Radial geodesics from one point reach every point under global exponential domain

We first prove the local sphere step used at both frontiers. Fix $x,y\in C_p$ with $D=d(x,y)>0$. Their component is positive-dimensional. Choose a coordinate basis of $T_{x}M$ and orthonormalize it by [F4]. By [F3], ${\exp }_x$ is a diffeomorphism from an open neighbourhood of $0_x$ onto an open neighbourhood of $x$. In the orthonormal coordinates, [F4] supplies $\rho >0$ whose open norm ball $B_{\rho }(0_x)$ lies in that source; after restriction, put $U={\exp }_x(B_{\rho }(0_x))$. Thus ${\exp }_x:B_{\rho }(0_x)\to U$ is a diffeomorphism and $U$ is open. By [F1], choose $a>0$ with the metric ball $B_d(x,a)\subseteq U$, and fix $0<\delta <\min \{\rho ,a,D\}\mathrm{.}$

Put ${\mathrm{\Sigma }}_{\delta }=\{w\in T_{x}M:\mathrm{\mid }w{\mathrm{\mid }}_{g_x}=\delta \}$ and $S_{\delta }(x)={\exp }_x({\mathrm{\Sigma }}_{\delta })$. Since $D>0$, the component is not zero-dimensional, so ${\mathrm{\Sigma }}_{\delta }$ is nonempty. It is closed and bounded in the orthonormal coordinates and hence compact by [F4]; the normal exponential is continuous, so [F4] makes $S_{\delta }(x)$ compact. The local distance formula in [F3], together with $B_d(x,a)\subseteq U$, gives the exact equality $S_{\delta }(x)=\{z\in C_p:d(x,z)=\delta \}\mathrm{.}$

The function $h:S_{\delta }(x)\to \mathbb{R}$, $h(z)=d(z,y)$, is continuous because the triangle inequality gives $\mathrm{\mid }h(z)-h(z^{\mathrm{\prime }})\mathrm{\mid }\le d(z,z^{\mathrm{\prime }})$. By [F4] it attains a minimum at some $z_0\in S_{\delta }(x)$. The triangle inequality and step 2.1 give $D=d(x,y)\le d(x,z_0)+d(z_0,y)=\delta +h(z_0),$ so $h(z_0)\ge D-\delta$.

Suppose $h(z_0)>D-\delta$ and put $\kappa =(h(z_0)-D+\delta )\mathrm{/}2>0$. The infimum definition of $D$ in [F1] supplies one piecewise $C^1$ curve $c:[b_0,b_1]\to C_p$ from $x$ to $y$ with $L(c)<D+\kappa$. By [F5], the continuous function $t\mapsto d(x,c(t))$, whose endpoint values are $0$ and $D$, takes the value $\delta$ at some $\tau$. Step 2.1 puts $c(\tau )$ in $S_{\delta }(x)$. By [F2], $L(c{\mathrm{\mid }}_{[b_0,\tau ]})\ge \delta ,d(c(\tau ),y)\le L(c{\mathrm{\mid }}_{[\tau ,b_1]})=L(c)-L(c{\mathrm{\mid }}_{[b_0,\tau ]})<D-\delta +\kappa <h(z_0),$ contradicting the minimality of $z_0$. Therefore $d(x,y)=\delta +d(z_0,y)\mathrm{.}$ This proves the local sphere step without choosing a sequence of approximate minimizers.

Return to $p,q$. If $q=p$, take $v=0_p$; the radial curve is constant, has length and distance zero, and is minimizing. Suppose $q\mathrm{\ne }p$ and put $\mathrm{\ell }=d(p,q)>0$. Apply steps 1.1--4.1 with $x=p$, $y=q$, and a sufficiently small $\epsilon$ in place of $\delta$. There are $z_{\epsilon }\in S_{\epsilon }(p)$ and a unit vector $u\in T_{p}M$ with $z_{\epsilon }={\exp }_p(\epsilon u),\mathrm{\ell }=\epsilon +d(z_{\epsilon },q)\mathrm{.}$

Because ${\mathcal{E}}_p=T_{p}M$, every $su$ belongs to ${\mathcal{E}}_p$. By [F6], $\gamma (s)={\exp }_p(su)$ is therefore defined for every real $s$ and is the geodesic with initial velocity $u$. Its image is connected and contains $p$, so it lies in $C_p$ and the distance $d$ used below is defined on it. Its speed is constantly one, so [F2] gives $L(\gamma {\mathrm{\mid }}_{[r,s]})=s-r$ whenever $0\le r\le s$.

Define $A=\{t\in [0,\mathrm{\ell }]:\mathrm{\ell }=t+d(\gamma (t),q)\}\mathrm{.}$ Step 5.1 says $\epsilon \in A$. If $t\in A$ and $0\le s\le t$, then [F1]--[F2] give $s+d(\gamma (s),q)\le s+d(\gamma (s),\gamma (t))+d(\gamma (t),q)\le t+d(\gamma (t),q)=\mathrm{\ell },$ whereas $\mathrm{\ell }=d(p,q)\le d(p,\gamma (s))+d(\gamma (s),q)\le s+d(\gamma (s),q)\mathrm{.}$ Thus equality holds throughout, $s\in A$, and $d(p,\gamma (s))=s$; in particular every prefix ending at a parameter in $A$ is minimizing.

