$\rho$ factorizes into $K \times X \to K \times X \xrightarrow{\mathrm{pr}_2} X$, where $\alpha(s, x) = (s, s.x)$. $\alpha$ is a homeomorphism, for $\alpha^{-1} : (s, y) \to (s, s^{-1}.y)$ is continuous. Since K is quasi-compact, $\mathrm{pr}_2$ is proper (Chapter I, § 10, no. 2, Theorem 1, Corollary 5); hence $\rho$ is proper (Chapter I, § 10, no. 1, Proposition 5).

The assertion concerning closed sets follows from Proposition 1 and the fact that a proper mapping is closed (Chapter I, § 10, no. 1, Proposition 1). The assertion concerning compact sets is trivial.

It should be noted that if L is a compact subset of X, and F a closed subset of G, then F.L is not necessarily closed in X ($§ 2$, Exercise 29; cf. no. 5, Corollary to Proposition 12).

The first assertion follows from Corollary 1 applied to G acting by right translations on itself, and the second assertion then follows from Chapter I, , no. 2, Theorem 1

Since x⁻¹y ∈ K is a closed equivalence relation (Corollary 2), the Corollary follows from Chapter I, § 5, no. 2, Proposition 4.

a) K operates properly on X. b) The mapping (s, x) → s.x of K × X into X is proper. c) The canonical mapping of X onto X/K is proper.

b) is a consequence of Proposition 1. As to a), since K is compact, pr₂ : (s, x) → x is proper (Chapter I, § 10, no. 2, Theorem 1, Corollary 5); hence, X being Hausdorff, (s, x) → (x, s.x) is proper (Chapter I, § 10, no. 1, Proposition 5, Corollary 3). It remains to prove c). By Corollary 1 to Proposition 1, the canonical mapping φ : X → X/K is closed. If Z is any topological space and if we make K operate trivially on Z, then K operates continuously on X × Z and hence the canonical mapping X × Z → (X × Z)/K is closed. But (X × Z)/K can be canonically identified with (X/K) × Z (§ 2, no. 4, Lemma 2 and Chapter I, § 5, no. 3, Corollary to Proposition 8); hence the canonical mapping X × Z → (X × Z)/K can be identified with φ × 1, and since it is closed for all Z it follows that φ is proper.

This follows from the fact that the canonical mapping X → X/K is proper, in view of Proposition 9 of Chapter I, § 10, no. 4.

Apply Corollary 1 to K operating on G by right translations.

Let $C \subset X \times X$ be the graph of the equivalence relation $R$ defined by $G$ on $X$; then $C$ is the image of $G \times X$ under the mapping $\theta : (s, x) \to (x, s.x)$. Since $\theta$ is proper, $C$ is closed in $X \times X$ (Chapter I, § 10, no. 1, Proposition 1). Since the relation $R$ is open ($§ 2$, no. 4, Lemma 2), it follows that $X/G$ is Hausdorff (Chapter I, § 8, no. 3, Proposition 8).

Now suppose that $G$ is Hausdorff. Then the mapping $x \to (e, x)$ of $X$ into $G \times X$ is a homeomorphism of $X$ onto a closed subspace of $G \times X$, and is therefore proper (Chapter I, § 10, no. 1, Proposition 2). If we compose this mapping with the mapping $(s, x) \to (x, s.x)$ of $G \times X$ into $X \times X$, which by hypothesis is proper, we obtain a proper mapping of $X$ into $X \times X$, namely the diagonal mapping $x \to (x, x)$. Hence the diagonal $\Delta$ of $X \times X$ is closed in $X$, and therefore $X$ is Hausdorff.

a) The mapping $s \to s.x$ of $G$ into $X$ is proper.

b) $K_x$ is quasi-compact.

c) The canonical mapping of $G/K_x$ onto $G.x$ is a homeomorphism.

d) The orbit $G.x$ is closed in $X$.

