Exercise 53.5

Question

Show that the map of Example $3$ is a covering map. Generalize to the map $p(z) = z^n$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Note that $p(z)$ when considered as a map on $S^1$ as a subset of
  $\mathbb{R}^2$ is given by $p(\cos 2\pi x,\sin 2\pi x) = (\cos 2\pi nx,\sin
  2\pi nx)$. Letting $q(x) = (2\cos\pi x,\sin 2\pi x)$ be the covering
  $\mathbb{R} 	o S^1$ in Theorem $53.1$, and $r\colon \mathbb{R} 	o \mathbb{R}$ be the multiplication by $n$, we then have the commutative diagram
  \begin{center}
    \begin{tikzcd}
      \mathbb{R}^2 \arrow{r}{r} & \mathbb{R}^2\arrow{d}{q}\
      S^1\arrow{r}{p}\arrow{u}{q^{-1}} & S^1
    \end{tikzcd}
  \end{center}
  Now let $x \in S^1$, and consider $U = S^1 \setminus \{x\}$. Taking $s \in
  q^{-1}(x)$, we see that we have a partition $q^{-1}(U) = \bigsqcup_{i \in
    \mathbb{Z}} W_i$ where $W_i = (s+i,s+i+1)$, and we also have $r^{-1}(W_i) =
    ((s+i)/n,(s+i+1)/n) \eqqcolon Z_i$. Letting $V_i = q(Z_i)$, we see that $p^{-1}(U) = \bigsqcup_{0\le i<n} V_i$. Thus, for all $i$,
  \begin{center}
    \begin{tikzcd}
      W_i \arrow{r}{r|W_i} & Z_i\arrow{d}{q|Z_i}\
      V_i \arrow{r}{p|V_i}\arrow{u}{(q|W_i)^{-1}} & U
    \end{tikzcd}
  \end{center}
  commutes, where the top and vertical arrows are homeomorphisms, and so we have a homeomorphism between $V_i$ and $U$ for all $i$.
\end{proof}

Exercise 53.5

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Exercise 53.5

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