that $s(D_1)_e=+$ and $s(D_2)_e=-$. (The other case is similar.) Then $\overline{D}_1\subset\overline{H_{\alpha_e}^+}$ and $\overline{D}_2\subset\overline{H_{\alpha_e}^-}$, so $C \subset \overline{D}_1\cap\overline{D}_2
\subset \overline{H_{\alpha_e}^+} \cap \overline{H_{\alpha_e}^-}=
H_{\alpha_e}$, which implies that $e\in E(C)$.

The first statement of (iii) follows easily from the definitions: the composition $D\circ D'$ is defined on the sign vectors of $D$ and $D'$, and the isomorphism $\iota_{C}$ just forgets the coordinates outside of $E(C)$ in these sign vectors.

We prove the second statement of (iii). Let $D\in{\mathscr{L}}_{\geq C}$, and let $D'$ be the unique face of ${\mathscr{L}}_{{\mathcal{H}}(C)}$ containing $D$. We clearly have $\dim(D)\leq\dim(D')$. If $\dim(D')>\dim(D)$ then there exists $e\in E$ such that $D\subset H_{\alpha_e}$ and $D'\not\subset H_{\alpha_e}$. But $C\subset\overline{D}$, so this implies that $e\in E(C)$. As $D'$ is not included in $H_{\alpha_e}$, it must be contained in one of the open half-spaces $H^{\pm}_{\alpha_e}$, contradicting the fact that $D'$ contains $D$.

\[rmk\_star\_convex\] Let $C\in{\mathscr{L}}$ and let $F'=\{e\in E :  C\not\subset H_{\alpha_e}\}$. Then the set ${\mathscr{T}}\cap{\mathscr{L}}_{\geq C}$ is equal to $\{T\in{\mathscr{T}}: \forall e\in F',\ s(T)_e=s(C)_e\}$, so it is a $T$-convex subset of ${\mathscr{T}}$ in the sense of [@OM Definition 4.2.5]; see [@OM Proposition 4.2.6]. In other words, it contains every shortest path in the chamber graph between any two of its elements, so it 