Local Cauchy neighborhoods in a conically bounded product chart

Since $\ker ds$ is spacelike, its one-dimensional Lorentz-orthogonal complement is timelike, and $\mathrm{\nabla }s$ lies in that complement. Since $ds({\mathrm{\partial }}_s)=1$ and ${\mathrm{\partial }}_s$ is future timelike, $\mathrm{\nabla }s$ is past timelike. Thus [L1] gives strict increase of $s$ along every future causal curve. On any compact parameter interval $[t_1,t_2]$, (C2) and the absolutely continuous fundamental theorem give $\mathrm{\mid }\sigma (\gamma (t_2))-\sigma (\gamma (t_1))\mathrm{\mid }\le {\int }_{t_1}^{t_2}\mathrm{\mid }(\sigma \circ \gamma {)}^{\mathrm{\prime }}\mathrm{\mid }\le K{\int }_{t_1}^{t_2}(s\circ \gamma {)}^{\mathrm{\prime }}=K(s(\gamma (t_2))-s(\gamma (t_1)))$. Hence the curve image is the graph $(s,z(s))$ of a $K$-Lipschitz function on its time interval, without requiring a smooth inverse parameter map.

For the generic construction, the global smooth future timelike field $N$ is transverse to $\mathrm{\Sigma }$, since no tangent vector to a spacelike hypersurface is timelike. By [L4] its flow $F(s,x)={\mathrm{\Phi }}_s^N(x)$ embeds an open zero-section neighborhood in $\mathbb{R}\times \mathrm{\Sigma }$ onto an open neighborhood of $\mathrm{\Sigma }$. At $(0,x)$ the inverse time covector annihilates $T_x\mathrm{\Sigma }$ and has value $1$ on $N_x$, so its gradient is past timelike. Shrink the flow domain to the open subset on which this remains true. Take all pairs $(V_j,a_j)$ with $V_j$ open in $\mathrm{\Sigma }$, $a_j$ a positive rational, and $(-a_j,a_j)\times V_j$ contained in that temporal domain; their spatial sets cover $\mathrm{\Sigma }$. A subordinate smooth locally finite partition $({\rho }_j)$ exists by [L4]. Define $\delta (x)=\frac{1}{2}{\sum }_j{a}_j{\rho }_j(x)$. This is smooth and positive. At each $x$ only finitely many terms are nonzero; one with maximal $a_j$ satisfies $x\in V_j$ and $\delta (x)\le a_j\mathrm{/}2$. Hence the entire strip $\mathrm{\mid }s\mathrm{\mid }<\delta (x)$ lies in the temporal flow domain. Restricting $F$ to this strip gives the diffeomorphism and temporal coordinate stated in (d), with future timelike $F_{\ast }{\mathrm{\partial }}_s=N$. Countable Choice is used for the flowout and partition suppliers; the causal calculus retains its stated Dependent Choice assumption.

The time graph is itself an admissible causal parametrization. At a differentiability point $s_0$ of this Lipschitz graph, choose fixed Euclidean coordinates and a positive-definite form $H_0$. For sufficiently small $ϵ>0$, the constant Lorentz form $g_0-ϵH_0$ has a closed convex future cone containing all future $g$-causal vectors based sufficiently near the limiting point: continuity makes $\mathrm{\mid }(g-g_0)(v,v)\mathrm{\mid }<ϵH_0(v,v)$ there, and orientation is fixed by the common temporal covector. The graph increment divided by its positive time increment is the integral of original causal velocities divided by that increment, so lies in this convex wider cone. Let the increment tend to zero and then $ϵ\downarrow 0$; the tangent $(1,z^{\mathrm{\prime }}(s_0))$ lies in the original future causal cone and is nonzero. This holds almost everywhere because the graph is Lipschitz. If it has a limit in $U$ at a finite time endpoint, adjoining that endpoint preserves the Lipschitz bound and the almost-everywhere causal tangent condition; hence it gives a causal segment from that endpoint to each interior point. The same reasoning holds in any relatively compact temporal coordinate box with a finite causal speed bound.

If $s(p)<0$, the past vertical fiber $(s^{\mathrm{\prime }},\sigma (p))$, $\inf I<s^{\mathrm{\prime }}\le s(p)$, avoids $S$ and has no limit in $U$ at its past end: a finite endpoint time is excluded from $I$, and an infinite endpoint has no finite coordinate limit. It is therefore past-inextendible in $U$, proving $p\mathrm{\notin }{D}_U^{+}(S)$. Reversing time proves $D_U^{-}(S)\subseteq \{s\le 0\}$, so (a) holds.

