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?
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.