The entire target segment stays on the chronological boundary

For $r=\gamma (s)$, causal restriction gives $p{\le }_{E}r{\le }_{E}c$. If $r\in I_E^{-}(C)$, choose $c^{\mathrm{\prime }}\in C$ with $r{\ll }_E{c}^{\mathrm{\prime }}$; push-up would imply $p{\ll }_E{c}^{\mathrm{\prime }}$, contradicting [F3]. Hence every such $r$ is outside $I_E^{-}(C)$.

Fix $r=\gamma (s)$ and use [F7] in $E$ to choose a smooth geodesic $\eta$ with $\eta (0)=r$ and $\dot{\eta }(0)=-N_r$. At $0$ one has $g(\dot{\eta },\dot{\eta })<0$ and $g(\dot{\eta },N)>0$. Smoothness of $g$, $N$ and $\eta$ makes these scalar functions continuous, so both strict inequalities persist on some $[0,\delta ]$ within its domain. Thus $\dot{\eta }$ is past timelike there. For $0<\epsilon <\delta$, put $r_{\epsilon }=\eta (\epsilon )$. The reversed segment $t\mapsto \eta (\epsilon -t)$, $0\le t\le \epsilon$, is smooth with future timelike tangent; [F8] makes it a future timelike curve in the stipulated class, so $r_{\epsilon }{\ll }_{E}r$, and continuity gives $r_{\epsilon }\to r$. Push-up and $r{\le }_{E}c$ imply $r_{\epsilon }{\ll }_{E}c$, including when $r=c$. Thus $r$ is in the closure of $I_E^{-}(C)$ and outside this open set, which places it in its boundary. This argument applies to each $s$, and [F4] supplies the particular null segment.