= Solution
Assume $M>0$ and use geometric units. The stationary <Killing vector field> $K=\partial_t$ gives the conserved <Killing energy> $E=-g(K,U)=f\dot t$. For radial motion, normalization of the <four-velocity> gives $-f\dot t^2+f^{-1}\dot r^2=-1$. Combining these identities yields
$$
\boxed{f\dot t=E,\qquad \dot r^2+f=E^2.}
$$
At rest at infinity, $f\to1$ and $\dot r\to0$, so $E=1$. Choose the inward branch $\dot r=-\sqrt{2M/r}$. Its <proper time> from the <Schwarzschild event horizon> to the <curvature singularity> is
$$
\boxed{\Delta\tau=\int_0^{2M}\sqrt{\frac r{2M}}\,dr=\frac{4M}{3}.}
$$
In ordinary units this is $4GM/(3c^3)$. The endpoint $r=0$ is not a regular centre of the <spacetime>. This is the <Schwarzschild horizon-to-singularity proper time for fall from infinity>, rather than a universal time for all infall energies.
For this <geodesic congruence>, set $w(r)=\sqrt{2M/r}$. Its <four-velocity> and metric dual are
$$
U=f^{-1}\partial_t-w\partial_r,
\qquad U_a\,dx^a=-dt-f^{-1}w\,dr.
$$
The right side is an exact one-form because $w/f$ depends only on $r$. Thus $U_a=-\partial_a\tau$, with <free-fall proper time in Schwarzschild spacetime> determined by
$$
\boxed{d\tau=dt+\frac{w}{f}\,dr,\qquad
\tau=t+2\sqrt{2Mr}+2M\log\left|\frac{\sqrt r-\sqrt{2M}}{\sqrt r+\sqrt{2M}}\right|+\text{constant}.}
$$
Indeed $U^a\partial_a\tau=-U^aU_a=1$, so this scalar advances by the actual <proper time> on every member of the infalling <geodesic congruence>, with a synchronized choice of additive offsets.
Substitution of $dt=d\tau-wf^{-1}dr$ in the <Schwarzschild metric> gives
$$
-f\,dt^2+f^{-1}dr^2=-f\,d\tau^2+2w\,d\tau\,dr+dr^2.
$$
Using $f=1-w^2$, the <Painlevé–Gullstrand coordinates> therefore have
$$
\boxed{ds^2=-d\tau^2+(dr+w\,d\tau)^2+r^2d\Sigma^2.}
$$
All coefficients are finite at $r=2M$; the determinant of the $\tau,r$ block is $-f-w^2=-1$. Hence the horizon is a <coordinate singularity> of <Schwarzschild time>, not a degeneracy of the metric. The new spatial slices have $d\ell^2=dr^2+r^2d\Sigma^2$, the flat Euclidean metric, and $g^{ab}\partial_a\tau\partial_b\tau=-1$ makes their normals the freely falling <four-velocities>. By contrast, the curvature scalar $R_{abcd}R^{abcd}=48M^2/r^6$ diverges at $r=0$; this genuine singularity is not removed.
For <ingoing Eddington-Finkelstein coordinates>, use $v=t+r_*$ with $dr_*/dr=f^{-1}$, for example $r_*=r+2M\log|r/(2M)-1|$. Then
$$
ds^2=-f\,dv^2+2\,dv\,dr+r^2d\Sigma^2.
$$
Both constructions cross the future horizon. Their time functions have different meanings: $v=\text{constant}$ is an ingoing <null hypersurface>, whereas $\tau=\text{constant}$ is a flat spacelike hypersurface orthogonal to the infallers. Thus the former is adapted to ingoing <null geodesics>, and the latter to a timelike free-fall clock congruence.
For the <Kruskal diagram>, choose <Kruskal–Szekeres coordinates> with $U_K=-e^{-(t-r_*)/(4M)}$, $V_K=e^{(t+r_*)/(4M)}$ in the right exterior and extend them across its horizon. They obey
$$
U_KV_K=\left(1-\frac r{2M}\right)e^{r/(2M)},\qquad
ds^2=-\frac{32M^3}{r}e^{-r/(2M)}dU_K\,dV_K+r^2d\Sigma^2.
$$
The four regions of <Kruskal spacetime> are I, the right exterior $(U_K<0,V_K>0)$; II, the future <black hole> $(U_K>0,V_K>0)$; III, the other exterior $(U_K>0,V_K<0)$; and IV, the past <white hole> $(U_K<0,V_K<0)$. The null axes are horizons, their intersection is the bifurcation two-sphere, and the spacelike boundaries $U_KV_K=1$ are the future and past curvature singularities. A point in this radial diagram represents a symmetry two-sphere.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-52-kruskal.png]
{title=Kruskal extension with both exteriors, black hole and white hole; the ingoing Painlevé-Gullstrand domain is shaded blue}
The <domain of the ingoing Painleve-Gullstrand chart> is \b[region I together with region II and their common future horizon]. More precisely, it is $V_K>0$, $U_KV_K<1$, or $r>0$ with finite $\tau$. To verify the boundary rather than infer it from the locally regular metric, put $\rho=\sqrt{r/(2M)}$ and write
$$
\tau=v+S(r),\qquad S(r)=4M\rho-2M\rho^2-4M\log(1+\rho).
$$
This follows by subtracting $r_*$ from the preceding radial primitive, and $S$ is finite at $r=2M$. Thus $V_K=\exp[(\tau-S(r))/(4M)]>0$ throughout the chart. Conversely every point with $V_K>0$ and $r>0$ gives a finite $\tau$. The past horizon $V_K=0$, the bifurcation sphere, region III and region IV are absent. Reversing the sign of the radial term produces the outgoing chart through the <white hole> instead.
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