Theorem 3

Let G and H be Lie groups and h a continuous morphism of L(G) into L(H).

(i) There exist an open subgroup G' of G and a Lie group morphism $\phi$ of G' into H such that $h = L(\phi)$.

(ii) Let G_1, G_2 be open subgroups of G and $\phi_i$ a morphism of G_i into H such that $h = L(\phi_i)$. Then $\phi_1$ and $\phi_2$ coincide on an open subgroup of G.

By Proposition 1, this follows from § 4, no. 1, Theorem 1.