Conjugacy, the exponential differential and the local proper-time maximum before the first conjugate instant

(Invertibility of the exponential differential along the geodesic.) By [L1] and the hypothesis, for every $s\in (0,b]$ there is no nonzero Jacobi field vanishing at $0$ and $s$, so $d({\exp }_p{)}_{sT(0)}$ is injective for every $s\in (0,b]$; at $s=0$ the differential is the identity by [L3] and hence injective. This proves (a), and together with [L2] proves (c).

(Gauss identity throughout the exponential domain.) The identity in [L4] holds at each lift value even if it is outside the small injective normal domain of that supplier's Statement. Indeed fix $v\in {\mathcal{E}}_p$ and $w\in T_{p}M$. Openness and star-shapedness of ${\mathcal{E}}_p$ let us form $F(t,r):={\exp }_p(t(v+rw))$ for $0\le t\le 1$ and all sufficiently small $r$. Put $A:={\mathrm{\partial }}_{t}F$ and $J:={\mathrm{\partial }}_{r}F$. By [L3], $D_{t}A=0$ and $g(A,A)=g_p(v+rw,v+rw)$. Torsion freedom gives $D_{t}J=D_{r}A$ by the symmetric coordinate connection formula, and metric compatibility gives ${\mathrm{\partial }}_{t}g(A,J)=g(A,D_{r}A)=\frac{1}{2}{\mathrm{\partial }}_{r}g(A,A)=g_p(v+rw,w)\mathrm{.}$ Since $J(0,r)=0$, integrating in $t$ and taking $r=0$ yields $g_{{\exp }_{p}v}(d({\exp }_p{)}_v(v),d({\exp }_p{)}_v(w))=g_p(v,w)$, with no injectivity hypothesis. The metric compatibility and torsion identities used here are supplied by Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields and Metric compatibility, torsion freedom and the metric-selected connection class.

(Finite local-inverse data.) For each $s\in [0,b]$, $d({\exp }_p{)}_{sT(0)}$ is invertible, so by [L7] there is an open $D_s\subseteq T_{p}M$ containing $sT(0)$ such that ${\exp }_p{\mathrm{\mid }}_{D_s}$ is a diffeomorphism onto $V_s:={\exp }_p(D_s)$. In particular, $d({\exp }_p{)}_v$ is invertible for every $v\in D_s$. Since $\{sT(0):s\in [0,b]\}$ is compact, finitely many of these sets $D_{s_1},\dots ,D_{s_N}$ cover the segment; write ${\psi }_j:V_j\to D_j$ for the inverse of each diffeomorphism. Thus every selected $D_j$ lies in the locus where $d({\exp }_p)$ is invertible.

(Compatible local inverse branches and the lift.) Refine the fixed chart partition of (b) by a finite partition $0=t_0<\cdots <t_M=b$ so that each radial subsegment $\{sT(0):t_{k-1}\le s\le t_k\}$ is contained in one $D_{j(k)}$. Such a refinement exists by compactness and a Lebesgue number for the open cover of the parameter interval induced by the $D_j$. Write $D_k^{\ast }:=D_{j(k)}$, $V_k^{\ast }:=V_{j(k)}$, and ${\psi }_k^{\ast }:={\psi }_{j(k)}$. At a junction $t_k$, the point $z_k:=t_{k}T(0)$ lies in both $D_k^{\ast }$ and $D_{k+1}^{\ast }$, and both inverses send $\gamma (t_k)$ to $z_k$. Therefore $O_k:=V_k^{\ast }\cap V_{k+1}^{\ast }\cap ({\psi }_k^{\ast }{)}^{-1}(D_k^{\ast }\cap D_{k+1}^{\ast })\cap ({\psi }_{k+1}^{\ast }{)}^{-1}(D_k^{\ast }\cap D_{k+1}^{\ast })$ is an open neighborhood of $\gamma (t_k)$. On $O_k$ both inverse values belong to $D_k^{\ast }$ and have the same exponential image, so injectivity on $D_k^{\ast }$ makes the inverses agree. Choose the tolerance $\delta$ so small that $\alpha ([t_{k-1},t_k])\subseteq V_k^{\ast }$ for each $k$, and $\alpha (t_k)\in O_k$ at every junction. These requirements follow from $C^0$ closeness in the fixed charts: each reference subsegment is compact in its corresponding open target, each junction lies in its open $O_k$, and there are only finitely many such conditions. Define $v(s):={\psi }_k^{\ast }(\alpha (s))$ on $[t_{k-1},t_k]$. The definitions agree at junctions, giving a continuous piecewise-$C^1$ lift with ${\exp }_p(v(s))=\alpha (s)$. Injectivity of the initial and final branches gives $v(0)=0$ and $v(b)=bT(0)$. This construction allows distinct inverse branches at a repeated spacetime event; compatibility is imposed only at adjacent parameter junctions.

(Uniform timelike control of the lift.) On every refined interval the inverse branch and its derivative are continuous near the compact reference image, so the fixed-chart $C^1$ closeness implies uniform closeness of $v$ to $sT(0)$ and, on each smooth piece, of $v^{\mathrm{\prime }}$ to $T(0)$; the chain rule gives $v^{\mathrm{\prime }}=d{\psi }_k^{\ast }{}_{\alpha (s)}({\alpha }^{\mathrm{\prime }}(s))$. To control the endpoint $s=0$ uniformly, choose a convex norm ball $B$ about $T(0)$ whose closure is inside the future timelike cone of $T_{p}M$, and first shrink the tolerance so that $v^{\mathrm{\prime }}(s)\in B$ throughout the initial refined interval, including one-sided values. Since $v(0)=0$, piecewise integration gives $v(s)\mathrm{/}s=s^{-1}{\int }_0^s{v}^{\mathrm{\prime }}(t)dt\in B$ there; hence $v(s)$ is future timelike for every positive $s$ in this interval. On the remaining compact radial subsegment, bounded away from $0$, uniform $C^0$ control keeps $v(s)$ in the open future timelike cone. Thus one tolerance works for all $s\in (0,b]$. Every $v(s)$ lies in its $D_k^{\ast }$, so $d({\exp }_p{)}_{v(s)}$ is invertible by step 2.1.

