Proof. Choose a generator \(\tau\) of the deck group and put \(R=\mathbb C[z,z^{-1}]\). There is a canonical deck action on the pulled-back local system: in the fiber description of \(p^*L\), it is \[ (\widehat x,v)\longmapsto(\tau\widehat x,v), \qquad v\in L_{p(\widehat x)}. \tag{4}\] The equality \(p\tau=p\) makes this action compatible with parallel transport. Let \(z\) act by this deck transformation. Choosing one lift of each cell of \(A\) and a basis in its coefficient fiber identifies \(C_j(\widehat A;p^*L)\) with a finite free \(R\)-module, of rank equal to the number of \(j\)-cells of \(A\).

Taking coinvariants of this deck action identifies translated cell lifts and their coefficient fibers. It therefore gives an isomorphism of chain complexes \[ C_*(\widehat A;p^*L)\otimes_R R/(z-1) \cong C_*(A;L). \tag{5}\] In the chosen bases, let \(D(z)\) be the matrix of the second boundary upstairs, and let \(m\) be its number of columns. The matrix \(D(1)\) is the second boundary downstairs, so it is injective by the hypothesis. If \(m=0\), there is nothing to prove. Otherwise, some \(m\times m\) minor of \(D(1)\) is nonzero. The corresponding minor of \(D(z)\) is consequently a nonzero Laurent polynomial. Thus \(D(z)\) has full column rank over \(\mathbb C(z)\) and is injective over \(R\), since \(R^m\) embeds in \(\mathbb C(z)^m\). There are no cells above dimension two, and hence \(H_2(\widehat A;p^*L)=\ker D(z)=0\). ◻

Proof. Suppose that such a map exists. We will repeatedly lift it to an infinite cyclic cover and replace that cover by the image of the lift. At every step the image will contain strictly more simplices, although the domain retains one fixed finite triangulation. For a finite simplicial complex \(A\), write \(s(A)\) for its number of nonempty simplices.

We first justify working simplicially without increasing the dimension of \(Y\). Its one-skeleton is a finite graph. Subdivide that graph to make it simplicial and homotope each two-cell attaching map to an edge path. After further subdivision if necessary, each nonconstant attaching path can be represented by a map from a polygonal circle with at least three edges, with every edge mapping homeomorphically onto a graph edge. Replace the attachment by the mapping cylinder of this map, capped by a disk along its free boundary circle. This merely inserts a collar into the attached disk. The resulting cylinder has one rectangle for each traversal of a graph edge. Each rectangle is embedded, with its two upper vertices in the separate domain circle and its two lower vertices at distinct endpoints of the graph edge. After compatible subdivisions of boundary edges, triangulate each rectangle from its own interior vertex and each capping disk from a new interior vertex. This yields a finite simplicial complex even when different rectangles traverse the same graph edge: their interiors and upper edges remain distinct. A constant attaching map contributes a triangulated two-sphere joined to the graph at one vertex. Homotoping the attaching maps preserves the homotopy type, so this construction gives a finite simplicial model of dimension at most two.

Transfer \(L\) and \(f\) across this homotopy equivalence. Homotopy invariance preserves both \(H_2(Y;L)=0\) and the nontrivial pullback monodromy. After a sufficiently fine subdivision of \(F\), simplicial approximation makes \(f\) simplicial. Fix this triangulation of \(F\) for the rest of the proof. Let \(A_0=f(F)\), let \(f_0:F\to A_0\) be the corestriction, and let \(L_0=L|_{A_0}\). The simplicial image \(A_0\) is a finite connected subcomplex, and (3) gives \(H_2(A_0;L_0)=0\).

Here is the inductive step. Suppose that \(A_i\) is a finite connected simplicial complex of dimension at most two, \(f_i:F\to A_i\) is a simplicial map onto its image \(A_i\), and \(L_i\) satisfies \[ H_2(A_i;L_i)=0, \qquad f_i^*L_i\text{ has nontrivial monodromy}. \tag{6}\] The second condition implies that \(L_i\) itself has nontrivial monodromy, so \(H_0(A_i;L_i)=0\). Since rank-one cellular chains have the same dimensions as ordinary complex cellular chains, the Euler characteristic gives \[\chi(A_i)=-\dim_\mathbb CH_1(A_i;L_i)\leq0.\] On the other hand, \[\chi(A_i)=1-b_1(A_i;\mathbb Q)+b_2(A_i;\mathbb Q).\] It follows that \(b_1(A_i;\mathbb Q)\geq1\). The finitely generated abelianization of \(\pi_1(A_i)\) therefore has a nonzero free part, and there is an epimorphism \(\phi_i:\pi_1(A_i)\twoheadrightarrow\mathbb Z\).

