Smooth transverse horizon branches are the null axes of the normal tube

The two null normal lines at $s\in S$ are spanned by $n_s+{\nu }_s$ and $n_s-{\nu }_s$. The radical of each branch is transverse to $S$ by [F1]; its flow coordinates cover a branch neighborhood by its generator curves from $S$. These curves are geodesics up to reparametrization. Uniqueness with the specified initial null directions and [F2] identify the two branches, as an unordered pair locally along $S$, with the images of the axes $a=b$ and $a=-b$. Compactness gives a common shrinking of the tube.

For each $s\in S$, the collar velocity lies in $T_s\mathrm{\Sigma }$ and decomposes uniquely as ${\mathrm{\partial }}_{t}c(s,0)=w_s+\lambda (s){\nu }_s$, where $w_s\in T_{s}S$. Its normal coefficient is positive: the collar differential is an isomorphism at the boundary, so the velocity is transverse to $S$, and its positive parameter points into the specified side ${\mathrm{\Sigma }}_{+}$ in a boundary chart. Thus $\lambda (s)>0$. Write ${\mathrm{\Xi }}^{-1}(c(s,t))=(q(s,t),a(s,t),b(s,t))$. The zero-section differential in [F2] gives $a(s,0)=b(s,0)=0$, ${\mathrm{\partial }}_{t}a(s,0)=0$ and ${\mathrm{\partial }}_{t}b(s,0)=\lambda (s)$. Smoothness gives $a(s,t)=O(t^2)$ and $b(s,t)=\lambda (s)t+O(t^2)$ uniformly on compact $S$. Since the smooth positive function $\lambda$ has a positive minimum there, for all sufficiently small $t>0$ one has $b>0$, $a=O(b^2)$, $\alpha =a+b>0$ and $\beta =a-b<0$. Compactness also makes this restricted collar lie in $O$.

An open past set with smooth boundary occupies the local past side of that boundary. Indeed, if $p\in \mathrm{\partial }G$, a point $x_{-}\ll p$ lies in $G$ by the closure property in [F3]. A point $p\ll x_{+}$ cannot lie in $\bar{G}$, since the same property would then put $p$ in the open set $G$, contradicting $p\in \mathrm{\partial }G$. Choose these points on a short timelike segment crossing the smooth null boundary. Each connected local side has constant membership: off $\mathrm{\partial }G$, the disjoint open sets $G$ and $M\setminus \bar{G}$ exhaust the neighborhood. This identifies the entire past side with $G$ and the entire future side with its exterior. Time reversal gives the corresponding future-side description of $F_{+}$. At $S$ the future timelike direction $n$ increases both $a-b$ and $a+b$; by continuity this fixes the future coorientation of either candidate axis near $S$.

The inward collar is in $G\cap F_{+}$ by hypothesis and has $\alpha >0,\beta <0$ by step 1.2. If $H_{+}$ were the axis $\alpha =0$ at any point of $S$, step 2.1 would place $G$ on its $\alpha <0$ side, contradicting these collar points. Thus $H_{+}$ is the axis $\beta =0$ everywhere along $S$, and $H_{-}$ is the other axis $\alpha =0$. Their respective past and future sides give $G\cap O=\{\beta <0\}$ and $F_{+}\cap O=\{\alpha >0\}$ after a common shrinking. Their intersection is the open four-dimensional spacetime quadrant; the positive-parameter hypersurface collar lies in it with the estimates from step 1.2. This proves all the asserted equations and the collar inclusion.