Finite Chains of Two-Complexes and Acyclic Covers

Introduction

A two-complex is aspherical if its second homotopy group vanishes. Whitehead asked whether every subcomplex of an aspherical two-complex is aspherical [17]; see [1] for a survey. Suppose that

is a chain of two-complexes. Every map $S^2\to\bigcup_i K_i$ has compact image, hence lands in some $K_i$, and is therefore null-homotopic in $K_{i+1}$. So $L=\bigcup_i K_i$ is aspherical and contains the non-aspherical subcomplex $K_0$, a negative answer to Whitehead’s question [9 §1]. Conversely, Howie Howie [11] showed that if the answer is negative, there is a counterexample that either is finite or is the union of an infinite chain of finite non-aspherical subcomplexes as in (1.1), with every inclusion null-homotopic.

This paper produces examples of two-complexes that admit a chain as in (1.1) of arbitrary finite length, via the following characterization:

Recall that a two-complex $K$ is Cockcroft [4] if its Hurewicz map $h_K:\pi_2(K)\to H_2(K)$ is zero. By Hopf’s exact sequence $\pi_2(K)\to H_2(K)\to H_2(\pi_1K)\to0$ [3 Ch. II, §5], this holds if and only if $H_2(K)\to H_2(\pi_1K)$ is an isomorphism. Moreover $K$ is Cockcroft if and only if there is a single strict inclusion $K\subsetneq L$ of two-complexes that is zero on $\pi_2$ (Lemma 2.1(a)). In particular, in a chain as in (1) the complexes $K_0,\ldots,K_{n-1}$ are Cockcroft, and truncating a chain of length $n+1$ shows that $K_n$ can be taken Cockcroft too. Theorem A thus describes the complexes that are “Cockcroft to every Theorem A does not produce an infinite chain (1): the chains constructed for different $n$ are not extensions of one another.

Proof. $K$ has one vertex, two edges and two 2-cells, so its cellular boundary $\partial_{1}$ vanishes and $\partial_{2}$ is the exponent-sum matrix $\begin{pmatrix}1&2\\2&3\end{pmatrix}$, of determinant $-1$. Hence $K$ is acyclic, so it is its own connected acyclic regular cover, and Theorem A gives the chains, with finite terms. A coset enumeration shows that $G=\pi_{1}(K)$ has order 120; being perfect, $G$ is $\operatorname{SL}(2,5)$. The universal cover $\widetilde{K}$ is simply connected with Euler characteristic $120\chi(K)=120$, so $\pi_{2}(K)\cong H_{2}(\widetilde{K})\cong\mathbb{Z}^{119}$.

Previous work

The second author [14 Theorem 1.1] constructed chains $K_{0}\subset K_{1}\subset K_{2}$ of finite presentation complexes with $\pi_{2}(K_{0})\ne0$ and both inclusions zero on $\pi_{2}$. Corollary B provides such chains of every finite length.

Outline of the proof

• (1)$\Rightarrow$(2) (Section 3). A chain of length $m$ gives $m+1$ quotients of $G=\pi_{1}(K)$. To every 2-chain $v$ on the universal cover of $K$ we associate a “cycle test”, and show that $v$ fails the test at no more than one of these quotients. Hence any $m$ tests are passed simultaneously by some quotient. A compactness argument in the space of normal subgroups of $G$ then gives a normal subgroup $N$ passing all tests, which means $H_{2}(K_{N})=0$ with all coefficients. If $N$ is minimal with this property, a Strebel-type lemma shows that $N$ is perfect, so the cover $K_{N}$ is acyclic.

• (2)$\Rightarrow$(1) (Sections 4–6). In Section 4 we show that it suffices to build the chain over the acyclic cover $D$ itself, with $\pi_{1}(D)$ dying after the first step; the chain descends to $K$ by a pushout. In Section 5 we build such chains over an acyclic $D$, by iterating an operation that turns generators into commutators and replaces relators by “blocks”, and then appending a terminal extension. The blocks are used through three properties (B1)–(B3). In Section 6 we construct the blocks from a truncated non-positively curved cube complex, using constructions of Davis [5] and of Davis–Januszkiewicz–Weinberger [6].

Conventions

A two-complex is a connected CW-complex of dimension at most 2, possibly with infinitely many cells. An inclusion $K\subsetneq L$ means that $K$ is a proper subcomplex of $L$. Homology has integer coefficients unless stated otherwise. A connected space is acyclic if its reduced integral homology vanishes. For a connected complex $K$ with $G=\pi_{1}(K)$ and a normal subgroup $N\trianglelefteq G$, we write $K_{N}$ for the connected regular cover with $\pi_{1}(K_{N})=N$; its deck group is $G/N$. We write $\widetilde{X}$ for the universal cover of $X$, write $\pi_{2}(X)=H_{2}(\widetilde{X})$ for the Hurewicz map, and we identify $\pi_{2}(X)=H_{2}(\widetilde{X})$ as $\mathbb{Z}[\pi_{1}X]$-modules (Hurewicz). A map is zero on $\pi_{2}$ if it induces the zero map on second homotopy groups.

AI usage and formalization

A variety of AI tools, including ChatGPT 6 Astra, Claude 5.5 Fable, and Aristotle, were used to assist with the proof of the main results and the redaction of the manuscript. The interaction with the AI involved a large amount of back-and-forth and experimentation during approximately 5 months. The arguments and text have been reviewed, checked, and substantially revised by the authors, who are responsible for the final manuscript.

Theorem 1 has been formalized in Lean/Mathlib and will shortly be available on the Palomar repository.