We record past heredity above the zero slice. If $r\in D_U^{+}(S)$ and $p{\le }_{U}r$ with $s(p)>0$, then $p\in D_U^{+}(S)$. Otherwise a past-inextendible avoiding curve through $p$, restricted to its past half ending at $p$, can be concatenated with a causal segment $p\to r$. Strict temporal increase keeps this segment at positive times, and finite concatenation preserves local absolute continuity and the causal tangent condition; its past end is unchanged. The resulting past-inextendible curve through $r$ avoids $S$, contradiction. The same argument works for any temporal collar whose zero slice is its initial hypersurface.

Around each $x\in \mathrm{\Sigma }$ choose spatial coordinates ${\sigma }_i:V_i\to B_{R_i}$ and $a_i>0$ so that the closure of the product box $(-a_i,a_i)\times B_{R_i}$ lies in a larger temporal coordinate box inside the strip. This is possible by openness after shrinking a slice chart and its time interval. On each such compact enlarged box, the causal vectors normalized by $ds(v)=1$ have uniformly bounded spatial components. Indeed, an unbounded sequence, divided by its Euclidean norms and passed to a convergent subsequence over the compact box, would limit to a nonzero future causal vector annihilated by $ds$, impossible since $\ker ds$ is spacelike. Closedness follows from continuity of the metric and orientation. Thus there is $K_i\ge 1$ with $\mathrm{\mid }d{\sigma }_i(v)\mathrm{\mid }\le K_{i}ds(v)$ throughout the local box. The corresponding intrinsic open funnels are $Q_i=F\{(s,x):x\in V_i,\mathrm{\mid }s\mathrm{\mid }<a_i,\mathrm{\mid }{\sigma }_i(x)\mathrm{\mid }+K_i\mathrm{\mid }s\mathrm{\mid }<R_i\}$. Their zero slices cover $\mathrm{\Sigma }$. These constants are local; no common width, radius or speed constant is assumed.

Suppose a past-inextendible curve through $q$ with $s(q)>0$ avoids $S$. Its time values on the past half remain positive, and their infimum is $a\in [0,s(q))$. Its Lipschitz spatial graph has a limit $z_{\ast }$ in the closed Euclidean ball as $s\downarrow a$. Since $a\in I$, a limit with $\mathrm{\mid }{z}_{\ast }\mathrm{\mid }<R$ would be a limit point in $U$, contradicting inextendibility; hence $\mathrm{\mid }{z}_{\ast }\mathrm{\mid }=R$ and $R-\mathrm{\mid }\sigma (q)\mathrm{\mid }\le \mathrm{\mid }{z}_{\ast }-\sigma (q)\mathrm{\mid }\le K(s(q)-a)\le Ks(q)$. This is incompatible with the strict defining inequality of $Q$. The negative-time case reverses time, and zero-time points already lie on $S$. Thus $Q\subseteq D_U^{+}(S)\cup D_U^{-}(S)$. It is open and contains every point $(0,\sigma )$ with $\mathrm{\mid }\sigma \mathrm{\mid }<R$, so is a neighborhood of the entire slice and satisfies $Q\subseteq W$.

If $q\in Q_i$ has positive time, a past-inextendible curve in $U_{\mathrm{\Sigma }}$ through $q$ avoiding $\mathrm{\Sigma }$ has only positive times. On the component of its past half remaining in the local box, the integrated speed estimate of step 1.1 holds with $K_i$. It cannot reach the spatial boundary at time $s\ge 0$, since that would give $R_i-\mathrm{\mid }{\sigma }_i(q)\mathrm{\mid }\le K_i(s(q)-s)\le K_{i}s(q)$, contradicting the funnel inequality. It cannot exit through a time boundary while proceeding to the past with $0<s\le s(q)<a_i$. If it never exits, its spatial graph has a limit at its nonnegative past time infimum, and its norm is at most $\mathrm{\mid }{\sigma }_i(q)\mathrm{\mid }+K_{i}s(q)<R_i$. That endpoint lies in $U_{\mathrm{\Sigma }}$, contradicting inextendibility there. Consequently $q\in D_{U_{\mathrm{\Sigma }}}^{+}(\mathrm{\Sigma })$. The negative case reverses time; zero-time points lie on $\mathrm{\Sigma }$ already. Therefore ${\bigcup }_i{Q}_i\subset D_{U_{\mathrm{\Sigma }}}(\mathrm{\Sigma })$, and $W_{\mathrm{\Sigma }}$ contains the entire hypersurface. This proves the inclusion in the intrinsic domain of the whole collar, not merely in domains computed in individual boxes.

