Let $G$ be a reductive group, $B$ a Borel subgroup, $P$ a minimal parabolic subgroup having a Levi decomposition $P = UL$, let $\alpha$ be one of the two roots of $L$ relative to $T$, and $U_\alpha, U_{-\alpha}$ the two unipotent subgroups corresponding to these roots. Then $B = TU_\alpha U$ and $B' = T U_{-\alpha} U$ are two Borel subgroups in $P$ containing $T$. Let $w$ be the unique nontrivial element of the Weyl group of $L$ relative to $T$.
I am reading an article which says that "Bruhat's lemma" (I suppose Bruhat's decomposition) implies that $P$ is the union of the open subset $BB' = (U_\alpha T U_{- \alpha})U$ and the closed subset $Bw = wB'$.
Can someone help me prove this?
Since all Borel subgroups of a reductive group are conjugate, and there is only one Weyl group element, we get the relation $B w = w B'$.
Now Bruhat decomposition implies that $P = B \dot{\cup} B w B$. Hence $P = Pw = B w \dot{\cup} B w B w = B w \dot{\cup} B B' w w = B w \dot{\cup} B B'$.