How is $\Bbb{C}$ topologically the same as $\Bbb{C}/\Bbb{Z}_n$? This article on Orbifolds says that $\Bbb{C}$ is topologically the same as $\Bbb{C}/\Bbb{Z}_n$, where $\Bbb{Z}_n$ are the $n$th roots of unity.
What does it mean to say that these two spaces are topologically the same? Are they homeomorphic (I don't think they are) or homotopic?
 A: The two spaces are homeomorphic: we define the map $\phi\colon \mathbb{C}/\mathbb{Z}_{n}\to \mathbb{C}$ such that $\phi[z]=z^{n}$. Firstly, the map is well defined, since given two equivalence classes $[z]=[w]$ we have that $z=w\xi$, where $\xi$ is an $n$-th root of unity. Therefore, $z^{n}=w^{n}\xi^{n}=w^{n}$, so the equation $\phi[z]=z^{n}$ makes sense. Furthermore, the same argument can be reversed to show that $z^{n}=w^{n}$ if and only if $[z]=[w]$, which implies that $\phi$  is injective.
$\phi$ is also continuous since the map $\tilde{\phi}\colon z\in \mathbb{C}\mapsto z^{n}$ is continous, and $\tilde{\phi}=\phi \circ \pi$, where $\pi\colon \mathbb{C}\to \mathbb{C}/\mathbb{Z}_{n}$ is the projection. Continuity now follows from the fact that $\pi$ is a quotient map.
We show that $\phi$ is surjective: given $w\in \mathbb{C}$, it is well known that there exists $n$ (complex) roots to the equation $z^{n}=w$. For any root $z$ of that equation, we have $\phi[z]=z^{n}=w$, so $\phi$ is surjective.
Now we know that $\phi$ is a continous and bijective map, so to prove that it's a homeomorphism we need to prove that it is open or closed, which is equivalent to proving that $\tilde{\phi}$ is open or closed. I propose two methods of doing this:
Method 1: $\tilde{\phi}$ is an open map: this is a consequence of the Open Mapping Theorem from Complex Analysis: $\tilde{\phi}$ is an entire and nonconstant function, so it has to be open.
Method 2: $\tilde{\phi}$ is a closed map: let $F\subseteq \mathbb{C}$ be closed. We'll show that $\tilde{\phi}(F)$ is also closed. Take a convergent sequence $(w_{i})$ contained in $\tilde{\phi}(F)$ and let $w=\lim w_{i}$. By definition, for every $i$ there exists some $z_{i}\in F$ such that $\tilde{\phi}(z_{i})=z_{i}^{n}=w_{i}$. In particular, $(|z_{i}|^{n})$ converges, which means that $(z_{i})$ is a bounded sequence. The Bolzano-Weierstrass Theorem guarantees that there is a convergent subsequence of $(z_{i})$, which we'll call $(z_{j})$. Then, if $z=\lim z_{j}\in F$ (since $F$ is closed), we have $\tilde{\phi}(z)=\tilde{\phi}(\lim z_{j})=\lim \tilde{\phi}(z_{j})=\lim w_{j}=w$, so $w\in \tilde{\phi}(F)$. We conclude that $\tilde{\phi}(F)$ is closed, so $\tilde{\phi}$ is a closed map.
In any case, we finally arrive to the conclusion that $\phi$ is an open/closed map, so it has to be a homeomorphism.
