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For $d{s}^2=-fd{t}^2+d{r}^2/f+R^{2}d{\Omega }^2$, suppose $f$ and $R$ are analytic at $r_H$, $f(r_H)=0$, $f^{\prime }(r_H)=2\kappa >0$ and $R(r_H)>0$. The tortoise coordinate obeys the displayed expansion with analytic $h$. For $U=-e^{-\kappa (t-r_{\ast })}$ and $V=e^{\kappa (t+r_{\ast })}$ in the right exterior, $-UV=(r-r_H)e^{2\kappa h(r)}$ after fixing an additive constant. The inverse function theorem makes $r$ analytic in $UV$ through zero, while $d{s}^2=fdUdV/({\kappa }^{2}UV)+R^{2}d{\Omega }^2$ has a finite nonzero radial coefficient there. This supplies the local analytic extension and its bifurcate Killing horizon.

Set $r_H=2m$. Since $f^{\prime }(r_H)=2\kappa$, the integrand defining the tortoise coordinate has a simple pole:where $h$ is analytic near $r_H$. Define the Kruskal–Szekeres coordinates in the right exterior by $U=-e^{-\kappa u}$ and $V=e^{\kappa v}$. ThenThe right-hand side has positive nonzero derivative at $r_H$. Continuing this relation analytically, rather than retaining an absolute value, makes $r$ an analytic function of $UV$ through both signs. The Kruskal extension across a simple static horizon has metricIts radial coefficient tends to $-2e^{-2\kappa h(r_H)}/\kappa$, finite and nonzero, while $R(r_H{)}^2=4m^2\cosh \alpha$. Therefore both null surfaces $U=0$ and $V=0$ are regular horizon branches, intersecting at a smooth two-sphere $U=V=0$.

The stationary Killing vector field becomesIt is null, tangent and normal on each horizon branch and vanishes on their intersection. This proves that $r=2m$ is a bifurcate Killing horizon, with bifurcation surface of area $\boxed{A_H=16\pi m^2\cosh \alpha }$. The exponential definitions specify one exterior; allowing all signs of $U,V$ supplies the remaining quadrants of the extension.