an homeomorphism in $\Bbb C^2$ so we can suppose that $A$ and $B$ intersect in the origin of $\Bbb C^2$.
Now I have four lines $A, B, C, D$ intersecting at the origin of $\Bbb C^2$, so we can consider building a homeomorphism : $X\to \Bbb C^2\setminus (C \cup D) $ using a rotation, but then A and B won't rotate with the same angle to map respectively to C and D?
Thank you for any help, comments.
You're also allowed to shear, in addition to translating and rotating. So you will be done if you can find a linear isomorphism $\phi \in GL(\mathbb{C}^2)$ that takes the lines $D$ and $C$ to the lines $A$ and $B$. This is easy to do. Let $A$ be the span of $(a_1, a_2)$ and $B$ be the span of $(b_1, b_2)$. Then the transformation $$\phi = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}$$ takes $D$, which is the span of $(1,0)$, to $A$ and $C$ to $B$. Moreover, $\phi$ is invertible since it takes a basis to a a basis.