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Write $f=1-r^2/a^2$ and $r_{\ast }=a\mathrm{artanh}(r/a)$. Since $d{r}_{\ast }/dr=f^{-1}$, $dt=du+dr/f$. Substitution gives the outgoing Outgoing Eddington-Finkelstein coordinatesThe 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 $x_i{x}_j/[a^2(1-r^2/a^2)]$. Thus the origin is not a spacetime singularity either.

Define Outgoing Eddington-Finkelstein coordinates by $u=t-r_{\ast }$. The radial Schwarzschild metric becomesOutgoing radio rays have constant $u$. Two nearby wave crests emitted at Alice's proper times $\tau$ and $\tau +d\tau$ have separation $du={\dot{u}}_{A}d\tau$. At Bob's fixed radius $r_B$, $d{t}_B=du$ and $d{\tau }_B=\sqrt{f_B}du$, where $f_B=1-2M/r_B$. Thus the exact local redshift isHere the wavelength ratio equals the proper time period ratio in geometric optics. Equivalently, the phase gradient $k_a=-{\omega }_{\infty }(du{)}_a$ gives Alice's measured frequency ${\omega }_A=-k_a{U}_A^a={\omega }_{\infty }{\dot{u}}_A$ and Bob's ${\omega }_B={\omega }_{\infty }/\sqrt{f_B}$.

In the asymptotically distant-observer limit $f_B\to 1$, this becomes ${\lambda }_B/{\lambda }_A=d{u}_A/d{\tau }_A$. “Far away” is the approximation behind the printed formula; at finite $r_B$ the factor $\sqrt{f_B}$ remains.