Axial averaging and positive boost at the pole

The circle flow preserves $g$ and $T$, and by [F2] preserves $D_0$ and $W_0$; hence it preserves $g_0$. It also preserves $S$ and, being connected to the identity, preserves the future cone. The original spacelike section supplies a smooth future $g$-unit normal $n$ along $S$. Because $T$ is tangent to $S$, $A_0$ maps $TS$ to itself. Thus $A_0^{-1}n$ is $g_0$-normal to $S$ and is timelike by the pullback formula. Choose its future sign continuously using the supplied Harrison family, which joins $g_0$ to the original metric. Applying [F3] gives the smooth invariant future $g_0$-unit normal $\bar{n}$.

Differentiate the equivariance of $\bar{n}$ along the circle flow. This gives $[Z,\bar{n}]=0$ along $S$, with the bracket computed using any local extension of $\bar{n}$; it is independent of the extension because $Z$ is tangent to $S$. At the fixed point $p$, ${\mathrm{\nabla }}_Z^{(0)}\bar{n}=0$. Torsion-freeness then gives ${\mathrm{\nabla }}_{\bar{n}}^{(0)}Z=0$ there as well.

Since $K(p)=0$, the difference of the two connections applied to $K$ vanishes at $p$. By [F4] the two metrics are positive scalar multiples there, so their normal planes and future/outward orientations agree. The boost in [F1] consequently gives ${\mathrm{\nabla }}_{\bar{n}}^{(0)}K=\kappa \bar{\nu }$ at $p$. The relation $K=T+{\mathrm{\Omega }}_{H}Z$ and step 2.1 give ${\mathrm{\nabla }}_{\bar{n}}^{(0)}T={\mathrm{\nabla }}_{\bar{n}}^{(0)}K$ there.

Apply [F4] to $(\bar{n},\bar{\nu })$ and use $g=(4\mathrm{/}{d}_0)g_0$ at $p$. Step 3.1 yields the displayed chain of equalities. Its last term is strictly positive because both $\kappa$ and $d_0$ are positive. This proves the claim.