The normal to a surface of constant $r$ has squared norm
$$
g^{ab}(\partial_ar)(\partial_br)=g^{rr}=f(r),
$$
which vanishes at $r=r_+$. Thus this surface is a <null hypersurface>. The stationary <Killing vector field>
$$
K=\partial_t
$$
has $K^2=g_{tt}=-f$, so it becomes null there and generates a <Killing horizon>. Since $f$ has a simple zero, its <surface gravity> is
$$
\boxed{\kappa=\frac12f'(r_+)=\frac1{r_+}}.
$$
A horizon cross-section has topology $S^3\times S^1$, where the circle is the periodic $z$ direction. Including a complete generator, the null hypersurface has topology
$$
\boxed{\mathbb R\times S^3\times S^1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 1 / c / Solution

{wiki}

A Killing horizon is a <null hypersurface> whose null generators are orbits of a <Killing vector field> $K$. On the horizon, $K^b\nabla_bK^a=\kappa K^a$ defines its <surface gravity>.

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