Global geodesic-polar coordinates from a boundary point on a simple manifold

Choose a smooth boundary defining function $\rho$ positive inside. At the boundary, $\mathrm{\nabla }\rho$ is a negative multiple of ${\nu }_{out}$; hence for a tangent unit vector $\omega$, strict convexity gives $(\rho \circ {\gamma }_{\omega }{)}^{\mathrm{\prime }\mathrm{\prime }}(0)=\mathrm{Hess}\rho (\omega ,\omega )<0$. Compactness of the tangent sphere from [F7] makes this bound uniform near tangency. For a nearby inward direction, put $a=d{\rho }_{x_0}(\omega )>0$. Taylor's formula factors the boundary equation as $\rho ({\gamma }_{\omega }(s))=s(a+sB(\omega ,s))$, with $B<0$ uniformly near tangency. The implicit function theorem gives the first positive exit $\tau (\omega )$ there, with $\tau \mathrm{/}a$ extending continuously and positively to tangent directions. Every other exit is transverse: a first exit with zero normal derivative would have $\rho =0$, ${\rho }^{\mathrm{\prime }}=0$, ${\rho }^{\mathrm{\prime }\mathrm{\prime }}<0$, forcing $\rho <0$ immediately before the exit. Thus all inward exits vary smoothly, and $\tau \to 0$ at the equator.

These exit times are uniformly bounded. Otherwise take inward unit directions ${\omega }_j$ and increasing lengths tending to infinity with no exit before those lengths. Compactness [F7] gives a convergent subsequence of directions, and continuity of $\omega \mapsto g(\omega ,{\nu }_{out})$ puts the limit in the closed inward hemisphere. Step 1.1 excludes a tangent limit. Smooth dependence then gives a limiting inward geodesic that remains in $M$ for every positive time. It enters $M^{\circ }$ immediately. If it had a first later boundary contact at $t_0$, then ${\rho }^{\mathrm{\prime }}(t_0)>0$ would force $\rho <0$ just before $t_0$, ${\rho }^{\mathrm{\prime }}(t_0)<0$ would force $\rho <0$ just after $t_0$, and ${\rho }^{\mathrm{\prime }}(t_0)=0$ together with strict convexity would force $\rho <0$ just before $t_0$. Each contradicts that the limit stays in $M$. Thus a positive-time shift is a geodesic starting in $M^{\circ }$ that never reaches $\mathrm{\partial }M$, contrary to nontrapping. Define $B_{x_0}=\{s\omega :\omega \text{ inward unit},0\le s\le \tau (\omega )\},$ including $0$ and tangent endpoints only as $0$. Identify the inward normal with the last Euclidean coordinate and compare this radial set with the closed ball of radius one centred at the inward normal, whose radial length is $2{\omega }_n$ on the inward hemisphere. The radial map multiplies by $\tau (\omega )\mathrm{/}(2{\omega }_n)$. Step 1.1 makes this ratio extend positively and continuously to the equator, so the radial map and its inverse are continuous also at zero. Hence $B_{x_0}$ is homeomorphic to a closed $n$-ball and compact by [F7].

In the complete exterior of [F2], the exponential map $E(v)={\exp }_{x_0}(v)$ is smooth on a neighborhood of $B_{x_0}$. Fix $v=s\omega \mathrm{\ne }0$ in that set and let $\alpha (t)={\exp }_{x_0}(tv)$, $0\le t\le 1$. By [F6], if $d{E}_v(w)=0$, the Jacobi field $J$ with $J(0)=0$ and $D_{t}J(0)=w$ has $J(1)=0$. Decompose $w=cv+w_{\perp }$ with $g(w_{\perp },v)=0$. The radial Jacobi field is $J_{\parallel }(t)=ct\dot{\alpha }(t)$, so $J_{\parallel }(1)$ is nonzero when $c\mathrm{\ne }0$. The Jacobi field $J_{\perp }$ from $w_{\perp }$ remains orthogonal to $\dot{\alpha }$: for $f(t)=g(J_{\perp }(t),\dot{\alpha }(t))$, the Jacobi equation gives $f^{\mathrm{\prime }\mathrm{\prime }}(t)=-g(R(J_{\perp },\dot{\alpha })\dot{\alpha },\dot{\alpha })=0$, since metric compatibility makes $R(J_{\perp },\dot{\alpha })$ skew-adjoint in its last two slots; also $f(0)=f^{\mathrm{\prime }}(0)=0$. Thus $J_{\perp }(1)$ is orthogonal to the radial vector. If $J(1)=0$, orthogonality forces $c=0$ and $J_{\perp }(1)=0$. For $w\mathrm{\ne }0$ this is a nonzero Jacobi field with endpoint zeros; reparametrizing $\alpha$ by arclength gives a unit-speed geodesic segment in $M$, contradicting [F1]. Hence $d{E}_v$ has trivial kernel. At $v=0$, [F6] on the constant geodesic gives $d{E}_0=I$. On the radial boundary $s=\tau (\omega )>0$, transversality from step 1.1 identifies $B_{x_0}$ locally with the inverse image of $M$ under this inverse chart, so the restricted map is a local homeomorphism of manifolds with boundary. The same is true at $0$: in a small normal neighborhood the inverse image of $M$ is precisely the short inward radial segments; tangent and outward rays immediately leave by strict convexity, and inward directions with very small exit time have already left. The exterior no-return property excludes extra portions of a ray in this local description. Thus $E:B_{x_0}\to M$ is a local homeomorphism everywhere.

Apply [F3] to this compact radial ball [F7]. It is a finite covering of $M$. The ball model contracts to its centre by straight-line homotopy, so $B_{x_0}$ is simply connected and [F3] makes the covering regular. If a fibre had two points, transitivity would supply a nonidentity deck transformation of $B_{x_0}$. Conjugate it by the ball homeomorphism of step 2.1. By [F4] it has a fixed point, and by uniqueness of lifts in [F3] a deck transformation with a fixed point is the identity, a contradiction. Every fibre therefore has one point. Interior inverse charts show that $E$ restricted to ${\mathcal{D}}_{x_0}$ is a smooth diffeomorphism onto $M^{\circ }$. This proves global injectivity without a path-space Morse theorem.

A minimizing exterior geodesic from $x_0$ to $y\in M^{\circ }$ exists by [F5]. It cannot leave $M$ and return, by [F2]; it must start inward, since an outward or tangent launch never enters $M$. It is therefore one of the radial segments just shown unique. Its length equals exterior distance and, being contained in $M$, equals intrinsic distance as well. To compute the metric in polar coordinates, vary the unit initial direction. The resulting Jacobi field $J$ has $J(0)=0$ and $g(D_{s}J(0),\dot{\gamma }(0))=0$. Metric compatibility and the Jacobi equation give $(g(D_{s}J,\dot{\gamma }){)}^{\mathrm{\prime }}=0$, and hence $(g(J,\dot{\gamma }){)}^{\mathrm{\prime }}=0$ with zero initial value. Thus the radial vector has unit length and is orthogonal to every angular vector. Consequently $r_{x_0}=s$ is smooth in the interior and $\mathrm{\mid }\mathrm{\nabla }{r}_{x_0}{\mathrm{\mid }}_g=1$. Completeness and no return are exactly [F2].