fibers of $BS_{\vec\alpha} f : BS_{\vec\alpha} C\to Y$ have the same number of components as the general fibers of $f$.

Consider the diagram $$\begin{array}{cccc}
    (BS_{\vec\alpha} \cdot pt) \times C
    &\widetilde{\longrightarrow}&  BS_{\vec\alpha} C&  \\
    \downarrow  & & \downarrow& \\
    C & \stackrel{f}{{\longrightarrow}} & Y&
  \end{array}$$ The left vertical map is projection onto the second factor, and the right vertical map is $BS_{\vec\alpha} f$. If the top map is $([p_k,\ldots,p_1],c) \mapsto [p_k,\ldots,p_1,p_1^{-1}\cdots p_k^{-1} c]$, which is easily seen to be well-defined and an isomorphism, then the diagram commutes.

We can now study the right-hand map $BS_{\vec\alpha} f$ by reversing the isomorphism on the top of the diagram. The fibers of the map from the northwest corner to the southeast are just products of the fibers of $f$ with Bott-Samelson manifolds, which are connected.

\[prop:BSshrinkdim\] Let $G$ act on a scheme $Y$, and $\iota:Z{\operatorname*{\hookrightarrow}}Y$ be the inclusion of a $B$-invariant subvariety. (In fact we may as well replace $Y$ by the subvariety $G\cdot Z$.) Let $\mu : G\times^B Z \to G\cdot Z$ be the projective map $[g,z] \mapsto g\cdot z$.

Then there exists a sequence of simple roots $(\alpha_1,\ldots,\alpha_k)$, such that $BS_{\vec\alpha}\iota$ is surjective and generically finite, and its degree is the number 