Let $\Delta$ be the unit disk in the complex plane and $0<\epsilon <<1$. My purpose is to compute the fundamental group of the following. $X :=\{ (z,w)\in \Delta \times \Delta \; | \; zw\neq \epsilon \}$. This is the complementary of the analytic hypersurface $\{ zw=\epsilon \}$ in $\Delta^2$.
I used Van Kampen's Theorem by using the following method:
$z=re^{i\alpha}$ and $w=se^{i\beta}$. Then $zw \in X $ if and only if $rs \neq \epsilon$ or $\alpha + \beta \neq 0 [2\pi]$.
It follows that
$X= X_1 \cup X_2$ where $X_1= \{ rs < \epsilon \; or \; \alpha + \beta \neq 0 [2\pi] \}$ and $X_2= \{ rs > \epsilon \; or \; \alpha + \beta \neq 0 [2\pi] \} $.
Is this the good way to solve my problem ?
Thanks
Your curve is a small perturbation of $\{zw=0\}$. The complement of the latter is $(\Delta\smallsetminus0)^2$, and its group $\Bbb Z^2$ is abelian. By Zariski's theorem, a small perturbation results in a quotient of $\pi_1$; hence, the group in question is also abelian, hence equals $H_1$, hence is $\Bbb Z$. (Poincare duality and such.) Of course, this particular group is just known to be $\Bbb Z$; what's written above is merely a brief explanation.