An infinite cap-center trajectory has recurrent common plaque-interior patches

Compactness of the ambient manifold and countable choice give an accumulation point $z$ of the sequence $q(s)$ as $s\to \mathrm{\infty }$. Choose a small box around $z$ in which the transverse derivative of the fixed negative flow has a uniform nonzero bound. If $q(s_n)\to z$, adjust each $s_n$ by a time tending to $0$ so that the adjusted point lies on the central plaque through $z$: solve the strictly monotone transverse-coordinate equation by the intermediate value theorem [F2]. The infinite trajectory permits both small time directions once $s_n$ is large, and the adjusted points all lie in the same leaf $B$ by construction.

The accumulation point $z$ cannot lie on $\gamma$. Otherwise choose the box inside the intrinsic neighbourhood $S^{\mathrm{\prime }}$ of the original leaf provided by the avoidance hypothesis; the same small-time adjustment places $q(s_n)$ on its central original-leaf plaque inside $S^{\mathrm{\prime }}$, contradicting that $q(s_n)$ is in the interior of a cap. Hence $z\mathrm{\notin }\gamma (S^1)$, and a fixed smaller box about $z$ has closure disjoint from $\gamma$.

Every late boundary ${\gamma }_{t(s_n)}$ avoids that smaller box by uniform convergence to $\gamma$, and each adjusted cap-centre is interior to its Jordan lifted disk by [F1]. In the lifted central plaque rectangle containing that centre, membership in the disk interior cannot change along a path without crossing the disk boundary; since the entire projected rectangle is boundary-free and connected, it lies inside the lifted disk. Shrinking to a fixed smaller rectangle around $z$ and taking $n$ sufficiently large gives one common intrinsic plaque patch in $B$ whose lifts lie in the interiors of all sufficiently late caps. The argument concerns actual central-plaque hits, rather than replacing ambient convergence by an assertion that the convergent points already share a leaf, and it uses one box, one rectangle and the cited flow and root facts, hence only the standing countable choice from [F3].