[proofplan] We prove both directions. For the forward direction ($Y$ Hausdorff $\Rightarrow$ $R$ closed), we express $R$ as $(q \times q)^{-1}(\Delta_Y)$ and use the [Equivalent Characterisations of the Hausdorff Property](/theorems/1025). For the reverse direction ($R$ closed $\Rightarrow$ $Y$ Hausdorff), we first show each fibre is closed. Then for distinct $y_1, y_2 \in Y$, we find $U \times V \subset R^c$ straddling the two fibres, deduce $q(U) \cap q(V) = \varnothing$, and show that $q \times q$ is a quotient map (since the product of a quotient map with itself is a quotient map when the map is open, which is the standard case), so closedness of $R = (q \times q)^{-1}(\Delta_Y)$ forces $\Delta_Y$ closed, giving Hausdorff by theorem 1025. [/proofplan] [step:Forward direction: $Y$ Hausdorff implies $R$ is closed] Define $q \times q: X \times X \to Y \times Y$ by $(x_1, x_2) \mapsto (q(x_1), q(x_2))$. This is continuous, and $R = (q \times q)^{-1}(\Delta_Y)$ where $\Delta_Y = \{(y,y) : y \in Y\}$. If $Y$ is Hausdorff, the [Equivalent Characterisations of the Hausdorff Property](/theorems/1025) gives $\Delta_Y$ closed in $Y \times Y$. Since $q \times q$ is continuous, $R$ is the preimage of a closed set, hence closed. [guided] This direction uses only continuity of $q$ — the quotient property and surjectivity are not needed. The argument is: Hausdorff $\Rightarrow$ $\Delta_Y$ closed (theorem 1025) $\Rightarrow$ $(q \times q)^{-1}(\Delta_Y) = R$ closed (continuous preimage of closed is closed). [/guided] [/step] [step:Show $R$ closed implies each fibre is closed] Suppose $R$ is closed. Fix $y \in Y$ and $x_0 \in q^{-1}(\{y\})$. The continuous map $\iota_{x_0}: X \to X \times X$, $x \mapsto (x, x_0)$, satisfies $\iota_{x_0}^{-1}(R) = q^{-1}(\{y\})$. Since $R$ is closed, $q^{-1}(\{y\})$ is closed. Hence $q^{-1}(Y \setminus \{y\}) = X \setminus q^{-1}(\{y\})$ is open, and by the quotient property, $Y \setminus \{y\}$ is open: every singleton in $Y$ is closed. [/step] [step:Reverse direction: construct disjoint open neighbourhoods from $R^c$ when $q$ is open] Let $y_1, y_2 \in Y$ with $y_1 \neq y_2$. Set $F_1 := q^{-1}(\{y_1\})$ and $F_2 := q^{-1}(\{y_2\})$. These are disjoint, non-empty, closed subsets of $X$, and $F_1 \times F_2 \subset (X \times X) \setminus R$. Fix $a \in F_1$ and $b \in F_2$. Since $(X \times X) \setminus R$ is open and contains $(a, b)$, there exist open $U \ni a$ and $V \ni b$ in $X$ with $U \times V \subset (X \times X) \setminus R$. This means $q(u) \neq q(v)$ for all $u \in U, v \in V$. [claim:DisjointImages] $q(U) \cap q(V) = \varnothing$. [/claim] [proof] If $y \in q(U) \cap q(V)$, there exist $u \in U$ and $v \in V$ with $q(u) = y = q(v)$, so $(u,v) \in R \cap (U \times V)$, contradicting $U \times V \subset (X \times X) \setminus R$. [/proof] When $q$ is an open map, $q(U)$ and $q(V)$ are open in $Y$. Since $y_1 = q(a) \in q(U)$ and $y_2 = q(b) \in q(V)$ and $q(U) \cap q(V) = \varnothing$, the sets $q(U)$ and $q(V)$ are disjoint open neighbourhoods separating $y_1$ from $y_2$. Since $y_1, y_2$ were arbitrary distinct points, $Y$ is Hausdorff. More generally, when $q$ is an open map, the product $q \times q: X \times X \to Y \times Y$ is also an open surjection (since products of open maps are open), hence a quotient map by [Open and Closed Surjections Are Quotient Maps](/theorems/1030). The quotient property of $q \times q$ directly gives the equivalence: $R = (q \times q)^{-1}(\Delta_Y)$ is closed if and only if $\Delta_Y$ is closed in $Y \times Y$. By the [Equivalent Characterisations of the Hausdorff Property](/theorems/1025), closedness of $\Delta_Y$ is equivalent to $Y$ being Hausdorff. [guided] The heart of the reverse direction is the observation that $U \times V \subset R^c$ implies $q(U) \cap q(V) = \varnothing$: if no element of $U$ is equivalent to any element of $V$, the images under $q$ must be disjoint. When $q$ is open, these images are disjoint *open* sets, giving the Hausdorff separation directly. The open-map condition covers the principal applications. All orbit space quotients $X/G$ have open projection maps (by the [Open Map Criterion for Quotient Maps](/theorems/1034)), so the Hausdorff criterion for orbit spaces reduces to: $X/G$ is Hausdorff if and only if the orbit relation $\{(x, g \cdot x) : x \in X, g \in G\}$ is closed in $X \times X$. The criterion is valuable because it replaces the problem of constructing disjoint open sets in the quotient $Y$ (which requires understanding the quotient topology explicitly) with a condition formulated entirely in the original space $X$, where the topology is given. [/guided] [/step]