Motivating Quotient Topologies

Munkres Topology

Given two topological spaces $X$ and $Y$, we say a map $p:X\to Y$ is a quotient map when $p$ is surjective and a subset $U\subset Y$ is open iff $p^{-1}(U)$ is open in $X$. An equivalent definition would be that $p$ is a quotient map precisely when $p$ is surjective, continuous, and maps all saturated open sets to open sets in $Y$.

We can also create a “quotient” structure by manipulating topologies instead of maps. Given a topological space $X$ and an arbitrary set $A$ with a surjective map $p:X\to A$, there is a unique topology $\mathcal{T}$ on $A$ such that $p$ becomes a quotient map; we refer to $\mathcal{T}$ here as the quotient topology induced by $p$. Clearly, $\mathcal{T}$ is the collection of subsets $U\subset A$ such that $p^{-1}(U)$ is open in the topology of $X$.

The topological quotient structure seems quite contrived with these definitions alone. We show that quotient maps, in fact, neatly fit in the conceptual picture of quotient sets $X/\sim$ generated by equivalence relations $\sim$. We first define the canonical topology on quotient sets.

Let $X^{\ast }$ be a partition of $X$ defined by an equivalence relation $\sim \subset X\times X$ (i.e. $X^{\ast }=X/\sim$). Define the canonical topology $\mathcal{T}$ on $X^{\ast }$ as follows: a set $U\subset X^{\ast }$ is open in $\mathcal{T}$ precisely when the union of equivalence classes $p^{-1}(U)={\bigcup }_{u\in U}{p}^{-1}(u)$ is open in the original space $X$; we can see this is a valid topology, and we denote the resulting topological space as the quotient space $X^{\ast }$. Note that this is essentially the most intuitive way to define a topology on equivalence classes.

These notes show the following: for any quotient map $q:X\to Y$, there is an associated equivalence relation $\sim \subset X\times X$ such that we can represent $q$ as $q=f\circ p$. $p:X\to X/\sim$ is the “identity” map that maps each element into its corresponding equivalence class in the quotient space $X/\sim$ and $f$ is a homeomorphism between $X/\sim$ and $Y$. In the reverse direction, we also show that if we take an arbitrary equivalence relation $\sim \subset X\times X$ and compose the “identity” map onto the resulting quotient space with a homeomorphism, we end up with a quotient map. Therefore, there is a very neat conceptual correspondence between quotient maps and equivalence classes on topological spaces. We show the reverse direction first, going from equivalence classes to quotient maps.

(Reverse Direction) Let $\sim \subset X\times X$ be an arbitrary equivalence relation inducing a quotient space $X^{\ast }$. It is not difficult to see that the “identity” map $p:X\to X^{\ast }$ is a quotient map. As homeomorphisms are automatically quotient maps and the composition of quotient maps are also quotient maps, if we take an arbitrary homeomorphism $f:X^{\ast }\to Y$, the resulting function $q=f\circ p$ is a quotient map.

(Forward Direction) Let $q:X\to Y$ be an arbitrary quotient map. We decompose $q$ into a canonical map onto a quotient space followed by a homeomorphism.

Define $\sim$ such that $x\sim y$ whenever $q(x)=q(y)$ for $x,y\in X$. Let $p:X\to X/\sim$ be the canonical quotient map that projects onto the equivalence classes. Then, $q$ induces a function $f:X/\sim \to Y$, which by Munkres 22.2 is a homeomorphism. Thus, we can represent $q$ as $q=f\circ p$ where $p$ is a canonical map onto a quotient space and $f$ is a homeomorphism. $q$ is, therefore, essentially a visualization of the quotient space structure of $X$.

We can, in fact, generalize this decomposition to represent any surjective continuous map $q:X\to Y$ into $q=f\circ p$ where $p$ is a quotient map onto the quotient space induced by $q$ and $p$ is a bijective continuous map that is a homeomorphism iff $q$ is a quotient map. See Munkres 22.3 for more. Note that one could see this decomposition as a corollary of the intuitive picture of quotient maps rather than a generalization.