Corollary 1

Let G be a simply connected topological group and let $G'$ be a group. Let V be a connected neighbourhood of the identity element in G and let $f: V\rightarrow G'$ be a local homomorphism. There exists a unique group homomorphism $h: G\rightarrow G'$ extending $f$. If $G'$ is a topological group and if the mapping $f$ is continuous, the homomorphism $h$ is continuous.

The group G is connected. Let H$,\varphi$ and $p$ be as in proposition 2. Since G is simply connected, H is a trivializable covering of G (I, p. 124, def. 3); since H is connected and nonempty, the mapping $p$ is an isomorphism of topological groups. Put $h=\varphi \circ p^{-1}$; the mapping $h$ is a group homomorphism and there exists an open neighbourhood U of the identity element $e$ of G contained in V such that $f|U =h|U$. It follows from lemma 1 that $f$ and $h$ coincide on V. In other words, the mapping $h$ extends the mapping $f$.

If $G'$ is a topological group and if the mapping $f$ is continuous, the homomorphism $\varphi$ is continuous, hence so is $h$.

Let us prove the uniqueness of such an extension. The set of points of G where two homomorphisms of G into $G'$ coincide is a subgroup of G. Since the group G is connected, every subgroup of G containing a neighbourhood of the identity element is equal to G (TG, III, p. 8, prop. 6). The uniqueness of $h$ follows.