Proof. Start with any two-dimensional \(K(\Gamma,1)\). Collapse a maximal tree in its connected one-skeleton and replace the two-cell attaching loops by finite edge words. These operations preserve the homotopy type: a contractible CW subcomplex can be collapsed, and homotopic attaching maps give homotopy equivalent adjunction spaces (Hatcher 2002, Propositions 0.17 and 0.18). These facts apply to infinite CW complexes as well; a loop in a graph is homotopic to a finite edge path. The resulting presentation complex has one vertex, with some sets of generators and relators. Its universal cover is acyclic and has no cells above dimension two. Consequently its relator boundary map is an isomorphism onto the cycle module. The relator words also normally generate the kernel of the map from the free generator group to \(\Gamma\).

Adjoin the prescribed letters \(S\), with one definition relator for each letter. Each new relator boundary has coefficient \(1\) on its own new edge and coefficient \(0\) on the other new edges. Subtracting these boundaries reduces any cycle uniquely to an old cycle, so the boundary basis property persists. For each former generator \(a\), choose a word \(w_a(S)\) with value \(a\) in \(\Gamma\), and replace \(a\) by \(t_a w_a(S)\). This is an invertible change of free generators: its inverse sends \(t_a\) to \(a w_a(S)^{-1}\) and fixes \(S\). The new letters \(t_a\) all have value \(1\). On edge chains the substitution fixes the \(S\)-edges and sends \(e_a\) to \(e_{t_a}+[w_a]\), so it is a chain isomorphism and preserves the boundary basis property. Let \(T\) be the set of these new letters, \(R\) the current relator index set, and \(I=\mathcal D\amalg T\). Equation (3) identifies the cycle module with \(\Lambda^{(I)}\). In these coordinates the relator boundary map is an isomorphism \[A:\Lambda^{(R)}\longrightarrow\Lambda^{(I)}.\]

We next arrange that the prescribed cycles themselves occur as relator boundaries. An auxiliary copy of the cycle module allows the three shears below to move this basis into a new relator block. Adjoin letters \(U=(u_i)_{i\in I}\) and defining relators \(u_i\). The stabilized boundary map is \[D_0=\begin{pmatrix}A&0\\0&1\end{pmatrix}: \Lambda^{(R)}\oplus\Lambda^{(I)} \longrightarrow\Lambda^{(I)}\oplus\Lambda^{(I)},\] where the second target summand consists of the \(U\)-edges. On its domain put \[P=\begin{pmatrix}1&A^{-1}\\0&1\end{pmatrix}, \qquad Q=\begin{pmatrix}1&0\\-A&1\end{pmatrix}.\] These are blocks of homomorphisms of left modules; all products below mean composition. Direct multiplication gives \[ PQP=\begin{pmatrix}0&A^{-1}\\-A&0\end{pmatrix}, \qquad D_0PQP=\begin{pmatrix}0&1\\-A&0\end{pmatrix}. \tag{5}\]

We realize each of the successive precompositions \(P,Q,P\) by actual relator words. A shear leaves one block of words unchanged and adds to each boundary in the other block a finite sum \(\sum_k n_k g_k[b_k]\) of translates of unchanged boundaries. Choose a generator word \(v_k\) representing \(g_k\) and multiply the changing relator by copies of \(v_k b_k^{\pm1}v_k^{-1}\), with the indicated multiplicities and signs. All these words represent \(1\) in \(\Gamma\); the added boundary is precisely the stated sum. The unchanged words belong to both old and new relator lists, so the old changing word is recovered from the new one by a finite inverse multiplication. Thus the old and new lists have exactly the same normal closure. Every column of \(A\) and \(A^{-1}\) has finite support because their domains and targets are direct sums. Hence this construction uses finite words for every relator even when the index sets are uncountable. It does not require an infinite product of words or a limiting sequence of presentation moves.

In the final second block, the relator indexed by \(i\in I\) has boundary \((e_i,0)\). Designate those indexed by \(z\in\mathcal D\) as \(\rho_z\). Set \(\mathcal T=T\amalg U\). The remaining second-block relators are indexed by \(T\), and the first-block relators are indexed by \(R\); these are the ordinary relators, with \(\mathcal R=T\amalg R\). Their projected boundary map is, in these orders, \[ \begin{pmatrix}1_T&0\\0&-A\end{pmatrix}: \Lambda^{(T)}\oplus\Lambda^{(R)} \xrightarrow{\ \cong\ } \Lambda^{(T)}\oplus\Lambda^{(I)}. \tag{6}\] Equation (5) also shows that the complete boundary list remains a basis. Normal closure was preserved at every step, so these are the relators of an actual presentation of \(\Gamma\). ◻

