Let $G$ be a reductive group and $T$ a maximal torus of $G$. Let $B$ be a Borel subgroup of $G$ containing $T$. Let $U$ be the unipotent radical of $B$. Let $g \in G$. Suppose that $g \in B w B$ for some $w \in W$, $W$ is the Weyl group of $G$. Define a rational map $f: G \to U$ by $g \mapsto x^{-1} y$, where $g=x \overline{w} t y$, $x, y \in U$, $t \in T$. I think that the image of $f$ is open. Usually how to show that the image of a rational map is open? Thank you very much.
Edit: how to show that the image of $f|_{B^-}: B^- \to U$ given by $g =x \overline{w} t y \mapsto x^{-1} y$ is open? Let $g=x \overline{w} t y \in B^-$. Then $y$ depends on $x$. If $y$ does not depend on $x$, then it is easy to see that the image of $f|_{B^-}$ is $U$ and hence the image of $f|_{B^-}$ is open. But now $y$ depends on $x$. How to show that the image of $f|_{B^-}$ is open? Thank you very much.