Relative free action, normal cover and exact sequence

By [F1] and [F2], the relative cone is convex, being the intersection of the convex full Tits cone with the linear subspace $E$. If $a,b\in T$, choose $r>0$ such that $B_E(a,r)$ and $B_E(b,r)$ are contained in the relative cone. For $0\le s\le 1$ and $\mathrm{\parallel }h\mathrm{\parallel }<r$, convexity puts $(1-s)(a+h)+s(b+h)=(1-s)a+sb+h$ in that cone. Thus the ball $B_E((1-s)a+sb,r)$ lies in the relative cone, so $(1-s)a+sb\in T$. Hence $T$ is convex. The coordinate face $C_J$, consisting of functionals vanishing on $J$ and strictly positive on $S\setminus J$, is a nonempty open subset of $E$ contained in the closed fundamental chamber and hence in the relative cone. Thus $T$ is nonempty. Local finiteness implies that the union of the hyperplanes in $T$ is relatively closed: near each point only finitely many need be considered. A nonempty open subset of a finite-dimensional real vector space cannot be covered by finitely many proper hyperplanes, so a locally finite family of such hyperplanes has nonempty complement in $T$. In particular $X$ is nonempty. The same local-finiteness argument shows that the deleted union in $T\times T$ is closed there; consequently $X$ is open in the Euclidean space $E\times E$ and is locally path-connected. These statements include the zero-dimensional case, where there are no proper hyperplanes.

Let $(u,v)\in X$. Choose a neighbourhood of $u$ meeting only finitely many members of $\mathcal{A}$. For sufficiently small positive $t$, $u+tv$ is in that neighbourhood and in $T$. If a hyperplane in the finite list contains $u$, it does not contain $v$, since $(u,v)\in X$; therefore it does not contain $u+tv$. For each remaining hyperplane in the list, at most one value of $t$ makes $u+tv$ lie on it. Choose $t$ avoiding those finitely many values. Then $u+tv$ belongs to a chamber $C$.

Let $w\in {\mathrm{Norm}}_W(W_J)$. For $\alpha =u{\alpha }_i\in \mathrm{\Phi }$, the equality $w\alpha =(wu){\alpha }_i$ shows $w\alpha \in \mathrm{\Phi }$; applying this also to $w^{-1}$ gives $w\mathrm{\Phi }=\mathrm{\Phi }$. For $H_{\alpha }=\{f:f(\alpha )=0\}$, the dual formula gives $(wf)(w\alpha )=f(\alpha )$, hence $w{H}_{\alpha }=H_{w\alpha }$. Since $wE=E$, it follows that $w(H_{\alpha }\cap E)=H_{w\alpha }\cap E$, and a trace is proper exactly when its image is proper. Thus $w$ permutes the relative hyperplanes. The restricted linear homeomorphism preserves the relative cone and therefore its interior $T$, so it permutes precisely those relative hyperplanes meeting $T$, namely $\mathcal{A}$. Its diagonal map consequently restricts to a homeomorphism of $X$. This action factors through $N(W,J)$ because $W_J$ fixes $E$ pointwise. In particular the quotient acts by homeomorphisms on the arrangement complement in $T$ and permutes its connected components, the chambers.

The arrangement complement in $T$ is open by step 1.1, and each of its components is open because sufficiently small Euclidean balls are connected. Hence $U=\{(a,b)\in X:a+tb\in C\}$ is an open neighbourhood of $(u,v)$. If $gU\cap U\mathrm{\ne }\mathrm{\varnothing }$, linearity gives $gC\cap C\mathrm{\ne }\mathrm{\varnothing }$. The homeomorphism $g$ permutes chambers, so $gC=C$; by chamber freeness $g$ is the identity. Thus every point has a disjoint-translate neighbourhood, which also proves freeness.

Let $p,q\in X$. Their straight segment lies in the open convex set $T\times T$. By compactness of that segment and local finiteness in each coordinate, a sufficiently small open tubular neighbourhood of the segment lies in $T\times T$ and meets only finitely many sets $H\times H$. It can be chosen convex. For each such $H$, the set of intermediate points $z$ for which either segment $[p,z]$ or $[z,q]$ meets $H\times H$ is contained in the union of the affine spans of $H\times H$ with $p$ and with $q$. Because $p,q\mathrm{\notin }H\times H$ and $H\times H$ has codimension two, these spans are proper affine subspaces. A finite union of proper affine subspaces cannot cover the nonempty tube. Choose $z$ outside that union. Both segments remain in the tube and avoid every deleted set, so they form a path in $X$ from $p$ to $q$. If $E$ has dimension zero, $X$ is a point and the conclusion is immediate.

Steps 1.1, 2.1 and 2.2 verify all hypotheses of [F5]. Applying it gives the normal covering, identifies its deck group with $N(W,J)$, and yields the exact sequence with the stated endpoint-lift map.