Remark

Instead of saying that a topological group $G$ operates continuously on a topological space $X$, it is often said that $G$ operates continuously on the left on $X$. When the topological group $G^0$ opposite to $G$ operates continuously on $X$, we say that $G$ operates continuously on the right on $X$. It comes to the same thing to say that $X$ has a continuous external law of composition $(s, x) \to s.x$ with $G$ as set of operators, such that $s.(t.x) = (ts).x$ and $e.x = x$. Such a law is often written on the right : $(s, x) \to x.s$ (whence the terminology), and we have then $(x.t).s = x.(ts)$. If $G$ operates continuously on the right on $X$ by the law $(s, x) \to x.s$, then also $G$ operates continuously on the left on $X$ according to the external law $(s, x) \to x.s^{-1}$, by virtue of axiom (GT$_{II}$).

Let $X$ (resp. $X'$) be a set with a group of operators $G$ (resp. $G'$) and let $f : G \to G'$ be a homomorphism and $g : X \to X'$ a mapping. $f$ and $g$ are said to be compatible if $g(s.x) = f(s).g(x)$ for all $s \in G$ and all $x \in X$. If $X''$ is a third set with a group of operators $G''$, if $f' : G' \to G''$ is a homomorphism, $g' : X' \to X''$ a mapping, and if $f'$ and $g'$ are compatible, then $f' \circ f$ and $g' \circ g$ are compatible. When $X, X'$ are topological spaces and $G, G'$ are topological groups operating continuously on $X, X'$ respectively, $(f, g)$ is said to be a morphism of the space with operators $X$ into the space with operators $X'$, provided that $f$ and $g$ are continuous and compatible. Passing to the quotients, $g$ then induces a continuous mapping $X/G \to X'/G'$ (Chapter I, , no. 4, Corollary to Proposition 6)

Let $G$ be a topological group operating continuously on a topological space $X$, and let $\varphi$ be the canonical mapping of $X$ onto the orbit space $X/G$. Let $A$ be any subset of $X$, and let $A'$ be the subspace of $X$ which is the saturation of $A$ with respect to the equivalence relation $R$ defined by $G$ (thus $A'$ is the union of the orbits of points of $A$, and is said to be the saturation of $A$ with respect to $G$). $G$ operates continuously on $A'$ by the restriction of $(s, x) \to s.x$ to $G \times A'$. Moreover, since $R$ is open (Lemma 2) and $A'$ is saturated, it follows from Proposition 4 of Chapter I, § 5, no. 2, and from the relation $\varphi(A) = \varphi(A')$, that: