Skip bases, cover roots, greatest-sortable projections, and the chamber union of each cone

Clause (1): [F1] gives that ${\pi }_c$ is well defined, independent of the initial-letter choices, idempotent, descent detecting and equal to the restriction of ${\pi }_{c{\mathrm{\mid }}_J}$ on parabolics; [F2] gives that ${\pi }_c$ is order preserving and that ${\pi }_c(w)$ is the unique greatest $c$-sortable element below $w$. Clause (1) is exactly the conjunction of these statements.

Clause (2): [F3] states that $C_c(v)$ is a basis of $V$ independent of the reduced Coxeter word for $c$ and of the recursion choices, and that the negative elements of the basis are exactly the negatives of the positive roots of the cover reflections, ${\mathcal{A}}_c(v)=\{-{\beta }_t:t\in \mathrm{C}\mathrm{o}\mathrm{v}(v)\}$; since the map $t\mapsto {\beta }_t$ is injective, the number of negative skip roots equals $\mathrm{\mid }\mathrm{C}\mathrm{o}\mathrm{v}(v)\mathrm{\mid }$, the number of cover reflections.

Clause (3), inclusion $\supseteq$: if ${\pi }_c(w)=v$ then $wC\subseteq {\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}_c(v)$ by [F2] (full criterion), so each such closed chamber is contained in the cone.

Clause (4): the parabolic compatibility ${\pi }_{c{\mathrm{\mid }}_J}(w_J)={\pi }_c(w{)}_J$ is [F2] (parabolic compatibility), and the abstention clause is a statement about what the proof does not use: no lattice congruence, no forcing of oriented rank-two contractions and no noncrossing-partition bijection is invoked anywhere in clauses (1)-(4), whose inputs are the recursion [F1], the cone criterion and monotonicity [F2], the skip basis [F3], the chamber tiling [F4] and the cover-root dictionary [F6].

Clause (3), reverse inclusion. Since the skip normals form a basis, their nonnegative halfspaces define a full-dimensional cone. A chamber whose interior meets its interior is contained in it: each bounding root hyperplane has constant sign on that open chamber, and closure preserves its inequalities. Choose one interior point $y$ avoiding all root hyperplanes; it exists because a finite union of proper hyperplanes cannot contain an open ball. For any $x$ in the cone, the points $x+\lambda (y-x)$ lie in its interior for $0<\lambda \le 1$, and each root hyperplane excludes at most one value of $\lambda$ because it does not contain $y$. For each integer $n\ge 1$, let $k_n$ be the least integer $k>n$ such that $x+k^{-1}(y-x)$ avoids all root hyperplanes. Finitely many values are excluded, so $k_n$ exists; these explicitly chosen points approach $x$ without any countable choice principle. Every such point lies in an open chamber contained in the cone, whose label projects to $v$ by [F2]. Finitely many chambers occur, so one such closed chamber contains a subsequence approaching $x$ and therefore contains $x$. Together with step 1.3 this proves the union equality. Interior points of the cone which happen to lie on additional arrangement hyperplanes require this generic approximation; they are not asserted to be in open chambers.

Clauses (1)-(4) are proved. No Choice is used: the approximation points are specified by least integers, and the remaining choices are single existential instantiations.