Proof. The boundary of a conjugate \(u r^{\sigma}u^{-1}\) is \(\sigma\bar u[r]\): the two conjugating paths cancel as chains because \(r\) has group value \(1\). Every factor in Equation (7) also has value \(1\), so multiplying the factors introduces no further prefix translations in its chain. Applying the complete relator boundary isomorphism to the left side of Equation (8) therefore gives \([z]\). The right side has the same image, namely \([\rho_z]=[z]\). Injectivity gives the asserted equality, coefficient by coefficient. ◻

Proof. For unit quaternions, telescoping products and bi-invariance give \[ \left|\prod_{j=1}^n a_j-\prod_{j=1}^n b_j\right| \le\sum_{j=1}^n|a_j-b_j|, \qquad |ab-ba|\le2|a-1|\,|b-1|. \tag{11}\] The second inequality follows by expanding \(ab-ba=(a-1)(b-1)-(b-1)(a-1)\). Inverses and conjugation preserve distance from \(1\).

Evaluate the fixed expression for each \(t\in\mathcal T_0\) at any vertex. Ordinary relators disappear, and every distinguished factor has distance at most \(h\) from \(1\), regardless of its conjugator. Thus \[|L(t,v)-1|\le M h\qquad(t\in\mathcal T_0)\] for one finite constant \(M\), independent of the vertex and the group quotient. No bound on the conjugators in these auxiliary expressions is required. Each \(u_{z,j}\) uses only letters of \(S\amalg\mathcal T_0\), so another fixed constant \(K\) gives \[|L(u_{z,j},v)-1|\le K(\epsilon+h).\] On evaluating Equation (7), discard its ordinary factors. Removing the conjugation from a remaining factor costs at most \[|aba^{-1}-b|=|ab-ba| \le 2K(\epsilon+h)h,\] including when the relator occurs with exponent \(-1\). There are at most \(n_z\) such factors.

The resulting factors are values of distinguished relators, or their inverses, at specified vertices. Any interchange of two adjacent factors costs at most \(2h^2\), by Equation (11). Group together occurrences having the same relator index and the same vertex; at most \(n_z(n_z-1)/2\) interchanges are needed. Their signed multiplicities are prescribed exactly by Equation (8), translated from \(1\) to \(v\). On a finite quotient, coincident vertices merely add the corresponding coefficients, so the same identity still applies. After cancellation within each group, the product is exactly \(L(\rho_z,v)\). The total error is bounded by \[2n_zK(\epsilon+h)h+n_z(n_z-1)h^2.\] Since \(\mathcal D\) is finite, this proves Equation (9) with one \(C\). The word \(z\) contains only \(S\)-letters, so \(|L(z,v)-1|\le\operatorname{length}(z)\epsilon\). Increasing \(C\) proves Equation (10) as well. If the distinguished list is empty, the two assertions are vacuous and any \(C\ge1\) suffices. ◻

Proof. Tensor the isomorphism (4) with the right \(\Lambda\)-module \(\mathbb Z[F]\). This gives an isomorphism \[ \mathbb Z[F]^{(\mathcal R)}\xrightarrow{\ \cong\ } \mathbb Z[F]^{(\mathcal T)}. \tag{14}\] No flatness claim is involved: tensoring the inverse gives an inverse to the displayed map. As an abelian-group basis, the source has indices \((v,r)\in F\times\mathcal R\). Its image consists of the projected boundary chains of \(r\) based at \(v\). Thus the chains for \(F\times\mathcal B\) are linearly independent over \(\mathbb Q\).

Give an unknown \(x_{v,t}\in H\) to each extra edge \((v,t)\). In the equation \(L(r,v)=1\), the exponent count of each unknown is exactly its coefficient in the projected boundary chain. This holds even though the fixed \(S\)-labels need not commute: the path traversed is determined by the generator values in \(F\), and all \(t\in\mathcal T\) have value \(1\). Write the \(m=|F|\,|\mathcal B|\) exponent vectors as rows. Their union of supports is finite, and their row rank is \(m\). Choose \(m\) unknowns giving a nonsingular \(m\times m\) integer minor \(B\), and set every other extra label equal to \(1\).

