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Identify the Lie algebra $\mathfrak{g}$ with the tangent space $T_{e}G$. For $X\in \mathfrak{g}$, its left-invariant vector field is $X^L(g)=(d{L}_g{)}_{e}X$. Let ${\gamma }_X$ be its integral curve with ${\gamma }_X(0)=e$. Uniqueness of integral curves and left invariance give ${\gamma }_X(s+t)={\gamma }_X(s){\gamma }_X(t)$ wherever initially defined. Repetition of a local curve extends it to all real times. Thus ${\gamma }_X$ is the unique one-parameter subgroup with derivative $X$ at zero, and the Exponential map of a Lie group is defined bySmooth dependence of solutions of differential equations on their initial data and parameters makes this map smooth.

The derivative at zero is particularly simple:It is the identity linear map from $\mathfrak{g}$ to $T_{e}G$. The inverse function theorem therefore proves that the Exponential map of a Lie group is a local diffeomorphism at zero, producing a local exponential chart. This proves the requested local assertion.

There is a necessary qualification to the other assertion: the image of every one-parameter subgroup lies in the identity component of a Lie group $G^{\circ }$. The correct statement without a connectedness hypothesis isIndeed, if $\exp$ is a group homomorphism, its image is an abelian subgroup. It contains an identity neighborhood by the local result. An open subgroup of a connected group is the entire group: every coset is open, so its complement is also open. Thus $\exp (\mathfrak{g})=G^{\circ }$, proving that this component is abelian.

Conversely, if $G^{\circ }$ is abelian, then $\exp (tX)\exp (tY)$ is a one-parameter subgroup: its group law follows by commuting the factors. Its derivative at zero is $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$ and, at $t=1$, additivity. This proves the additivity of the Lie exponential map criterion. For connected $G$ it is exactly the stated equivalence with an abelian group. For disconnected $G$ the unrestricted claim is false: $S_3\times \mathbb{R}$ is a nonabelian Lie group, but its exponential is the homomorphism $x\mapsto (e,x)$. The original PDF does not impose connectedness in that sentence, so that qualification cannot be omitted.

Now suppose $G$ is connected and abelian, of dimension $n$. The exponential is a surjective group homomorphism, and its kernel $\Lambda$ is discrete by local injectivity at zero. The induced mapis a bijective Lie group homomorphism and a local diffeomorphism, hence a Lie group isomorphism. To identify the quotient, take linearly independent elements ${\lambda }_1,\dots ,{\lambda }_b\in \Lambda$ spanning the real linear span of $\Lambda$, and let ${\Lambda }_0={\sum }_i\mathbb{Z}{\lambda }_i$. Every coset of ${\Lambda }_0$ in $\Lambda$ has a representative in the bounded fundamental parallelepiped of these vectors. A discrete additive subgroup has finite intersection with a compact set: otherwise differences of arbitrarily close elements would approach zero, contradicting discreteness at zero. Thus $\Lambda /{\Lambda }_0$ is finite.

It follows that $\Lambda$ is a finitely generated torsion-free abelian group of rank $b$, hence has a $\mathbb{Z}$-basis of $b$ elements. These basis elements also span its $b$-dimensional real span and are linearly independent over $\mathbb{R}$. Extending them to a real basis identifies $\Lambda$ with ${\mathbb{Z}}^b\times \{0{\}}^{n-b}$. Consequently the classification of connected abelian Lie groups isThe compact circle factors record the periods of the one-parameter subgroups.