Proposition 9

Let $X$ be a locally compact space, $R$ an equivalence relation on $X$, $C$ the graph of $R$ in $X \times X$, $f$ the canonical mapping $X \to X/R$; let $X'$ be the compact space obtained by adjoining a point at infinity $\omega$ to $X$, and let $R'$ be the equivalence relation on $X'$ whose graph is $C' = C \cup \{(\omega, \omega)\}$. Then the following conditions are equivalent:

a) $f$ is proper.

b) The saturation of each compact subset of $X$ with respect to $R$ is compact.

c) $R'$ is closed.

d) The restriction of $\mathrm{pr}_2$ to $C$ is proper.

e) $R$ is closed and the equivalence classes with respect to $R$ are compact.

Furthermore, if these conditions are satisfied, $X/R$ is locally compact.

a) $\Longrightarrow$ b): Since $X/R = f(X)$ and $f$ is proper, $X/R$ is Hausdorff (no. 1, Proposition 5, Corollary 2); hence the image under $f$ of every compact subset $K$ of $X$ is compact ($\S$ 9, no. 4, Theorem 2, Corollary 1).

The saturation of $K$ with respect to $R$ is $\overline{f}^{-1}(f(K))$ and is therefore compact by no. 2, Proposition 6.

b) $\Longrightarrow$ c) : If $F'$ is closed in $X'$ and does not contain $\omega$, then $F'$ is a compact subset of $X$; hence its saturation with respect to $R'$, which is the same as its saturation with respect to $R$, is compact and a fortiori closed in $X'$. If $\omega \in F'$ and if $F = F' \cap X = F' - \{ \omega \}$, then the saturation of $F'$ with respect to $R'$ is the union of $\{ \omega \}$ and the saturation $H$ of $F$ with respect to $R$; hence it is enough to show that $H$ is closed in $X$ (i.e. that $R$ is a closed relation). For this, it is enough to show that if $K$ is any compact subset of $X$ then $H \cap K$ is compact ($§ 9$, no. 7, Proposition 11). Now the saturation $L$ of $K$ with respect to $R$ is compact by hypothesis, and $H \cap L$ is the saturation of $F \cap L$, which is also compact; a fortiori $H \cap K = (H \cap L) \cap K$ is compact.

c) $\Longrightarrow$ d) : Since $X'$ is regular ($§ 9$, no. 2, Corollary to Proposition 1), $C'$ is closed in $X' \times X'$ ($§ 8$, no. 6, Proposition 14) and therefore compact. It follows that $C'$ is the one-point compactification of $C$ ($§ 9$, no. 8, Theorem 4). Since the restriction to $C'$ of $pr_2 : X' \times X' \to X'$ is continuous at $\omega$, the result follows from no. 3, Corollary to Proposition 7.

d) $\Longrightarrow$ e) : If $F$ is any closed subset of $X$, then $C \cap (F \times X)$ is closed in $C$, whence the saturation of $F$ with respect to $R$, which is equal to $pr_2(C \cap (F \times X))$, is closed in $X$ (no. 1, Proposition 1). Also the equivalence class of $x \in X$ mod $R$ is homeomorphic to the inverse image of $\{ x \}$ under the restriction of $pr_2$ to $C$ and is therefore compact [no. 2, Theorem 1 b)].

e) $\Longrightarrow$ a) : If $R$ is closed, then by definition $f$ is closed, and for each $z \in X/R$, $f^{-1}(z)$ is an equivalence class mod $R$ and is therefore compact; hence $f$ is proper by Theorem 1 b) of no. 2.

Finally we have to prove that $X/R$ is locally compact. $X'/R'$ is compact by c) and Proposition 8; the relation $R$ is that induced on $X$ by $R'$; $X$ is open in $X'$ and is saturated with respect to $R'$; hence $X/R$ is homeomorphic to the image $f'(X)$ of $X$ under the canonical mapping $f' : X' \to X'/R'$ ($§ 3$, no. 6, Proposition 10, Corollary 1). Now $f'(X)$ is open in $X'/R'$, and hence is a locally compact subspace of $X'/R'$.

Q.E.D.