Quotient of space and a group of maps, Riemann surfaces I've been attempting to study Riemann surfaces, and I have continuously run into this notion which eludes me. I see people write things like $
\mathbb H / <z\mapsto z+1>$ or $\mathbb D / PSL$. I know what all these things are individually, but I don't understand how someone takes a quotient of a space and some maps (as opposed to say, treating the space as a group and taking the quotient over a normal subgroup). The end result of this is hopefully understanding how people come up with the "isomorphism maps" that are discussed in this context. 
Can someone spell this out a little bit for me? The whole issue feels opaque to me.
 A: Say $G$ is a group of bijections of the set $X$. For $x,y\in X$ say $x\sim y$ if $x=gy$ for some $g\in G$. Then $\sim$ is an equivalence relation on $X$, and $X/G$ is the same thing as the set $X/\sim$ of equivalences classes.
A: Suppose a group $G$ acts (on the left) on a set $X$, so we have a map
\begin{align*}
G \times X &\to X\\
(g,x) &\mapsto gx \, .
\end{align*}
Given $x \in X$, consider its orbit $Gx = \{gx : g \in G\}$.  These orbits partition $X$, so we can define the quotient $G \backslash X = \{Gx : x \in X\}$ as the set of orbits.  (I write $G \backslash X$ rather than $X/G$ because $G$ is acting on the left.)
If $X$ is a topological space and $G$ is a topological group and acting continuously on $X$, i.e., the map
\begin{align*}
G \times X &\to X\\
(g,x) &\mapsto gx
\end{align*}
is continuous, then we can upgrade $G \backslash X$ to a topological space by giving it the quotient topology as follows.  We have a quotient map
\begin{align*}
\pi: X &\to G \backslash X\\
x &\mapsto Gx
\end{align*}
that sends $x$ to its orbit, and we define the topology on $G \backslash X$ to be the strongest topology such that $\pi$ is continuous.  That is, $U \subseteq G \backslash X$ is open iff $\pi^{-1}(U) \subseteq X$ is open.
Let's look at a geometric-flavored example.  The group $\text{PSL}_2(\mathbb{R})$ acts on the upper half-plane by Möbius transformations:
\begin{align*}
\text{PSL}_2(\mathbb{R}) \times \mathcal{H} &\to \mathcal{H}\\
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix} \cdot
z &= \frac{az + b}{cz + d} \, .
\end{align*}
Then
$$
\begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix} \cdot z = z + 1
$$
so it acts as $T(z) = z+1$.  I claim that $\langle T \rangle \backslash \mathcal{H}$ is homeomorphic to an infinite cylinder.  To see this, we find a fundamental domain for the action, i.e., a set of representative points in $\mathcal{H}$ so that each orbit has exactly one representative.  In this case $\{z \in \mathcal{H} : 0 \leq \Re(z) < 1\}$, is such a set.  Since $T$ identifies the left and right edges of this domain, then we get a cylinder as the quotient $\langle T \rangle \backslash \mathcal{H}$.
For another example, take the subgroup $\text{PSL}_2(\mathbb{Z}) \leq \text{PSL}_2(\mathbb{R})$ and form $\text{PSL}_2(\mathbb{Z})\backslash \mathcal{H}$.  This has a nice fundamental domain as shown here.  This yields a funny looking quotient that is homeomorphic to a sphere minus a point.
