The exponential domain is open and the exponential map is smooth

The map $j:TM\to \mathbb{R}\times TM$, $j(v)=(1,v)$, is smooth. By [F1] and [F2], $\mathcal{E}=j^{-1}(\mathcal{G})$, so $\mathcal{E}$ is open in $TM$. Every zero vector $0_p$ lies in $\mathcal{E}$ because its maximal geodesic is the constant curve on $\mathbb{R}$.

The restriction $j{\mathrm{\mid }}_{\mathcal{E}}:\mathcal{E}\to \mathcal{G}$ is smooth, and [F1] gives $\exp =G\circ j{\mathrm{\mid }}_{\mathcal{E}}$. Hence $\exp$ is smooth. For fixed $p$, ${\mathcal{E}}_p$ is the inverse image of the open set $\mathcal{E}$ under the smooth linear inclusion $T_{p}M\hookrightarrow TM$ and is therefore open in $T_{p}M$; the restriction ${\exp }_p$ is smooth.

If $M$ is empty, then $TM$, $\mathcal{E}$, and the zero section are empty, and openness and smoothness are vacuous. In dimension zero, $TM$ is the zero section and $\mathcal{E}=TM$; in dimension one the same slice and composition arguments apply unchanged. The zero-vector case was checked in step 1.1, and time $1$ is an interior point of each relevant open interval. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is used only through [F1] and [F2] to obtain the global smooth tangent-bundle/geodesic construction; taking a preimage and restricting a map require no further choice.