There seem to be two formulations of the Bott–Samelson resolution flowing around. For concreteness, let $ G = \mathrm{GL}\_{n} ( \mathbb{C} ) $ with the Borel subgroup $ B \subset G $ of upper triangular matrices, and denote by $ P\_i \subset G $ the $i$th parabolic. Fix a reduced expression $ w = s\_{i\_1} \cdots s\_{i\_k} $.


Denoting by $ \times^B $ the cross product modulo the diagonal $ B $-action, the first formulation says that multiplication defines a map
$$
P\_{i\_1} \times^B \dotsb \times^B P\_{i\_k}
\xrightarrow{\textrm{mult}}
\overline{BwB}
\subset
G
$$
which is a resolution of singularities since it restricts to an isomorphism of open subsets
$$
(P\_{i\_1}-B) \times^B \dotsb \times^B (P\_{i\_k}-B)
\xrightarrow{\;\sim\;}
BwB
.
$$


On the other hand, we can consider the flag variety $ G / B $ and define a map
$$
G/B \times\_{G/P\_{i\_1}}\dotsb \times\_{G/P\_{i\_k}} G/B
\xrightarrow{\mathrm{first\times last}}
\overline{ G ( wB , B ) }
\subset
G/B\times G/B
$$
which drops all coordinates except the first and last one. This is also a resolution of singularities since it restricts to an isomorphism of open subsets
$$
\bigl\{\,
( g\_0 B , \dotsc , g\_nB ) \bigm|
g\_i B \neq g\_{i+1} B \text{ for all $i$}
\,\bigr\}
\xrightarrow{\;\sim\;}
G(wB,B).
$$


We note that the RHSs of these maps differ by $ B \backslash G / B \cong (G/B \times G/B)/G $, so the LHSs should differ the same way.
For now, let us ignore all stack-related stuff and just pretend that these are honest quotient spaces.


The varieties $ G / P\_i $ can be thought of as the varieties of partial flags $ 0 = V\_0 \subset V\_1 \subset \dotsb \subset V\_{i-1} \subset V\_{i+1} \subset\dotsb\subset V\_n = \mathbb{C}^{n} $ where $ \dim (V\_j / V\_k) = j - k $. Thus an element in the fibre product $ G/B \times\_{ G / P\_{i} } G/B $ is of the form $ (gpB,gB) $ for some $ p \in P\_i $.
Therefore, in passing from one formulation to the other, it is natural to try to write a map like
\begin{align\*}
B\backslash P\_{i\_1} \times^B \dotsb \times^B P\_{i\_k}/B
& \longrightarrow
\bigl( G/B \times\_{G/P\_{i\_1}}\dotsb \times\_{G/P\_{i\_k}} G/B \bigr)/G \\
(p\_1 , \dotsc , p\_k )
& \longmapsto
\bigl(
p\_1 \dotsm p\_k B , p\_2 \dotsm p\_k B , \dotsc, p\_k B , B
\bigr)
.
\end{align\*}
However, **how can this map be well-defined**? Any $ (p\_1 , \dotsc , p\_k ) $ is equivalent to $ (p\_1 , \dotsc , p\_i b^{-1} , b p\_{i-1} , \dotsc , p\_k ) $,
but I don’t see how $ ( p\_1 \dotsm p\_k B , p\_2 \dotsm p\_k B , \dotsc, p\_k B , B ) $ can be equivalent to $ ( p\_1 \dotsm p\_k B , p\_2 \dotsm p\_k B , \dotsc , b p\_{i-1} \dotsm p\_{k} B, \dotsc, p\_k B , B ) $. What am I doing wrong?



Two different formulations of the Bott–Samelson resolution
I was being stupid. Of course, as @LSpice also [remarks](https://mathoverflow.net/questions/378660/two-different-formulations-of-the-bott-samelson-resolution#comment960977_378660), the real map is
\begin{align\*}
B\backslash P\_{i\_1} \times^B \dotsb \times^B P\_{i\_k}/B
& \longrightarrow
\bigl( G/B \times\_{G/P\_{i\_1}}\dotsb \times\_{G/P\_{i\_k}} G/B \bigr)/G \\
(p\_1 , \dotsc , p\_k )
& \longmapsto
\bigl(
B , p\_1 B , p\_1 p\_2 B , \dotsc, p\_1 \dotsm p\_k B
\bigr)
.
\end{align\*}