Preliminaries

Cockcroft complexes and zero maps

Proof. (a) Let $i:K \subset L$ with $L$ a two-complex. Since there are no 3-cells, $H_2=\ker\partial_2$ on cellular chains, so $H_2(i)$ is injective. If $i$ is zero on $\pi_2$, then $H_2(i)\circ h_K=h_L\circ i_*=0$, hence $h_K=0$. Conversely, let $K$ be Cockcroft. If $\pi_1(K)\ne1$, attach 2-cells along loops representing a generating set of $\pi_1(K)$ (finite if $K$ is finite). The resulting $L$ is simply connected, so $h_L$ is an isomorphism, and $h_L\circ i_*=H_2(i)\circ h_K=0$ gives $i_*=0$. If $\pi_1(K)=1$, then $h_K$ is an isomorphism and zero, so $\pi_2(K)=0$, and we may take $L=K\cup e^1\cup e^2$ with $e^2$ attached along the loop $e^1$.

(b) By the lifting criterion, $f$ lifts to $\widetilde{f}:A\to\widetilde{Y}$, and $f_*=p_*\widetilde{f}_*$ where $p:\widetilde{Y}\to Y$ is the covering. Now $h_{\widetilde{Y}}\widetilde{f}_*=H_2(\widetilde{f})h_A=0$, and $h_{\widetilde{Y}}$ is an isomorphism, so $\widetilde{f}_*=0$.

(c) The Hurewicz map is invariant under the action of $\pi_1$, so $h_X(\gamma\cdot f_*s)=h_X(f_*s)=H_2(f)h_A(s)=0$ for $\gamma\in\pi_1X$, $s\in\pi_2A$. Similarly $g_*(\gamma\cdot f_*s)=g_*(\gamma)\cdot g_*f_*s=0$.

(d) We have $H_2(g)\circ h_X=h_Y\circ g_*=0$ and $H_2(g)$ is injective, so $h_X=0$. For a wedge $W=\bigvee_r X_r$ with retractions $\rho_r:W\to X_r$, the map $H_2(W)\to\prod_r H_2(X_r)$ with coordinates $H_2(\rho_r)$ is injective, and $H_2(\rho_r)h_W=h_{X_r}\rho_{r*}=0$. □

Regular covers

Let $K$ be a two-complex and $G=\pi_1(K)$. Write $I$ and $J$ for the sets of 1-cells and 2-cells of $K$, and fix an orientation and a lift to $\widetilde{K}$ of every cell. For $N\trianglelefteq G$, the images of these lifts in $K_N$ are lifts of the cells of $K$, and the cells of $K_N$ are their translates under the deck group $G/N$. Hence, for $k=\mathbb{Z}$ or $k=\mathbb{F}_p$, the cellular chain module $C_2(K_N;k)$ is identified with $k[G/N]^{(J)}$, and $C_1(K_N;k)$ with $k[G/N]^{(I)}$ (direct sums, left modules). Let

be the cellular boundary map in these bases. For $v\in k[G]^{(J)}$ we write $v_N\in k[G/N]^{(J)}$ for its image; then $\partial_N(v_N)=(\partial_1v)_N$, since the covering $\widetilde{K}\to K_N$ is cellular and maps chosen lifts to chosen lifts.

Proposition 2.2. $H_2(K_N;k)=\ker\partial_N$ (over $k[G/N]$), and $H_1(K_N)=N/[N,N]$.

Proof. $K_N$ has no 3-cells, so $H_2=\ker\partial_2$; and $H_1(K_N)=\pi_1(K_N)^{\mathrm{ab}}=N/[N,N]$. □

Remark 2.3. If $K$ has a single vertex, it is the presentation complex of a presentation $\langle x_i\ (i\in I)\mid r_j\ (j\in J)\rangle$ of $G$, and $\partial_N$ is right multiplication by the matrix $(\partial r_j/\partial x_i)$ of Fox derivatives, reduced to $\mathbb{Z}[G/N]$ [7], [13 Ch. II, §3]. For $N=G$ this is the exponent-sum matrix. Every two-complex is homotopy equivalent to such a complex, by collapsing a maximal tree of its 1-skeleton.

Proof of (1) implies (2)

Throughout this section $K$ is a two-complex satisfying (1), $G = \pi_1(K)$, and we use the notation of Section 2.2. The letter $k$ denotes $\mathbb{Z}$ or $\mathbb{F}_p$ for a prime $p$.

We begin with a special case of a theorem of Strebel [16, Cor. 1.8(ii)]; we include a proof for completeness.

Cycle tests

Definition 3.2. A test is a pair $(k,v)$ with $k \in\{\mathbb{Z}\} \cup\{\mathbb{F}_p : p\ \text{prime}\}$ and $v \in k[G]^{(J)}$. A normal subgroup $M \trianglelefteq G$ passes the test $(k,v)$ if $\partial_M(v_M) = 0$ implies $v_M = 0$; otherwise it fails it. We call $M$ good if it passes every test.

Since every element of $k[G/M]^{(J)}$ is of the form $v_M$, a normal subgroup $M$ is good if and only if $\partial_M$ is injective over $\mathbb{Z}$ and over every $\mathbb{F}_p$, i.e. (Proposition 2.2) $H_2(K_M;k) = 0$ for all such $k$.

Consequently each test is failed by at most one $M_r$, and any $m$ tests are all passed by some $M_r$ with $0 \le r \le m$.

Proof. Fix $r < m$ and write $X = X_r$, $X' = X_{r+1}$, $M = M_r$, $M' = M_{r+1}$. By Lemma 2.4, the components of the preimages of $K$ in $\widetilde{X}$ and $\widetilde{X}'$ through the base points are copies of $K_M$ and $K_{M'}$, and the based lift $\widetilde{X} \to\widetilde{X}'$ of the inclusion restricts to the covering map $K_M \to K_{M'}$. On 2-chains, the latter is the reduction

Assume $\partial_M(v_M) = 0$. Then $v_M$ is a 2-cycle of $K_M \subset\widetilde{X}$, and its image $v_{M'}$ is a 2-cycle of $\widetilde{X}'$. The map $H_2(\widetilde{X};k) \to H_2(\widetilde{X}';k)$ is zero: for $k = \mathbb{Z}$ it is the map $\pi_2(X) \to\pi_2(X')$; for $k = \mathbb{F}_p$, since $H_1(\widetilde{X}) = H_1(\widetilde{X}') = 0$, the universal coefficient theorem identifies it naturally with $(\pi_2(X) \to\pi_2(X')) \otimes\mathbb{F}_p = 0$. Thus $v_{M'}$ is a null-homologous 2-cycle in a complex without 3-cells, so $v_{M'} = 0$.

