Proposition 5

Let $G$ (resp. $G'$) be a topological group operating continuously on a topological space $X$ (resp. $X'$). Let $\varphi$ be a continuous homomorphism of $G$ into $G'$ and let $\psi$ be a continuous mapping of $X$ into $X'$ compatible with $\varphi$ (§ 2, no. 4). Then:

(i) If $\varphi$ is surjective and $\psi$ is surjective and proper, and if $G$ operates properly on $X$, then $G'$ operates properly on $X'$.

(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).