Let \(p_i:\widehat A_i\to A_i\) be the connected cover corresponding to \(\ker\phi_i\). The composite \(\phi_i(f_i)_*\) is zero: every homomorphism \(\pi_1(F)\to\mathbb Z\) factors through \(H_1(F;\mathbb Z)\), which is finite because \(F\) is finite and \(b_1(F;\mathbb Q)=0\). Thus \(f_i\) lifts to \(\widetilde f_i:F\to\widehat A_i\). Give the cover the lifted triangulation. The lift is simplicial on the already fixed triangulation of \(F\), since on each domain simplex it is the lift of a simplicial map into a single target simplex. No new subdivision of \(F\) is needed.

Set \[A_{i+1}=\widetilde f_i(F), \qquad L_{i+1}=(p_i^*L_i)|_{A_{i+1}},\] and corestrict the lift to \(f_{i+1}:F\to A_{i+1}\). Lemma 2, followed by (3), gives \(H_2(A_{i+1};L_{i+1})=0\). Moreover, \(f_{i+1}^*L_{i+1}\cong f_i^*L_i\), so the nontrivial monodromy persists. These are precisely the conditions (6) needed to continue. Figure 1 shows one step.

It remains to show that \(s(A_i)\) increases strictly. The restriction \(p_i|_{A_{i+1}}\) maps onto \(A_i\) and maps each simplex isomorphically onto a simplex. Hence \(s(A_{i+1})\geq s(A_i)\). If equality held, there would be exactly one simplex above each simplex of \(A_i\), compatibly with all faces. The restriction would then be a simplicial isomorphism. Its inverse, followed by the inclusion into \(\widehat A_i\), would give a section of \(p_i\). Such a section would make \((p_i)_*\) surjective, contrary to \[\operatorname{im}(p_i)_*=\ker\phi_i\ne\pi_1(A_i).\] Thus \(s(A_{i+1})>s(A_i)\). But every simplex of \(A_i\) is the image of a simplex in the fixed triangulation of \(F\), so \[s(A_0)<s(A_1)<s(A_2)<\cdots\leq s(F).\] This is impossible, and the theorem follows. ◻

Proof. Suppose that \(g\) has finite order \(n\) and \(\rho(g)\ne1\); in particular, \(n\geq2\). Let \[M_n=S^1\cup_{\deg n}D^2\] be the Moore complex obtained by attaching a disk along a map of degree \(n\). A loop in \(Y\) representing \(g\) extends to a map \(M_n\to Y\), because its \(n\)th power is nullhomotopic. The cellular boundary \(C_2(M_n;\mathbb Z)\to C_1(M_n;\mathbb Z)\) is multiplication by \(n\), so \(H_1(M_n;\mathbb Z)=\mathbb Z/n\) and \(b_1(M_n;\mathbb Q)=0\). The pullback character sends the circle generator to \(\rho(g)\ne1\).

Use the finite simplicial model for \(M_n\) constructed in the proof of Theorem 3, and compose its homotopy equivalence to \(M_n\) with the map to \(Y\). This gives a finite connected simplicial domain with vanishing first rational Betti number and nontrivial pullback monodromy, contradicting that theorem. For \(n=2\) the standard model \(M_2\) is \(\mathbb{RP}^2\), the case needed below. ◻

Proof. Attach two-dimensional cells along loops representing \(t_1,\ldots,t_r\), and call the resulting complex \(B'\). The van Kampen theorem gives \(\pi_1(B')=G\). Let \(\overline B\) be the full preimage of \(B\) in the universal cover \(\widetilde{B'}\). Since \(P\to G\) is onto, \(\overline B\) is connected; it is the covering of \(B\) corresponding to \(K\). Hence \[H_1(\overline B;\mathbb Z)=K_{\mathrm{ab}}=0.\]

Write \(R=\mathbb ZG\) and regard cellular chains of these covers as left \(R\)-modules. The relative cellular complex \(C_*(\widetilde{B'},\overline B)\) consists of \(R^r\) in degree two and zero in every other degree. The homology exact sequence of the pair therefore gives a surjection \[ H_2(\widetilde{B'};\mathbb Z) \longrightarrow H_2(\widetilde{B'},\overline B;\mathbb Z)=R^r. \tag{7}\] Choose classes \(\alpha_1,\ldots,\alpha_r\) mapping to the standard basis vectors, where the basis records chosen lifts and orientations of the new cells. The space \(\widetilde{B'}\) is simply connected, so the degree-two Hurewicz theorem represents each \(\alpha_j\) by a map \(S^2\to\widetilde{B'}\). Project these maps to \(B'\) and attach one three-dimensional cell along each projected map. This defines \(X\). Only finitely many cells have been added, and attaching the last cells does not change the fundamental group.

