A complete stationary flow preserves its compact tangent boundary

If $S$ is empty the conclusion holds immediately. Otherwise tangency and [F4] give a smooth vector field $Y=X{\mathrm{\mid }}_S$ on the boundaryless manifold $S$. By [F5], the open subsets of $S$ admitting a solution map on some interval $(-h,h)$ form an open cover of $S$. Use intrinsic compactness in [F3] to take a finite subcover $U_1,\dots ,U_N$, with corresponding maps ${\mathrm{\Psi }}_j:(-h_j,h_j)\times U_j\to S$ and $h_j>0$, and put $h={\min }_j{h}_j>0$. Every point of $S$ therefore has a solution for $Y$ on $(-h,h)$.

Fix $p\in S$ and let $\gamma :(a,b)\to S$ be its maximal integral curve for $Y$, with $\gamma (0)=p$, supplied by [F6]. If $b<\mathrm{\infty }$, choose $t_0\in (a,b)$ with $b-t_0<h$ and a $U_j$ containing $\gamma (t_0)$. The curve $t\mapsto {\mathrm{\Psi }}_j(t-t_0,\gamma (t_0))$ is defined on $(t_0-h_j,t_0+h_j)$ and agrees with $\gamma$ on the overlap by uniqueness in [F5], applied at their common time $t_0$. Gluing extends $\gamma$ beyond $b$, contradicting maximality. If $a>-\mathrm{\infty }$, choose $t_0$ with $t_0-a<h$ and use the same map for negative time to extend past $a$. Thus $a=-\mathrm{\infty }$ and $b=\mathrm{\infty }$.

Viewed in $M$, each such restricted curve solves the ambient equation: in a slice chart its tangential derivatives are the components of $Y$, while its normal coordinates are zero, and tangency makes these derivatives exactly the components of $X$. It is therefore an ambient integral curve through $p$ by [F1]. Ambient uniqueness in [F5] and [F6] identifies it with $t\mapsto {\varphi }_t(p)$ for every real $t$. Hence ${\varphi }_t(S)\subseteq S$. Applying the same inclusion to $-t$ and using the inverse identity in [F1] gives $S\subseteq {\varphi }_t(S)$, and consequently ${\varphi }_t(S)=S$.