Structure of corresponding boundary points: achronality of the future boundary, equality of pasts, and uniqueness

(Corresponding points have the same side.) Since the initial surface is contained in the open set $U$, a boundary point $p$ is outside it and therefore lies in $I^{+}(\mathrm{\Sigma })$ or $I^{-}(\mathrm{\Sigma })$ by [L1]. A point $x\in U\cap I^{+}(\mathrm{\Sigma })$ is timelike reachable from $\mathrm{\Sigma }$ along a curve lying in $U$ by [L3], so $\phi (x)\in I^{\mathrm{\prime }+}({\mathrm{\Sigma }}^{\mathrm{\prime }})$ by [L4]; likewise for the past, using time reversal and [L3]. If $p\in I^{+}(\mathrm{\Sigma })$, restrict all source neighborhoods in the correspondence definition to that open set. The corresponding point $p^{\mathrm{\prime }}$ cannot lie in $I^{\mathrm{\prime }-}({\mathrm{\Sigma }}^{\mathrm{\prime }})$, because this is open and would give a paired witness with inconsistent sides; and it cannot lie in ${\mathrm{\Sigma }}^{\mathrm{\prime }}\subseteq \phi (U)$ because it is a boundary point of the open image. Thus $p^{\mathrm{\prime }}\in I^{\mathrm{\prime }+}({\mathrm{\Sigma }}^{\mathrm{\prime }})$. The past case is identical after time reversal.

(Actual interior past at a future boundary.) Let $p\in \mathrm{\partial }U\cap I^{+}(\mathrm{\Sigma })$ and $x\in I^{-}(p)\cap I^{+}(\mathrm{\Sigma })$. The open set $I^{+}(x)\cap I^{+}(\mathrm{\Sigma })$ contains $p$ and therefore meets $U$ at some $y$. A future timelike curve from $\mathrm{\Sigma }$ to $x$ followed by one from $x$ to $y\in U\cap I^{+}(\mathrm{\Sigma })$ lies entirely in $U$ by [L3]; hence $x\in U$. This proves $I^{-}(p)\cap I^{+}(\mathrm{\Sigma })\subseteq U$. Short timelike geodesic segments ending at $p$, with all earlier points sufficiently close to $p$, lie in this set by [L5] and the openness of $I^{+}(\mathrm{\Sigma })$. Thus there are actual interior past points tending to $p$. Apply the same proof in $M^{\mathrm{\prime }}$ to $\phi (U)$ and reverse time on the past side.

(Achronality.) If distinct $x,y\in \mathrm{\partial }U\cap I^{+}(\mathrm{\Sigma })$ satisfied $x\ll y$, step 1.2 at $y$ would put $x\in U$, contradicting $x\in \mathrm{\partial }U$. Thus this boundary is achronal. The identical argument on $\phi (U)$ proves the second assertion of (a).

(A closed past inclusion from paired witnesses.) Let $(p,p^{\mathrm{\prime }})$ be a corresponding future pair and $q\in I^{-}(p)\cap U$. Take a short future timelike geodesic through $p^{\mathrm{\prime }}$ and points $t_n^{\mathrm{\prime }}$ on it with $p^{\mathrm{\prime }}\ll t_n^{\mathrm{\prime }}$ and $t_n^{\mathrm{\prime }}\to p^{\mathrm{\prime }}$. For every $n$, $I^{+}(q)$ is an open neighborhood of $p$, and $I^{\mathrm{\prime }-}(t_n^{\mathrm{\prime }})$ is an open neighborhood of $p^{\mathrm{\prime }}$. Correspondence supplies $r_n\in U\cap I^{+}(q)$ with $\phi (r_n)\in I^{\mathrm{\prime }-}(t_n^{\mathrm{\prime }})$. The timelike curve from $q$ to $r_n$ lies in $U$ by [L3], so $\phi (q)\ll \phi (r_n)\ll t_n^{\mathrm{\prime }}$. Closedness of the causal relation in $M^{\mathrm{\prime }}$ gives $\phi (q)\le p^{\mathrm{\prime }}$. Notice that the neighborhood of $p^{\mathrm{\prime }}$ used here is the past of the strictly later point $t_n^{\mathrm{\prime }}$.

