Lie Subgroups, Actions, and Homogeneous Spaces

Since $i$ is an immersion by [F1], its differential $d{i}_e:T_{e}H\to T_{e}G$ is injective by [F2]. It is therefore a linear isomorphism from $T_{e}H$ onto the linear subspace $\mathfrak{h}:=d{i}_e(T_{e}H)\subseteq T_{e}G$.

Because $i$ is also a smooth Lie-group homomorphism, [F3] gives $d{i}_e([X,Y{]}_H)=[d{i}_{e}X,d{i}_{e}Y{]}_G$ for all $X,Y\in T_{e}H$. Hence the bracket of any two vectors in $\mathfrak{h}$ again lies in $\mathfrak{h}$.

Thus $\mathfrak{h}$ is a bracket-closed linear subspace of $T_{e}G$ and so is a Lie subalgebra by [F4]. Step 1.1 identifies $\mathrm{Lie}(H)$ with it, and step 2.1 shows that this identification respects Lie brackets. This includes the zero-dimensional and full-dimensional cases. The only choice used is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited by [F3].

Choose one finite basis $E_1,\dots ,E_r$ of $\mathfrak{h}$, using the empty basis if $r=0$, and let $E_i^L(g)=d(L_g{)}_e{E}_i$. By [F1], the fields $E_1^L,\dots ,E_r^L$ form a global smooth frame for ${\mathcal{D}}^{\mathfrak{h}}$.

By [F3], $[E_i^L,E_j^L]$ is left invariant. Its value at $e$ is the Lie-algebra bracket $[E_i,E_j]$, which belongs to $\mathfrak{h}$ because $\mathfrak{h}$ is a subalgebra. If $[E_i,E_j]={\sum }_k{c}_{ij}^k{E}_k$, left invariance gives $[E_i^L,E_j^L]={\sum }_k{c}_{ij}^k{E}_k^L$, a section of ${\mathcal{D}}^{\mathfrak{h}}$.

The frame criterion [F2] applied to step 2.1 proves involutivity. When $r=0$, every local section is zero and the same conclusion is vacuous; when $r=\dim G$, the distribution is $TG$. The hypothesis [A1] is used through [F1]'s smooth tangent-bundle trivialization and [F3]'s invariant-bracket result; choosing the single finite basis in step 1.1 adds no choice.

By [F1] and [F2], let $H$ be the maximal connected integral leaf of $\mathcal{D}$ through $e$, with its intrinsic leaf manifold structure and injective immersion $i:H\hookrightarrow G$. Its tangent space at $e$ is ${\mathcal{D}}_e=\mathfrak{h}$.

For every $h\in H$, [F3] makes $L_h$ a diffeomorphism preserving $\mathcal{D}$ in both directions. It therefore carries the maximal connected integral leaf $H$ to a maximal connected integral leaf $L_h(H)$: any larger connected integral manifold containing $L_h(H)$ would pull back under $L_{h^{-1}}$ to one properly containing $H$. But $L_h(H)$ contains $h=L_h(e)$, which belongs to $H$, so uniqueness of the maximal leaf through $h$ in [F2] gives $L_h(H)=H$. Hence $h{h}^{\mathrm{\prime }}\in H$ for $h,h^{\mathrm{\prime }}\in H$, and because $e\in L_h(H)$ there is $h^{\mathrm{\prime }}\in H$ with $h{h}^{\mathrm{\prime }}=e$, so $h^{-1}\in H$. Thus the leaf is a subgroup of $G$.

The ambient division map $\delta :G\times G\to G$, $\delta (a,b)=a{b}^{-1}$, is smooth, and by step 2.1 its restriction to the smooth manifold $H\times H$ has image setwise in $H$. Setwise inclusion alone would not prove smoothness for the intrinsic leaf topology. We use the countable-plaque construction in the proof of [F2]. Fix a flat-chart domain $U$ for $\mathcal{D}$. The leaf $H$ has a countable plaque atlas by [F2]; each atlas plaque meets $U$ in at most countably many connected components, and each such component lies in one $U$-plaque by the local plaque lemma used in [F2]. Under [A1], $H\cap U$ is therefore a countable union of $U$-plaques. Near any $(a,b)\in H\times H$, choose a connected source chart $C$ whose ambient division image is contained in $U$. The transverse coordinate of $\delta (C)$ is a continuous image of connected $C$ into the countable set of transverse coordinates of those plaques. A connected countable subset of Euclidean space is a singleton, so this transverse coordinate is constant and $\delta (C)$ lies in the single plaque through $a{b}^{-1}$. Its longitudinal coordinates are smooth as ambient coordinates of $\delta$, giving a smooth $H$-valued factor in that plaque chart. Hence division is smooth locally everywhere; inversion $b^{-1}=\delta (e,b)$ and then multiplication $ab=\delta (a,b^{-1})$ are smooth. Thus $H$ is a Lie group and $i$ is a smooth injective homomorphism and immersion.

