$\exp$ map gives a Lie group isomorphism for connected abelian Lie groups

This question comes from the proof of theorem 9.1.11 of this document: http://www.cds.caltech.edu/~marsden/volume/ms/2000/Supplement/ms_internet_supp.pdf

Consider $G$ an abelian connected Lie group with Lie algebra $\mathfrak{g}$. Then we have that $\exp: \mathfrak{g} \rightarrow G$ is a surjective Lie group homomorphism that is a local diffeomorphism from a neighbourhood of zero to the identity element in $G$ and its kernel is a discrete subgroup.

In the link mentioned above, the author then states that $\mathfrak{g}/\ker \exp$ is isomorphic to $G$ and "diffeomorphic to $G$ as manifolds by working in a small neighbourhood of the origin in $\mathfrak{g}$ where $\exp$ is a diffeomorphism with an open neighbourhood of the identity element in $G$. Thus, $\mathfrak{g}/\ker \exp$ and $G$ are isomorphic Lie groups."

What I don't understand is how the conclusion arises, i.e. how does the Lie group isomorphism appear? More specifically, I don't understand how does this imply that we have a diffeomorphism between $\mathfrak{g}/\ker \exp$ and $G$, since we only have the diffeomorphism guaranteed in a small neighbourhood...

• The idea is that you can use the fact that multiplication in $G$ is smooth and that $G$ is generated by an open neighborhood of the identity to "spread the diffeomorphism around," but I'll let someone whose Lie theory isn't as rusty give a more precise answer. – Stahl May 29 '17 at 1:11