The preimage of \(B'\) in \(\widetilde X\) is its universal cover, so we may retain the notation \(\widetilde{B'}\) and \(\overline B\). In the chosen cellular bases, \[C_3(\widetilde X)=R^r, \qquad C_2(\widetilde X)=C_2(\overline B)\oplus R^r.\] Let \(q:C_2(\widetilde X)\to R^r\) be projection onto the second summand. The choice in (7) says exactly that \[ q\,\partial_3=\operatorname{id}_{R^r}. \tag{8}\] Indeed, the relative cellular boundary of the \(j\)th three-dimensional cell is the \(j\)th basis vector; in the two-dimensional complex \(\widetilde{B'}\), a homology class is represented by its cellular two-cycle without any ambiguity from three-boundaries.

Equation (8) proves that \(\partial_3\) is injective. There are no cells above dimension three, so \(H_i(\widetilde X;\mathbb Z)=0\) for all \(i>2\). It also gives the cohomological assertion directly. For any \(R\)-linear map \(\varphi:C_3(\widetilde X)\to M\), its extension \(\varphi q:C_2(\widetilde X)\to M\) satisfies \[(\varphi q)\partial_3=\varphi.\] Thus the cochain boundary \(\operatorname{Hom}_R(C_2(\widetilde X),M)\to\operatorname{Hom}_R(C_3(\widetilde X),M)\) is onto, and \(H^3(X;M)=0\).

Finally, the relative cellular complex of \((\widetilde X,\overline B)\) is \[ 0\longrightarrow R^r\xrightarrow{\,\operatorname{id}\,}R^r \longrightarrow0, \tag{9}\] in degrees three and two. It is contractible. The short exact sequence of absolute and relative cellular complexes is split in each degree, since the cells of \(\overline B\) form subsets of the chosen cellular bases. Tensoring with any right \(R\)-module \(N\) therefore leaves a short exact sequence of complexes whose relative term is still contractible. Its homology exact sequence gives \[H_*(N\otimes_R C_*(\overline B)) \cong H_*(N\otimes_R C_*(\widetilde X)).\] The left side computes \(H_*(B;N)\): the coefficient action of \(P\) factors through the deck group \(G\) of \(\overline B\to B\). The right side computes \(H_*(X;N)\), proving the last assertion. ◻

Proof. The exponents are matched by \(4^2-1=15=3\cdot5\) and \(3^3-1=26=2\cdot13\), with \(\gcd(15,26)=1\). Conjugation by \(t_1=x_1^2\) sends \(a_1\) to \(a_1^{16}\), while conjugation by \(t_2=x_2^3\) sends \(a_2\) to \(a_2^{27}\). Squaring \(x_1=a_2^{13}\) and cubing \(s x_2s^{-1}=a_1^5\) now give \[ t_1=a_2^{26}=[t_2,a_2], \qquad s t_2 s^{-1}=a_1^{15}=[t_1,a_1]. \tag{11}\] Write \(K_{\mathrm{ab}}\) additively and let \(u_j\) be the image of \(t_j\). Conjugation makes \(K_{\mathrm{ab}}\) a left \(\mathbb ZG\)-module: every element of \(K\) acts trivially on its own abelianization. For \(t\in K\), the image of \([t,a]\) is \((1-a)[t]\). Consequently \[ u_1=(1-a_2)u_2, \qquad s u_2=(1-a_1)u_1. \tag{12}\]

Since \(t_1=a_2^{26}\), the element \(a_2\) fixes \(u_1\) and \(a_2^{26}\) acts trivially on all of \(K_{\mathrm{ab}}\). Applying \(1+a_2+\cdots+a_2^{25}\) to the first equality gives \[26u_1=(1-a_2^{26})u_2=0.\] Likewise, \(a_1\) fixes \(s u_2\), because \(s t_2s^{-1}=a_1^{15}\); and \(a_1^{15}\) acts trivially on \(K_{\mathrm{ab}}\). Applying \(1+a_1+\cdots+a_1^{14}\) to the second equality gives \[15s u_2=(1-a_1^{15})u_1=0.\] The action of \(s\) is invertible, so \(15u_2=0\), and the first equality of (12) then gives \(15u_1=0\). Since \(15\) and \(26\) are relatively prime, \(u_1=0\). The second equality gives \(s u_2=0\), hence \(u_2=0\).

