Theorem 1

Tag 0AE9, permanent. Cite it as [Bourbaki, Tag 0AE9].

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

Let $G$ and $H$ be Lie groups and $h$ a continuous morphism of $L(G)$ into $L(H)$ . Suppose that $G$ is simply connected. Then there exists one and only one Lie group morphism $\phi$ of $G$ into $H$ such that $h = L(\phi)$ .

The existence of $\phi$ follows from Lemma 1 and § 4, no. 1, Theorem 1 (i). The uniqueness of $\phi$ follows from § 4, no. 1, Theorem 1 (ii) and the fact that $G$ is connected.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Corollary, § 6
Corollary 4, § 6
Proposition 14, § 6
Proposition 7, § 6
Proposition 8, § 2

Cites

§ 4, no. 1, Theorem 1
Lemma 1