If a test fails at $r$, then $v_{M_{r+1}} = 0$, hence $v_{M_s} = 0$ for all $s > r$ (as $M_{r+1} \subseteq M_s$), so it passes at every $s > r$. Applying this to the smallest index where the test fails shows that it fails at most once. Since $m$ tests have at most $m$ failures among the $m+1$ indices, some index is failed by none. □

Proof. Identify subsets of $G$ with points of the compact space $\{0,1\}^G$; the normal subgroups form a closed subset $\mathcal{N}$. Fix a test $(k,v)$, and let $S \subset G$ be a finite set containing the supports of all coordinates of $v$ and of $\partial_1v$. Since $v_M$ and $\partial_M(v_M) = (\partial_1v)_M$ are obtained by summing coefficients over left cosets $sM$, and $sM = s'M$ if and only if $s^{-1}s' \in M$, whether $M$ passes $(k,v)$ depends only on $M \cap S^{-1}S$. Hence the set $\mathcal{N}_{(k,v)}$ of normal subgroups passing $(k,v)$ is clopen in $\mathcal{N}$. By hypothesis (1) and Lemma 3.3, applied to a chain of length $m$, any $m$ of the sets $\mathcal{N}_{(k,v)}$ have a common point. By compactness, $\mathcal{G} = \bigcap_{(k,v)} \mathcal{N}_{(k,v)}$ is nonempty, and it is closed.

If $C \subseteq\mathcal{G}$ is totally ordered by inclusion, then $\bigcap C$ is the limit in $\{0,1\}^G$ of the net $C$, ordered by reverse inclusion: for each $g \in G$, membership of $g$ is eventually constant and equal to membership in $\bigcap C$. Hence $\bigcap C \in\mathcal{G}$, and Zorn’s lemma gives a minimal element. □

Proof of (1) $\Rightarrow$ (2). Let $N$ be a minimal element of $\mathcal{G}$. Then $H_2(K_N;k) = 0$ for $k = \mathbb{Z}$ and every $k = \mathbb{F}_p$. By the universal coefficient theorem, $\operatorname{Tor}(H_1(K_N),\mathbb{F}_p) \subseteq H_2(K_N;\mathbb{F}_p) = 0$ for every $p$, so $H_1(K_N) = N/[N,N]$ is torsion-free.

Apply Lemma 3.1 with $Q = G/[N,N]$, $H = N/[N,N]$ and $f = \partial_{[N,N]}$ over $k$. Since $C_*(K_N;k) = k[G/N] \otimes_{k[Q]} C_*(K_{[N,N]};k)$, we have $k[Q/H] \otimes_{k[Q]} f = \partial_N$, which is injective. Hence $\partial_{[N,N]}$ is injective over every $k$, i.e. $[N,N] \in\mathcal{G}$. Minimality gives $[N,N] = N$. Thus $H_1(K_N) = 0$ and $H_2(K_N) = 0$, and $K_N$ is a connected acyclic regular cover of $K$. □

Reduction to the acyclic cover

Let $p\colon D \to K$ be a connected acyclic regular cover, $G = \pi_1(K)$, $N = p_*\pi_1(D)$ and $H = G/N$. Section 5 builds chains over $D$ in which $\pi_1(D)$ dies immediately. This section transfers them to $K$.

two-complexes and all inclusions are zero on $\pi_2$. If $K$ and all $L_i$ are finite, so are all $E_i$.

Proof. Cells and $\pi_1$. The cells of $E_i$ are those of $K$ and those of $L_i \setminus D$, so the $E_i$ are strictly nested two-complexes. Put $\Pi_i=\pi_1(L_i)$. By van Kampen’s theorem $\pi_1(E_i)$ is the pushout of $G \leftarrow\pi_1(D) \to\Pi_i$. The first map has normal image $N$ and the second is trivial, so

$\pi_2$ is generated from $L_i$. Write $\Lambda=\Lambda_i$ and $\Pi=\Pi_i$. By Lemma 2.4, in $\widetilde{E_i}$: the preimage of $K$ is a disjoint union of copies of $\widetilde{K}_N=D$ indexed by $\Lambda/H$ (the kernel of $G\to\Lambda$ is $N$, since $H$ embeds in $H*\Pi$); the preimage of $L_i$ is a disjoint union of copies of $\widetilde{L_i}$ indexed by $\Lambda/\Pi$; and the preimage of $D$ is a disjoint union of copies of $D$ indexed by $\Lambda$, as $\pi_1(D)$ dies in $\Lambda$. These are subcomplexes, the first two cover $\widetilde{E_i}$ and intersect in the third, so we have a short exact sequence of cellular chain complexes

Since $H_1(D)=H_2(D)=0$, the long exact homology sequence gives an isomorphism

that is,

induced by $L_i\to E_i$. In particular $\pi_2(E_i)=\mathbb{Z}[\Lambda]\cdot\operatorname{im}\pi_2(L_i)$.