The identity differential of $i$ has image $T_{e}H=\mathfrak{h}$ by step 1.1, so the constructed immersed subgroup has the required Lie algebra. This also covers $\mathfrak{h}=0$, when the connected leaf is $\{e\}$, and $\mathfrak{h}=\mathfrak{g}$, when the leaf is the identity component of $G$.

Let $j:K\to G$ be any connected immersed Lie subgroup whose tangent algebra is the same $\mathfrak{h}$. Translation in $K$ shows $dj(T_{k}K)={\mathcal{D}}_{j(k)}$, so $j$ is a connected integral immersion through $e$. The factorization clause of [F2] gives a unique smooth map $f:K\to H$ with $i\circ f=j$; injectivity of $i$ and the homomorphism law for $j$ make $f$ a homomorphism. Its identity differential is an isomorphism because both tangent images are $\mathfrak{h}$, so [F4] makes $f(K)$ contain an open identity neighborhood in $H$.

The image $f(K)$ is a subgroup. Since it is open, all its left cosets are open, so its complement is open as well; connectedness of $H$ forces $f(K)=H$. Injectivity of $i$ and $i\circ f=j$ forces $f$ to be injective. Translation of the isomorphism $d{f}_e$ shows that $df$ is invertible everywhere, so [F4] makes $f$ a bijective local diffeomorphism and hence a Lie-group isomorphism; uniqueness follows again from injectivity of $i$.

The only countable construction is [F2], whose proof spends [A1] on a countable flat-chart cover and countable unions. The countable-plaque argument in step 3.1 uses the same premise and is precisely what makes a setwise leaf-valued smooth map intrinsically smooth. The remaining selections are single finite-dimensional or local choices and use no stronger choice principle. Steps 1.1–4.1 give existence and steps 4.2–5.1 give uniqueness, establishing the stated correspondence.

Apply [F1] to the common subalgebra $\mathfrak{h}$. Its uniqueness clause supplies a Lie-group isomorphism $\varphi :H_1\to H_2$ satisfying $i_2\circ \varphi =i_1$.

The equality $i_2\circ \varphi =i_1$ gives $i_1(H_1)=i_2(H_2)$ as subsets of $G$. Because $\varphi$ and ${\varphi }^{-1}$ are smooth, they identify the intrinsic manifold structures, not merely the underlying image. The zero-dimensional and full-dimensional cases are included, and the stated countable-choice assumption is exactly the one inherited from [F1].

Fix $X\in \mathfrak{g}$. By [F2], ${\mathrm{ad}}_X$ restricts to an endomorphism of $\mathfrak{h}$. For $Y\in \mathfrak{h}$, solve $u^{\mathrm{\prime }}={\mathrm{ad}}_{X}u$, $u(0)=Y$ inside the finite-dimensional space $\mathfrak{h}$. Its inclusion into $\mathfrak{g}$ solves the same ambient initial-value problem, so uniqueness in [F4] gives $e^{t{\mathrm{ad}}_X}Y\in \mathfrak{h}$. Applying this with $-t$ proves equality $e^{t{\mathrm{ad}}_X}\mathfrak{h}=\mathfrak{h}$.

By [F3] and step 1.1, ${\mathrm{Ad}}_{\exp X}\mathfrak{h}=\mathfrak{h}$ for every $X\in \mathfrak{g}$. Define $K=\{g\in G:{\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}\}\mathrm{.}$ The representation law in [F3] makes $K$ a subgroup. The local exponential neighborhood in [F4] lies in $K$, so $K$ is open; every other coset is open as well, and therefore $K$ is also closed. Since $K$ contains $e$, connectedness puts the identity component $G^0$ inside $K$.

