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Use $\hslash =c=k_B=1$ and normalize the time-translation Killing vector field at infinity. The simple outer zero and positivity on its exterior imply $b=V^{\prime }(r_0)>0$. Near the horizon, write $x=r-r_0$; then $V=bx+O(x^2)$. The Wick rotation $t=-i{t}_E$ gives the radial Euclidean metricDefine $\rho =2\sqrt{x/b}$, so $x=b{\rho }^2/4$. Its leading near-horizon form isThus $b{t}_E/2$ is a polar angle. Smoothness at the origin requires period $2\pi$, rather than a cone or a multiple cover with a branched origin. The Euclidean black-hole regularity condition isA thermal quantum state has imaginary-time period $\beta =1/T$. Since $V\to 1$, this time coordinate is normalized to the clocks at infinity, and the Hawking temperature isA static clock at radius $r>r_0$ measures the redshifted local temperature $T/\sqrt{V(r)}$; that local quantity is not the temperature requested with the asymptotic normalization.

The static coordinates are singular on the horizon, so calculate its surface gravity in a regular ingoing chart. With $v=t+\int dr/V$, the metric and the same normalized Killing vector field areThe inverse radial block has $g^{vr}=1$, $g^{rr}=V$, $g^{vv}=0$. Direct calculation givesSince $k$ has constant coordinate components, its acceleration isOn the future horizon, $V=0$ and $k$ is null, so the defining equation gives $\kappa =b/2$. In the outgoing chart regular on the past branch, the radial cross-term is $-2dudr$ and the same calculation gives $\kappa =-b/2$ for $k={\partial }_u$. Hence the branch-independent physical statement isThis proves the Euclidean temperature and signed horizon surface gravity relation directly. Rescaling the Killing vector field would rescale its surface gravity; the condition at infinity fixes that otherwise arbitrary normalization.