Proposition 21

Let $G$ be a topological group, $G_0$ a dense subgroup of $G$, $H_0$ a normal subgroup of $G_0$, $H$ the closure of $H_0$ in $G$ and $\varphi$ the canonical mapping $G \to G/H$. Then the canonical bijection $G_0/H_0 \to \varphi(G_0)$ is an isomorphism of the topological group $G_0/H_0$ onto a dense subgroup of $G/H$.

Since $H_0 = H \cap G_0$, it is enough to show that if $U_0$ is any open subset of $G_0$ which is saturated with respect to the relation $x^{-1}y \in H_0$, then $U_0$ is the intersection of $G_0$ and an open subset of $G$ which is saturated with respect to the relation $x^{-1}y \in H$ (Chapter I, § 3, no. 6, Proposition 10). Let $U$ be an open subset of $G$ such that $U_0 = U \cap G_0$. Since $U_0 = U_0 H_0$, it is easily seen that $U_0 = U H_0 \cap G_0$; but $U H_0$ is open in $G$, so we may suppose that $U = U H_0$. The set $U H$ is open in $G$ and is saturated with respect to the relation $x^{-1}y \in H$; hence the proposition will be proved if we show that $U H \cap G_0 = U_0$. Now, if $u \in U$ and $h \in H$ are such that $u h \in G_0$, then there is a symmetric neighbourhood $V$ of $e$ in $G$ such that $u V \subset U$; since $V h$ is a neighbourhood of $h$ in $G$, there exists $z \in V$ such that $z h \in H_0$. But then we have $u z^{-1} \in U$, and $u h = (u z^{-1})(z h)$; therefore $U H \cap G_0 \subset U H_0$. But $U H_0 = U$, hence $U H \cap G_0 \subset U \cap G_0 = U_0$. This completes the proof.

Let $G$ be a topological group operating continuously on a topological space $X$, and let $K$ be a normal subgroup of $G$ which is contained in the stabilizer of each point of $X$. For each $x \in X$, the relation $s \equiv t \pmod{K}$ therefore implies $s.x = t.x$, and passing to the quotient we define a mapping $s \to s.x$ of $G/K$ into $X$. It is immediately verified that with respect to the external law $(s', x) \to s'.x$, X has G/K as a group of operators. Moreover, G/K operates continuously on X with respect to this law; for the relation of equality on X and the relation $s \equiv t \pmod{K}$ on G are both open equivalence relations, and therefore the result follows from the continuity of the mapping

of $G \times X$ into X (Chapter I, § 5, no. 3, Corollary to Proposition 8, and § 3, no. 4, Proposition 6).

Now let G be a topological group operating continuously on a topological space X, and let H be any normal subgroup of G; then H operates continuously on X. Let S be the equivalence relation defined by H on X; then S is open (no. 4, Lemma 2). The relation S is compatible with the group G (no. 4); for if $y \equiv x \pmod{S}$ then there exists $t \in H$ such that $y = t.x$; hence for all $s \in G$ we have $s.y = (sts^{-1}).s.x$, and $sts^{-1} \in H$ since H is normal in G; thus $s.y \equiv s.x \pmod{S}$. If $\psi$ is the canonical mapping of X onto X/S, the group G therefore operates continuously on X/S with respect to the external law

(no. 4, Proposition 11). Moreover, the group H is contained in the stabilizer of each of the points of X/S; as we have seen above, G/H operates continuously on $X/S = X/H$ with respect to the external law $(s', \psi(x)) \to \psi(s.x)$. If R denotes the equivalence relation on X defined by G, then the relation S implies R, and the equivalence relation R/S on X/S is that defined by the group G/H. Hence (Chapter I, § 3, no. 4, Proposition 7):