An arc with one height extremum is boundary-parallel

Let $m$ be the unique maximum. The two components of $a\setminus \{m\}$ are strictly monotone in height, so each intermediate regular level meets $a$ in exactly two points. In a Morse chart on $a$ near $m$, the height is $h(m)-t^2$; a short level arc joining the two nearby points and the subarc through $m$ bound a small embedded cap disk. Under [F1], the two points on all lower regular levels form a smooth path of distinct ordered pairs in $S^2$.

Apply [F2] to the two-point isotopy of step 1.1, starting at the cap level. The extended sphere isotopy transports the short cap arc through simple arcs on the intervening levels. Their union is an embedded strip because distinct levels are disjoint; adjoining the cap disk gives an embedded disk with boundary $a\cup {\beta }_0$ for a simple arc ${\beta }_0\subset S_c$.

Let ${\gamma }_0$ be the initial cap arc on the reference $S^2$, and let $f_r$ be the ambient isotopy from [F2] carrying its endpoint pair to the two intersection points on level $S_r$. The arc $f_c^{-1}(\beta )$ is a simple arc with the same endpoints as ${\gamma }_0$. Bigon elimination gives an isotopy $g_t$ of $S^2$, fixed at those endpoints, with $g_1({\gamma }_0)=f_c^{-1}(\beta )$. Precompose the transported cap arc by $g_{\lambda (r)}$ before applying $f_r$, where $\lambda$ is zero near the cap level and increases smoothly to $1$ at $r=c$. Every $g_t$ fixes the reference endpoints, so the joining arcs still have the required endpoint tracks; at $r=c$ the joining arc is exactly $\beta$. The union of this family with the cap disk is therefore an embedded disk bounded by $a\cup \beta$. Reversing height proves the lower-ball statement.