Past access to the far tail by exact time reversal

Reverse the spacetime time orientation and replace $T$ by $T^{\mathrm{\prime }}=-T$ and its stationary coordinate $t$ by $t^{\mathrm{\prime }}=-t$. The future unit normal is $n^{\mathrm{\prime }}=-n$. Hence $T^{\mathrm{\prime }}=N{n}^{\mathrm{\prime }}+X^{\mathrm{\prime }}$ with $X^{\mathrm{\prime }}=-X$ and unchanged positive $N$. The induced metric is unchanged; $k^{\mathrm{\prime }}=-k$ is still trace-free. Boundary tangency and smooth extension are unchanged, as are the complete stationary flow and product topology. The end components satisfy $g_{00}^{\mathrm{\prime }}=g_{00}$, $g_{0i}^{\mathrm{\prime }}=-g_{0i}$, $g_{ij}^{\mathrm{\prime }}=g_{ij}$, so all end bounds and future timelikeness of $T^{\mathrm{\prime }}$ in the reversed orientation hold. The set $U_f$ is the same subset because its time factor is all of $\mathbb{R}$. Thus every hypothesis of [F1] is preserved, not merely the maximal-slice condition.

Apply [F1] in this reversed setting. It gives $E\subset I_{E,\mathrm{r}\mathrm{e}\mathrm{v}}^{-}(U_f)$. Reversing a future timelike curve in the reversed orientation gives a future timelike curve in the original orientation from $U_f$ to its original starting point. Therefore this is precisely $E\subset I_E^{+}(U_f)$. Combining it with the original [F1] gives both chronological accesses. No assertion about a deformed metric is used.