Spacelike normal projection is locally invertible with positive boundary jet

The kernel of $d{\pi }^{\perp }$ is the one-dimensional timelike Gaussian fibre direction. Its intersection with the candidate's positive-definite tangent three-plane is zero, so $d{\pi }^{\perp }$ restricted to that plane is an isomorphism onto the three-dimensional reference tangent space. By the flow-graph definition, the candidate graph map $\mathrm{\Phi }$ is an embedding on the reference interior; thus $d\mathrm{\Phi }$ identifies its tangent space with the candidate tangent plane. The chain rule gives $dP=d{\pi }^{\perp }\circ d\mathrm{\Phi }$, an isomorphism. The inverse theorem proves both claimed interior local diffeomorphisms.

Write $\bar{\nu }={\mathrm{\partial }}_{\rho }c(0,y)$ for the inward collar normal and $e=-\bar{\nu }$ for the exterior normal. The infinitesimal boost rate in $(\bar{n},e)$ is $\lambda =-\kappa$. For $\kappa <0$ the positive-rate fixed-boundary supplier applies to $(K,e,\lambda )$; for $\kappa >0$ apply it to $(-K,e,\kappa )$ at time $-s$, using ${\varphi }_{-s}^{-K}={\varphi }_s^K$. For the variable-time graph the latter replacement also changes $u$ to $-u$, leaving $\mathrm{\Phi }$ unchanged and preserving its embedding and spacelikeness hypotheses. Thus both cases give $d{\varphi }_s^K(e)=\cosh (\kappa s)e-\sinh (\kappa s)\bar{n}$, and hence $d{\varphi }_s^K(\bar{\nu })=\cosh (\kappa s)\bar{\nu }+\sinh (\kappa s)\bar{n}$, with no $TS$ component. For $\kappa =0$, the zero-rate differential calculation in F2's Gaussian-cone supplier, Proof, phase three, first step, gives the identity on the normal plane, so the same formula holds. In every case the differential is the identity on $TS$. The flow fixes $S$, so $P(0,y)=(0,y)$. By the chain rule, $d\mathrm{\Phi }(X)=d{\varphi }_{u(0,y)}^K(X)+du(X)K_y$ on $S$, and the last term vanishes since $K_y=0$. The Gaussian projection kills $\bar{n}$ and keeps reference tangent vectors. Consequently $d{P}_{(0,y)}(a,Y)=(\cosh (\kappa u(0,y))a,Y)$ for $Y\in T_{y}S$. This proves the positive radial coefficient, tangential identity and absence of a tangential first-order radial component.

Fix a boundary point and use source and target collar charts with nonnegative radial coordinate on the chosen side. By F4 the coordinate representative of $P$ has a smooth extension $f=(f_0,f^{\mathrm{\prime }})$ to a Euclidean-open neighborhood across the source boundary. On its boundary patch $f_0(0,z)=0$, while step 1.2 gives ${\mathrm{\partial }}_r{f}_0(0,z_0)=\cosh (\kappa u(0,z_0))>0$ and an invertible full derivative. Shrink to a product neighborhood on which ${\mathrm{\partial }}_r{f}_0>0$. Since $f_0(0,z)=0$, strict increase along each radial line shows that $f_0(r,z)$ has the same sign as $r$. Apply F3 to $f$ and shrink its inverse neighborhoods within this product neighborhood. The sign identity shows that this Euclidean inverse restricts bijectively to the two nonnegative half-neighborhoods, taking boundary to boundary and interior to interior. Its restriction and that of $f$ are smooth in the local-extension sense. This proves a smooth half-collar inverse near each boundary point.

Fix the candidate. The union of the source inverse neighborhoods in step 2.1 is relatively open and contains $\{0\}\times S$. Product neighborhoods $[0,{\epsilon }_y)\times V_y$ contained in that union cover the boundary; a finite subcover in F6 and the minimum of its finitely many positive widths give a source collar $[0,{\epsilon }_0)\times S$. If no smaller source collar were injective, A1 selects, for each integer $j\ge 1$, distinct colliding points $(r_j,y_j)$ and $(r_j^{\mathrm{\prime }},y_j^{\mathrm{\prime }})$ with $0\le r_j,r_j^{\mathrm{\prime }}<\min ({\epsilon }_0\mathrm{/}2,1\mathrm{/}j)$. Apply F6 first to $(y_j)$ and then to $(y_j^{\mathrm{\prime }})$ along that subsequence, giving a common subsequence with limits $y$ and $y^{\mathrm{\prime }}$. Continuity and the boundary identity in step 1.2 imply $(0,y)=(0,y^{\mathrm{\prime }})$, so $y=y^{\mathrm{\prime }}$. Both preimages then eventually lie in the same source inverse neighborhood from step 2.1, contradicting its injectivity. Thus some source collar is injective. Its image is a relative open neighborhood of the entire target boundary, using step 2.1 at boundary points and step 1.1 at interior points. For each target boundary point choose a product neighborhood contained in this image; a finite subcover of $S$ and the minimum of its positive widths give a positive-width target collar contained in the image. The local smooth inverses agree there by injectivity. This proves exactly the asserted individual collar germ and coverage; every shrinking may depend on the fixed candidate.