A connected manifold $N$ can be identified with its universal covering quotient a discrete group

Question

In my Holomorphic dynamics course, we covered the following corollary:

Every Riemann surface is the quotient of either $\mathbb{D},\mathbb{C}$ or $\hat{\mathbb{C}}$ by a discrete group $\Gamma$ of automorphisms of $\mathbb{D}, \mathbb{C}$ or $\hat{\mathbb{C}}$.

I don't have much experience with Algebraic [..] so I'm having a lot of trouble understanding the proof of this.

For the proof, we first need to identify an arbitrary Riemann surface $X$ with ($\tilde{X}$ quotient $\Gamma$) where $\tilde{X}$ is the universal cover and $\Gamma$ is the discrete group. I'm having trouble with this identification.

I believe $\Gamma$ is the group of all deck transformations from $\tilde{X}$ to $\tilde{X}$ and every universal covering is a normal covering. My book then says that given two points $x$ and $x'$ in $\tilde{X}$, there exists one and only one deck transformation which takes $x$ to $x'$. My book then says that it follows $X$ can be identified with the quotient ($\tilde{X}$ quotient $\Gamma$).

However, I still don't quite understand the identification that follows. I believe it may be due to my inexperience with some of the group-related notions above so any help is widely appreciated.

Also, I'm not sure what to tag this because I don't really know what field this falls under so any help there would also be appreciated!

P.P.S this probably should go as another question but what does the notation $\Gamma <$ Aut$(X)$ mean ?

Answer 1

Let $p : \tilde{X} \to X$ be the universal covering map.

My book then says that given two points $x$ and $x'$ in $\tilde{X}$, there exists one and only one deck transformation which takes $x$ to $x'$.

This is false. If $x$ and $x'$ are in the same fiber (i.e. $p(x) = p(x')$), then there is a unique deck transformation which maps $x$ to $x'$.

Recall that $\tilde{X}/\Gamma$ is the collection of equivalence classes given by the relation $x \sim y$ if there is $\gamma \in \Gamma$ such that $\gamma(x) = y$. Note, if $x \sim y$, then $p(y) = p(\gamma(x)) = p(x)$. Conversely, if $p(x) = p(y)$, then as mentioned above, $x \sim y$. That is, the equivalence classes of $\sim$ are precisely the preimages of the points in $X$. So we can identify $\tilde{X}/\Gamma$ with $X$ via the map $\varphi : \tilde{X}/\Gamma \to X$ given by $[x] \mapsto p(x)$.