For $g\in K$, the composite $C_g\circ i:H\to G$ is an injectively immersed homomorphism with connected source, and [F5] says that its identity tangent image is ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Uniqueness in [F1] therefore identifies its image $gH{g}^{-1}$ with $H$. Thus every $g\in K$ normalizes $H$, and step 2.1 gives $H⊴G^0$.

If $\mathfrak{h}$ is invariant under every ${\mathrm{Ad}}_g$, then $K=G$ by definition, and step 3.1 gives $H⊴G$. The cases $\mathfrak{h}=0$ and $\mathfrak{h}=\mathfrak{g}$ are included: their connected integral subgroups are respectively $\{e\}$ and $G^0$. Disconnected $G$ is allowed, and the stronger global conclusion uses exactly the separately stated full $\mathrm{Ad}$-invariance. No closedness conclusion follows. The only choice use is [A1], inherited through [F1], [F3], and [F4].

Choose a smooth chart $\phi :W\to \phi (W)\subseteq {\mathbb{R}}^n$ with $\phi (e)=0$. On a smaller neighborhood of $0$, the coordinate form of the squaring map is $s(v)=\phi ({\phi }^{-1}(v{)}^2)$. The differential of multiplication at $(e,e)$ sends $(X,Y)$ to $X+Y$: its restrictions to the two coordinate axes are the identity because $ge=g$ and $eh=h$, and the differential is linear. Therefore $d{s}_0(X)=2X$.

Fix a Euclidean norm. Differentiability at $0$ gives $r>0$ such that $B_r(0)\subseteq \phi (W)$, the coordinate squaring map is defined there, and $\parallel s(v)-2v\parallel \le \frac{1}{2}\parallel v\parallel$ whenever $\parallel v\parallel <r$. Hence $\parallel s(v)\parallel \ge \frac{3}{2}\parallel v\parallel$ throughout that ball.

Put $U={\phi }^{-1}(B_r(0))$. Suppose a subgroup $K\subseteq U$ contains $h\mathrm{\ne }e$, and write $v_j=\phi (h^{2^j})$. Because every power belongs to $K\subseteq U$, all $v_j$ lie in $B_r(0)$, while $v_{j+1}=s(v_j)$. Step 2.1 gives $\parallel v_j\parallel \ge (3\mathrm{/}2{)}^j\parallel v_0\parallel \mathrm{.}$ Since $h\mathrm{\ne }e$, $\parallel v_0\parallel >0$, so the right side eventually exceeds $r$, contradicting $v_j\in B_r(0)$.

Thus every subgroup contained in $U$ is $\{e\}$. In dimension zero, the same argument reduces to the open singleton identity chart. No choice principle is used: only one chart, one norm, and one radius are fixed.

Put $\mathfrak{g}=T_{e}G$ and define $\mathfrak{h}=\{X\in \mathfrak{g}:\exp (tX)\in H\text{ for every }t\in \mathbb{R}\}\mathrm{.}$ This set contains $0$ and is closed under real scalar multiplication. If $X,Y\in \mathfrak{h}$ and $t\in \mathbb{R}$, then for all sufficiently large positive integers $n$, [F2] gives $\exp (tX\mathrm{/}n)\exp (tY\mathrm{/}n)=\exp Z_n,Z_n=\mathrm{BCH}(tX\mathrm{/}n,tY\mathrm{/}n)\mathrm{.}$ Both factors lie in $H$. The homogeneous expansion and uniform convergence in [F2] give $n{Z}_n\to t(X+Y)$. By [F3], $\exp (n{Z}_n)=(\exp Z_n{)}^n\in H$; closedness of $H$ gives $\exp (t(X+Y))\in H$. Since $t$ was arbitrary, $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a linear subspace.

Choose a complement $\mathfrak{b}$ with $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{b}$ by [F4]. The smooth map $\mathrm{\Psi }:\mathfrak{h}\times \mathfrak{b}⟶G,\mathrm{\Psi }(X,Y)=\exp X\exp Y,$ has differential $(X,Y)\mapsto X+Y$ at $(0,0)$, an isomorphism. By [F4], after shrinking around $(0,0)$ it is a diffeomorphism onto an identity neighborhood.