Zero maps. The covering $p$ is bijective on $\pi_2$, and the composite $D \xrightarrow{p} K \subset E_1$ equals $D\subset L_1\to E_1$, which is zero on $\pi_2$; hence $K\subset E_1$ is zero on $\pi_2$. For $i\ge1$, the square formed by $L_i\subset L_{i+1}$ and $E_i\subset E_{i+1}$ commutes, its top arrow is zero on $\pi_2$, and $\pi_2(E_i)$ is generated by the image of $\pi_2(L_i)$; so $E_i\subset E_{i+1}$ is zero on $\pi_2$ by Lemma 2.1(c). □

Proof of (2) $\Rightarrow$ (1), given Proposition 5.1. Apply Proposition 5.1 to the acyclic cover $D$ and then Proposition 4.1. If $K$ is finite and acyclic, take $D=K$ and $p$ the identity; then $E_i=L_i$, which are finite by Proposition 5.1. □

The chain over an acyclic complex

Proposition 5.1. Let $D$ be an acyclic two-complex and $n\ge1$. There are two-complexes

with all inclusions zero on $\pi_2$ and $\pi_1(D)\to\pi_1(L_i)$ trivial for $i\ge1$. If $D$ is finite, all $L_i$ can be chosen finite.

Presentations extending $D$

Fix a maximal tree $T_D$ of the 1-skeleton of $D$ and a base vertex. Collapsing $T_D$ gives a presentation $\langle x_i\ (i\in I)\mid r_j\ (j\in J)\rangle$, where the $x_i$ are the edges of $D$ outside $T_D$; we fix a total order on $I$. Every word in the $x_i$ determines a based edge loop in $D$, going through $T_D$. A presentation extending $D$ is a complex $P\supseteq D$ obtained by attaching loops at the base vertex (extra generators) and 2-cells along words in the $x_i$ and the extra generators (extra relators). The $x_i$ and $r_j$ are the core generators and relators. Relators are labelled: equal words with different labels are different 2-cells. We write $G(P)=\pi_1(P)$, and $K(P)$ for the presentation complex of a presentation $P$. Since $P\simeq P/T_D$, and $K(P)$ is the presentation complex of the corresponding presentation, we freely pass between the two.

Since $D$ is acyclic and $D/T_D$ has a single vertex, its cellular boundary (Remark 2.3)

is an isomorphism, since its kernel is $H_2(D)$ and its cokernel is $H_1(D)$. Note that $\Theta^{-1}$ maps finitely supported vectors to finitely supported vectors.

If $I=\varnothing$, then $D/T_D$ is a wedge of $|J|$ spheres, so $J=\varnothing$ since $H_2(D)=0$, and $D$ is contractible. We then take for $L_i$ the complex $D$ with $i$ pairs $(e^1,e^2)$ attached, $e^2$ along the loop $e^1$; all $\pi_2$ vanish. Henceforth $I\ne\varnothing$.

Outline

We write $[u,v]=uvu^{-1}v^{-1}$. The chain is built by induction on $n$. We start from a pair $D\subset Y_D$ that kills $\pi_1(D)$ (Lemma 5.2). Given a chain $D=L_0\subset\cdots\subset L_m$, we apply to each term an operation $T$ that turns every extra generator $z$ into a commutator $[a_z,b_z]$ of two new generators, corrects every extra relator by core relators so that it becomes a product of commutators, and then replaces it by a block. By the defining property (B2) of blocks, $\pi_2$ of $T(P)$ is generated by the image of $\pi_2(P)$, so $T$ preserves inclusions that are zero on $\pi_2$ (Lemma 5.4). We then append the terminal extension $T^+(L_m)$, obtained by capping all generators $a_z,b_z$. It kills $\pi_1(L_m)$, and its second homology comes from “capped blocks”, which are Cockcroft by (B3) (Lemma 5.5). Each such pass lengthens the chain by one.

The initial pair

Define

a presentation extending $D$.

Proof. The group. Eliminate $x_i=U_i:=[a_i,b_i]$. The cross-commutators make the subgroups $\langle a_i,b_i\rangle$ commute pairwise, so the $U_i$ commute, and $r_j$ becomes $r_j(U)=\prod_i U_i^{\Theta_{ij}}$. For $c=\Theta^{-1}e_i\in\mathbb{Z}^{(J)}$ we get

So $x_i=1$ and all $a_i,b_i$ commute. Conversely, sending $a_i,b_i$ to a basis of the free abelian group on $\{a_i,b_i\}$ and $x_i$ to $0$ satisfies every relator.

Cockcroft. Eliminating the $x_i$ by the relators $x_iU_i^{-1}$ is a Tietze transformation, hence a homotopy equivalence [10 Ch. I], from $Y_D/T_D$ to the presentation complex $Y'$ with generators $a_i,b_i$ and relators $r_j(U)$ and the cross-commutators. All these relators have zero exponent sums, so the cellular boundary $\partial_2$ of $Y'$ vanishes and $H_2(Y')$ is free on the $2$-cells. Let $\mathbb{T}$ be the union of the finite subtori of $\prod_{i\in I}(S_{a_i}^1\times S_{b_i}^1)$, with its product cell structure. It is a $K(\mathbb{Z}^{(2I)},1)$, and $H_2(\mathbb{T})$ is free on the coordinate squares. Map $Y'\to\mathbb{T}$ by sending generators to the corresponding circles, each cross-commutator disk onto its square with degree one, and the disk of $r_j(U)$, whose boundary is a concatenation of loops $U_i^{\pm1}$, onto the concatenation of the corresponding $(a_i,b_i)$-squares. On $H_2$:

A class in the kernel has zero cross-commutator coefficients, and its vector $\lambda\in\mathbb{Z}^{(J)}$ of remaining coefficients satisfies $\Theta\lambda=0$, so $\lambda=0$. Thus $H_2(Y')\to H_2(\mathbb{T})$ is injective, and $Y'\simeq Y_D$ is Cockcroft by Lemma 2.1(d).

The inclusion. $D\subset Y_D$ is trivial on $\pi_1$, as $\pi_1(D)$ is generated by the $x_i$, and $D$ is Cockcroft since $H_2(D)=0$; apply Lemma 2.1(b). ∎

Blocks

For $q \ge1$ let $\Sigma_q$ be the closed oriented surface of genus $q$ with its standard one-vertex cell structure: edges $s_1,t_1,\ldots,s_q,t_q$ and one 2-cell attached along $\omega_q=\prod_{h=1}^{q}[s_h,t_h]$.

