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

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?

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  • $\begingroup$ Can you see what the complement of $Y$ in $C^2$ is? $\endgroup$ Commented Jan 17, 2017 at 2:30
  • $\begingroup$ @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. $\endgroup$
    – darkgbm
    Commented Jan 17, 2017 at 2:48
  • $\begingroup$ Well... a geometrical description of that set would surly be of use. What you answered was a restatement of the definition of the set :-\ $\endgroup$ Commented Jan 17, 2017 at 2:56

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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 $z^4-1=0$.

The triple cover of the plane branching at four points can be explicitly constructed by doing some cut and paste. See for example Kauffman's book "On knots" for an exposition of this method.

Here a pictorial description of your space (the blue points should be removed).

enter image description here

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 $\mathbb{Z}$.

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