= Solution
Write $f=1-r^2/a^2$ and $r_*=a\operatorname{artanh}(r/a)$. Since $dr_*/dr=f^{-1}$, $dt=du+dr/f$. Substitution gives the outgoing <Outgoing Eddington-Finkelstein coordinates>
$$
\boxed{ds^2=-f\,du^2-2\,du\,dr+r^2d\Omega^2.}
$$
The radial block has determinant $-1$, independent of $f$, and all coefficients are smooth at $r=a$. This is a <coordinate singularity> of the static chart, not a spacetime singularity. The transformation was defined inside the horizon, but the new metric itself extends smoothly across it.
At $r=0$ the spherical angular coordinates degenerate as usual. Cartesian spatial coordinates make the original static metric regular there: $f=1+O(r^2)$ and the extra spatial radial correction is $x_ix_j/[a^2(1-r^2/a^2)]$. Thus \b[the origin is not a spacetime singularity either].
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