Section 6 proves this theorem. Property (B2) is what makes $T$ preserve inclusions that are zero on $\pi_2$, and (B3) is what makes the terminal extension Cockcroft.

The operations $T$ and $T^+$

Let $P$ be a presentation extending $D$. Fix once and for all, for every word $w$ with zero exponent sum in every generator, an expression $w=\prod_{h=1}^{q}[u_h,v_h]$ in the free group with $q\ge1$. Such expressions exist since such $w$ lie in the commutator subgroup; for instance, repeatedly use $p\ell mt=[p\ell p^{-1},pmp^{-1}]\cdot pm\ell t$ to sort the letters of $w$ according to a fixed order on the generators, and use $[1,1]$ if no commutator occurs. Define $T(P)$ by the following three rules.

1. Generators become commutators. For each extra generator $z$ of $P$ take a fresh pair of generators $a_z,b_z$ and substitute $z\mapsto[a_z,b_z]$ in every extra relator. Core generators and core relators are untouched. Call $r^*$ the substituted form of an extra relator $r$.

2. Correction by core relators. Let $e\in\mathbb{Z}^{(I)}$ be the vector of exponent sums of $r^*$ in the $x_i$, and $c=-\Theta^{-1}e\in\mathbb{Z}^{(J)}$. Replace $r^*$ by

which has zero exponent sum in every generator: in the $x_i$ by the choice of $c$, and in the $a_z,b_z$ because they only occur inside the commutators $[a_z,b_z]$.

3. Blocks. Write $r^0=\prod_{h=1}^{q}[u_h,v_h]$ using the fixed expression, and replace the relator $r^0$ by the block (5.3), with fresh internal generators $Z_r$ for each label $r$.

The core is unchanged, so $T(D)=D$. The words $u_h,v_h$ involve only the $x_i$ and the $a_z,b_z$. Composing the maps of the three rules (the third being the structural map of (B2)) gives a structural map $\eta_P:P\to T(P)$, with $x_i\mapsto x_i$ and $z\mapsto[a_z,b_z]$. The terminal extension is

internal generators of blocks are not capped.

Naturality. All choices in $T$ (the names $a_z,b_z$, the corrections, the factorizations, the internal generators) depend only on the labelled generator or relator concerned. Hence a labelled inclusion $j:P\subset P'$ induces a labelled inclusion $T(j):T(P)\subset T(P')$ with $T(j)\circ\eta_P=\eta_{P'}\circ j$. A new extra generator yields a new pair $a_z,b_z$, and a new extra relator yields a new block, which has $m_q\geq1$ relators; so $T$ preserves strict inclusions.

Proof. If $f:A\to B$ and $g:B\to C$ satisfy the generation property $\pi_2(B)=\mathbb{Z}[\pi_1B]\cdot f_*\pi_2(A)$ and similarly for $g$, then so does $gf$, because $g_*(\gamma s)=g_*(\gamma)g_*(s)$. It therefore suffices to check generation for each rule.

Rule 1. Let $P'$ be obtained from $P$ by adding, for each extra generator $z$, a loop $b_z$, a loop $a_z$ and a 2-cell along $a_zb_za_z^{-1}(zb_z)^{-1}$. This relator says $z=[a_z,b_z]$, and eliminating each $z$ by it is a Tietze transformation [10 Ch. I], giving the presentation of rule 1; under this homotopy equivalence the inclusion $P\subset P'$ becomes the map of rule 1. Now $P'$ is a graph of spaces [15]: the vertex space is $P\vee\bigvee_z S^1_{b_z}$, and each new 2-cell together with $a_z$ is a cylinder $S^1\times[0,1]$ glued along $b_z$ and $zb_z$. Both loops have infinite order in $G(P)*F(\{b_z\})$, as is seen by killing $G(P)$. So $G(P')$ is an HNN extension along infinite cyclic subgroups, in which the vertex group embeds [13 Ch. IV, §2]. Consequently $\widetilde{P'}$ is a tree of spaces whose vertex spaces $V_\alpha$ are copies of the universal cover of $P\vee\bigvee_z S^1_{b_z}$ and whose edge spaces are strips $\mathbb{R}\times[0,1]$. Since $H_*(\mathbb{R}\times[0,1],\mathbb{R}\times\{0,1\})$ is concentrated in degree 1, excision gives $H_2(\widetilde{P'})/\bigoplus_\alpha V_\alpha=0$, so $\bigoplus_\alpha H_2(V_\alpha)\to H_2(\widetilde{P'})$ is onto. By the same argument applied to the wedge with circles (a tree of copies of $\widetilde P$ joined by lines), each $H_2(V_\alpha)$ is generated by translates of $\pi_2(P)$.

Rule 2. The attaching maps of $r^*$ and $r^0$ are homotopic in the complex obtained by removing all extra relator cells, since the correcting word is null-homotopic in $D$. Hence the map of rule 2 is a homotopy equivalence [8 Prop. 0.18].

Rule 3. This is exactly (B2).

The consequences follow from (5.7) and Lemma 2.1(c): if $P$ is Cockcroft, so is $T(P)$; and if $j_*=0$ on $\pi_2$, then $T(j)_*\eta_{P*}=\eta_{P'*}j_*=0$, so $T(j)_*=0$. □

Proof. The $x_i$ remain trivial because $\eta_P$ fixes them.

The inclusion. The composite $P\to T(P)\subset T^+(P)$ is trivial on $\pi_1$: the $x_i$ are trivial by hypothesis, and each extra generator $z$ maps to $[a_z,b_z]$, a commutator of two capped loops. Since $P$ is Cockcroft, the composite is zero on $\pi_2$ by Lemma 2.1(b). By (5.7) and Lemma 2.1(c), the inclusion $T(P)\subset T^+(P)$ is zero on $\pi_2$.

