Finite equivariant Salvetti and orbit homotopy models

Choose Euclidean coordinates on $E\times E\cong {\mathbb{R}}^d$. The removed set is a finite union of closed linear subspaces, so $M$ is open and Hausdorff. For $d>0$, choose a nonzero defining linear form ${\lambda }_H$ for each of the finitely many hyperplanes. Put $\delta (x,y)=\min (\{1\}\cup \{\sqrt{{\lambda }_H(x{)}^2+{\lambda }_H(y{)}^2}:H\in \mathcal{A}\})$. This finite minimum is continuous on the ambient space, positive exactly on $M$, and identically $1$ for an empty arrangement. On $M$ set $\rho (z)=\mathrm{\parallel }z\mathrm{\parallel }+1\mathrm{/}\delta (z)$. For $t\ge 1$, the sublevel $\{\rho \le t\}$ is a closed bounded subset of the closed region $\{\delta \ge 1\mathrm{/}t\}$, since the formula is continuous there and $\mathrm{\parallel }z\mathrm{\parallel }\le t$. Sublevels for $t<1$ are empty. They are compact by [F3]. Thus $B_n=\{n-1\le \rho \le n+1\}$ is compact and lies in the open set $V_n=\{n-2<\rho <n+2\}$ for each integer $n\ge 1$. These bands cover $M$, since $\rho \ge 1$.

Finite-dimensional linear maps and their linear inverses are continuous, so the action on $M$ is by homeomorphisms. For every compact $C\subseteq M$, the set $\{g\in G:gC\cap C\mathrm{\ne }\mathrm{\varnothing }\}$ is a subset of the finite group $G$ and hence finite. This proves proper discontinuity in the compact-set convention of [F5]. We also verify the disjoint-neighbourhood condition used for the comparison. Average any inner product on $E\times E$ over $G$ to obtain an invariant norm. If $G$ is nontrivial and $z\in M$, freeness under [F5] gives $a={\min }_{g\mathrm{\ne }1}\mathrm{\parallel }z-gz\mathrm{\parallel }>0$. Choose $r<a\mathrm{/}3$ so small that $B(z,r)\subset M$. Invariance gives $gB(z,r)=B(gz,r)$, disjoint from $B(z,r)$ for $g\mathrm{\ne }1$ by the triangle inequality. The condition is vacuous for trivial $G$.

Let $\mathcal{O}$ be any open cover of $M$ and $d>0$. Fix an enumeration of all balls with rational centers and positive rational radii in ${\mathbb{R}}^d$, using integer codes for the finitely many rational coordinates. These balls form a basis: rational grids approximate every coordinate, and one can choose a rational radius between the distance to the chosen center and the remaining radius of any prescribed ball. For each $n$, retain the rational balls contained in $V_n$ and in some member of $\mathcal{O}$. They cover $B_n$ because $M$, $V_n$ and the members of $\mathcal{O}$ are ambient-open. Compactness gives a finite subfamily covering $B_n$. Select the subfamily with least finite-subset code; for an empty band select the empty family. This specifies each finite family uniquely, without countable choice or a selection of containing members of $\mathcal{O}$. Their union covers $M$ and refines $\mathcal{O}$. Near $z$, continuity gives $\rho <\rho (z)+1$; a selected ball at level $n$ lies in $V_n$, so it can meet this neighbourhood only if $n<\rho (z)+3$. There are finitely many such levels and finitely many selected balls at each level, proving local finiteness. This proves paracompactness under [F3]. If $d=0$, there are no proper hyperplanes, $M$ is a singleton, and its one-member open refining cover proves paracompactness directly.

Apply [F1]'s facet-point supplier to choose $x(F)\in F$ for every facet, with $x(gF)=gx(F)$ for every $g\in G$. Use these points in the natural cover construction. By [F1] the cover is invariant and open with contractible nonempty finite intersections and disjoint nonidentity translates of each member. Steps 1.2 and 2.1 give both proper-discontinuity checks in [F5] and paracompactness; freeness in the sense of [F5] is assumed. Hence [F99] applies. Composing its map with the inverse of [F1]'s equivariant nerve homeomorphism yields $f:K\to M$, an equivariant ordinary homotopy equivalence under [F4]. Equivariance makes $\bar{f}([k])=[f(k)]$ well defined, and the quotient topology makes it continuous. Its homotopy-equivalence assertion and the free nerve action are also supplied by [F99].

