Riemann surface associated with complete analytic function of $(z^2-1)^{1/3}$ I'm trying to define an analytic function on '$\mathbb{C}$' of the form $f(z)=(z^2-1)^{1/3}$, i.e. I first remove two semi-infinite rays $l_1$ and $l_2$, one going from $1$ to $\infty$ along the positive reals, one from $-1$ to $\infty$ along the negative reals, ending up with the domain $U=\mathbb{C}\setminus(l_1\cup l_2)$.
Then I have three choices for $f$, and they will differ by factors of $e^{2\pi i/3}$ and $e^{4\pi i/3}$. Since $U$ is simply-connected, such a function is well-defined after making the initial choice.
First question: This is pretty vague. How can I justify this?
Second question: If I now want to describe the Riemann surface associated with the complete analytic function of $(z^2-1)^{1/3}$, how can I do so using sort of a gluing procedure? I start with three copies of $U$, but how can I glue them together?

A more direct way, without any glueing, is to look directly at the locus $E$ of $y^3=z^2-1$ in $\mathbf C^2$. This equation defines a smooth affine curve, and the projection $(y,z) \mapsto z$ is locally an isomorphism away from the points $(0,1)$ and $(0,-1)$. Moreover this projection is generically three-sheeted. On $E$, the function $y$ satisfies $y^3=z^2-1$, i.e., morally $y$ 'is' $(z^2-1)^{1/3}$, and the function $y$ restricted to each "sheet" gives the three branches of your function.
