An existing stationary graph supplies global projection before any estimate

The inverse of $G_{\ast }$ is $(t,x)\mapsto (t-u_{\ast }(x),j(x))$. Both maps are smooth in the graph smooth structure, and composing them in either order gives the identity. Likewise $j^{-1}$ on its image is the smooth spatial projection. Thus $G_{\ast }$ and $j:C\to {\mathrm{\Sigma }}_{\ast }^{\circ }$ are diffeomorphisms, and ${\mathrm{\Pi }}_{\ast }$ is smooth everywhere. Every stationary orbit meets the existing graph exactly once. This proves global existence of projection before any collar estimate.

Write $c_i(r,y)$ for the two radial graph collars, where $i=0,\ast$. They are smooth down to zero with transverse inward radial derivative by [F3]. At $p=c_i(0,y)$ let $n_i$ be the future unit section normal and let ${\nu }_i$ be the unit normal to $S$ tangent to the section and pointing into its collar. This ${\nu }_i$ is constructed by raising the nonzero collar covector $dr$ with the positive induced metric and normalizing; hence it is smooth. Decompose $V_i={\mathrm{\partial }}_r{c}_i(0,y)=W_i+b_i{\nu }_i$ with $W_i\in T_{p}S$. Since $dr(V_i)=1$ and $dr({\nu }_i)>0$, $b_i=1\mathrm{/}dr({\nu }_i)>0$. The signed boost hypothesis gives $d{K}_p(V_i)=\kappa b_i{n}_i$. In finite tube charts, $K{\mathrm{\mid }}_S=0$ and Newton–Leibniz give $r^{-1}K(c_i(r,y))={\int }_0^1{\mathrm{\partial }}_r(K\circ c_i)(sr,y)ds$. Continuity on compact collar domains makes this converge uniformly, in chart components, to $\kappa b_i{n}_i$ as $r\downarrow 0$: for a prescribed component error, finitely many neighborhoods of the compact zero-radius slice bound the integrand error uniformly, and [F6] bounds its integral by the same error. The positive continuous $b_i$ have positive minima on compact $S$. Consequently $g_i(K,K)\mathrm{/}{r}^2$ and $g_i(K,n_i)\mathrm{/}r$ have uniformly negative limits $-{\kappa }^2{b}_i^2$ and $-\kappa b_i$ respectively. After shrinking both collars, $K$ is future timelike there. This verifies, rather than assumes, the extra sign condition needed for the crossing supplier.

Fix either section, pull its induced metric back by $c_i$, and call the resulting smooth positive metric $h$. Choose $\epsilon >0$ with the closed radial strip $S\times [0,\epsilon ]$ inside this smooth collar. It is compact by [F5], so [F4] bounds the continuous functions $\sqrt{h({\mathrm{\partial }}_r,{\mathrm{\partial }}_r)}$ and $\sqrt{h^{-1}(dr,dr)}$ above by finite positive constants $M,L$. Here $h^{-1}$ is the inverse positive matrix in collar coordinates; the contractions are coordinate independent. The fixed-$y$ radial path from zero to $r$ has length at most $Mr$ by [F6]. For any tangent vector $v$, put $w=h^{-1}dr$, so $dr(v)=h(w,v)$; Cauchy–Schwarz gives $\mathrm{\mid }dr(v)\mathrm{\mid }\le L\sqrt{h(v,v)}$. Newton–Leibniz on each smooth piece of a curve in this strip, followed by the pointwise integral comparison and addition over the pieces, therefore gives $\mathrm{\mid }r(\gamma (b))-r(\gamma (a))\mathrm{\mid }\le L{\mathrm{length}}_h(\gamma )$ for every piecewise $C^1$ subcurve there.

By the local field formula in [F2], a formal stationary orbit is $(r,\tau +t,{\psi }_{-\mathrm{\Omega }t}y)$, so $r$ is fixed. The positive stationary lapses in the hypotheses and step 1.2 verify all sign hypotheses of the bounded-rapidity supplier for both graphs. Each such orbit therefore meets each graph exactly once for every small $r>0$ and every orbit label. The original graph supplies the complete saturation in [F2], since $T$ is complete by [F1] and each section metric is stationary with the local Killing fields as stipulated. Shrink the radius once more so the bounded rapidities of both graphs, and segments between them, lie in the actual tube. The saturation agrees there with the actual tube and everywhere with actual stationary flow. Thus its formal crossings are actual crossings. Global uniqueness from step 1.1 identifies the crossing of the existing section with ${\mathrm{\Pi }}_{\ast }$. Applying the crossings in both directions proves that the entire two small collars correspond, with the same $r$, not merely that a subset of one collar enters the other. The correspondence and its inverse are smooth as restrictions of the global stationary projections.

Let $p=c_i(r,y)$ with $0<r<\epsilon \mathrm{/}2$ and consider any piecewise $C^1$ curve in the whole section joining $p$ to any point of $S$. If it reaches $S$ before leaving the strip, step 1.3 bounds its length below by $r\mathrm{/}L$. If it leaves first, its initial segment must reach radius $\epsilon$: the collar is an embedding, and the compact closed strip has no other exit except its zero and outer faces. The length of that segment is at least $(\epsilon -r)\mathrm{/}L\ge r\mathrm{/}L$. Reaching the zero face is already reaching $S$. These cases cover curves that later leave and return, and all possible boundary endpoints. Taking the infimum over all such curves and using the radial competitor from step 1.3 proves $r\mathrm{/}L\le {\rho }_i(p)\le Mr$. This is the intrinsic distance on the whole section, not a distance restricted to collar paths. Applying this argument separately to both sections, and using the radius preservation in step 2.1, proves all the distance comparisons.

For $\eta >0$ and $R>1+\eta$, the base annulus $A_{\eta ,R}=\{1+\eta \le \mathrm{\mid }x\mathrm{\mid }\le R\}$ is compact by [F5], and its continuous graph image $j(A_{\eta ,R})$ is compact by [F4]. By step 2.1 a full small original collar $Q_0$ maps under $j={\mathrm{\Pi }}_{\ast }{\mathrm{\mid }}_C$ onto the full existing collar $Q_{\ast }$. Since $Q_0$ is a neighborhood of the compact boundary sphere in the usual compactification of $C$, compactness of that sphere gives an $\eta >0$ with $\{1<\mathrm{\mid }x\mathrm{\mid }<1+\eta \}\subset Q_0$. Explicitly, pull this neighborhood back by $(y,s)\mapsto (1+s)y$; open product neighborhoods of each $(y,0)$ have a finite subcover on the zero slice, and the minimum of their positive widths works. Remove $Q_{\ast }$ and the open stationary graph end $j(\{\mathrm{\mid }x\mathrm{\mid }>R\})$ from ${\mathrm{\Sigma }}_{\ast }^{\circ }$. The remainder is closed, is contained in $j(A_{\eta ,R})$, and is therefore compact by [F5]. Letting the collars shrink and the end thresholds grow gives the asserted annular truncations: compact escape can occur only toward the boundary collar or the specified stationary graph end.