 Unipotent quotient in a Deligne-Lusztig variety

Choose rational $B=TU$, a representative $\dot{w}$, and $x$ with $x^{-1}F(x)=\dot{w}$. Set $T^{\prime }=xT{x}^{-1}$ and $U^{\prime }=F(xU{x}^{-1})$. Then $T^{\prime }$ is rational, andThe map sends $h$ to $hxB$. Quotienting only by the finite group $T^{\prime F}$ leaves fibres isomorphic to the positive-dimensional unipotent intersection when that intersection is nontrivial. For $w=1$ in ${\mathrm{SL}}_2$, this is an affine-line fibre over each rational flag, so the missing quotient cannot be ignored in a variety isomorphism.