of components in a general fiber of the map $\mu$.

We will show there exist [*two*]{} sequences of simple roots $(\alpha_1,\ldots,\alpha_k)$, $(\beta_1,\ldots,\beta_j)$ and a natural commutative diagram $$\begin{array}{ccccc}
    BS_{\vec\alpha} Z &\stackrel{BS_{\vec\alpha} \iota} {\longrightarrow}&
    G\cdot Z   &\stackrel{\mu}{\longleftarrow}& G\times^B Z \\
    &&\uparrow&&\uparrow \\
    && BS_{\vec\beta} (G\cdot Z) &{\longleftarrow}& BS_{\vec\beta} BS_{\vec\alpha} Z
  \end{array}$$ in which all maps are onto, the map $BS_{\vec\beta} BS_{\vec\alpha} Z \to G\times^B Z$ is generically $1$:$1$, and the map $BS_{\vec\alpha}\iota$ is generically finite to one. From this diagram we will derive the conclusions of the proposition.

Let $Z_0 = Z$. Since the subgroups $\{P_\alpha\}$ generate $G$, for each $i$ either $Z_i$ is $G$-invariant or we may pick a simple root $\alpha_i$ such that $Z_i$ is not $P_{\alpha_i}$-invariant. Define $Z_i := P_{\alpha_i} \cdot Z_{i-1}$.

Since $Z_i$ is the image of the proper map $P_{\alpha_i}\times^B Z_{i-1} {\mathop{\twoheadrightarrow}}Z_i$, and by inductive assumption $Z_{i-1}$ is closed, reduced, and irreducible, we find $Z_i$ is too. Since $Z_{i-1}$ was not $P_{\alpha_i}$-invariant, $Z_i \supset Z_{i-1}$ and $\dim Z_i = \dim Z_{i-1} + 1 = \dim Z + i$. Obviously $G\cdot Z_i = G\cdot Z$, so $Z_i \subseteq G\cdot Z$ is only $G$-invariant if $Z_i = G\cdot Z$. Hence this process stops when $\dim G\cdot Z = 