Move all the prescribed \(S\)-labels from \(1\) to their final values along the short quaternion arcs, with common parameter \(\tau\in[0,1]\). These arcs stay at distance less than \(\eta\) from \(1\). The selected equations now define a continuous family of smooth maps \[f_\tau:H^m\longrightarrow H^m.\] For an assignment \(x\in H^m\), let \[h(\tau,x)=\max_{z\in\mathcal D,\ v\in F} |L_\tau(\rho_z,v;x)-1|, \qquad \mathcal O_\tau=\{x:h(\tau,x)<\delta\}.\] Only finitely many words occur here, so \(h\) is continuous. Every root \(f_\tau(x)=\mathbf 1\) satisfies all equations from \(\mathcal A\). Equation (13) therefore implies \[ f_\tau(x)=\mathbf 1\quad\Longrightarrow\quad h(\tau,x)\ne\delta. \tag{15}\] If \(m=0\), the domain and target are points. The unique assignment starts with \(h=0\), and continuity together with Equation (15) keeps \(h<\delta\) throughout. This proves the assertion in that case; henceforth assume \(m\ge1\).

We justify degree invariance with these moving open sets. For fixed \(\tau_0\), the roots with \(h(\tau_0,x)<\delta\) form a compact set: by (15) they are also the roots with \(h(\tau_0,x)\le\delta\). Choose an open neighborhood \(V\) containing exactly these roots, with \(\overline V\subset\mathcal O_{\tau_0}\). If there are no such roots, take \(V=\varnothing\). Compactness of \(H^m\) and continuity show that, for all \(\tau\) sufficiently close to \(\tau_0\), every root with \(h(\tau,x)<\delta\) lies in \(V\), every root in \(\overline V\) has \(h(\tau,x)<\delta\), and there is no root on \(\partial V\). For example, a sequence of selected roots outside \(V\) with parameters tending to \(\tau_0\) would have a limit root with \(h\le\delta\), hence \(h<\delta\), outside \(V\), a contradiction. The other assertions follow in the same way, or from \(\overline V\subset\mathcal O_{\tau_0}\). Excision and homotopy invariance of ordinary degree now give \[\deg(f_\tau,\mathcal O_\tau,\mathbf 1) =\deg(f_\tau,V,\mathbf 1) =\deg(f_{\tau_0},V,\mathbf 1)\] near \(\tau_0\). Thus the selected degree is locally constant, and hence constant on \([0,1]\).

At \(\tau=0\), the map \(f_0\) is a pure word map with exponent matrix \(B\): all fixed coefficients are \(1\). The set \(\mathcal O_0\) is invariant under simultaneous conjugation, contains the identity assignment, and its boundary contains no root. Lemma 2 therefore gives \[\deg(f_0,\mathcal O_0,\mathbf 1)\ne0.\] The same holds at \(\tau=1\), so a root exists in \(\mathcal O_1\), as required. ◻

Proof of Theorem 4. Let \(a\) be any prescribed labeling of the \(S\)-edges on \(\Gamma\) satisfying the theorem’s hypotheses, with the fixed \(\eta\) of Equation (12). In the compact product \[\mathscr P=H^{\Gamma\times\mathcal T}\] consider all the following conditions on the unknown extra-edge labels: \[\begin{align*} L(r,g)&=1 &&(r\in\mathcal R,\ g\in\Gamma), \tag{16}\\ |L(\rho_z,g)-1|&\le\delta &&(z\in\mathcal D,\ g\in\Gamma). \tag{17}\end{align*}\] Each condition is closed and depends on finitely many coordinates. We verify the finite intersection property, keeping the original labeling \(a\) fixed throughout.

First choose an arbitrary finite collection \(\mathscr C\) of the conditions (16)–(17). Let \(P\subset\Gamma\) be the finite set of starting vertices of all positively oriented \(S\)-edges used when evaluating the words in \(\mathscr C\). An inverse occurrence traversing the edge from \(g\) to \(gs^{-1}\) uses the positively oriented edge starting at \(gs^{-1}\), and this starting vertex is included in \(P\). Let \(\mathcal B\) contain \(\mathcal A\) and all ordinary relator indices appearing in \(\mathscr C\). These choices are made before choosing a finite quotient.

Residual finiteness supplies a finite quotient \(q:\Gamma\twoheadrightarrow F\) that is injective on \(P\): separate the finitely many nonidentity elements \(p^{-1}p'\) for distinct \(p,p'\in P\), and let \(F\) be the image of \(\Gamma\) in the product of the resulting finite quotients. For each \(S\)-edge actually used in \(\mathscr C\), assign to its image edge in \(F\) its prescribed label from \(a\). Injectivity on \(P\) ensures that no two inconsistent prescriptions are placed on the same edge. Give every remaining \(S\)-edge of \(F\) label \(1\). All these labels have distance less than \(\eta\) from \(1\).

Apply Lemma 8 to this finite quotient, this assignment, and \(\mathcal B\). Pull the resulting extra-edge labels back by \(q\), and combine them with the original \(S\)-labels \(a\) on \(\Gamma\). In every word from \(\mathscr C\), each traversed \(S\)-edge has the same label as its image in \(F\), by construction; the same is true for each extra edge by pullback. Its path product therefore agrees with the quotient path product. The selected ordinary equations hold, and the selected distinguished products even have distance strictly less than \(\delta\). This supplies a point of \(\mathscr P\) satisfying \(\mathscr C\).

