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 $X$?
Best Answer
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 $(\mathbb{R}^3/\sim) \approx \mathbb{R}$.