Every element of \(K\) is a product of \(P\)-conjugates of \(t_1^{\pm1}\) and \(t_2^{\pm1}\). Thus \(u_1,u_2\) generate \(K_{\mathrm{ab}}\) under addition and the \(P\)-action. Their vanishing proves \(K_{\mathrm{ab}}=0\). ◻

Proof. Proposition 5 reduces the computation to \(B\). Its twisted cellular groups in degrees two and one are \(\mathbb C^4\) and \(\mathbb C^5\), respectively. We compute the second boundary by the lifted edge-path sum, equivalently by Fox differentiation (Fox 1953). In this convention the derivative of a word obeys \[\frac{\partial(uv)}{\partial g} =\frac{\partial u}{\partial g} +u\frac{\partial v}{\partial g},\qquad \frac{\partial g}{\partial g}=1,\qquad \frac{\partial g^{-1}}{\partial g}=-g^{-1},\] and the derivatives with respect to other generators are zero. Evaluate their group-ring coefficients at \(\rho\).

Use the relators \[\begin{split} r_1&=x_1a_1x_1^{-1}a_1^{-4},\qquad r_2=x_2a_2x_2^{-1}a_2^{-3},\\ r_3&=x_1a_2^{-13},\qquad r_4=sx_2s^{-1}a_1^{-5}. \end{split}\] With the edges ordered as \(x_1,a_1,x_2,a_2,s\), the four boundary vectors, written as rows, are \[ D=\begin{pmatrix} 1-\zeta & -2 & 0 & 0 & 0\\ 0&0&2&\zeta^2-1&0\\ 1&0&0&-1&0\\ 0&-(1+\zeta)&1&0&1-\zeta^2 \end{pmatrix}. \tag{15}\] Here is a direct check of the entries. For a relator \(xax^{-1}a^{-n}\), the evaluated entries in the \(x\) and \(a\) coordinates are \[1-\rho(a)^n, \qquad \rho(x)-\sum_{j=0}^{n-1}\rho(a)^j.\] For \(r_1\), the latter sum is \(1+\zeta+\zeta^2+\zeta^3=1\); for \(r_2\), it is \(1-1+1=1\). For \(r_3\), the alternating sum of thirteen terms is \(1\). For \(r_4\), the \(a_1\) entry is \(-\sum_{j=0}^4\zeta^j=-(1+\zeta)\), the \(x_2\) entry is \(\rho(s)=1\), and the \(s\) entry is \(1-\rho(x_2)=1-\zeta^2\). These give (15).

The minor using columns \(x_1,a_1,x_2,s\) has determinant \[ -4(1-\zeta^2)\ne0. \tag{16}\] Thus the four rows are independent, so the boundary \(C_2(B;L_\rho)\to C_1(B;L_\rho)\) is injective. Since \(B\) has no three-dimensional cells, \(H_2(B;L_\rho)=0\), as required. ◻

Proof of Theorem 1. The complex \(X\) constructed above is finite and satisfies (13). In \(G\), the element \(x_1\) has square one, and it is nonidentity because \(\rho(x_1)=-1\). Thus its order is exactly two.

Suppose that \(X\) were homotopy equivalent to a finite connected complex \(Y\) of dimension at most two. Transfer \(\rho\) along the induced isomorphism of fundamental groups, giving a local system \(L\) on \(Y\). Homotopy invariance and Lemma 7 give \(H_2(Y;L)=0\). But the transferred character takes the value \(-1\) on an element of order two, contrary to Corollary 4.

Concretely, a loop representing that element extends to a map \(\mathbb{RP}^2\to Y\): the attaching loop of the two-dimensional cell of \(\mathbb{RP}^2\) is its generator traversed twice. The pulled-back local system has monodromy \(-1\), whereas \(b_1(\mathbb{RP}^2;\mathbb Q)=0\). This is precisely the map excluded by Theorem 3. ◻