In the notation of the preceding exercise, give an example for which the covering \(U_0\to U_1\) is normal (or Galois), i.e. for which \[ f_*(\pi_1(U_0))\triangleleft \pi_1(U_1). \]
Solution
1. Take\[ f(z)=z^d \]for any integer \(d\ge2\).
We have \[ f'(z)=dz^{d-1}, \] so the only critical point is \(0\) and the only critical value is \(0\). Hence, in the notation of E-HM213A17-05, \[ V_1=\{0\}, \qquad V_0=f^{-1}(V_1)=\{0\}, \] and therefore \[ U_0=U_1=\mathbb C^*. \] Thus the covering is \[ f:\mathbb C^*\longrightarrow\mathbb C^*, \qquad z\longmapsto z^d. \]
2. On fundamental groups, \(f_*\) is multiplication by \(d\).
Use the standard generator \[ \gamma(t)=e^{2\pi i t}, \qquad 0\le t\le1, \] of \[ \pi_1(\mathbb C^*,1)\cong\mathbb Z. \] Then \[ (f\circ\gamma)(t)=e^{2\pi i d t}, \] which winds \(d\) times around the origin. Hence \[ f_*:\mathbb Z\longrightarrow\mathbb Z, \qquad n\longmapsto dn, \] and so \[ f_*(\pi_1(U_0))=d\mathbb Z. \]
3. The covering is normal.
The group \[ \pi_1(U_1)\cong\mathbb Z \] is abelian, so every subgroup is normal. In particular, \[ d\mathbb Z\triangleleft\mathbb Z. \] Therefore the covering \(z\mapsto z^d\) of \(\mathbb C^*\) is normal (Galois). Its deck group is equivalently the cyclic group of \(d\)th roots of unity, acting by \[ z\longmapsto\zeta z, \qquad \zeta^d=1. \]