The complex $K$ is finite by [F2]. By [F1] a simplex is a finite chain and the group acts by order-preserving simplicial automorphisms. If $g$ maps a simplex to itself, its order-preserving permutation of the finite chain is the identity, so it fixes the vertices and the simplex pointwise. The free action from step 3.1, in the sense of [F5], forces $g=1$. Thus distinct translates of an open simplex are disjoint, and the orbit projection is injective on each open simplex.

Write $Q=K\mathrm{/}G$ and $q:K\to Q$. The space $K$ is Hausdorff by [F2]. Two distinct orbits are finite disjoint sets. For each pair of points in the two sets take disjoint open neighbourhoods, intersect these for each fixed point, and then take the two respective unions. This produces disjoint open neighbourhoods of the two finite sets, using only finitely many choices. Intersect each neighbourhood with all its $G$-translates; it remains an open neighbourhood of its orbit and is invariant. Their images in $Q$ are open and disjoint. Indeed $q$ is open because $q^{-1}(q(O))={\bigcup }_{g\in G}gO$ for open $O$, by [F4]. Hence $Q$ is Hausdorff.

Let $K^n$ be the simplicial $n$-skeleton, put $Q^n=q(K^n)$ with the subspace topology, and set $Q^{-1}=\mathrm{\varnothing }$. Closed simplices are compact by [F2] and [F3], with a zero-simplex compact directly. Finite unions are compact by taking finitely many finite subcovers. Continuous images are compact by pulling covers back. Thus $K^n$ and $Q^n$ are compact. For $n\ge 1$, form the attachment space $A_n$ from $Q^{n-1}$ and one closed $n$-simplex for each simplex orbit, identifying boundary points with their images in $Q^{n-1}$, as in [F2]. Boundaries lie in the preceding skeleton, and their maps are continuous restrictions of $q$. The finite disjoint union and its quotient $A_n$ are compact. Its induced continuous map $A_n\to Q^n$ is bijective: boundaries have exactly the specified identifications in $Q^{n-1}$, while step 4.1 gives precisely one injective interior per orbit. This bijection is a homeomorphism by the following elementary argument. A closed subset of a compact space is compact, since one may add its open complement to a cover. A compact subset of a Hausdorff space is closed: for an outside point, the open sets admitting a disjoint neighbourhood of that point cover the compact subset; a finite subcover and the intersection of the corresponding finitely many neighbourhoods separate it from that point. Therefore a continuous bijection from a compact space to a Hausdorff space maps closed sets to closed sets. Apply this to $A_n\to Q^n$, which is Hausdorff as a subspace of $Q$. This proves the disk-attachment topology on each skeleton.

The finite Hausdorff zero-skeleton $Q^0$ is discrete, by separating any one of its points from each of the finitely many other points. In each attachment quotient, the interior of a disk summand is open and saturated, and undergoes no identifications. Its restriction is therefore a homeomorphism onto its image, by [F4], and step 6.1 transports this to $Q^n$. Since $Q^n$ has the subspace topology from $Q$, this also gives the required homeomorphism onto the cell in $Q$. Its characteristic map is the closed-simplex map followed by $q$, with boundary in $Q^{n-1}$. These open cells partition $Q$ by step 4.1.

Each characteristic image is compact and hence closed in $Q$ by step 6.1. Density of the simplex interior and continuity show that this image is precisely the closure of its open cell. The characteristic map onto this closed cell is a closed surjection by the same compact-to-Hausdorff argument, hence is a quotient map: a subset with closed preimage is the image of that closed preimage and is closed. Thus a subset $A\subseteq Q$ meets each closed cell in a closed set exactly when its preimage under every characteristic map is closed. By [F4], $A$ is closed exactly when $q^{-1}(A)$ is closed in $K$. The weak topology in [F2] tests the latter on closed simplices; one representative per orbit suffices, since all translates have the same composite with $q$. This is the CW weak topology. There are finitely many simplex orbits, so every cell closure meets finitely many cells. Together with steps 5.1–7.1 this proves that $Q$ is a finite CW complex under [F2]. The same finiteness for $K$ was already supplied by [F2], completing all assertions.