 Solution

Put $f(r)=1-2M/r$. An incoming radial null ray has $dt/dr=-1/f$. The tortoise coordinate $r_{\ast }=r+2M\log |r/(2M)-1|$ obeys $d{r}_{\ast }/dr=1/f$, so $v=t+r_{\ast }$ is constant along such a ray. For $r>2M$ this is exactly the printed logarithm. Substituting $dt=dv-dr/f$ gives the metric in ingoing Eddington-Finkelstein coordinatesThe radial null directions satisfy either $dv=0$ or $dr/dv=f/2$. Future-directed ingoing rays have decreasing $r$ on the first family. Outgoing rays increase $r$ outside the horizon, stay at $r=2M$ on its generators, and decrease $r$ inside. Thus inside the horizon every future timelike trajectory also moves toward smaller $r$.

The $(v,r)$ metric block has determinant minus one and is nonsingular at $r=2M$. The horizon is a null surface, not a curvature singularity. For example the Kretschmann scalar $48M^2/r^6$ is finite there and diverges at $r=0$.

Inside, set $A=2M/r-1>0$. For radial timelike motion, the Schwarzschild expression becomes $d{\tau }^2=A^{-1}d{r}^2-Ad{t}^2$, henceAngular motion would only reduce the available proper time further. Integrating from the horizon to zero, with $r=2M{\sin }^2\theta$, givesThe bound is sharp for the zero-energy radial geodesic with constant Schwarzschild $t$ in the interior. For a geodesic arriving from the exterior with strictly positive conserved energy, the proper time is strictly smaller; $\pi M$ is the supremum as that energy tends to zero. Thus the universally available horizon-to-singularity upper bound is $\pi M$, rather than a time attainable by every exterior infaller. Restoring units gives $\pi GM/c^3$.