The inverse image of $\{ x \} \times X$ under $\theta : (s, x) \to (x, s.x)$ is $G \times \{ x \}$; hence, by Proposition 3 of Chapter I, § 10, no. 1, the restriction of $\theta$ to $G \times \{ x \}$ is a proper mapping of $G \times \{ x \}$ into $\{ x \} \times X$, whence a) follows. Since $K_x$ is the inverse image of $x$ under $s \to s.x$, b) follows from Chapter I, § 10, no. 2, Theorem 1. c) and d) are consequences of a), by virtue of Chapter I, § 10, no. 1, Propositions 2 and 5 b).

(ii) If $\varphi$ is proper, if $G'$ operates properly on $X'$ and if $X$ is Hausdorff, then $G$ operates properly on $X$.

To prove (i), consider the commutative diagram

where $\alpha = \varphi \times \psi$ and $\beta = \psi \times \psi$. By hypothesis, $\theta$ is proper; so is $\beta$ [Chapter I, § 10, no. 1, Proposition 4 a)]; hence $\beta \circ \theta = \theta' \circ \alpha$ is proper [Chapter I, § 10, no. 1, Proposition 5 a)]. Since $\alpha$ is surjective it follows that $\theta'$ is proper [Chapter I, § 10, no. 1, Proposition 5 b)].

To prove (ii) consider an ultrafilter $U$ on $G \times X$, such that the mappings

converge with respect to $U$ to $y_0$ and $x_0$ respectively. It follows that $(s, x) \to \varphi(s).\psi(x)$ and $(s, x) \to \psi(x)$ converge with respect to $U$. Since $G'$ operates properly on $X'$, this implies (no. 1) that $(s, x) \to \varphi(s)$ converges with respect to $U$ to a point $s'_0 \in G'$. Since $\varphi$ is proper, we deduce (Chapter I, § 10, no. 2, Theorem 1) that $(s, x) \to s$ converges with respect to $U$ to a point $s_0 \in G$. The uniqueness of the limit in $X$ then shows that $y_0 = s_0 x_0$, and hence that $G$ operates properly on $X$ (no. 1).

(FP) The graph $C$ of the equivalence relation defined by $G$ is closed in $X \times X$, and the canonical mapping $\varphi : C \to G$ is continuous.

The set $C$ is the image of the mapping $\theta : (s, x) \to (x, s.x)$ of $G \times X$ into $X \times X$. We know (Chapter I, § 10, no. 1, Proposition 2) that $\theta$ is proper if and only if $C$ is closed in $X \times X$ and (if $\theta'$ denotes the mapping $\theta$ considered as a mapping of $G \times X$ into $C$) $\theta'$ is a homeomorphism. Now the hypothesis implies that $\theta'$ is bijective and that its inverse is the mapping $(x, y) \to (\varphi(x, y), x)$. Hence $\theta'$ is a homeomorphism if and only if $\varphi$ is continuous.

Let $F$ be the compact space obtained by adjoining a point at infinity $\omega$ to $G$, and let $\Gamma$ be the graph of $\rho : (s, x) \to s.x$ considered as a subset of $F \times X \times X$. Let us show that if the restriction of $\mathrm{pr}_{23}$ to $\Gamma$ is proper, then $\Gamma$ is closed in $F \times X \times X$. Indeed, this hypothesis implies that the map $u : (t, s, x, y) \to (t, x, y)$ of $F \times \Gamma$ into $F \times X \times X$ is closed. If $\Gamma'$ is the set of points $(s, s)$ in $F \times G$, where $s \in G$, then $\Gamma'$ is closed in $F \times G$, since it is the graph of the canonical injection $G \to F$ (Chapter I, § 8, no. 1, Corollary 2 to Proposition 2); hence the intersection $(\Gamma' \times X \times X) \cap (F \times \Gamma)$ is closed in $F \times \Gamma$, and it is immediately seen that its image under $u$ is the set $\Gamma$ considered as a subset of $F \times X \times X$; hence $\Gamma$ is closed in $F \times X \times X$. Now, we have $(\{\omega\} \times X \times X) \cap \Gamma = \emptyset$. From the definition of $F$, therefore, for every point $(x, y) \in X \times X$ there is a neighbourhood $W$ of $(x, y)$ in $X \times X$ and a compact subset $K$ of $G$ such that

