Proposition 10

*Let $G$ and $H$ be Lie groups and $\phi$ a morphism of $G$ into $H$. (i) $\phi \circ \exp_G = \exp_H \circ L(\phi)$. (ii) If $G$ is an integral subgroup of $H$, then $\exp_G = \exp_H | L(G)$.

EXponential mapping

The two sides of the equality (i) are analytic mappings of $L(G)$ into $H$ which coincide in a neighbourhood of 0 ($§ 4$, Proposition 8, no. 4) and hence are equal. Assertion (ii) is a special case of (i).