Outgoing Kerr null curves and the selected future null infinity

Substitute $dt=du+(r^2+a^2)dr\mathrm{/}\mathrm{\Delta }$ and $d\varphi =d\phi +adr\mathrm{/}\mathrm{\Delta }$ into the exterior tensor. The resulting formula is the ingoing expression with $v,\psi$ replaced by $u,\phi$ and with both terms linear in $dr$ reversed in sign. In particular its $d{r}^2$ coefficient is zero. A curve at fixed $u,\theta ,\phi$ is therefore null. At large $r$ its $t$ derivative is positive and it is future null, so continuity keeps it future null along the entire exterior.

On multiplying by $R^2$ and setting $dr=-R^{-2}dR$, all coefficients extend smoothly to $R=0$: the denominators become $1+a^2{R}^2{\cos }^2\theta$. The boundary value is $2dudR-2a{\sin }^2\theta dRd\phi +{\gamma }_{S^2}$. This is nondegenerate, its restriction to $R=0$ is the round sphere metric with kernel ${\mathrm{\partial }}_u$, and its inverse has ${\bar{g}}^{-1}(dR,dR)=0$ there. Thus this is a smooth null boundary, including at the axes. The curves of step 1.1 have constant finite $u,\theta ,\phi$ and reach $R=0$; they supply a future causal path to the selected ${\mathcal{I}}^{+}$. Smooth conformal rescaling preserves causal directions in the interior, which fixes the boundary's future designation.