Proposition 3
Tag 0AEM, permanent. Cite it as [Bourbaki, Tag 0AEM].
Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups
Let G be a Lie group, H an integral subgroup of G, M a non-empty connected analytic manifold, f a mapping of M into G and r \in NK. Consider the following conditions:* (i) f is of class C^r and f(M) \subset H; (ii) f(M) \subset H and f, considered as a mapping of M into H, is of class C^r; (iii) f is of class C^r, f(M) meets H and the image of Tm(M) is contained in f(m).L(H) for all m \in M*.
Then (ii) (iii) (i). If the topology on H admits a countable base, the three conditions are equivalent. (ii) (i) and (ii) (iii): obvious. (iii) (ii): suppose that condition (iii) holds. By Differentiable and Analytic Manifolds, R, 9.2.8, f is a morphism of class C^r of M into the left foliation associated with L(H). As M is connected, f(M) \subset H.
If the topology on H admits a countable base, condition (i) implies that f is a mapping of class C^r of M into H (Differentiable and Analytic Manifolds, R, 9.2.8); hence (i) (ii).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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