We have proved: for each finite collection of constraints, one can choose a finite quotient, copy the finitely many required prescribed labels, solve a finite system there, and pull back a solution to that collection. The quotient may depend on the collection; the original assignment \(a\) is never assumed to descend to any one finite quotient. Compactness of \(\mathscr P\) now gives an extension satisfying all of (16)–(17) simultaneously.

For this extension, \(h\le\delta\) and all the equations in \(\mathcal A\) hold. The original hypothesis says \(L(z,g)=1\) for every \(z\) and \(g\). Taking suprema in Equation (9), with \(\epsilon=\eta\), gives \[h\le C(h+\eta)h\le C(\delta+\eta)h<\tfrac12h \quad\text{if }h>0.\] Thus \(h=0\). All distinguished relators, as well as all ordinary relators, have product \(1\) at every vertex. Finally, a closed edge path spells a word representing \(1\) in \(\Gamma\). Because Lemma 5 gives an actual presentation, that word is a finite product of conjugates of final relators and their inverses. Evaluating at the starting vertex makes every factor \(1\). Hence the original closed path has product \(1\), as claimed. ◻

Proof. Start with the presentation complex \(K\) of \[ \langle x,y\mid x^2y^{-5},\ x^2(xy^{-1})^{-3}\rangle. \tag{18}\] The presented group also occurs in Hatcher (Hatcher 2002, Example 2.38), with relations \(a^5=b^3=(ab)^2\): take \(a=y\) and \(b=y^{-1}x\). Indeed, \(x^2=y^5\) is central and \(b\) is conjugate to \(xy^{-1}\). We verify the acyclicity of \(K\) and construct the required \(\operatorname{SU}(2)\) representation directly.

Its cellular boundary \(C_2(K;\mathbb Z)\to C_1(K;\mathbb Z)\) has exponent-sum rows \((2,-5)\) and \((-1,3)\). Their determinant is \(1\), so this boundary is an isomorphism. Since \(K\) has one vertex and no cells above dimension two, it is integrally acyclic.

Identify \(H\) with the unit quaternions. Put \(\theta=\pi/5\) and choose imaginary unit quaternions \(u,v\) with \[u\mathbin{\cdot}v=\frac{1}{2\sin\theta};\] this is possible because \(\sin\theta>1/2\). Send \(x\) to \(u\) and \(y\) to \(\cos\theta+v\sin\theta\). Then \(x^2\) and \(y^5\) both map to \(-1\). Moreover, the real part of \(u(\cos\theta-v\sin\theta)\) is \(1/2\). A unit quaternion with real part \(1/2\) has cube \(-1\), so both relations in (18) hold. The homomorphism is nontrivial because the image of \(x\) is \(u\ne1\).

Here is a flag triangulation that avoids any assumption that a presentation complex is regular. Subdivide each circle of its one-skeleton into at least three edges. Express the two attaching maps as polygonal edge paths, and cone the boundary of each polygon to a new interior vertex before making the boundary identifications. Keep the radial edges indexed by the boundary occurrences: distinct occurrences of the same graph vertex give distinct radial edges. Every sector is now a closed triangle with three distinct vertices, and its closure is embedded. Indeed, its one outer edge is embedded in the subdivided graph, while its two radial edges have disjoint interiors. Thus this is a finite regular CW structure on \(K\).

Take the order complex of its nonempty cells, ordered by inclusion of closures. This triangulates \(K\): inductively triangulate cell boundaries and cone each such triangulation from an interior point of the cell. Regularity makes these constructions compatible on common faces. The result is two-dimensional, and it is flag, since pairwise comparable cells form a chain. Call it \(L\). The resulting homeomorphism transfers both acyclicity and the nontrivial homomorphism to \(L\). ◻

Proof. We induct on the number of vertices. Delete one vertex, with generator \(t\), and let \(P'\) be the group on the remaining graph. The neighbors of the deleted vertex generate the graph group \(C\). Both inclusions into \(P'\) and into the original group are injective: killing the other generators gives retractions. Let \(r:P'\to C\) be the first of these retractions. There is an isomorphism \[ P\cong F(P'/C)\rtimes P', \tag{20}\] where \(P'/C\) denotes left cosets, \(F(P'/C)\) is the free group on symbols \(b_{qC}\), and \(P'\) acts by left translation of their indices. To verify the formula, map \(P'\) identically to the second factor and \(t\) to \(b_C\). This respects the relations \([t,c]=1\) for \(c\in C\). Conversely, send \(b_{qC}\) to \(qtq^{-1}\). This is independent of the representative of \(qC\) and respects the semidirect-product action. The maps are inverse on the indicated generators.

