The exterior is the interior domain of dependence in the glued ambient

Suppose $p\in E$ were outside $D_{M_{\ast }}^{+}(A)\cup D_{M_{\ast }}^{-}(A)$. Choose a past-inextendible causal half-curve ending at $p$ and a future-inextendible causal half-curve starting at $p$, both avoiding $A$. Concatenating them gives an inextendible causal curve $\gamma$ in $M_{\ast }$ avoiding $A$. Restrict $\gamma$ to the connected component of ${\gamma }^{-1}(E)$ containing its visit to $p$. This restriction is inextendible in $E$: at a parameter end internal to the domain of $\gamma$, continuity and openness of $E$ would extend that component if its endpoint were in $E$; at an original parameter end, a limit in $E$ would also be a limit in $M_{\ast }$, contradicting inextendibility of $\gamma$. The supplied Cauchy property in $E$ forces the restricted causal curve to meet $C_0\subset A$, a contradiction. Therefore $E\subset D_{M_{\ast }}(A)$; openness of $E$ gives the forward interior inclusion. For the separately defined timelike-curve convention, take any inextendible timelike curve through $p\in E$. Its component inside $E$ is inextendible there by the same endpoint argument, and is causal, so the Cauchy property forces it to meet $C_0$. Thus $E\subset D_{M_{\ast }}^{\mathrm{t}\mathrm{i}\mathrm{m}}(A)$, and openness gives the corresponding interior inclusion.

Every added open non-right quadrant has an inextendible timelike curve avoiding the right section. In the left wedge, the boost field is timelike with the opposite orientation from the right wedge. For either time direction add a sufficiently small outward radial component, proportional to $r$, to the appropriate signed boost field. It stays timelike and reaches the artificial boundary $r=\delta$ in finite boost time, because $dr\mathrm{/}dt=ar$ with fixed sufficiently small $a>0$; compactness of $S$ and a smaller uniform $\delta$ give a common cone margin. Hence both curve ends exit the band through its excluded left cutoff, without entering the right wedge.

Fix $p\in E$ and its original witness paths $a{\ll }_{M}p{\ll }_{M}b$ from F3. Choose $a_{-},b_{+}\in M^{\mathrm{e}\mathrm{n}\mathrm{d}}$ with short future timelike segments $a_{-}{\ll }_{M}a$ and $b{\ll }_M{b}_{+}$ entirely in the end. For this local construction, take a coordinate chart in the open end about each endpoint, choose a future timelike vector there, and hold its coordinate components constant: continuity of the smooth metric and time orientation keeps that vector future timelike on a sufficiently small chart neighborhood, and sufficiently short straight coordinate segments remain in that neighborhood. The piecewise $C^1$ clause of F4 makes them timelike curves in the adopted class. The same construction on both sides of any $x\in M^{\mathrm{e}\mathrm{n}\mathrm{d}}$ shows $M^{\mathrm{e}\mathrm{n}\mathrm{d}}\subset E$. Concatenate these segments with the two witness paths. For every point $q$ on either original witness path, including $a,p,b$, restriction and timelike concatenation give $a_{-}{\ll }_{M}q{\ll }_M{b}_{+}$. Thus $q\in I_M^{+}(M^{\mathrm{e}\mathrm{n}\mathrm{d}})\cap I_M^{-}(M^{\mathrm{e}\mathrm{n}\mathrm{d}})=E$. Both original witness paths, including their end endpoints, therefore lie entirely in $E$.

For an added future timelike quadrant, translate the point by the abstract stationary action to $(U,V)=(-r,r)$. A short past timelike segment reaches the left section $U=V=-r\mathrm{/}4$: its normal displacement has strictly timelike Minkowski limit, uniformly for small $\delta$. Continue to the excluded left cutoff as above. In the future direction the outward timelike normal ray reaches $\mathrm{\mid }UV\mathrm{\mid }={\delta }^2$ in that timelike quadrant. Translate this whole curve back; quadrants and excluded cutoff are invariant. The past timelike quadrant is handled by time reversal. The resulting curves have both endpoints on excluded non-right cutoff and avoid $C_0\cup S$, so they are inextendible within $M_{\ast }$. Their past and future halves are causal and avoid $A$, witnessing nonmembership in both one-sided causal domains. Points on the non-right axes or at $S$ cannot lie in the interior of the domain of dependence, since every neighborhood contains such added points outside it. This proves the reverse interior inclusion. These full inextendible witnesses are timelike and avoid $A$, so they also exclude every added open quadrant from $D_{M_{\ast }}^{\mathrm{t}\mathrm{i}\mathrm{m}}(A)$. Every non-right axis point, including $S$, is approached by such points. Therefore none of those axis points is interior to this timelike-curve domain either, proving $\mathrm{int}{D}_{M_{\ast }}^{\mathrm{t}\mathrm{i}\mathrm{m}}(A)=E$.

The gluing embeds $E$ with its original metric and time orientation in $M_{\ast }$. Smooth composition and the velocity chain rule in F4 show that the two witness paths remain future timelike after this embedding, since its differential preserves their metric and time orientation. Their end endpoints $a,b$ are also in $E$ by step 1.3, so $a{\ll }_{M_{\ast }}p{\ll }_{M_{\ast }}b$ gives $E\subset I_{M_{\ast }}^{-}(M^{\mathrm{e}\mathrm{n}\mathrm{d}})\cap I_{M_{\ast }}^{+}(M^{\mathrm{e}\mathrm{n}\mathrm{d}})$.

By step 1.3 the original end is contained in the embedded $E$. Use its unchanged coordinates and the compatible action in F6. For every $x$ with $\mathrm{\mid }x\mathrm{\mid }>R_0$, the orbit through $(0,x)\in {\mathrm{\Sigma }}_1$ is ${\mathrm{\Phi }}_s(0,x)=(s,x)$ for all real $s$, entirely in that end. Thus every point of $M^{\mathrm{e}\mathrm{n}\mathrm{d}}$ lies in the stationary saturation of ${\mathrm{\Sigma }}_1$, and every point in that saturation lies in $M^{\mathrm{e}\mathrm{n}\mathrm{d}}$. The CW convention in F5 uses exactly this one end of the chosen section, so $M_{\mathrm{e}\mathrm{x}\mathrm{t}}={\bigcup }_{s\in \mathbb{R}}{\mathrm{\Phi }}_s({\mathrm{\Sigma }}_1)=M^{\mathrm{e}\mathrm{n}\mathrm{d}}$ in $M_{\ast }$.

Substitute the identity from step 2.3 into the chronological inclusion of step 2.2. It gives $E\subset I_{M_{\ast }}^{-}(M_{\mathrm{e}\mathrm{x}\mathrm{t}})\cap I_{M_{\ast }}^{+}(M_{\mathrm{e}\mathrm{x}\mathrm{t}})$. Apply the exclusion equivalence in F5 with ambient $M_{\ast }$ and $U=E$: $(B_{\ast }\cup W_{\ast })\cap E=\mathrm{\varnothing }$. Since steps 1.1 and 2.1 identify both domain interiors with $E$, this is the asserted exclusion on each interior, including the exact timelike-curve domain used by CW. All chronological sets and complements in this application are taken in $M_{\ast }$; no global exterior-identification claim for $\widehat{M}$ is required.