Identifying the spherical Cauchy portion without a stationary lapse

By [F6], the choice assumptions of the suppliers are satisfied. Put $P=I_D^{-}(U)$ and $Q=I_D^{+}(U)$. By [F4] both are unions of open chronological sets, so $E=P\cap Q$ is open. To check $U\subset E$, fix $u\in U$, choose a coordinate chart $y$ with $y(u)=0$, and put $v=d{y}_u(N_u)$. The smooth arc $\beta (t)=y^{-1}(tv)$ has future timelike tangent at $0$. Continuity of the metric, tangent and orientation field makes it future timelike on a sufficiently small interval $[-\epsilon ,\epsilon ]$, whose image lies in $U$ by openness. Its two halves are admissible by [F5], and show $\beta (-\epsilon ){\ll }_{D}u{\ll }_D\beta (\epsilon )$, hence $u\in P\cap Q$. Thus $E$ is nonempty. In particular, the specified far-out section $\chi (\{\mathrm{\mid }x\mathrm{\mid }>\rho \})=e(\{\mathrm{\mid }x\mathrm{\mid }>\rho \})$ lies in $\mathrm{\Sigma }\cap U\subset C$.

For $p\in E$, choose $u,v\in U$ with $u{\ll }_{D}p{\ll }_{D}v$, and let $\gamma$ be a timelike curve from $u$ to $p$. At any interior point $z$ of $\gamma$, restriction as in [F5] gives $u{\ll }_{D}z{\ll }_{D}p$. Concatenating the latter segment with a curve from $p$ to $v$, using [F5], gives $z{\ll }_{D}v$. Thus $z\in Q\cap P$. Both endpoints also lie in $E$ by step 1.1 and the choice of $p$, so the whole connecting curve lies in $E$. After an increasing affine reparametrization as in [F5], this is a path on $[0,1]$ in the sense of [F12]. Its image is connected by the connectedness of $[0,1]$ in [F12] and the continuous-image theorem [F13]. If $E$ had a separation in the sense of [F11], connectedness of $U\subset E$ would put $U$ on one side, while any point on the other side would have this connecting path to $U$, contradicting connectedness of a continuous path image. Hence $E$ is connected. Applying [F2] now gives connectedness of $C$ and a diffeomorphism $\mathbb{R}\times C\to E$.

Since $E$ is open, $C$ is open in $\mathrm{\Sigma }$. A point of its relative boundary in $\mathrm{\Sigma }$ is therefore outside $C$; since it lies in $\mathrm{\Sigma }$, it is outside $E$, but is an ambient limit point of $C$. By [F3] it lies in $S$. No point of $S$ lies in $E$, because a boundary point of either open set $P$ or $Q$ is outside that set. Consequently $C$ is contained in $\mathrm{\Sigma }\setminus S$ and is both open and closed there. Since $\chi$ is a homeomorphism by [F8], connectedness is preserved by [F13]. In the domain of $\chi$ the complement of the unit sphere has the two connected components, in the sense of [F11], $1\mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }<1$ and $\mathrm{\mid }x\mathrm{\mid }>1$: each radial annulus is path connected by radial paths and paths on the unit sphere supplied by [F7], hence connected by [F12]. The continuous radius function separates the two annuli by the omitted value $1$, so they are complementary clopen subsets of that complement. A connected subset cannot meet both, since the two intersections would separate it; thus each annulus is a largest connected subset through any of its points, as required by [F11]. Step 1.1 puts the nonempty far-out subset $\chi (\{\mathrm{\mid }x\mathrm{\mid }>\rho \})$ in $C$, so $C$ meets the outer component in a nonempty clopen subset. The consequence in [F11] makes this intersection the whole connected outer component, so that component lies in $C$. Connectedness of $C$ from step 2.1 excludes the inner component. Thus $C=\chi (\{\mathrm{\mid }x\mathrm{\mid }>1\})$.

Equip $S^2$ with [F9]'s atlas and all products with [F10]'s product structure. On $\{\mathrm{\mid }x\mathrm{\mid }>1\}$ the polar map $P(x)=(\mathrm{\mid }x\mathrm{\mid },x\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$ is a bijection with inverse $J(r,\omega )=r\omega$, for $r>1$. To check smoothness in the product charts, put $r=\mathrm{\mid }x\mathrm{\mid }=\sqrt{x_1^2+x_2^2+x_3^2}$. In the north and south stereographic charts the coordinate expressions of $P$ are respectively $(r,(x_1,x_2)\mathrm{/}(r-x_3))$ and $(r,(x_1,x_2)\mathrm{/}(r+x_3))$. The denominators are nonzero on the respective chart preimages, and $r>1$, so both expressions are smooth there. Conversely, writing $u\in {\mathbb{R}}^2$ and $q=\mathrm{\mid }u{\mathrm{\mid }}^2$, the expressions of $J$ in the two source product charts are $r(2u\mathrm{/}(1+q),(q-1)\mathrm{/}(1+q))$ and $r(2u\mathrm{/}(1+q),(1-q)\mathrm{/}(1+q))$, which are smooth since $1+q>0$. Thus $P$ and $J$ are smooth mutually inverse maps in the sense of [F8]. Step 3.1 and the given diffeomorphism $\chi$ identify $C$ with $(1,\mathrm{\infty })\times S^2$. Taking the product with the identity on $\mathbb{R}$ and composing with the diffeomorphism in step 2.1 gives $E\cong \mathbb{R}\times (1,\mathrm{\infty })\times S^2$; these compositions and the reassociation of product factors are smooth in [F10]'s product charts.