The differential of exp at zero is the identity

Fix $w\in T_{p}M$. By [F1], the straight line $c(s)=sw$ lies in ${\mathcal{E}}_p$ for all sufficiently small $s$, and $c^{\mathrm{\prime }}(0)$ corresponds to $w$ under $T_{0_p}(T_{p}M)\cong T_{p}M$. The curve definition of the differential and [F2] give $d({\exp }_p{)}_{0_p}(w)={\frac{d}{ds}\mid }_0{\exp }_p(sw)={\frac{d}{ds}\mid }_0{\gamma }_{p,w}(s)=w,$ because ${\gamma }_{p,w}^{\mathrm{\prime }}(0)=w$. Hence the differential is the identity.

For $w=0_p$, both sides in step 1.1 are zero. In dimension zero the identity is the unique map on the zero vector space; dimension one is the same one-vector computation. If $M$ is empty there is no point $p$, so the statement is vacuous. Only an arbitrarily small open parameter interval around zero is used, not an endpoint of the exponential domain. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1]--[F2], and differentiating a fixed curve introduces no choice.