By induction \(P'\) is residually finite. For \(g\notin C\), we have \(g\ne r(g)\), so some finite quotient \(p\) of \(P'\) satisfies \(p(g)\ne p(r(g))\). The map \((p,p\circ r)\) separates \(g\) from \(C\): every element of \(C\) has equal coordinates, whereas \(g\) does not. Taking a product of finitely many such maps shows that any finite list of distinct left cosets in \(P'/C\) remains distinct in \(D/\overline C\) for a suitable finite quotient \(D\) of \(P'\), where \(\overline C\) is the image of \(C\).

Consider a nonidentity element \((w,q)\) of (20). If \(q\ne1\), a finite quotient of \(P'\) separates it from the identity. If \(q=1\), choose \(D\) preserving the distinct indices appearing in the nonempty reduced free word \(w\). Its image is still nontrivial in \(F(D/\overline C)\rtimes D\). A free group is residually finite: for a nonempty reduced word of length \(n\), prescribe its successive steps on distinct points \(0,\ldots,n\). For each free generator these steps prescribe an injective partial permutation, because the word has no adjacent inverse letters. Complete each partial permutation to a permutation of this finite set. The resulting finite permutation representation sends \(0\) to \(n\) along the word. Apply this fact to obtain a finite-index normal subgroup \(N\) of \(F(D/\overline C)\) not containing the image of \(w\). Intersect the finitely many translates of \(N\) under \(D\). The intersection is normal, has finite index, is \(D\)-invariant, and still omits \(w\). Its quotient, semidirect \(D\), is the required finite quotient. This proves the induction. Residual finiteness passes to subgroups, so it also holds for \(G\). ◻

Proof. First we prove acyclicity of the universal cover for every finite commutation graph, by induction on its number of vertices. Use \(t,P',C\) as in the preceding proof. Downstairs, the cubical complex is obtained from \(S_{P'}\) by attaching \(S_C\times[0,1]\), identifying both ends with the coordinate subcomplex \(S_C\subset S_{P'}\). Here \(S_{P'}\) and \(S_C\) denote the same coordinate-torus construction for those graphs.

Upstairs, the components over \(S_{P'}\) are its universal covers, indexed by left cosets \(P/P'\), and the lifted cylinders are products of the universal cover of \(S_C\) with \([0,1]\), indexed by \(P/C\). These covering assertions follow from the injectivity of the subgroup inclusions. The cylinder indexed by \(gC\) joins the vertex spaces indexed by \(gP'\) and \(gtP'\). Its incidence graph is a tree. Indeed, under (20), write \(g=wq\) with \(w\in F(P'/C)\) and \(q\in P'\). The vertex cosets are identified with \(w\), and this cylinder runs from \(w\) to \(wb_{qC}\). The incidence graph is therefore the Cayley tree of the free group \(F(P'/C)\).

By induction every vertex space and every cylinder cross-section is acyclic. The union over a finite subtree is acyclic: adjoining a leaf amounts to gluing in its vertex space and incident cylinder along one acyclic cylinder end, and the assertion follows from Mayer–Vietoris. Any cellular cycle in \(E\) meets only finitely many vertex spaces and cylinders, hence lies in such a finite-subtree union. It bounds there. This proves acyclicity of \(E\). The base case, the empty graph, is a point.

As a universal cover, \(E\) is simply connected. A simply connected acyclic CW complex is contractible: a first nonzero homotopy group would, by the Hurewicz theorem, give a nonzero homology group, and the Whitehead theorem then applies to the map to a point. These are the usual CW forms of the two theorems; see (Hatcher 2002, secs. 4.1–4.2).

For completeness, consider the cubical structure itself. A cube is specified by its initial vertex \(p\) and a clique \(\sigma\); its vertices are \[p\prod_{q\in\tau}a_q,\qquad \tau\subseteq\sigma.\] They are distinct by the abelianization map \(P\to\mathbb Z^V\). Each lifted face is determined by its initial vertex and its direction subset; distinct faces of one cube have distinct such data, again by abelianization. Thus every closed cube is embedded. Suppose two cubes contain a vertex \(p\). Seen from \(p\), each has a set of distinct signed directions \(a_q^{\varepsilon_q}\), with mutually commuting underlying generators. Abelianization shows that any common vertex is obtained from \(p\) using only the equally signed directions shared by the two cubes. These directions span the same face in each cube, by uniqueness of lifts. It contains all common vertices and hence every common cell. Since the intersection is a union of common cells, it is exactly this face. Every nonempty intersection contains such a vertex. Finally, only finitely many signed cliques are available at a vertex, proving local finiteness. The three-dimensional assertion follows from the existence of two-simplices in \(L\). ◻

Proof. Triangulate each cube of \(E\) by chains in its Boolean poset of vertices: a simplex is a chain under the order of increasing coordinates. These triangulations agree on faces. Heights are strictly increasing along every such chain; in particular, no edge of this triangulation joins vertices of equal height.

