# Quotients from Equivalence Relations Defined by Functions from Compact to Hausdorff Spaces

Recall from awhile back on the Topological Quotients from Equivalence Relations Defined by a Function page that if $X$ and $Y$ are topological spaces and if $f : X \to Y$ then for the equivalence relation $\sim_f$ on $X$ defined for all $x_1, x_2 \in X$ by $x_1 \sim_f x_2$ if and only if $f(x_1) = f(x_2)$ we have two important results.

First off, if $g : X \: / \sim_f \to Y$ is defined for all $[x] \in X \: / \sim_f$ by $g([x]) = f(x)$ then the function $g$ is injective and continuous.

Secondly, if $f$ is additionally an open or closed map then $X \: / \sim_f$ is homeomorphic to $f(X)$, and the function $g$ above a homeomorphism between the two spaces.

Also recall that a map $f : X \to Y$ is said to be closed if for every closed set $C$ in $X$ we have that the image, $f(C)$ is closed in $Y$.

We first look at a very nice lemma which says that if $X$ is compact, $Y$ is Hausdorff, and $f : X \to Y$ is a continuous map then we can moreover conclude that $f$ is a closed map.

Lemma 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. If $X$ is compact, $Y$ is Hausdorff, and $f$ is continuous, then $f$ is a closed map. |

**Proof:**Let $C$ be a closed set in $X$. Then $C$ is compact in $X$ since $X$ is compact. So $f(C)$ is a compact set in $Y$ from the continuity of $f$. But every compact set in a Hausdorff space is closed, so $f(C)$ is closed in $Y$.

- This shows that $f$ is a closed map. $\blacksquare$

We are now ready to state a very nice theorem regarding topological quotients from equivalence relations defined by functions from compact spaces to Hausdorff spaces.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. If $X$ is compact, $Y$ is Hausdorff, and $f : X \to Y$ is continuous and surjective, then if $\sim_f$ is the equivalence relation defined for all $x_1, x_2 \in X$ by $x_1 \sim_f x_2$ if and only if $f(x_1) = f(x_2)$ then $X \: / \sim_f$ is homeomorphic to $Y$. |

**Proof:**From Lemma 1 we see that since $X$ is compact, $Y$ is Hausdorff, and $f$ is continuous that then $f$ is a closed map.

- Since $f$ is a surjective map we see that $f(X) = Y$. So by the theorem mentioned at the top of this page we have that $X \: / \sim_f$ is homeomorphic to $f(X) = Y$. $\blacksquare$