Quotient topology vs quotient space vs identifications? The book I'm reading isn't very clear and doesn't provide any concrete examples or definitions as to what a quotient topology is besides this paragraph.

Let $(X,T)$ be any topological space and ~ any equivalence relation on $X$.  Let $Y$ be the set of all equivalence classes of ~.  We can denote $Y$ by $X/$~.  The natural topology to put on the set $Y = X/$~ is the quotient toplogy under the map which identifies the equivalence classes; that is, maps each equivalence class to a point.

I don't understand this, what exactly is a quotient topology?  What does it mean by a map that identifies equivalence classes?  If a function maps an equivalence class to a point, then isn't it a function from $Y$ to $X$?
 A: If $X$ is a topological space, and we define an equivalence relation $\sim$ on $X$, then we can construct the quotient topology for $X$ as such.  Let $X/\sim$ be the quotient space (i.e. space of equivalence classes), and let $\pi: X \to X/\sim$ be the map $\pi(p) = [p]$ be the map that sends each element of $X$ to its equivalence class in $X/\sim$.  We define the quotient topology on $X/\sim$ to be the collection of subsets $U \subseteq X/\sim$ such that $\pi^{-1}(U)$ is open in $X$.  Observe that this makes $\pi$ a continuous map.
To illustrate this, imagine Euclidean space $\mathbb{R}^3$.  We can define an equivalence relation on it by saying that $(x,y,z) \sim (x',y,',z')$ if and only if $z=z'$.  Observe that this automatically satisfies the reflexive, symmetric, and transitive properties.  We then see that equivalence classes are of the form $[z_0] = \{(x,y,z) \in \mathbb{R} \; | \; z=z_0\}$, and thus each plane parallel to the $xy$-plane, in a sense, collapses to a single point along the $z$-axis.  Therefore we can say that $(\mathbb{R}^3/\sim) \approx \mathbb{R}$.
A: For topological spaces $X$ and $Y$, a map (of sets) $f\colon X\to Y$ is an identification if $f$ is surjective and $Y$ carries the coinduced topology, i.e. $U \subset Y$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. Note that this topology on $Y$ makes the map $f$ continuous.
Even more is true: If $f$ as above is an identification, then for each topological space $T$ a map $g\colon Y\to T$ is continuous if and only if the composition $gf\colon X\to T$ is continuous.
In particular, for an equivalence relation $\sim$ on $X$ you may form the set $Y=X/\sim$ and endow it with the coinduced topology with respect to the projection $\pi\colon X\to X/\sim$. Since this map is surjective, it is an identification. In this particular case, the topology on $X/\sim$ is also called quotient topology. 
I guess the name "identification" comes from these sort of application, where you make points equal in the sense that points with $x\sim y$ are mapped to the same point $[y]=[x] \in X/\sim$.
