= 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 $T'=xTx^{-1}$ and $U'=F(xUx^{-1})$. Then $T'$ is rational, and
$$
\mathcal L_F^{-1}(U')/[(U'\cap F^{-1}(U'))T'^F]\cong X(w).
$$
The map sends $h$ to $hxB$. Quotienting only by the finite group $T'^F$ 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.
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