# Proof - $\mathbb S^1 \cong \mathbb S^1/ \mathbb Z_n$.

if possible, analyze my proof. (I don't know if it's correct).

Exercise is here

proof: Type $$\xi = e^{\frac{2 \pi i}{n}}$$. Note that for all $$k \in \mathbb Z_n$$ \begin{align*} \varphi_k(z) = \left[e^{\frac{2 \pi i}{n}} \right]^k z = e^{\frac{2k \pi i}{n}} z, \end{align*} furthermore, it is easily verified that $$\varphi_k$$ is an action as follows

i) $$\varphi_0(z) = e^0z = z$$

ii) $$\varphi_{k + l}(z) = e^{\frac{2\pi i (k + l)}{n}} z = e^{\frac{2\pi ik}{n}} e^{\frac{2\pi il}{n}} z =\varphi_k(\varphi_l(z))$$

Therefore $$\varphi_k$$ is a action of $$\mathbb Z_n$$ in $$\mathbb S^1$$. Furthermore, the equivalence relation induced by the action is given by \begin{align*} x\mathcal{R}y & \Leftrightarrow \exists k \in \mathbb Z_n: \xi^k x = y \\ &\Leftrightarrow e^{\frac{2k\pi i}{n}} x = y \\ &\Leftrightarrow e^{\frac{2k\pi i}{n}} e^{i \theta} = e^{i \lambda} \end{align*} Let $$\pi: \mathbb S^1 \rightarrow \mathbb S^1 / \mathbb Z_n$$ be the projection, and $$p: \mathbb S^1 \rightarrow \mathbb S^1$$, $$z \mapsto z^n.$$ Note that $$p$$ is continuous, and in particular satisfies \begin{align*} p(t) = p(t') \Leftrightarrow t^n = t'^n &\Leftrightarrow e^{i \theta} = e^{i \lambda} \end{align*} but then there is $$k \in \mathbb Z_n$$ such that \begin{align*} e^{\frac{2k\ pi i}{n}} e^{i \theta} = e^{i \lambda} \Leftrightarrow \xi^k t = t ' \end{align*} therefore there is a continuous bijection $$h: \mathbb S^1 / \mathbb Z_n \rightarrow \mathbb S^1$$. But being $$\mathbb S^1 / \mathbb Z_n$$ compact (compact image by continuous application), and $$\mathbb S^1$$ Hausdorff, $$h$$ is a homeomorphism.

• the proof is right
– ali
Oct 24, 2020 at 21:02
• Thank you very much @ali. Oct 25, 2020 at 4:17