The exponential map is globally injective on every connected Lie group

Addition and negation descend to smooth operations in the quotient charts, so $G$ is a one-dimensional Lie group. The quotient map $\pi :\mathbb{R}\to G$ is continuous and surjective; since $\mathbb{R}$ is connected, [F3] makes $G$ connected.

For $X\in T_{0}G\simeq \mathbb{R}$, the curve ${\gamma }_X(t)=[tX]$ is a one-parameter subgroup with initial velocity $X$. Hence [F1] gives ${\exp }_G(X)=[X]$.

In particular, ${\exp }_G(0)=[0]=[1]={\exp }_G(1)$ although $0\mathrm{\ne }1$. Thus the exponential is not globally injective.

The witness is nonempty, connected, and one-dimensional, so it also covers the lowest positive dimension. No metric, degeneracy, endpoint, choice, or biconditional occurs. The zero-dimensional connected case is harmless but cannot rescue the universal claim.