Stationary coefficients in a transverse collar retain the tangential motion

On $s=0$, ${\mathrm{\partial }}_s=Y=n_{\ast }$, so the coefficient $a(0,\rho ,y)$ is exactly $N_{\ast }$. Tangency of $T$ to $S$ gives $a(0,0,y)=b(0,0,y)=0$. Smoothness and compactness therefore give $\mathrm{\mid }a(s,\rho ,y)-a(0,\rho ,y)\mathrm{\mid }\le C\mathrm{\mid }s\mathrm{\mid }$ and $\mathrm{\mid }b(s,\rho ,y)\mathrm{\mid }\le C(\rho +\mathrm{\mid }s\mathrm{\mid })$ on a common coordinate collar; the tangential coefficient has a uniform norm bound $M$ in the chosen smooth Riemannian metric on $S$. Choose $\eta$ with $C\eta \le c\mathrm{/}2$. The two asserted coefficient inequalities follow, with $B=C(1+\eta )$. These estimates are uniform over the whole compact $S$, allowing chart changes as $y$ moves.

Let $\nu$ be the inward unit normal to $S$ within the section. We first prove $d{\rho }_p({\nu }_p)=1$ for every $p\in S$, without assuming a normal-geodesic extension of $y$. Choose a boundary half-chart $(r,z)$ centered at $p$, with $r\ge 0$, $S:r=0$, ${\mathrm{\partial }}_r{\mathrm{\mid }}_p={\nu }_p$ and induced metric $h_p=d{r}^2+\mathrm{\mid }dz{\mathrm{\mid }}^2$. For each $0<\delta <1$, continuity gives a smaller half-ball where $(1-\delta {)}^2(d{r}^2+\mathrm{\mid }dz{\mathrm{\mid }}^2)\le h\le (1+\delta {)}^2(d{r}^2+\mathrm{\mid }dz{\mathrm{\mid }}^2)$. The radial segment to $(r,0)$ has length at most $(1+\delta )r$. Every path from $(r,0)$ to $S$ that stays in the half-ball has length at least $(1-\delta )r$, since its total $r$-variation is at least $r$. A path leaving the half-ball first has a fixed positive lower length bound for all sufficiently small $r$, from the same metric comparison applied up to its first exit, so it cannot violate that lower bound. Taking the infimum over all paths gives $(1-\delta )r\le \rho (r,0)\le (1+\delta )r$. Taking $r\downarrow 0$ and then $\delta \downarrow 0$ proves the derivative claim. Since $\rho {\mathrm{\mid }}_S=0$ and ${\mathrm{\partial }}_{\rho }\rho =1$, it follows that ${\mathrm{\partial }}_{\rho }{\mathrm{\mid }}_S=\nu +V$ for some $V\in TS$; $V$ need not vanish.

Assume now the exterior hypotheses. Shrink the flow collar so that $Y$ is timelike. A null tangent hyperplane contains no timelike vector, so $Y$ is transverse to each null boundary. The implicit function theorem gives their smooth graphs $s=h_{\pm }(\rho ,y)$, with $h_{\pm }(0,y)=0$; local graphs agree on overlaps by uniqueness, and compactness gives a common smaller collar. At $S$, their null radical lines are orthogonal to $TS$, hence are the two lines spanned by $n_{\ast }+\nu$ and $n_{\ast }-\nu$ in the normal plane of $S$. Their tangent hyperplanes are the orthogonal complements of those lines. The graph's radial tangent vector there is $({\mathrm{\partial }}_{\rho }{h}_{\pm })n_{\ast }+\nu +V$. Pairing it with $n_{\ast }\pm \nu$ gives $-{\mathrm{\partial }}_{\rho }{h}_{\pm }\pm 1=0$, independently of $V$. After labeling by slope on each component of $S$, this proves ${\mathrm{\partial }}_{\rho }{h}_{\pm }(0,y)=\pm 1$.

Smoothness and compactness in step 2.1 give $h_{-}(\rho ,y)\le -3\rho \mathrm{/}4$ and $h_{+}(\rho ,y)\ge 3\rho \mathrm{/}4$ for all sufficiently small positive $\rho$. Shrink $\eta$ from step 1.1 to at most $1\mathrm{/}2$ and shrink ${\rho }_0$ so that the whole wedge is in the common neighborhood where these graphs are the full event boundaries. For each wedge point, the vertical segment from $(0,\rho ,y)$ to $(s,\rho ,y)$ stays strictly between the graphs and avoids both boundaries. For either $A=I_D^{-}(U)$ or $A=I_D^{+}(U)$, a connected segment avoiding ${\mathrm{\partial }}_{D}A$ has constant membership in $A$, because the two complementary regions in [F4] are disjoint open sets. Its initial point belongs to both sets by the reference-collar hypothesis, so its endpoint belongs to $\mathcal{E}$. All coefficient estimates survive the shrinking. This proves the asserted uniform exterior wedge without identifying its side from transversality alone.