At a vertex \(p\), the full subcomplex of its link spanned by vertices of height greater than \(\lambda(p)\) is the barycentric subdivision of \(L\). Indeed, a higher vertex adjacent to \(p\) in the triangulation is obtained by multiplying by the positive generators of a nonempty clique, and a simplex of these vertices corresponds exactly to a chain of such cliques. This description remains valid when \(p\) is an intermediate corner of a cube, since any comparable higher corner changes only positive directions. We call this the ascending link. The descending link is described identically using negative directions, and is also a subdivision of \(L\). Both links are integrally acyclic.

Let \(K_n\) be the full subcomplex of the triangulation spanned by vertices of height at least \(-n\), for \(n\ge0\). To pass from \(K_n\) to \(K_{n+1}\), add simultaneously the vertices \(p\) of height \(-n-1\) and the cones on their ascending links. No simplex contains two of these new vertices. Consequently the relative cellular chain complex is the direct sum, over the new vertices, of \[C_*\bigl(\operatorname{cone}(\operatorname{Asc}(p)), \operatorname{Asc}(p);\mathbb Z\bigr).\] Each summand has zero homology, since its base is acyclic and nonempty. Hence \(K_n\mathrel{% \pdfliteral direct{/Span << /ActualText <FEFF21AA> >> BDC}% \hookrightarrow \pdfliteral direct{EMC}}K_{n+1}\) induces an isomorphism on all homology groups. This argument explicitly allows infinitely many vertices at each height. The union of the \(K_n\) is \(E\), and cellular chains have finite support, so homology commutes with this union. Thus \(K_0\mathrel{% \pdfliteral direct{/Span << /ActualText <FEFF21AA> >> BDC}% \hookrightarrow \pdfliteral direct{EMC}}E\) is a homology isomorphism. Using descending links in the same way, the full subcomplex on vertices of height at most zero is also acyclic.

The closed halfspace \(E_+=\lambda^{-1}([0,\infty))\) deformation retracts onto \(K_0\). To see this directly, write a point in a triangulated simplex as \(\sum_i t_iv_i\) and discard the weights of vertices of negative height, renormalizing the others: \[R_+\left(\sum_i t_iv_i\right) =\frac{\sum_{\lambda(v_i)\ge0}t_iv_i} {\sum_{\lambda(v_i)\ge0}t_i}.\] The denominator is positive on \(E_+\): otherwise all vertices with positive weight would have negative height, contrary to the affine height of the point. The linear homotopy from the point to \(R_+\) stays in the same simplex and in \(E_+\). It fixes \(K_0\), and its formulas agree on faces. Local finiteness gives a continuous global homotopy. The corresponding construction retracts \(E_-=\lambda^{-1}(( -\infty,0])\) onto the full subcomplex on nonpositive-height vertices. Thus both closed halfspaces are acyclic.

Cut the triangulation along height zero, and subdivide the resulting polytopes if desired. The spaces \(E_+\), \(E_-\), and their intersection \(X\) are then subcomplexes, so the reduced Mayer–Vietoris sequence applies to \(E=E_+\cup E_-\). Since \(E\) and the two halfspaces are acyclic and \(X\) contains the vertex \(1\), that sequence proves \(\widetilde H_*(X;\mathbb Z)=0\), including connectedness.

For the final cell structure on \(X\), return to the original cubes. The integer slices of \([0,1]^2\) by \(x_1+x_2=m\) are a vertex or, for \(m=1\), the diagonal joining \((1,0)\) to \((0,1)\). The nondegenerate integer slices of \([0,1]^3\) are the triangles at sums \(1\) and \(2\). Slices of edges introduce no new vertices. Formula (21) therefore gives exactly the asserted vertices, edges, and triangles. The face-intersection property in Lemma 11 makes them a simplicial complex. Triangles occur by taking a three-cube whose initial vertex has height \(-1\), so its dimension is two. Figure 1 illustrates the positive-dimensional slices. ◻

Proof. The first two lemmas supply \(L\), \(\alpha\), and residual finiteness. The cubical deck action restricts to \(G\) on \(X\). It preserves the cell structure just described, and no nonidentity element stabilizes a cell: such an element would preserve its unique ambient open cube in \(E\), hence also that cube’s initial vertex. The action is free, being the restriction of a deck action. Vertices of \(X\) are exactly the elements of \(G\), so there is one vertex orbit.

