The exponential map scales geodesic time

Suppose first that $t\mathrm{\ne }0$. By [F1], $1\in I_{p,tv}$ iff $t\in I_{p,v}$, and on that domain ${\gamma }_{p,tv}(1)={\gamma }_{p,v}(t)$. Applying [F2] proves both the domain equivalence and the exponential identity.

If $t=0$, then $0\in I_{p,v}$, while [F1] makes ${\gamma }_{p,0}$ the constant curve on $\mathbb{R}$; hence $0v\in {\mathcal{E}}_p$ and ${\exp }_p(0)=p={\gamma }_{p,v}(0)$. Thus the equivalence and identity also hold at zero.

Let $v\in {\mathcal{E}}_p$ and $s\in [0,1]$. Then $1\in I_{p,v}$ by [F2]. Since $I_{p,v}$ is an interval containing $0$, it contains $s$, so step 1.1 or 1.2 gives $sv\in {\mathcal{E}}_p$. Thus every fibre domain is star-shaped about zero. Negative parameters are covered by step 1.1 whenever the corresponding geodesic time lies in the maximal interval.

On an empty manifold there are no initial vectors. In dimension zero only step 1.2 occurs, and in dimension one the proof is unchanged. Zero velocity and zero scale were handled explicitly; membership is always at the interior time $1$ of an open maximal interval, including when the original time is a finite endpoint candidate. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1]--[F2], and no additional selection is made.