We claim that some exponential neighborhood $U\subseteq \mathfrak{g}$ satisfies $H\cap \exp U=\exp (U\cap \mathfrak{h})\mathrm{.}$ The inclusion from right to left follows from the definition of $\mathfrak{h}$. If no such neighborhood existed, take a nested sequence of coordinate balls $U_n$ shrinking to $0$, all inside the injectivity domain in [F1] and with $\exp U_n$ inside the image in step 1.2. By [A1], select $h_n\in (H\cap \exp U_n)\setminus \exp (U_n\cap \mathfrak{h})\mathrm{.}$ Write $h_n=\exp X_n\exp Y_n$ using the inverse in step 1.2. Then $(X_n,Y_n)\to (0,0)$, $X_n\in \mathfrak{h}$, and $\exp Y_n\in H$. For all sufficiently large $n$, $Y_n\mathrm{\ne }0$: otherwise injectivity of the common exponential chart would put $h_n$ in $\exp (U_n\cap \mathfrak{h})$.

Fix a norm on $\mathfrak{b}$, put $c_n=\parallel Y_n\parallel$, and discard the finitely many zero terms. The unit vectors $c_n^{-1}{Y}_n$ have a convergent subsequence by [F5]; relabel it so that $c_n^{-1}{Y}_n\to Y\in \mathfrak{b}$ with $\parallel Y\parallel =1$. For arbitrary $t\in \mathbb{R}$, choose the integer $k_n=\lfloor t\mathrm{/}{c}_n\rfloor$. Then $k_n{c}_n\to t$, so $k_n{Y}_n\to tY$. By [F3], $\exp (k_n{Y}_n)=(\exp Y_n{)}^{k_n}\in H\mathrm{.}$ Closedness gives $\exp (tY)\in H$. Since this holds for every $t$, $Y\in \mathfrak{h}\cap \mathfrak{b}=\{0\}$, contradicting $\parallel Y\parallel =1$. The claim in step 2.1 follows.

Choose a linear coordinate isomorphism $E:\mathfrak{g}\to {\mathbb{R}}^m$ carrying $\mathfrak{h}$ to ${\mathbb{R}}^k\times \{0\}$. By step 2.1, the chart $E\circ \log$ on $\exp U$ sends $H\cap \exp U$ to the coordinate slice $E(U)\cap ({\mathbb{R}}^k\times \{0\})$. For each $h\in H$, left translation carries this chart to a slice chart at $h$ because $L_h(H)=H$. Hence [F6] makes $H$ an embedded submanifold with its subspace topology.

Ambient multiplication and inversion preserve $H$. In the slice charts from step 3.2 their restrictions have smooth coordinate representatives, so they make $H$ a Lie group and its inclusion into $G$ a smooth embedded homomorphism. Its tangent space at $e$ is $\mathfrak{h}$, and [F6] confirms that this space is bracket closed.

Any other smooth structure making the same subset $H$ an embedded Lie subgroup has the same subspace topology by definition. In every ambient slice chart from step 3.2, both intrinsic structures use the restriction to the same Euclidean slice, so the identity between them is locally a diffeomorphism and hence globally a diffeomorphism. This proves uniqueness. The cases $H=\{e\}$, $H=G$, dimensions zero and one, and disconnected $H$ are included. A subgroup contains the identity, so the empty case cannot occur; no metric, nondegeneracy, manifold-boundary, or interval-endpoint hypothesis occurs. Choice is used exactly as stated in [A1] and through [F1]–[F3].

Suppose $\mathrm{\Gamma }$ is discrete in the subspace topology. There is an identity neighborhood $U$ with $U\cap \mathrm{\Gamma }=\{e\}$. Choose a symmetric identity neighborhood $V$ with $V^{-1}V\subseteq U$. Every translate $gV$ contains at most one point of $\mathrm{\Gamma }$: two such points ${\gamma }_1,{\gamma }_2$ would satisfy ${\gamma }_1^{-1}{\gamma }_2\in U\cap \mathrm{\Gamma }$.

If $g$ lies in the closure of $\mathrm{\Gamma }$, then $gV$ contains some $\gamma \in \mathrm{\Gamma }$. If $g\mathrm{\ne }\gamma$, Hausdorffness makes $(gV)\setminus \{\gamma \}$ an open neighborhood of $g$ disjoint from $\mathrm{\Gamma }$, contradicting closure. Thus $g=\gamma$, so $\mathrm{\Gamma }$ is closed.

By [A1], $\mathrm{\Gamma }$ has its unique embedded Lie-subgroup structure. Its embedded topology is its discrete subspace topology, so every singleton is an open coordinate neighborhood; hence its manifold dimension is zero.