Cockcroft. Since $D$ is acyclic, the quotient map $\kappa:T^+(P)\to T^+(P)/D$ is an isomorphism on $H_2$ (long exact sequence of the pair). In $T^+(P)/D$, all $x_i$ and core relators have disappeared, and the remaining relators are the loops $a_z,b_z$ with their capping disks, and the blocks. Eliminating the $a_z,b_z$ by their capping disks is a homotopy equivalence [10 Ch. I]. As the words $u_h,v_h$ only involve the $x_i$ and the $a_z,b_z$, every block relator $\beta_{q,j}(u,v,Z_r)$ becomes $\beta_{q,j}(1,\ldots,1,Z_r)$, and distinct blocks have disjoint internal generators. Therefore

the wedge being over the extra relators $r$ of $P$. Each wedge summand is Cockcroft by (B3), hence so is the wedge (Lemma 2.1(d)), and $T^{+}(P)$ is Cockcroft by Lemma 2.1(d) applied to $\kappa$. $\square$

Proof of Proposition 5.1

We construct chains $D=L_0\subsetneq L_1\subsetneq\cdots\subsetneq L_m$ of presentations extending $D$, related by labelled inclusions, such that

By Lemma 5.2, $(D,Y_D)$ is such a chain with $m=1$. Given such a chain, set

By naturality, this is a strictly increasing chain of labelled inclusions; note that $T^{+}(L_m)\ne T(L_m)$ since $L_m\supseteq Y_D$ has extra generators. By Lemma 5.4, the inclusions $D=T(D)\subset T(L_1)$ and $T(L_i)\subset T(L_{i+1})$ are zero on $\pi_2$, and every $T(L_i)$ is Cockcroft. By Lemma 5.5, applicable because $L_m$ satisfies (10) and $m\geq1$, the $x_i$ are trivial in $G(T(L_i))$ and $G(T^{+}(L_m))$, the complex $T^{+}(L_m)$ is Cockcroft, and $T(L_m)\subset T^{+}(L_m)$ is zero on $\pi_2$. So the new chain satisfies (10). Applying (11) $n-1$ times to $(D,Y_D)$ gives a chain of length $n$; as $\pi_1(D)$ is generated by the $x_i$, it is killed in every $L_i$ with $i\geq1$.

If $D$ is finite, then $I$ and $J$ are finite, and each pass introduces finitely many generators and finitely many finite blocks; so all $L_i$ are finite. $\square$

Construction of the blocks

This section proves Theorem 5.3. We combine two constructions. Davis’s real moment-angle complexes realize a prescribed finite flag complex as every vertex link of a cube complex [5 Example 2.2]. The relative hyperbolization of Davis–Januszkiewicz–Weinberger [6 §2] cones off boundary components, hyperbolizes, and truncates the cone points. We adapt two of their arguments, the radial retraction [6 Lemma 2.2] and the CAT(0) property of a branched cover [6 Lemma 2.6], to show that $\pi_2$ does not grow when an arbitrary two-complex is glued into the truncated corner (Lemma 6.2).

The cube complex and its truncation

A simplicial complex is flag if every finite set of pairwise adjacent vertices spans a simplex. For a finite flag complex $L$ on the vertex set $\{1,\ldots,d\}$ put

a cube complex whose cubes are indexed by a simplex $\sigma$ and a choice of signs on the remaining coordinates. Every vertex link of $C(L)$ is isomorphic to $L$. We use Gromov’s link condition: a cube complex whose vertex links are flag is CAT(0), so its universal cover is CAT(0) and in particular contractible [2 Thms. II.5.20 and II.4.1(2)]. For cube complexes that are not locally finite, which occur in the proof of Lemma 6.2, see [12, Thm. C.9 and its proof; Thms. C.4, A.7].

Choose a finite triangulation of $\Sigma_q$ in which the $2q$ standard loops are edge paths, and let $L_q$ be its barycentric subdivision. It is flag, since its simplices are the chains of faces of the original triangulation, and pairwise comparable faces form a chain.

Number the vertices of $L_q$, set $C_q = C(L_q)$ and $v = (1,\ldots,1)$, and cut off the corner at $v$:

Since every other vertex of $C_q$ is at $\ell^1$-distance at least 2 from $v$, only the corner at $v$ is removed. The cut surface $C_q \cap\left\{\sum_h(1-x_h)=\frac{1}{2}\right\}$ is a copy of the link of $v$, hence of $\Sigma_q$ (triangulated as $L_q$), and the removed part is the open cone on it. We regard $\Sigma_q$ as the cut surface, the standard loops being edge paths in it.

Proof. (i) This follows from the discussion above, as $\dim L_q = 2$.

(iii) Since $L_q$ is a closed surface, every edge $\tau$ of $L_q$ lies in exactly two triangles; hence every square of $C_q$ lies in exactly two 3-cubes. Orient $\Sigma_q$, and let $o(\sigma)=\pm1$ compare the orientation of a triangle $\sigma$ with the one given by increasing order of its vertices. Orient each cube $[-1,1]^\sigma\times\{\epsilon\}$ by increasing order of coordinates, and give it the coefficient $o(\sigma)\prod_{h\notin\sigma}\epsilon_h$. Let a square $[-1,1]^\tau\times\{\eta\}$ lie in the cubes indexed by $\sigma_i=\tau\cup\{k_i\}$, $i=1,2$, and let $p_i$ be the number of elements of $\tau$ smaller than $k_i$. The square appears in the boundary of the $i$-th cube with coefficient $\eta_{k_1}\eta_{k_2}E\cdot o(\sigma_i)(-1)^{p_i}$, where $E=\prod_{h\notin\sigma_1\cup\sigma_2}\eta_h$; and $o(\sigma_1)(-1)^{p_1}+o(\sigma_2)(-1)^{p_2}=0$ because $\tau$ cancels in the fundamental cycle of $\Sigma_q$. So these coefficients define a 3-cycle $\zeta$ of $C_q$. Now consider the exact sequence

By excision $H_3(C_q,M_q)\cong H_3(\operatorname{Cone}\Sigma_q,\Sigma_q)\cong H_2(\Sigma_q)\cong\mathbb{Z}$, and under this identification $\delta$ becomes the map induced by $\Sigma_q\subset M_q$. The cubes containing $v$ are those with all signs $+1$, and their coefficients in $\zeta$ are the $o(\sigma)$; so $\zeta$ maps to a generator of $H_3(C_q,M_q)$. Hence $\delta=0$, and $[\Sigma_q]\mapsto0$.

