Lie subgroup–Lie subalgebra correspondence

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.