(Strict inclusion and equality of chronological pasts.) With $q$ as above, a sufficiently late point $r$ on the interior timelike geodesic ending at $p$ from step 1.2 satisfies $q\ll r\ll p$, because $I^{+}(q)$ is an open neighborhood of $p$. It lies in $U$, so step 2.2 gives $\phi (r)\le p^{\mathrm{\prime }}$, while [L3]-[L4] give $\phi (q)\ll \phi (r)$. Push-up yields $\phi (q)\ll p^{\mathrm{\prime }}$. Hence $\phi (I^{-}(p)\cap U)\subseteq I^{\mathrm{\prime }-}(p^{\mathrm{\prime }})\cap \phi (U)$. Apply the same strict inclusion to the reversed common subdevelopment and the corresponding pair $(p^{\mathrm{\prime }},p)$ to obtain equality. For the second equality in (b), if $r\in I^{+}(q)\cap I^{-}(p)\cap U$, both endpoints $q,r$ lie in $U$ and their joining timelike curve lies there by [L3]; consequently $\phi (q)\ll \phi (r)\ll p^{\mathrm{\prime }}$. Conversely if $r^{\mathrm{\prime }}\in I^{\mathrm{\prime }+}(\phi (q))\cap I^{\mathrm{\prime }-}(p^{\mathrm{\prime }})\cap \phi (U)$, the first equality puts $r={\phi }^{-1}(r^{\mathrm{\prime }})$ in $I^{-}(p)\cap U$, and the joining timelike curve between $\phi (q),r^{\mathrm{\prime }}$ lies in $\phi (U)$ by [L3], so its inverse gives $q\ll r$. This proves both equalities.

(Limit along an interior timelike curve.) Let $\gamma :(-\epsilon ,0)\to U$ be future timelike with limit $p$. For every $s<t<0$, $\gamma (s)\ll \gamma (t)$, and closedness applied to later values tending to $p$ gives $\gamma (t)\le p$; push-up gives $\gamma (s)\ll p$. If $p$ were to the past of $\mathrm{\Sigma }$, then for sufficiently late $s$ the point $\gamma (s)$ would lie in $I^{-}(\mathrm{\Sigma })$, and a concatenated timelike curve from $\gamma (s)$ through $p$ to $\mathrm{\Sigma }$ would have both endpoints in $U$, putting $p\in U$ by [L3], a contradiction. Therefore such a curve can converge only to a future boundary point. Fix an open neighborhood $V^{\mathrm{\prime }}$ of the corresponding $p^{\mathrm{\prime }}$. By [L6] choose an open causal-convex $W^{\mathrm{\prime }}$ with $p^{\mathrm{\prime }}\in W^{\mathrm{\prime }}\subseteq V^{\mathrm{\prime }}$. The interior past-geodesic construction of step 1.2 on the primed side supplies $q^{\mathrm{\prime }}\in W^{\mathrm{\prime }}\cap I^{\mathrm{\prime }-}(p^{\mathrm{\prime }})\cap \phi (U)$, with $q^{\mathrm{\prime }}\in I^{\mathrm{\prime }+}({\mathrm{\Sigma }}^{\mathrm{\prime }})$. Step 3.1 gives $q:={\phi }^{-1}(q^{\mathrm{\prime }})\in I^{-}(p)\cap U$. For all sufficiently late $s$, $\gamma (s)\in I^{+}(q)$ by chronology openness, and also $\gamma (s)\in I^{-}(p)$ by the first argument of this step. The second equality of (b) then gives $q^{\mathrm{\prime }}\ll \phi (\gamma (s))\ll p^{\mathrm{\prime }}$. Concatenate the ambient timelike curves from $q^{\mathrm{\prime }}$ to this point and from it to $p^{\mathrm{\prime }}$; both endpoints $q^{\mathrm{\prime }},p^{\mathrm{\prime }}$ are in $W^{\mathrm{\prime }}$, so causal convexity places $\phi (\gamma (s))$ in $W^{\mathrm{\prime }}\subseteq V^{\mathrm{\prime }}$. This proves $\phi (\gamma (s))\to p^{\mathrm{\prime }}$. Reversing time proves the corresponding limit statement for past-directed curves ending at a past boundary point.

(Uniqueness.) Suppose a future boundary point $p$ corresponds to $p_1^{\mathrm{\prime }},p_2^{\mathrm{\prime }}$. By step 1.1 both lie in the future of ${\mathrm{\Sigma }}^{\mathrm{\prime }}$, and by step 3.1 their chronological pasts intersected with $\phi (U)$ agree. Step 1.2 supplies points $y_n^{\mathrm{\prime }}\in I^{\mathrm{\prime }-}(p_1^{\mathrm{\prime }})\cap I^{\mathrm{\prime }+}({\mathrm{\Sigma }}^{\mathrm{\prime }})\subseteq \phi (U)$ on a short past geodesic, tending to $p_1^{\mathrm{\prime }}$. The past equality therefore gives $y_n^{\mathrm{\prime }}\ll p_2^{\mathrm{\prime }}$ for every $n$. Closedness of the causal relation implies $p_1^{\mathrm{\prime }}\le p_2^{\mathrm{\prime }}$. Interchanging the two points gives $p_2^{\mathrm{\prime }}\le p_1^{\mathrm{\prime }}$. Causality forces equality. For a past boundary point, apply this proof with the time orientation reversed; uniqueness holds on both sides. Together with step 4.1, this proves (c).