is empty; and since we may take $W$ to be a neighbourhood $V_x \times V_y$, where $V_x$ and $V_y$ are neighbourhoods of $x$ and $y$ respectively in $X$, the statement “((G — K) × W) ∩ Γ = ∅” becomes “if s ∉ K, then s · V_x ∩ V_y = ∅”. We have thus proved the necessity of the condition stated in the proposition. Conversely, suppose this condition is satisfied; let A be a set filtered by an ultrafilter $\mathfrak{F}$, and let α → (s_α, x_α) be a mapping of A into G × X such that lim_$\mathfrak{F}$ x_α = x and lim_$\mathfrak{F}$ s_α · x_α = y. Suppose that K, V_x and V_y satisfy the condition of the proposition. By hypothesis there is a set M ∈ $\mathfrak{F}$ such that if α ∈ M then x_α ∈ V_x and s_α · x_α ∈ V_y, hence s_α ∈ K. This shows that α → s_α converges with respect to $\mathfrak{F}$, and the proof is complete.

If G is compact, the condition of Proposition 7 is trivially satisfied; we thus retrieve Proposition 2 a).

Proposition 7 shows in particular that a discrete group G, operating continuously on a Hausdorff space X, operates properly on X if and only if, for each pair (x, y) of points of X, there is a neighbourhood V_x of x and a neighbourhood V_y of y such that the set of points s ∈ G for which s · V_x ∩ V_y ≠ ∅ is finite.

a) The subgroup K_x is finite and there is an open set U ⊂ X, containing x, which is stable under K_x, and on which the equivalence relation induced by the relation defined by G is the equivalence relation defined by K_x.

b) The canonical mapping U/K_x → X/G is a homeomorphism of U/K_x onto an open neighbourhood of the class of x in X/G.

By Proposition 7, K_x is finite. To construct an open set U which satisfies the required conditions, notice first that by Proposition 7 there is an open set U_0 containing x and such that the set K of s ∈ G for which s · U_0 ∩ U_0 ≠ ∅ is finite. Clearly K_x ⊂ K; let s_1, ..., s_n be the elements of K — K_x. If we put x_i = s_i · x (1 ≤ i ≤ n), then x_i ≠ x for each i; since X is Hausdorff, there is for each index i an open neighbourhood V_i of x and an open neighbourhood V'_i of s_i · x such that V_i ∩ V'_i = ∅. Let U_i = V_i ∩ s_i^{-1} · V'_i; then U_i is clearly open and contains x, and we have U_i ∩ s_i · U_i ⊂ V_i ∩ V'_i = ∅. Let U' = U_0 ∩ U_1 ∩ ... ∩ U_n; U' is open, contains x and is such that U' ∩ s · U' = ∅ for s ∉ K_x. Putting U = ⋂_{t ∈ K_x} t · U' we finally obtain an open set, stable under K_x, containing x and such that U ∩ s · U = ∅ for s ∉ K_x: U is the open set required.

The fact that the canonical mapping U/K_x → X/G is a homeomorphism of U/K_x onto an open set in X/G follows from Chapter I, § 5, no. 2, Proposition 4, since U is open and the equivalence relation defined by G is open (§ 2, no. 4, Lemma 2).

Since the equivalence relation on $X$ defined by $G$ is open ($§ 2$, no. 4, Lemma 2), the proposition results from Chapter I, $§ 10$, no. 4, Proposition 10.

Since the equivalence relation defined by $G$ is open ($§ 2$, no. 4, Lemma 2), the proposition is a particular case of Proposition 10 of Chapter I, $§ 10$, no. 4.

By hypothesis, the mapping $\theta : (s, x) \to (x, s.x)$ of $G \times X$ into $X \times X$ is proper; if $X \times X$ is compact (resp. locally compact) then the Corollary to Proposition 9 of Chapter I, $§ 10$, no. 4 shows that $G \times X$ is also compact (resp. locally compact), and therefore so is $G$ since $X \neq \emptyset$. Since $X/G$ is Hausdorff (no. 2, Proposition 3), the compactness (resp. local compactness) of $X$ implies the compactness (resp. local compactness) of $X/G$ [Chapter I, $§ 10$, no. 4, Proposition 8 (resp. Proposition 9)] (see $§ 2$, Exercise 29).

