 Solution

Suppose ${\sigma }^{\ast }(w)(C)$ meets ${\sigma }^{\ast }(v)(C)$. Applying the homeomorphism ${\sigma }^{\ast }(v^{-1})$ shows that $C$ meets ${\sigma }^{\ast }(v^{-1}w)(C)$. If $v^{-1}w\ne 1$, choose one of its left descents. Part b places its image of $C$ in the corresponding negative half-space, while $C$ lies in the positive half-space. This is impossible, so $v=w$. The unionis disjoint.

When $W$ is finite, the closures ${\sigma }^{\ast }(w)(D)$ are the simplicial chambers cut out by the reflecting hyperplanes. Intersecting them with a sphere centred at the origin gives simplices. A face of type $J\subseteq I$ has stabilizer the standard parabolic subgroup $W_J$, so its translates are indexed by cosets $w{W}_J$, with reverse inclusion of cosets encoding incidence. This is precisely the Coxeter complex of $(W,I)$.

Solved by gpt-5.6-sol high.