Proposition

Suppose $q: X \rightarrow Y$ is an open quotient map. Then $Y$ is Hausdorff if and only if the set $\mathscr{R} = \{ (x_1, x_2) \mid q(x_1) = q(x_2) \}$ is closed in $X \times X$.

Solution

First, we will suppose that $Y$ is Hausdorff. We will show that $\mathscr{R}$ is closed in $X \times X$. Let $(a, b) \in X \times X \setminus \mathscr{R}$. Then $q(a) \ne q(b)$.

Since $Y$ is Hausdorff, there must exist disjoint neighborhoods $U, V$ of $q(a), q(b)$, respectively. Then $a \in q^{-1}(U), b \in q^{-1}(V)$. $q$ is a quotient map, so $q^{-1}(U)$ and $q^{-1}(V)$ are both open. Thus $q^{-1}(U) \times q^{-1}(V)$ is a neighborhood of $(a, b)$. We claim that this neighborhood is contained in $X \times X \setminus \mathscr{R}$. Let $(c, d) \in q^{-1}(U) \times q^{-1}(V)$.

• $q(c) \in q(q^{-1}(U)) = U$ because $q$ is surjective.
• $q(d) \in q(q^{-1}(V)) = V$ because $q$ is surjective.

Since $U \cap V = \emptyset$, $q(c) \ne q(d)$. Therefore, $(c, d) \in X \times X \setminus \mathscr{R}$. Thus $X \times X \setminus \mathscr{R}$ is open, so $\mathscr{R}$ is closed.

On the other hand, suppose that the set $\mathscr{R} = \{ (x_1, x_2) \mid q(x_1) = q(x_2) \}$ is closed in $X \times X$. We will show that $Y$ is Hausdorff.

Let $y_1 \ne y_2 \in Y$. Since $q$ is surjective, there exist $x_1, x_2 \in X$ such that $q(x_1) = y_1, q(x_2) = y_2$. Since $q(x_1) = y_1 \ne y_2 = q(x_2)$, $(x_1, x_2) \notin \mathscr{R}$. Since $\mathscr{R}$ is closed, the complement is open. Therefore, $X \times X \setminus \mathscr{R}$ is a neighborhood of $(x_1, x_2)$. Thus there must exist a basis element $U_1 \times U_2$ such that $(x_1, x_2) \in U_1 \times U_2 \subset X \times X \setminus \mathscr{R}$ where $U_1, U_2$ are open in $X$. Then $(y_1, y_2) = (q(x_1), q(x_2)) \in q(U_1) \times q(U_2)$. Since $q$ is open, $q(U_1), q(U_2)$ are both open. Therefore, they are neighborhoods of $y_1, y_2$, respectively.

We claim that $q(U_1) \cap q(U_2) = \emptyset$. Suppose otherwise. Let $y \in q(U_1) \cap q(U_2)$. Then there exists an $a_1 \in U_1$ such that $q(a_1) = y$. Similarly, there exists an $a_2 \in U_2$ such that $q(a_2) = y$. Then $(a_1, a_2) \in \mathscr{R}$. However, since $(a_1, a_2) \in U_1 \times U_2 \subset X \times X \setminus \mathscr{R}$, $(a_1, a_2)$ cannot be an element of $\mathscr{R}$. Therefore, $q(U_1) \cap q(U_2) = \emptyset$.

Then $q(U_1), q(U_2)$ are disjoint neighborhoods of $y_1, y_2$, so $Y$ is Hausdorff.