We shall now give criteria which allow us to assert that a Hausdorff topological group $G$ operates properly on a locally compact space $X$. For each pair of subsets $K, L$ of $X$ we denote by $P(K, L)$ the set of all $s \in G$ such that $s.K \cap L \neq \emptyset$.

b) If G operates properly on X and if L is compact, then P(K, L) is compact.

c) Conversely, if X is locally compact and if, for each pair K, L of compact subsets of X, P(K, L) is relatively compact in G [and therefore compact by a)]; then G operates properly on X (and if X is not empty, G is locally compact by Proposition 11).

The mapping $(s, x) \to s.x$ of $G \times K$ into $X$ is continuous, and the inverse image $L'$ of $L$ under this mapping is therefore closed. Since $K$ is compact, the projection $\operatorname{pr}_1 : G \times K \to G$ is proper (Chapter I, § 10, no. 2, Corollary 5 of Theorem 1) and the image of $L'$ under $\operatorname{pr}_1$ is therefore closed. This image is $P(K, L)$, hence a) is proved.

To prove b): $X$ is Hausdorff (no. 2, Proposition 3). By hypothesis the mapping $\theta : (s, x) \to (x, s.x)$ of $G \times X$ into $X \times X$ is proper; since $K \times L$ is compact, so is $\overline{\theta}(K \times L)$ since $X$ is Hausdorff (Chapter I, § 10, no. 2, Proposition 6). Hence the projection $P(K, L)$ of $\overline{\theta}(K \times L)$ into $G$ is a compact set.

To prove c): since $K \times L$ is closed in $X \times X$, it follows that $\overline{\theta}(K \times L)$ is closed in $P(K, L) \times K$ and is therefore compact under the hypotheses of c). Since every compact subset of $X \times X$ is contained in a compact subset of the form $K \times L$, it follows that the inverse image of any compact subset of $X \times X$ under $\theta$ is compact, and since $X \times X$ is locally compact, this shows that $\theta$ is proper (Chapter I, § 10, no. 3, Proposition 7) [see Chapter IV, § 1, Exercise 4c)].

a) If G operates properly on X, $\rho_K$ is proper.

b) If X is locally compact and $\rho_K$ is proper for each compact subset K of X, G operates properly on X.

The mapping $\rho_K$ factorizes as $G \times K \xrightarrow{\theta_K} K \times X \xrightarrow{\mathrm{pr}_2} X$, where $\theta_K$ is the restriction to $G \times K$ of the mapping $\theta : (s, x) \to (x, s.x)$ of $G \times X$ into $X \times X$. Since $\theta^{-1}(K \times X) = G \times K$, $\theta_K$ is proper if $\theta$ is proper (Chapter I, § 10, no. 1, Proposition 3). On the other hand, since K is compact, the projection $\mathrm{pr}_2 : K \times X \to X$ is proper (Chapter I, § 10, no. 2, Theorem 1, Corollary 5), hence $\rho_K$ is proper (ibid., no. 1, Proposition 5).

Suppose conversely that $\rho_K$ is proper for each compact $K \subset X$. If L is a compact subset of X, then $\rho_K^{-1}(L)$ is a compact subset of $G \times K$, whose projection into G is $P(K, L)$; therefore $P(K, L)$ is compact. Hence if X is locally compact it follows from Theorem 1 that G operates properly on X.

This follows from Proposition 12 and from Chapter I, § 10, no. 1, Proposition 1.

