Proposition 4

Let G and H be Lie groups and f a homomorphism of the group G into the group H. For f to be analytic, it is necessary and sufficient that there exist a non-empty open subset U of G such that $f|_U$ is analytic.

The condition is obviously necessary. Suppose that it holds. For all $x_0 \in G$, $f(x_0x) = f(x_0)f(x)$ for all $x \in U$ and hence $f|_{x_0U}$ is analytic. But the sets $x_0U$, where $x_0 \in G$, form an open covering of G.