There are only finitely many cell orbits. An ambient cube is specified by its clique of directions and its initial vertex; a nonempty integer slice has only finitely many possible relative heights. Two initial vertices of the same height differ by left multiplication by an element of \(G\). Hence the finite list of cliques and possible relative heights accounts for all the orbits. The quotient \(Y=X/G\) is thus a finite complex with one vertex, and \(X\to Y\) is a regular \(G\)-cover. One can obtain this covering by restricting \(E\to E/G\).

Choose orientations for the edge orbits. For each orbit choose its oriented lift starting at \(1\), and call its endpoint \(s\in G\). Keep different edge orbits as different letters even if their values coincide; this defines \(S\). Left translation identifies every oriented edge with \(g\to gs\). Connectedness of \(X\) implies that its graph is connected, so \(S\) generates \(G\).

For each triangle orbit choose an orientation, a starting corner, and the lift having that corner at \(1\). Reading its boundary gives a word \(z\) on \(S\), necessarily trivial in \(G\). These indexed words form \(\mathcal D\). The cellular chains of \(X\), with the chosen cell orbit representatives, are precisely the free modules in (23). Acyclicity makes the augmented chain complex exact. In particular, \(H_2(X;\mathbb Z)=0\) and the absence of three-cells make \(\partial_2\) injective, which gives the asserted basis rather than only a spanning set.

This free resolution proves \(\operatorname{cd}_{\mathbb Z}(G)\le2\). To obtain the opposite inequality, choose a two-simplex of \(L\) with vertices \(q,r,s\). The three corresponding commuting generators form an embedded \(\mathbb Z^3\) in \(P\): their map from \(\mathbb Z^3\) is injective after abelianization in \(\mathbb Z^V\). Its height kernel is a subgroup \(\mathbb Z^2\subset G\). Cohomological dimension cannot increase upon restriction to a subgroup, because a free group-ring module remains free over the subgroup ring, and consequently projective modules remain projective. The torus resolution gives \(H^2(\mathbb Z^2;\mathbb Z)\cong\mathbb Z\), so \(\operatorname{cd}_{\mathbb Z}(G)\ge2\).

Finally, \(P\) is finitely generated, hence countable, and so is \(G\). The contractible three-dimensional complex \(E\) is also a free \(G\)-complex; the quotient \(E/G\) is a three-dimensional \(K(G,1)\). Thus \(\operatorname{gd}(G)\le3\), completing the proof. ◻

Proof. An inverse pair can be deleted exactly when every occurrence between its endpoints is independent of its direction. This criterion is independent of the representative: the order of an occurrence dependent on that direction relative to either endpoint cannot change under exchanges. Deletion gives the trace represented by removing the two occurrences.

We check the local confluence of deletion. For two disjoint deletable pairs, either deletion leaves the other available, and both orders remove exactly the same four occurrences. If two pairs share an occurrence, they cannot share their first endpoint or their last endpoint, because an intervening occurrence of the same direction would prevent deletion. The relevant subword consequently has the form \[x B x^{-1} C x\] or the same form with \(x\) replaced by \(x^{-1}\), where every letter of \(B\) and \(C\) is independent of \(x\). The two deletions leave \(BCx\) and \(xBC\), which are the same trace. Since deletion decreases length, induction on length using these local diamonds proves uniqueness of the terminal trace. Exchanges and adjacent inverse insertions or deletions preserve that terminal trace. These operations generate equality in the defining presentation of \(P\), so the terminal trace depends only on the group element. Every word representing that element reduces to it, proving the assertion about word length.

Write \(|g|_P\) for word length in the signed standard generators. Let \(w\) be reduced of length \(n\), and append a signed generator \(x\). If the result is reduced, its new occurrence is maximal. Otherwise an available deletion must involve the new last occurrence, since a deletion involving only old occurrences would already have been available in \(w\). Its inverse mate \(y\) has only independent occurrences after it in \(w\), and is therefore maximal in the old occurrence poset. Deleting \(y\) and the new \(x\) leaves a word of length \(n-1\). If \(g\) is the element represented by \(w\), the word-metric inequality gives \[|gx|_P\geq |g|_P-1=n-1.\] Thus the remaining word is reduced. Deleting a maximal occurrence also leaves exactly the induced order on the retained occurrences: no dependency chain between two retained occurrences can pass through a maximal occurrence. This proves the last assertion. ◻

Proof. By Lemma 15, one occurrence poset is obtained from the other by adjoining a maximal occurrence. Every retained occurrence has the same principal downset before and after this operation: a maximal occurrence cannot lie on a dependency chain ending at a different occurrence. The globally minimal predecessors of a retained occurrence are precisely the minimal elements of its principal downset. Its assigned direction is therefore unchanged. This also covers a changed occurrence that is both minimal and maximal: it is isolated and has no retained successors.