Since $G/H$ is Hausdorff (§ 2, no. 5, Proposition 13) it is locally compact, by Proposition 9 of no. 5 applied to H operating on G on the right. Thus it remains to show that $G/H$ is paracompact. Let V be a symmetric compact neighbourhood of e in G, and let $G_0 = V^\infty$ be the subgroup of G generated by V. $G_0$ is open (§ 2, no. 1, Corollary to Proposition 4) and operates continuously on $G/H$ (§ 2, no. 5, Proposition 12). If we can show that each of the orbits $G_0.z \ (z \in G/H)$ is an open subset of $G/H$ and a countable union of compact sets, then it will follow that $G/H$ is the topological sum of the distinct orbits $G_0.z$ and is therefore paracompact (Chapter I, , no. 10, Theorem 5). That $G_0 . z$ is open in $G/H$ follows from the facts that $G_0$ is open in $G$ and the equivalence relation $x^{-1} y \in H$ is open ($\S 2$, no. 4, Lemma 2). On the other hand, $G_0 z$ is the union of the sets $V^n . z$ ($n \geq 1$), and since $V^n$ is compact in $G$ and $G/H$ is Hausdorff, $V^n . z$ is compact. This completes the proof

$C$ is a closed normal subgroup of $G$ ($§ 2$, no. 2, Proposition 7), and hence $G/C$ is locally compact (Proposition 13) and totally disconnected (Chapter I, $§ 11$, no. 5, Proposition 9). Since the inverse image of an open subgroup of $G/C$, under the canonical mapping of $G$ onto $G/C$, is an open subgroup of $G$ containing $C$, we see that we can restrict ourselves to proving the proposition for the group $G/C$. In other words, we may assume that $G$ is totally disconnected. We know then (Chapter II, $§ 4$, no. 4, Corollary to Proposition 6) that every compact neighbourhood $V$ of $e$ contains a neighbourhood $U$ of $e$ which is both open and closed. Since $U$ is compact and $B = C U$ is closed, there is a symmetric open neighbourhood $W$ of $e$ such that $W \subset U$ and $UW \cap BW = \varnothing$ ($§ 3$, no. 1 and Chapter II, $§ 4$, no. 3, Proposition 4), and a fortiori $UW \subset U$. By induction on $n$ it follows that $W^n \subset U$ for every integer $n > 0$. Hence the subgroup $W^\infty = \bigcup_{n > 0} W^n$, generated by $W$, is contained in $U$; but $W^\infty$ is open in $G$ ($§ 2$, no. 1, Corollary to Proposition 4). This completes the proof.

We have also proved:

Let $C$ be the identity component of $G$. The components of $G$ are the sets $sC$, where $s \in G$ ($§ 2$, no. 2, Proposition 7); $\varphi(sC)$ is clearly connected, hence so is $\varphi(sC)$ (Chapter I, $§ 11$, no. 1, Proposition 1). But $\varphi(sC) = \varphi(sCH)$, and since $sCH$ is saturated with respect to the equivalence relation defined by $H$, and since this equivalence relation is open (§ 2, no. 4, Lemma 2), we have $\overline{\varphi(sCH)} = \varphi(\overline{sCH}) = \varphi(s.\overline{CH})$ (Chapter I, § 5, no. 3, Proposition 7).

Put $L = CH$. $L$ is a closed subgroup of $G$ which contains $C$ and $H$; hence to prove that the sets $\varphi(s.L) = s.\varphi(L)$ are the components of $G/H$ it is enough to show that the quotient space of $G/H$ by the equivalence relation whose classes are the sets $s.\varphi(L)$ is totally disconnected. Now this quotient space is homeomorphic to the homogeneous space $G/L$ (Chapter I, § 3, no. 4, Proposition 7); we are thus reduced to proving that when $C \subset H$, $G/H$ is totally disconnected. Since $G/H$ may be identified with $(G/C)/(H/C)$ (§ 2, no. 7, Proposition 22), we may even assume that $G$ itself is totally disconnected. Every neighbourhood of $\varphi(e)$ in $G/H$ contains a neighbourhood of the form $\varphi(V)$, where $V$ is a neighbourhood of $e$ in $G$, and therefore (Corollary 1) contains a neighbourhood of the form $\varphi(K)$, where $K$ is a compact open subgroup of $G$. $\varphi(K)$ is therefore both open and closed in $G/H$, and this shows that the component of $\varphi(e)$ in $G/H$ consists of $\varphi(e)$ alone. By translation, the same is true for the component of every point of $G/H$, and the corollary is proved.