By [F8], $T=\sup A$ exists and $\epsilon \le T\le \mathrm{\ell }$. We claim $T\in A$. Given $\eta >0$, the definition of supremum supplies $t\in A$ with $T-\eta <t\le T$; otherwise $T-\eta$ would be a smaller upper bound. The triangle inequality and step 6.1 give $\mathrm{\mid }d(\gamma (T),q)-d(\gamma (t),q)\mathrm{\mid }\le d(\gamma (T),\gamma (t))\le T-t<\eta \mathrm{.}$ Since $d(\gamma (t),q)=\mathrm{\ell }-t$, the difference between $d(\gamma (T),q)$ and $\mathrm{\ell }-T$ has absolute value less than $2\eta$. If that difference were nonzero, taking $\eta$ smaller than one third of its absolute value would be impossible. Hence $d(\gamma (T),q)=\mathrm{\ell }-T$ and $T\in A$.

Suppose, for contradiction, that $T<\mathrm{\ell }$, and put $x=\gamma (T)$. Carry out steps 1.1--4.1 for $x,q$, choosing $0<\delta <\min \{T,\mathrm{\ell }-T\}$ as well as smaller than the local normal and metric radii there. We obtain $z_0={\exp }_x(\delta w)$ for a unit $w\in T_{x}M$, with radial segment $\sigma (s)={\exp }_x(sw)$, $0\le s\le \delta$, and $d(x,q)=\delta +d(z_0,q)=\mathrm{\ell }-T\mathrm{.}$

Concatenate $\gamma {\mathrm{\mid }}_{[0,T]}$ with $\sigma$. By [F2], [F3], and step 6.1 its length is $T+\delta$. On the other hand, the triangle inequality and step 9.1 give $\mathrm{\ell }=d(p,q)\le d(p,z_0)+d(z_0,q)=d(p,z_0)+\mathrm{\ell }-T-\delta ,$ so $d(p,z_0)\ge T+\delta$. The concatenated curve therefore has length exactly $d(p,z_0)=T+\delta$ and is globally minimizing. Every one of its subarcs is also minimizing, since a shorter replacement would shorten the full curve by finite additivity.

The entire concatenation in step 10.1 is a piecewise smooth global minimizer, with the unit vectors ${\gamma }^{\mathrm{\prime }}(T)$ and $w$ as its nonzero one-sided velocities at its sole possible corner $x$. By [F7] these vectors are positive multiples of the same tangent vector; because both have norm one, they are equal. Initial-value uniqueness in [F6] now gives $\sigma (s)=\gamma (T+s)(0\le s\le \delta )\mathrm{.}$

Thus $z_0=\gamma (T+\delta )$, and step 9.1 becomes $\mathrm{\ell }=(T+\delta )+d(\gamma (T+\delta ),q)\mathrm{.}$ So $T+\delta \in A$, contradicting that $T$ is an upper bound of $A$. Therefore $T=\mathrm{\ell }$. Since $T\in A$, [F1] gives $d(\gamma (\mathrm{\ell }),q)=0$ and hence $\gamma (\mathrm{\ell })=q$.

Step 7.1 and $T=\mathrm{\ell }$ show $d(\gamma (s),q)=\mathrm{\ell }-s$ and $d(p,\gamma (s))=s$ for every $s\in [0,\mathrm{\ell }]$. Hence the unit-speed radial curve $\gamma {\mathrm{\mid }}_{[0,\mathrm{\ell }]}$ has length $\mathrm{\ell }=d(p,q)$ and is minimizing. Put $v=\mathrm{\ell }u$. By [F6], ${\exp }_p(v)=\gamma (\mathrm{\ell })=q$, and $t\mapsto {\exp }_p(tv)=\gamma (\mathrm{\ell }t)$ has length $\mathrm{\ell }=\mathrm{\mid }v{\mathrm{\mid }}_{g_p}$.

Every piecewise-$C^1$ curve from $p$ has connected image and therefore stays in $C_p$, so minimizing inside $C_p$ is equivalent to minimizing among such curves in $M$ with these endpoints. Empty $M$ supplies no $p$. In dimension zero each component is an open singleton, so only the constant case of step 5.1 occurs; dimension one is covered because its positive-radius tangent sphere has two points. Zero distance and zero velocity were handled in step 5.1, all normal and metric radii were chosen strictly below their open endpoints, and the parameter endpoints $0,\mathrm{\ell }$ were included in steps 8.1 and 12.1. No converse is claimed. Assumption [A1] is used exactly through [F3], [F6], and [F7] for the already-constructed normal/exponential, global-geodesic, and minimizer-regularity results. Each compact minimum, basis, radius, and near-minimizing curve is instantiated only at one of finitely many fixed stages; the sphere-crossing and supremum arguments select no sequence or arbitrary family, so no further choice is used.