Covering space of $C \backslash \{0,1,\lambda \}$

Question

Let $\lambda\in C \backslash \{0,1\}$, $E= \{(x,y) \in C^2 : y^2=x(x-1)(x-\lambda),\ x\neq 0,1,\lambda \}$. Prove that $E$ is a connected $2$-fold covering space of $C \backslash \{ 0,1,\lambda \}$ for the first projection. Describe another such covering space non isomorphic with $E$. Can anyone help me? Thank you.

Answer 1

Since the projection is continuous on $\mathbb{C}^2$, its restriction $p$ to $E$ certainly is continuous.

For each $z_0 \in \mathbb{C}\setminus \{0,\,1,\,\lambda\}$, on the disk $D$ with radius $r = \min \{\lvert z_0\rvert,\, \lvert z_0-1\rvert,\, \lvert z_0 -\lambda\rvert\}$ with centre $z_0$, the function $f(z) = z(z-1)(z-\lambda)$ has no zeros. Since the disk is simply connected, $f$ has a holomorphic square root $g$ on $D$. The two mappings

$$s_1(z) = (z,g(z)),\quad s_2(z) = (z, -g(z))$$

are continuous maps $D \to \mathbb{C}^2$, their images are disjoint and contained in $E$. Since each fibre of $p$ contains two points,

$$T_{z_0} \colon D\times \{-1,\,1\} \to p^{-1}(D);\quad T_{z_0}(z,\varepsilon) = (z, \varepsilon\cdot g(z))$$

is a local trivialisation of $p$, thus $p \colon E \to \mathbb{C}\setminus\{0,\,1,\,\lambda\}$ is a two-sheeted covering.

It remains to see that $E$ is connected. Let $\rho = \frac12\min \{1,\, \lvert\lambda\rvert\}$, and choose a square root $\omega$ of $\rho(\rho-1)(\rho-\lambda)$. The lift of the curve $c_\rho\colon t \mapsto \rho e^{it},\; t\in [0,\,2\pi]$ starting in $(\rho,\omega)$ ends in $(\rho,-\omega)$.

Now let $a,b \in E$. Let $\gamma_a$ be a curve in $\mathbb{C}\setminus\{0,\,1,\,\lambda\}$ connecting $p(a)$ with $\rho$, and $\gamma_b$ a curve in $\mathbb{C}\setminus\{0,\,1,\,\lambda\}$ connecting $p(b)$ with $\rho$. Let $\tilde{\gamma}_a$ and $\tilde{\gamma}_b$ be the lifts of $\gamma_a$ resp $\gamma_b$ starting in $a$ resp. $b$. If $\tilde{\gamma}_a$ and $\tilde{\gamma}_b$ end in the same point above $\rho$, the composition $\tilde{\gamma}_a \tilde{\gamma}_b^{-1}$ is a path in $E$ connecting $a$ and $b$. If the two curves end in different points above $\rho$, the lift of $\gamma_a c_\rho \gamma_b^{-1}$ starting in $a$ is a path in $E$ connecting $a$ and $b$. Thus $E$ is path-connected, hence connected.

Describe another such covering space non isomorphic with $E$.

Let $F = \{ (x,y) : y^2 = x\} \subset \mathbb{C}^2$. Arguments analogous to the above show that $F' = F \setminus\{(0,0)\}$ with the projection $p_1$ to the first component is a connected two-sheeted covering of $\mathbb{C}\setminus\{0\}$. Hence $F'' = F' \setminus p_1^{-1}(\{1,\,\lambda\})$ with the projection $p_2 = p_1\lvert_{F''}$ to the first component is a connected two-sheeted covering of $\mathbb{C}\setminus\{0,\,1,\,\lambda\}$.

To see that $F''$ is not isomorphic to $E$, note that a circle of small radius around $1$ (or $\lambda$) lifts to a closed curve in $F''$, but not in $E$.