(ii) $M_q$ is a polyhedral complex whose 3-cells are the cubes of $C_q$, those containing $v$ being truncated. Consider the graph whose vertices are these 3-cells, two being adjacent when they share a 2-cell. It is connected: the cubes containing a given vertex $w\in\{\pm1\}^d$ correspond to the triangles of $L_q$, and are connected because $L_q$ is a connected closed surface; and two vertices $w,w'$ differing in one coordinate $k$ lie in a common cube, namely any cube indexed by a triangle containing $k$. Choose a spanning tree, rooted at a truncated cube. A triangle of the cut surface lies in only one 3-cell, so we can collapse the root across it. Then collapse each other 3-cell, in order of distance to the root, across the 2-cell it shares with its parent: that 2-cell lies in exactly two 3-cells and the parent has been removed, so it is a free face. All 3-cells disappear. $\square$

The presentation $B_q$ and the marking (B1)

Let $S$ be the two-dimensional subcomplex of Lemma 6.1(ii), and let $\rho:M_q\to S$ be the deformation retraction given by the collapse. By cellular approximation [8 Thm. 4.8], $\rho|_{\Sigma_q}$ is homotopic to a cellular map $c:\Sigma_q\to S$, where $\Sigma_q$ carries its one-vertex structure. Contract a maximal tree of $S$, call $z_1,\ldots,z_k$ the remaining oriented edges, let $w_F(z)$ be the boundary word of each 2-cell $F$ of $S$, and let $\mu_h(z),\nu_h(z)$ be the words read along $c(s_h), c(t_h)$. Put

with $Z_q=(z_1,\ldots,z_k)$. The relators $s_h\mu_h^{-1},t_h\nu_h^{-1}$ form generator–relator pairs, so the inclusion $S/\text{tree}\subset K(B_q)$ is a homotopy equivalence, and $K(B_q)\simeq M_q$. Define $b_q$ on the 1-skeleton by $s_h\mapsto s_h$, $t_h\mapsto t_h$. The loop $\omega_q(s,t)$ is homotopic in $K(B_q)$, across the cells $s_h\mu_h^{-1},t_h\nu_h^{-1}$, to $\omega_q(\mu,\nu)$. This is the image under $c$ of the boundary of the 2-cell of $\Sigma_q$, so it bounds the image under $c$ of that 2-cell. Following the homotopy and then this disk defines $b_q$ on the 2-cell, and gives (B1). By construction, under $K(B_q)\simeq M_q$ the map $b_q$ corresponds to the inclusion $\Sigma_q\subset M_q$ up to homotopy.

Generation in the pushout: proof of (B2)

Let $E$, $r=\prod_h[u_h,v_h]$ and $E'$ be as in (B2), and let $a:\Sigma_q\to K(E)$ be the cellular map sending $s_h,t_h$ to the loops $u_h,v_h$ and the 2-cell of $\Sigma_q$ onto the 2-cell $r$ with degree one (through a collar realizing the free reduction of $\prod_h[u_h,v_h]$ to $r$). Consider the double mapping cylinder

In $W$, the 3-cell (polygon)$\times(0,1)$ has the 2-cell $r$ as a free face; removing both is an elementary collapse. The remaining cylinder cells are an edge (vertex)$\times(0,1)$ joining the two base points, and 2-cells (edge)$\times(0,1)$ identifying $s_h,t_h$ with $u_h,v_h$. Contracting the edge and eliminating $s_h,t_h$ by these 2-cells gives exactly the presentation complex of (5.3). Under this homotopy equivalence $W\simeq K(E')$, the inclusion $K(E)\subset W$ corresponds to the structural map $f$. Moreover, since double mapping cylinders are homotopy invariant [8 Ch. 0], and $b_q$ corresponds to the inclusion of the cut surface, $W$ is homotopy equivalent rel $K(E)$ to the space in the following lemma, with $X=K(E)$. Three-dimensional spaces occur only in this proof: every complex in our chains is a two-complex.

Proof. Collapsing $X$ to a point turns $\Sigma_q\times[0,1]$ into $\operatorname{Cone}(\Sigma_q)$, so $W/X=C_q$; let $\kappa:W\to C_q$ be the quotient map. It is a homeomorphism from $W\setminus X$ onto $C_q\setminus\{v\}$.

Step 1: $\pi_1$-injectivity (cf. [6 Lemma 2.2]). Let $\widehat{W}\to W$ be the pullback of the universal cover $\widetilde{C}_q\to C_q$ along $\kappa$. It is obtained from $\widetilde{C}_q$ by replacing each lift $v_\lambda$ of $v$, together with its open cone neighbourhood, by a copy $X_\lambda\cup_a(\Sigma_q\times[0,1])$; in particular it is connected. Fix a lift $v_0$. Since $\widetilde{C}_q$ is CAT(0), each point $y$ outside the open cone neighbourhood of $v_0$ is joined to $v_0$ by a unique geodesic, which crosses the cut surface $\Sigma_{(0)}$ at $v_0$ exactly once (inside each cube at $v_0$ geodesics from $v_0$ are straight segments, and the cut is an affine hyperplane). Mapping $y$ to this crossing point defines a retraction $\rho$ onto $\Sigma_{(0)}\cong\Sigma_q$, continuous because geodesics in CAT(0) spaces depend continuously on their endpoints. Define $R:\widehat{W}\to X_0=X$ as follows:

• on the lift of $M_q$, $R=a\circ\rho$;

• on $X_0$, $R$ is the identity, and on its cylinder $R(\sigma,t)=a(\sigma)$;

• at $\lambda\ne0$: $\rho$ is defined on the whole closed cone neighbourhood of $v_\lambda$, so its restriction to the cut surface $\Sigma_{(\lambda)}\cong\Sigma_q$ extends over a cone and is therefore homotopic to a constant map by a homotopy $H:\Sigma_q\times[0,1]\to\Sigma_q$ with $H_1=\rho|_{\Sigma_{(\lambda)}}$ and $H_0$ constant. Put $R(\sigma,t)=a(H(\sigma,t))$ on the cylinder and let $R$ be constant on $X_\lambda$.

