Proposition 13

Let $G$ be a Lie group, $H$ a normal Lie subgroup of $G$, $X$ a manifold of class $C^r$ and $(g, x) \mapsto gx$ a law of left operation of class $C^r$ of $G$ on $X$. Suppose that conditions (a) and (b) of Proposition 10 hold.

(i) The law of left operation $(h, x) \mapsto hx$ of $H$ on $X$ satisfies conditions (a) and (b) of Proposition 10 (so that we can consider the quotient manifolds $X/G$ and $X/H$).

(ii) The law of left operation of $G$ on $X$ defines on taking quotients a law of left operation of class $C^r$ of $G/H$ on $X/H$; this law satisfies conditions (a) and (b) of Proposition 10 (so that we can consider the quotient manifold $(X/H)/(G/H)$).

(iii) The canonical mapping of $X$ onto $X/H$ defines on taking quotients a bijection of $X/G$ onto $(X/H)/(G/H)$. This bijection is an isomorphism of manifolds of class $C^r$.

Clearly $H$ operates freely on $X$; it operates properly by General Topology, Chapter III, § 4, no. 1, Example 1. The orbital mappings of $H$ on $X$ are immersions since the canonical injection of $H$ into $G$ is an immersion. This proves (i).

The law of left operation of $G$ on $X$ obviously defines, on taking quotients, a law of left operation of $G/H$ on $X/H$. This law is of class $C^r$ by Differentiable and Analytic Manifolds, R, 5.9.6. Let $g \in G$ and $x \in X$ be such that $(Hg)(Hx) = Hx$; then $H(gx) = Hx$ and hence $gx \in Hx$ and $g \in H$; this proves that $G/H$ operates freely on $X/H$. The mapping $\theta : (g, x) \mapsto (x, gx)$ of $G \times X$ into $X \times X$ is closed; on the other hand, $\theta(Hg \times Hx) = Hx \times H(gx)$; it follows immediately that the mapping

of $(G/H) \times (X/H)$ into $(X/H) \times (X/H)$ is closed; as moreover $G/H$ operates freely on $X/H$, Theorem 1 (c) of General Topology, Chapter I, § 10, no. 2 proves that $G/H$ operates properly on $X/H$.

Let $\pi$ be the canonical mapping of $X$ onto $X/H$, $\sigma$ the canonical mapping of $G$ onto $G/H$, $x$ an element of $X$ and $y = \pi(x)$.

Then $\pi \circ \rho(x) = \rho(y) \circ \sigma$ and hence

Let $u \in T_e(G/H)$ be such that $T_e(\rho(y))u = 0$. There exists $v \in T_e(G)$ such that $u = T_e(\sigma)v$. Then $T_x(\pi)(T_e(\rho(x))v) = 0$, hence $T_e(\rho(x))v$ is tangent to $Hx$ (Differentiable and Analytic Manifolds, R, 5.10.5) and therefore is of the form $T_e(\rho(x) | H)v'$ for some $v' \in T_e(H)$. As $T_e(\rho(x))$ is injective, it follows that $v = v'$, whence $v \in T_e(H)$ and therefore $u = 0$. Thus $T_e(\rho(y))$ is injective. The image of $T_e(\rho(y))$ is equal to that of $T_x(\pi) \circ T_e(\rho(x))$; now the image of $T_e(\rho(x))$ admits a topological supplement in $T_x(X)$ and contains the kernel of $T_x(\pi)$. It is therefore seen that $\rho(y)$ is an immersion, which completes the proof of (ii).

Assertion (iii) follows from the above and Differentiable and Analytic Manifolds, R, 5.9.7.