(Length inequality.) Define $R(s):=\sqrt{-g_p(v(s),v(s))}>0$ on $(0,b]$. On each smooth piece, step 1.2 gives $R^{\mathrm{\prime }}=-g_p(v,v^{\mathrm{\prime }})\mathrm{/}R=-g_{\alpha }(N,{\alpha }^{\mathrm{\prime }})$, where $N:=d({\exp }_p{)}_v(v)\mathrm{/}R$ and ${\alpha }^{\mathrm{\prime }}=d({\exp }_p{)}_v(v^{\mathrm{\prime }})$. The radial tangent has constant squared norm $g_p(v,v)=-R^2$ by metric compatibility along its geodesic. It starts future timelike and remains so by continuity of its pairing with the time-orientation field; hence $N$ is a future unit timelike vector. By [L5], $R^{\mathrm{\prime }}\ge \sqrt{-g({\alpha }^{\mathrm{\prime }},{\alpha }^{\mathrm{\prime }})}$. Integrating over the pieces from $\epsilon >0$ to $b$ and letting $\epsilon \downarrow 0$, using $R\to 0$ and bounded piecewise-continuous length density, gives $L_g(\alpha )\le R(b)=\sqrt{-g_p(bT(0),bT(0))}=b$.

(Equality case.) If equality holds in step 5.1, the nonnegative function $R^{\mathrm{\prime }}-\sqrt{-g({\alpha }^{\mathrm{\prime }},{\alpha }^{\mathrm{\prime }})}$ has total limiting integral zero. Its integral over every compact subinterval of a smooth piece inside $(0,b]$ is therefore zero, since it is bounded above by that total integral. It is continuous on each such compact subinterval, so [L8] makes it vanish there. These subintervals cover the smooth interiors, and by the equality case of [L5] the vector ${\alpha }^{\mathrm{\prime }}(s)$ is a positive multiple of $N(s)$ at every interior parameter value; since ${\alpha }^{\mathrm{\prime }}=d({\exp }_p{)}_v(v^{\mathrm{\prime }})$ and $N=d({\exp }_p{)}_v(v)\mathrm{/}R$, and $d({\exp }_p{)}_{v(s)}$ is injective by step 4.1, $v^{\mathrm{\prime }}(s)$ is a positive multiple of $v(s)$: $v^{\mathrm{\prime }}(s)=\lambda (s)v(s)$ with $\lambda (s)>0$, piecewise continuous.

(Monotone reparametrization.) Under equality, step 6.1 gives $v^{\mathrm{\prime }}=\lambda v$ with $\lambda >0$ on the smooth pieces. Differentiating $R$ gives $R^{\mathrm{\prime }}=\lambda R$, so $(v\mathrm{/}R{)}^{\mathrm{\prime }}=0$ there. Continuity across the finitely many corners makes $v\mathrm{/}R$ one constant vector on $(0,b]$, and its terminal value is $v(b)\mathrm{/}R(b)=T(0)$. Thus $v(s)=R(s)T(0)$ and $\alpha (s)=\gamma (R(s))$. Put $\rho (s):=R(s)$ for $s>0$ and $\rho (0):=0$. One has $\rho (b)=b$ and ${\rho }^{\mathrm{\prime }}=\sqrt{-g({\alpha }^{\mathrm{\prime }},{\alpha }^{\mathrm{\prime }})}>0$ on every smooth piece, so $\rho$ is strictly increasing and surjects onto $[0,b]$. At $0$, $v(s)=s{\alpha }^{\mathrm{\prime }}(0)+o(s)$ and $v^{\mathrm{\prime }}(s)\to {\alpha }^{\mathrm{\prime }}(0)$ from the initial piece. Consequently $R(s)\mathrm{/}s\to \sqrt{-g_p({\alpha }^{\mathrm{\prime }}(0),{\alpha }^{\mathrm{\prime }}(0))}$ and $R^{\mathrm{\prime }}(s)=-g_p(v,v^{\mathrm{\prime }})\mathrm{/}R$ has the same limit, proving the required one-sided $C^1$ extension. Hence $\rho$ is piecewise $C^1$ on $[0,b]$ and gives the equality characterization.

(Converse and conclusion.) Conversely, if $\alpha =\gamma \circ \rho$ with $\rho$ a piecewise-$C^1$ strictly increasing surjection of $[0,b]$ onto itself, and $\alpha$ belongs to the comparison class of (b), then ${\alpha }^{\mathrm{\prime }}={\rho }^{\mathrm{\prime }}T\circ \rho$ on every piece and at its one-sided endpoints. Since $\alpha$ is future timelike and $T$ is future timelike, these tangents force ${\rho }^{\mathrm{\prime }}>0$ there. Its image interval is $[0,b]$, so [L6] gives $L_g(\alpha )=L_g(\gamma )=b$ and equality holds in the stated sense. Steps 5.1 and 7.1 give the inequality and the forward equality characterization for every $\alpha$ in the chosen neighborhood, and steps 1.1 and 2.1 give (a) and (c); the proof is complete.