These definitions agree on overlaps, so $R$ is a retraction of $\widehat{W}$ onto $X_0$. Hence $\pi_1(X) \to\pi_1(\widehat{W})$ is injective, and so is $\pi_1(\widehat{W}) \to\pi_1(W)$; their composite is induced by $X \subset W$.

Step 2: a CAT(0) quotient (cf. [6 Lemma 2.6]). Let $U \subset\widehat{W}$ be the preimage of $X$. By Step 1 and Lemma 2.4, its components $U_\alpha$ are copies of $\widehat{X}$. Collapse each $U_\alpha$ to a point $v_\alpha$ to obtain a space $V$, and let $V_0 = \{v_\alpha\}$. As $\widehat{W}$ is simply connected and the $U_\alpha$ are disjoint connected subcomplexes, $V \simeq\widehat{W} \cup_\alpha\operatorname{Cone}(U_\alpha)$ [8 Ch. 0] is simply connected by van Kampen's theorem. The cylinders attached to $U_\alpha$ become cones with apex $v_\alpha$, and each truncated cube together with the cone on its cut triangle becomes a full cube; so $V$ is a cube complex. The vertex links at old vertices are copies of $L_q$. The link at $v_\alpha$ is the preimage in $\widehat{W}$ of the cut surface adjacent to $U_\alpha$, that is, the pullback of $\widehat{X} \to X$ along $a$: a disjoint union of covers of $L_q$. Covers of $L_q$ are flag: the vertices of a clique in a cover project to pairwise adjacent, hence distinct, vertices of $L_q$, which span a simplex; this simplex lifts to a simplex through any vertex of the clique, and since a covering map is an isomorphism on the star of a vertex, the lift contains the whole clique. Hence $V$ is CAT(0) (by [12], as $V$ need not be locally finite), so contractible.

Step 3: homology. Since $(\widehat{W}, U) \to(V, V_0)$ is a relative homeomorphism of CW pairs and $V_0$ is discrete, $H_2(\widehat{W}, U) \cong H_2(V, V_0) \cong H_2(V) = 0$. Hence $H_2(U) \to H_2(\widehat{W})$ is surjective. By Lemma 2.4, $\pi_1(W)$ permutes the components of $U$ transitively, and the component through the base point is a copy of $\widehat{X}$ with $H_2 = \pi_2(X)$. Since $\pi_2(W) = H_2(\widehat{W})$, it is generated over $\mathbb{Z}[\pi_1 W]$ by the image of $\pi_2(X)$. □

Lemma 6.2 gives (5.4) for a single replacement, and by induction (generation is preserved under composition, as in the proof of Lemma 5.4) for finitely many. For an arbitrary set of replacements, let $E'$ be the full replacement and $s \in\pi_2(E')$. A representing sphere is compact, so it meets only finitely many blocks, coming from a finite set $\Phi$ of relators. Let $E'_\Phi$ be the presentation in which only the relators in $\Phi$ are replaced. The structural map $g : E'_\Phi\to E'$ replacing the remaining relators is the identity on the subcomplex containing the sphere, so $s = g_*(s')$ for some $s' \in\pi_2(E'_\Phi) = \mathbb{Z}[G(E'_\Phi)] \cdot f_{\Phi*}\pi_2(E)$. Since $g \circ f_\Phi= f$ and $g_*$ is equivariant, $s \in\mathbb{Z}[G(E')] \cdot f_*\pi_2(E)$. This proves (B2).

The capped block: proof of (B3)

Proof. By homotopy invariance, $C_q = M_q \cup_{\Sigma_q} \operatorname{Cone}(\Sigma_q) \simeq K(B_q) \cup_{b_q} \operatorname{Cone}(\Sigma_q)$. With the one-vertex structure on $\Sigma_q$, the cone consists of the cone point, an edge (the cone on the vertex), 2-cells (the cones on $s_h,t_h$) and one 3-cell $e^3$ (the cone on the polygon). Contracting the edge [8 Prop. 0.17] gives $C_q \simeq V_q \cup e^3$.

The cellular boundary of $e^3$ consists of the 2-chain $b_q\#[\Sigma_q]$ on $K(B_q)$, and the capping 2-cells with coefficients the exponent sums of $s_h$ and $t_h$ in $\omega_q$, which vanish. The chain $b_q\#[\Sigma_q]$ is a 2-cycle of $K(B_q)$ representing the image of $[\Sigma_q]$, which is zero by Lemma 6.1(iii) since $b_q$ corresponds to the inclusion $\Sigma_q \subset M_q$. A null-homologous 2-cycle in a two-complex is the zero chain, so $\partial e^3 = 0$, and $H_2(V_q) \to H_2(V_q \cup e^3)$ is an isomorphism. As $V_q \cup e^3 \simeq C_q$ is Cockcroft by Lemma 2.1(d). Finally, eliminating each $s_h,t_h$ by its capping 2-cell is a homotopy equivalence from $V_q$ to the presentation complex of the capped presentation $\langle Z_q \mid\beta_{q,j}(1,\ldots,1,Z_q)\rangle$. □

This proves (B3), and completes the proofs of Theorem 5.3, Proposition 5.1 and Theorem A. □

References