Conversely, an embedded zero-dimensional Lie subgroup has the subspace topology, and every point has a chart into the one-point space ${\mathbb{R}}^0$; it is therefore discrete. Its closedness is already part of the right-hand condition. This proves both directions. The trivial subgroup and discrete ambient groups are included. Choice is used only through [A1].

The graph ${\mathrm{\Gamma }}_F=\{(g,F(g)):g\in G\}$ is a subgroup of $G\times H$. It is closed: if $(g,h)$ is not on the graph, then $h\mathrm{\ne }F(g)$, and continuity of $F$ together with Hausdorffness of $H$ gives product neighborhoods of $(g,h)$ disjoint from the graph. By [A1], ${\mathrm{\Gamma }}_F$ is an embedded Lie subgroup.

The first projection $P:{\mathrm{\Gamma }}_F\to G$ is a smooth Lie-group homomorphism and a homeomorphism, with continuous inverse $g\mapsto (g,F(g))$. Manifold charts and [F1] therefore give $\dim {\mathrm{\Gamma }}_F=\dim G$.

We show that $d{P}_{(e,e)}$ is injective. If $X$ is in its kernel, [F3] gives $P({\exp }_{{\mathrm{\Gamma }}_F}(tX))={\exp }_G(t,d{P}_{(e,e)}X)=e$ for every $t$. The algebraic kernel of $P$ is the singleton $(e,e)$, so this one-parameter subgroup is constant; differentiating it at zero gives $X=0$. Equal dimensions from step 2.1 now make $d{P}_{(e,e)}$ an isomorphism.

Translation of the homomorphism identity makes $dP$ invertible everywhere. By [F2], $P$ has smooth local inverses around every point. Since the set-theoretic inverse is unique, these local inverses agree with the global continuous inverse $P^{-1}$, so $P^{-1}$ is smooth.

Let $Q:{\mathrm{\Gamma }}_F\to H$ be the second projection. Then $F=Q\circ P^{-1}$ is smooth. Zero-dimensional and disconnected groups are included; no injectivity or surjectivity of $F$ is assumed. The exponential argument in step 3.1 is the needed correction to the scaffold: a smooth homeomorphism between equal-dimensional manifolds need not have invertible differential. Countable choice is used through [A1] and [F3].

The homomorphism identity is $F\circ L_g=L_{F(g)}\circ F$. Differentiating it at $e$ and using [F2] gives $d{F}_g\circ d(L_g{)}_e=d(L_{F(g)}{)}_e\circ d{F}_e$.

Both translation differentials in step 1.1 are isomorphisms by [F1]. Therefore $d{F}_g=d(L_{F(g)}{)}_e\circ d{F}_e\circ d(L_{g^{-1}}{)}_g$, so composing with the two isomorphisms does not change rank. This proves the displayed equality. Dimension zero, disconnected groups, and the zero differential require no separate argument, and no choice principle is used.

The identity singleton in $H$ is closed, so $K=F^{-1}(e_H)$ is closed. The homomorphism law gives $gK{g}^{-1}\subseteq K$ for every $g\in G$, and applying it to $g^{-1}$ gives equality. Thus $K$ is a closed normal subgroup, and [A1] gives its unique embedded Lie-subgroup structure.

Let $r=\mathrm{rank}d{F}_e$. By [F1], the rank is constant. Choose constant-rank charts at $e$ and $e_H$ that send these points to zero and in which $F$ is $(u,v)\mapsto (u,0)$. In the source chart, the fibre $F^{-1}(e_H)=K$ is locally the slice $u=0$, whose tangent space at the origin is exactly the kernel of the displayed linear map. Because [A1] gives $K$ the embedded subspace structure, this slice tangent is $T_{e}K$. Hence $T_{e}K=\ker d{F}_e$.

By definition $\mathrm{Lie}(K)=T_{e}K$, so step 2.1 proves the formula. The trivial kernel, the constant map, disconnected groups, and ranks zero or full are included. Countable choice is used only through [A1].

Give $B=G\mathrm{/}K$ the quotient topology and let $q:G\to B$ be the coset map. It is open because $q^{-1}(q(O))=OK$ is a union of right translates of an open set $O$. It is Hausdorff: the equivalence relation is the closed set $R=\{(g,h):g^{-1}h\in K\}$, and if $(g,h)\mathrm{\notin }R$, a product neighborhood $U\times V$ disjoint from $R$ gives disjoint open quotient neighborhoods $q(U)$ and $q(V)$. Images under the open map $q$ of a countable basis of $G$ form a countable basis of $B$.

