Topological Quotients from Equiv. Relations Defined by a Function

# Topological Quotients from Equivalence Relations Defined by a Function

Recall from the Topological Quotients page that if $(X, \tau)$ is a topological space, $\sim$ is an equivalence relation on $X$, for all $x \in X$ we let $[x] = \{ y \in X : x \sim y \}$ be the equivalence class of $x$ and let $X \: / \sim$ be the set of all of these equivalence classes then the quotient topology on $X \: / \sim$ is the final topology induced by the canonical quotient map $q : X \to X \: / \sim$ defined for all $x \in X$ by $q(x) = [x]$ ($x \in X$ is sent to its equivalence class $[x] \in X \: / \sim$).

We also proved a simple proposition which said that $U \subseteq X \: / \sim$ is open in $X \: / \sim$ if and only if $\displaystyle{\bigcup_{[x] \in U} [x]}$ is open in $X$.

Now suppose that $f : X \ to Y$ is a function. Then we can define an equivalence relation on $X$ by saying that $x_1 \sim x_2$ if and only if $f(x_1) = f(x_2)$. The following theorem gives some nice properties of these topological quotients.

 Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Define an equivalence relation on $\sim_f$ on $X$ by $x_1 \sim_f x_2$ if and only if $f(x_1) = f(x_2)$. Let $g : X \: / \sim_f \to Y$ be defined for all $[x] \in X \: / \sim_f$ by $g([x]) = f(x)$. Then: a) $g$ is injective and continuous. b) If $f$ is an open (or closed) map then $X \: / \sim_f$ and $f(X)$ are homeomorphic and $g$ is a homeomorphism between $X \: / \sim_f$ and $f(X)$.
• Proof of a) We first show that $g$ is injective. Let $[x_1], [x_2] \in X \: / \sim_f$ and suppose that:
(1)
\begin{align} \quad g([x_1]) = g([x_2]) \end{align}
• Then $f(x_1) = f(x_2)$ which implies that $x_1 \sim_f x_2$. In other words, $x_1$ and $x_2$ belong to the same equivalence class so $[x_1] = [x_2]$. Therefore, $g$ is injective.
• We now show that $g$ is continuous. Consider the canonical quotient map $q : X \to X \: / \sim_f$ defined for all $x \in X$ by $q(x) = [x]$. Applying $g : X \: / \sim_f \to Y$ and we have that for all $[x] \in X \: / \sim_f$ that $g([x]) = f(x)$. Therefore the composition map $g \circ q : X \to Y$ defined for all $x \in X$ by $(g \circ q)(x) = g(q(x)) = g([x]) = f(x)$ is simply equal to $f$:
(2)
\begin{align} \quad f = g \circ q \end{align} • Let $U$ be an open set in $Y$. Since $f$ is continuous we have that $f^{-1}(U)$ is open in $X$. Equivalently:
(3)
\begin{align} \quad f^{-1}(U) = (g \circ q)^{-1}(U) = q^{-1}(g^{-1}(U)) \end{align}
• is open in $X$. But $g^{-1}(U)$ is simply a collection of equivalence classes and:
(4)
\begin{align} \quad q^{-1}(g^{-1}(U)) = \bigcup_{[x] \in g^{-1}(U)} [x] \end{align}
• From the position mentioned earlier, since the lefthand side of the equation above is open we have that the righthand side is open and furthermore this happens if and only if $g^{-1}(U)$ is open in $X \: / \sim_f$ and so $g : X \: / \sim_f \to Y$ is continuous.
• Proof of b) Let $f$ be an open map. From (a) we have already proven that $g$ is continuous, so we only need to show that $g$ is open.
• Let $U$ be an open subset of $X \: / \sim_f$. Then $q^{-1}(U)$ is open in $X$ by the quotient topology, and since $f$ is an open map, we have that $f(q^{-1}(U))$ is open in $Y$. From the diagram above we see that:
(5)
\begin{align} \quad g(U) = f(q^{-1}(U)) \end{align}
• Therefore $g(U)$ is open in $Y$. Since $g(U)$ is open in $Y$ and since $f(X)$ is open in $Y$ (from $f$ being an open map) we see that $g(U) \cap f(X)$ is open in $Y$, and so the map $g : X \: / \sim_f \to f(X)$ is a continuous open map, i.e., a homeomorphism. $\blacksquare$