Stationary end access and future orientation need no maximal slice

The open connected product end $U$ lies in $E$ by the selected-exterior supplier in [F1]. For each $p\in E$, the definition of $E$ gives $a,b\in U$ and future timelike curves $a{\ll }_{M}p{\ll }_{M}b$; causal convexity from the same supplier keeps both curves in $E$. Explicitly, an interior point $z$ of the first curve satisfies $a{\ll }_{M}z{\le }_{M}p{\ll }_{M}b$ by restriction, and the push-up input in [F1] gives $z{\ll }_{M}b$; on the second curve it applies to $a{\ll }_{M}p{\le }_{M}z{\ll }_{M}b$ to give $a{\ll }_{M}z$. The endpoints are already in $E$. Hence $a{\ll }_{E}p{\ll }_{E}b$. The chart in [F5] has the full real time coordinate and $T={\mathrm{\partial }}_t$, so stationary translations send $(t,x)$ to $(t+s,x)$ and preserve $U$. This proves its invariance and the stated two-sided access to the full selected end.

Completeness of the stationary orbits in $E$ and the flow laws in [F3] give smooth maps ${\mathrm{\Phi }}_s^T:E\to E$ for all real $s$, with inverses ${\mathrm{\Phi }}_{-s}^T$. They are isometries by [F3]. For a fixed future timelike $v\in T_{q}E$, the vector $(d{\mathrm{\Phi }}_s^T{)}_{q}v$ varies continuously with $s$ and remains timelike. Relative to the continuous time orientation at ${\mathrm{\Phi }}_s^{T}q$, its future or past component cannot change along the real parameter interval; at $s=0$ it is future. Thus each ${\mathrm{\Phi }}_s^T$ preserves time orientation. By the smooth-composition formula F2, it carries every locally absolutely continuous future timelike curve to another such curve, with almost-everywhere velocity $(d{\mathrm{\Phi }}_s^T){\gamma }^{\mathrm{\prime }}$. In particular, it preserves chronological comparisons inside $E$. Differentiating the flow group law in its orbit parameter also gives $(d{\mathrm{\Phi }}_s^T{)}_q{T}_q=T_{{\mathrm{\Phi }}_s^{T}q}$.

Fix $(t,x)\in U$ with $t\mathrm{\ne }0$. Choose $R>\mathrm{\mid }x\mathrm{\mid }$ sufficiently large that $x_{\ast }=Rx\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ lies in the future timelike tail. The continuous radial path $u\mapsto (t,((1-u)+uR\mathrm{/}\mathrm{\mid }x\mathrm{\mid })x)$, $0\le u\le 1$, stays in $U\subset E$ and avoids $C$ by [F5]. Its image is connected by [F7]. The disjoint open sides in [F6] cover this image, so it lies entirely in one side. At its far endpoint, the vertical orbit segment from $(0,x_{\ast })$ to $(t,x_{\ast })$ is future timelike if $t>0$, and the segment from $(t,x_{\ast })$ to $(0,x_{\ast })$ is future timelike if $t<0$, by [F5] and F2. Since $(0,x_{\ast })\in C$, this identifies the side throughout the radial path. Consequently every end point with $t>0$ belongs to $I_E^{+}(C)$ and every end point with $t<0$ belongs to $I_E^{-}(C)$. This uses no chronological connection between the radial path's endpoints.

Suppose $T$ were past timelike at $p\in E$. For some $ϵ>0$ the negative-flow local orbit segment is smooth future timelike, so $p{\ll }_E{\mathrm{\Phi }}_{-ϵ}^{T}p$ by F2. Choose $a,b\in U$ with $a{\ll }_{E}p{\ll }_{E}b$ by step 1.1. Choose $s$ so that $t(a)+s>0$ and set $A={\mathrm{\Phi }}_s^{T}a$, $P={\mathrm{\Phi }}_s^{T}p$, $B={\mathrm{\Phi }}_s^{T}b$. Step 1.2 gives $A{\ll }_{E}P{\ll }_{E}B$ and $P{\ll }_E{\mathrm{\Phi }}_{-ϵ}^{T}P$. Translating the latter comparison and concatenating finitely many curves using F2–(e) gives $P{\ll }_E{\mathrm{\Phi }}_{-nϵ}^{T}P$ for every positive integer $n$. Translating $P{\ll }_{E}B$ and concatenating yields $A{\ll }_E{\mathrm{\Phi }}_{-nϵ}^{T}B$.

Choose a positive integer $n$ with $nϵ>t(b)+s$. Then $A$ is an end point of positive time and ${\mathrm{\Phi }}_{-nϵ}^{T}B$ is an end point of negative time. Step 2.1 gives $c_{-},c_{+}\in C$ with $c_{-}{\ll }_{E}A$ and ${\mathrm{\Phi }}_{-nϵ}^{T}B{\ll }_E{c}_{+}$. Concatenating these comparisons with step 2.2 gives $c_{-}{\ll }_E{c}_{+}$. If $c_{-}\mathrm{\ne }{c}_{+}$ this contradicts acausality of $C$ in [F6]; if they coincide it is a closed future timelike curve, also excluded by [F6]. Thus $T$ cannot be past timelike anywhere in $E$, as claimed.