In a constant-rank product chart from [F1], choose the transverse slice $S$ obtained by setting the kernel coordinates to zero. The restriction $q{\mathrm{\mid }}_S$ is bijective onto $q(U)$: points have the same $F$-value exactly when they differ by an element of $K$, and the normal form makes each local fibre meet $S$ once. It is a homeomorphism because an open subset of $S$ thickens in the kernel coordinates to an open subset of $U$ with the same $q$-image. These charts make $B$ a Hausdorff second-countable smooth manifold and make $q$ locally the projection $(u,v)\mapsto u$, hence a surjective submersion. Their changes are smooth because each has the smooth local section supplied by its slice.

Normality of $K$ gives $B$ its quotient group law. Multiplication and inversion are smooth: near any arguments, choose the smooth local sections from step 2.1 and express the descended maps as $q(s_1(x)s_2(y))$ and $q(s(x{)}^{-1})$. Thus $B$ is a Lie group and $q$ is a smooth homomorphism.

Define $j:B\to H$ by $j(gK)=F(g)$. Algebraically this is a well-defined injective homomorphism with image $\mathrm{im}F$. In the local coordinates of step 2.1 and the target constant-rank chart, $j$ is $u\mapsto (u,0)$, so it is a smooth immersion. Therefore $j(B)$ with the transported intrinsic structure is an immersed Lie subgroup, and $F=j\circ q$.

At the identity, $d{F}_e=d{j}_{eK}\circ d{q}_e$. The differential $d{q}_e$ is surjective with kernel $T_{e}K=\ker d{F}_e$ by the local projection and [A1], while $d{j}_{eK}$ is injective. Hence $d{j}_{eK}(T_{eK}B)=\mathrm{im}d{F}_e$, which is the tangent algebra of the immersed image.

If another manifold structure on the same image makes the corestriction from $G$ a surjective submersion, its local smooth sections show that the identity map in either direction is locally a composite of that corestriction with a local section for the other structure. Thus the identity is a diffeomorphism and the structure is unique. Rank zero, trivial image, noninjective $F$, and disconnected groups are included. Nothing in the construction identifies the intrinsic topology with the subspace topology of $H$; no embeddedness or closedness conclusion is asserted. Choice is inherited only through [A1].

Let $\bar{F}:G\to \mathrm{im}F$ be the corestriction and let $j:\mathrm{im}F\hookrightarrow H$ be inclusion. Then $F=j\circ \bar{F}$ set-theoretically and as homomorphisms. By [F2], $\bar{F}$ is a surjective submersion and $j$ is an injective immersion with the canonical immersed-subgroup structure.

For $g,g^{\mathrm{\prime }}\in G$, $F(g)=F(g^{\mathrm{\prime }})⟺F(g^{-1}{g}^{\mathrm{\prime }})=e_H⟺g^{-1}{g}^{\mathrm{\prime }}\in \ker F⟺g^{\mathrm{\prime }}\in g\ker F\mathrm{.}$ Thus the fibres of both $F$ and $\bar{F}$ are precisely the left kernel cosets; [F1] also makes right cosets equal because the kernel is normal.

Steps 1.1 and 1.2 prove the asserted differential-geometric and algebraic factorization. The zero map, trivial kernel, nonclosed image, and disconnected groups are included. No embeddedness of the image is inferred. Countable choice is inherited through [F1]–[F2].

By [A1], put $\mathfrak{g}=T_{e}G$ and $\mathfrak{h}=T_{e}H$. By [F1], choose a linear projection of $\mathfrak{g}$ onto $\mathfrak{h}$ and put $\mathfrak{m}$ equal to its kernel, so $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{h}$.

Give $G\mathrm{/}H$ the quotient topology. The map $q$ is open, since $q^{-1}(q(O))=OH={\bigcup }_{h\in H}Oh$ for every open $O\subseteq G$. It is Hausdorff: the orbit relation $R=\{(g_1,g_2):g_1^{-1}{g}_2\in H\}$ is closed, and for two inequivalent points choose a product neighborhood $O_1\times O_2$ disjoint from $R$; the open sets $q(O_1)$ and $q(O_2)$ are then disjoint. Images of a countable basis of $G$ form a countable basis of $G\mathrm{/}H$.