Let $\gamma$ be a two-sided inextendible causal curve in $W$ avoiding $S$. Its times are either entirely positive or entirely negative; assume positive. By (a), every image point belongs to $D_U^{+}(S)$. If its past end has no limit in $U$, its past half is already a past-inextendible avoiding curve in $U$, contradicting this membership. Otherwise it converges to $q\in U$ with $s(q)=a\ge 0$. If $a=0$, then $q\in S\subseteq W$, contradicting inextendibility in $W$. It remains to exclude a positive-time endpoint $q$.

Suppose that endpoint has $s(q)>0$. Choose a small convex normal set $C⋐U$ about $q$ and a point $r$ on a sufficiently short tail of $\gamma$ contained in $C$. Step 2.1 supplies a causal segment from $q$ to $r$, so $v={\exp }_q^{-1}(r)$ is nonzero future causal by [L2]. Move $r$ a small positive distance along its future timelike coordinate fiber to $r_{ϵ}\in C\cap W$. If $v$ is timelike, continuity keeps ${\exp }_q^{-1}(r_{ϵ})$ future timelike. If $v$ is null, its radial tangent $P(q,r)$ is nonzero future null, and the Gauss differential gives ${\frac{d}{dϵ}{D}_q^2(r_{ϵ})\mid }_0=2g_r(P(q,r),{\mathrm{\partial }}_s)<0$. Therefore $D_q^2(r_{ϵ})<0$ for small positive $ϵ$, and future orientation persists because $v\mathrm{\ne }0$. Thus in either case $q{\ll }_C{r}_{ϵ}$. Smoothness of the two-point inverse and openness of the timelike cone give a neighborhood $V\subset C\cap \{s>0\}$ of $q$ such that $p{\ll }_C{r}_{ϵ}$ for every $p\in V$. Since $r_{ϵ}\in W$ has positive time, it belongs to $D_U^{+}(S)$ by (a), and step 2.3 gives $V\subset D_U^{+}(S)$. Hence $q\in W$, contradicting inextendibility in $W$. The negative case reverses time. Every inextendible $W$-curve therefore meets $S$, and strict temporal increase gives exactly one parameter visit. This proves (c)'s Cauchy property without asserting openness of a one-sided domain.

Negative-time vertical fibers of the variable strip avoid $\mathrm{\Sigma }$ and have no limit in $U_{\mathrm{\Sigma }}$ at their past endpoint $-\delta (x)$; the future counterparts handle positive times. Hence the intrinsic analogue of (a) holds. The heredity proof of step 2.3 applies on this temporal collar. An inextendible $W_{\mathrm{\Sigma }}$-curve avoiding $\mathrm{\Sigma }$ has one time sign; if its appropriate endpoint has no limit in $U_{\mathrm{\Sigma }}$, the domain definition contradicts its membership. If the endpoint lies in $U_{\mathrm{\Sigma }}$ at zero time, it is in $\mathrm{\Sigma }\subset W_{\mathrm{\Sigma }}$. If it lies at nonzero time, a relatively compact temporal chart around it has a finite speed bound by the normalization argument in step 2.4. Steps 2.1 and 5.1 then give the same local causal endpoint segment, radial timelike perturbation, and neighborhood forced into the appropriate one-sided domain by heredity. Thus that endpoint is in $W_{\mathrm{\Sigma }}$, again contradiction. Every inextendible $W_{\mathrm{\Sigma }}$-curve therefore meets $\mathrm{\Sigma }$ exactly once. This argument also handles curves escaping spatially at infinity: they have no collar endpoint and are already quantified over by the intrinsic domain definition.

The hypersurfaces in (c),(d) are smooth spacelike embedded hypersurfaces and have the proved Cauchy intersection property. Applying [L5] to their respective induced open manifolds establishes both stated global-hyperbolicity consequences. Clauses (a),(b),(c),(d) follow respectively from steps 2.2, 3.1, 4.1 and 5.1, and 1.2, 2.4, 3.2 and 6.1.