One-parameter subgroups are exactly exponentials

Suppose $\gamma$ is a one-parameter subgroup, and put $X={\gamma }^{\mathrm{\prime }}(0)$. By uniqueness in [F4], $\gamma ={\gamma }_X$, so [F2] gives $\gamma (t)={\exp }_G(tX)$ for every real $t$. If also $\gamma (t)={\exp }_G(tY)$ for every $t$, [F2] identifies this curve with ${\gamma }_Y$; differentiating at zero yields $Y={\gamma }^{\mathrm{\prime }}(0)=X$.

Conversely, fix $X\in \mathfrak{g}$. By [F4], ${\gamma }_X$ is a one-parameter subgroup, and [F2] gives $t\mapsto {\exp }_G(tX)={\gamma }_X(t)$. Hence every exponential curve is a one-parameter subgroup in the sense of [F3].

Lie groups are nonempty and boundaryless. If $\dim G=0$, then $X=0$ and the only exponential curve is constant; in dimension one the proof is unchanged. The curves have domain all of $\mathbb{R}$, so there is no finite endpoint. No metric or nondegeneracy condition occurs. The only choice use is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited through [F2] and [F4]; taking one derivative adds no choice. Step 1.1 proves the forward implication and uniqueness, and step 1.2 proves the reverse implication.