Consequently, if \(c\) is the count vector for the shorter trace, the longer trace has count vector \(c+e_v\) for some \(v\in V\). Since \(\left\lVert c/n\right\rVert_1=1\), \[\left\lVert\frac{c+e_v}{n+1}-\frac{c}{n}\right\rVert_1 =\frac{\left\lVert e_v-c/n\right\rVert_1}{n+1}\leq\frac{2}{n+1}.\] For the common-simplex assertion, adjoining a maximal occurrence preserves all old minimal occurrences. If the new occurrence is not minimal, the minimal set is unchanged. If it is also minimal, it is isolated, and its direction is independent of every old occurrence: dependent occurrences would be comparable. Thus the union of the two sets of minimal directions is a clique. Flagness of \(L\) completes the proof. ◻

Proof of Proposition 14. Choose a base vertex \(q_0\) for \(\alpha\) and a base lift of \(q_0\) in the universal cover \(\widetilde L\to L\). Choose the left deck action convention so that lifting a loop representing \(\gamma\) from that base lift ends at \(\gamma\) times the lift. There is a continuous map \[ D:\widetilde L\longrightarrow H, \qquad D(\gamma\xi)=\alpha(\gamma)D(\xi). \tag{30}\] To construct it, choose images for representatives of the vertex orbits and extend equivariantly. On representatives of the edge orbits, extend between the prescribed endpoints using path connectedness of \(H\). Extend over representatives of the two-cell orbits using \(\pi_1(H)=0\). The deck action is free on cells, so equivariance introduces no ambiguity; \(L\) has dimension two, so there are no further extensions. The CW topology gives continuity.

Fix \(\epsilon>0\). There is a finite open cover \((U_i)\) of \(L\) such that each \(U_i\) is evenly covered and \(D\) has image of diameter less than \(\epsilon\) on each sheet over \(U_i\). Indeed, around any lift of a point, continuity of \(D\) gives a sufficiently small such neighborhood on one sheet. All other sheets are its deck translates, and left multiplication by \(\alpha(\gamma)\) preserves quaternion distance. Compactness of \(L\) then gives a finite subcover. Let \(\ell>0\) be a Lebesgue number in the diameter form: every subset of \(L\) of diameter less than \(\ell\) lies in some \(U_i\). Choose \(N\geq2\) so large that \(8/N<\ell\), and construct \(f_N\) as above.

For an oriented edge \(e\) of \(X\), lift its assigned path \(f_N(e)\) from \(\xi\) to \(\xi'\) in \(\widetilde L\), and define \[ h_e=D(\xi)^{-1}D(\xi'). \tag{31}\] Changing the lift multiplies both \(D\) values on the left by the same \(\alpha(\gamma)\), so the label is independent of the lift. Reversing the path inverts the label. Its image lies in some \(U_i\) because its diameter is at most \(4/N<\ell\). The lift stays in one sheet, and therefore \[|h_e-1|=|D(\xi')-D(\xi)|<\epsilon.\]

The whole image of the boundary of a triangular two-cell lies in one \(U_i\) by (27). The inverse homeomorphism of a sheet over \(U_i\) lifts this boundary to a closed loop in that sheet. In multiplying the three edge labels, choose these consecutive lifts; the intermediate \(D\) values cancel, giving product \(1\).

Finally, choose a finite edge loop based at \(q_0\), \(q_0,q_1,\ldots,q_k=q_0\) in \(L\) whose class \(\gamma\) has \(\alpha(\gamma)\ne1\). Such a loop exists because \(\alpha\) is nontrivial and every element of the fundamental group of a simplicial complex is represented by a finite edge loop. For each successive pair, concatenate the path (28). The endpoint of each piece is \(d a_{q_{i+1}}^{-N}\), the initial vertex of the next piece, so this is a closed edge path in \(X\) based at \(d a_{q_0}^{-N}\). By (29), its image under \(f_N\) is precisely the chosen loop with each edge subdivided. Lift that image from the chosen base lift \(\xi\) of \(q_0\). The terminal lift is \(\gamma\xi\), and telescoping (31) gives the path product \[D(\xi)^{-1}D(\gamma\xi) =D(\xi)^{-1}\alpha(\gamma)D(\xi)\ne1.\] This proves all assertions of the proposition. ◻