Corollary

Let f be a local isomorphism of $\mathbf{R}$ with a topological group G. Then there is a unique strict morphism of $\mathbf{R}$ onto an open subgroup of G which coincides with f at all points of some neighbourhood of 0.

Let $\overline{f}$ be the continuous homomorphism of $\mathbf{R}$ into G which coincides with $f$ at all points of an open interval I, which contains 0 and is contained in the set on which $f$ is defined; $\overline{f}(\mathbf{R})$ by hypothesis contains a neighbourhood of the identity element of G, hence (Chapter III, § 2, no. 1, Corollary to Proposition 4) is an open subgroup of G; and $\overline{f}$ is a strict morphism of $\mathbf{R}$ onto $\overline{f}(\mathbf{R})$, by Chapter III, § 2, no. 8, Proposition 24.