# Fundamental group of the following subspace of $\mathbb{C}\times\mathbb{C}$

Consider the subspace of $\mathbb{C}\times\mathbb{C}$: $$Y = \{(w,z) \in \mathbb{C}\times\mathbb{C}| w^3 = z^4 - 1\}$$

The goal is to compute the fundamental group of $Y$. The hint is to use some kind of deformation retraction that turn it into lines. I can't seem to be able to imagine how it looks like.

From a previous problem, I can imagine $Y$ to exist as 2 complex planes, 1 for $w$ and 1 for $z$. For each value of $w$, there are 4 values of $z$ and for each $z$, there are 3 $w$. Each pair of $(w, z)$ is a point in $Y$.

How can we go from here?

• Can you see what the complement of $Y$ in $C^2$ is? Commented Jan 17, 2017 at 2:30
• @MarianoSuárez-Álvarez It consists of every pair $(w,z)$ that does not satisfy the equation. That seems like a lot of points... It is not obvious to me what this space looks like. Commented Jan 17, 2017 at 2:48
• Well... a geometrical description of that set would surly be of use. What you answered was a restatement of the definition of the set :-\ Commented Jan 17, 2017 at 2:56

Let $V \subset \mathbb{C}^2$ be the set of solutions of $w^3=z^4-1$. Notice that the projection on the second coordinate $(w,z) \mapsto z$ induces a triple cover $V \to \mathbb{C}$ branching at four points (namely $\{\pm i, \pm 1\}$). Roughly speaking this is because the equation $w^3=z^4-1$ has three distinct solutions in $w$ unless $z^4-1=0$.
This is a genus 2 orientable surface with 3 points removed and it has the homotopy type of a bouquet of $6$ circles. Consequently $\pi_1(V)$ is the free product of $6$ copies of $\mathbb{Z}$.