The finite plus construction

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Let $B$ be a finite complex of dimension two and $K$ a normal subgroup of its fundamental group $P$. Attaching cells of dimension two kills $K$, but it changes the homology. If $K$ is perfect, the change can be undone by cells of dimension three. The result is a complex $X$ of dimension three with the fundamental group $P/K$ and with the same homology as $B$, for all coefficients that come from $P/K$. This is Quillen's plus construction, in a finite form. This lesson proves that $X$ satisfies the condition D(2).

Perfect subgroups and the cover that belongs to $K$ (Lemma 1.1).
Cycles that pass once through the new cells (Lemma 2.1).
The construction and its three properties (Proposition 3.1).

The lesson follows [OpenAI-W, Section 3]. It uses chains of covers, coefficients in a module and the theorems TOP1 to TOP4 from Wall's D(2) problem and twisted homology, which are proved in the programme.

1. The setting

A group $K$ is perfect if it is generated by its commutators $ghg^{-1}h^{-1}$, that is, if its abelianization $K_{\mathrm{ab}}$ is trivial. The normal closure of elements $t_1,\dots,t_r$ of a group $P$ is the smallest normal subgroup of $P$ that contains them; it consists of the products of conjugates of the $t_j$ and their inverses.

Throughout the lesson:

$B$ is a finite connected CW complex of dimension at most two, and $P=\pi_1(B)$;
$t_1,\dots,t_r\in P$, $K$ is their normal closure, and $K$ is perfect;
$G=P/K$ and $R=\mathbb Z[G]$.

Let $B'$ be the complex $B$ with $r$ further cells $d_1,\dots,d_r$ of dimension two, attached along loops that represent $t_1,\dots,t_r$. By TOP1, $\pi_1(B')=G$, and $\pi_1(B)\to\pi_1(B')$ is the quotient map $P\to G$. Let $p:\widetilde{B'}\to B'$ be the universal cover and $\overline B=p^{-1}(B)\subseteq\widetilde{B'} .$

Lemma 1.1. $\overline B$ is a connected subcomplex, the restriction $\overline B\to B$ is the covering $B_K$ of $B$ that belongs to $K$, and $H_1(\overline B;\mathbb Z)=0$.

Proof. By TOP2, $\overline B$ is a subcomplex, and the group $G$ permutes the lifts of each cell of $B$ freely and transitively. All cells of $\widetilde{B'}$ outside $\overline B$ have the dimension two. So $\overline B$ contains all vertices and edges of the connected complex $\widetilde{B'}$, and it is connected. A loop of $B$ lifts to a closed path in $\widetilde{B'}$ exactly if its class in $\pi_1(B')=G$ is trivial, that is, if its class in $P$ lies in $K$. So $\overline B\to B$ is the connected covering with the group $K$. By TOP3, $H_1(\overline B;\mathbb Z)=K_{\mathrm{ab}}=0$. $\blacksquare$

By Section 3 of the first lesson, the cellular chain groups of $\overline B$ and $\widetilde{B'}$ are free left $R$-modules. Choose lifts $\tilde d_1,\dots,\tilde d_r$ of the new cells. Then $C_2(\widetilde{B'})=C_2(\overline B)\oplus R^r,$ where $R^r$ has the basis $\tilde d_1,\dots,\tilde d_r$; in the other degrees $C_j(\widetilde{B'})=C_j(\overline B)$. Let $q:C_2(\widetilde{B'})\to R^r$ be the projection onto the second summand. It is $R$-linear.

2. Cycles through the new cells

Lemma 2.1. For every $j$ there is a chain $c_j\in C_2(\overline B)$ such that $\alpha_j=\tilde d_j-c_j$ is a cycle of $C_2(\widetilde{B'})$. It satisfies $q(\alpha_j)=\tilde d_j$.

Proof. The chain $\partial\tilde d_j$ lies in $C_1(\overline B)$, and it is a cycle, because $\partial\partial=0$. Since $H_1(\overline B;\mathbb Z)=0$ and the homology of $\overline B$ is computed by its cellular chains (TOP3), there is $c_j\in C_2(\overline B)$ with $\partial c_j=\partial\tilde d_j$. $\blacksquare$

This is the place where the perfectness of $K$ is used. The cycle $\alpha_j$ consists of one new cell and of a chain in $\overline B$ that closes it up.

3. The construction

Proposition 3.1 ([OpenAI-W]; Quillen, Mannan). In the setting of Section 1 there is a finite complex $X$ that contains $B$ and has, besides the cells of $B$, exactly $r$ cells of dimension two and $r$ cells of dimension three, such that:

the inclusion $B\to X$ induces the quotient map $P\to G$ on the fundamental groups;
$H_i(\widetilde X;\mathbb Z)=0$ for all $i>2$, and $H^3(X;M)=0$ for every left $R$-module $M$: the complex $X$ satisfies the condition D(2);
for every right $R$-module $N$ and every $i$, the inclusion induces an isomorphism $H_i(B;N)\cong H_i(X;N),$ where on the left $P$ acts on $N$ through $P\to G$.

Proof. The cells of dimension three. The complex $\widetilde{B'}$ is simply connected and has the dimension two, so $H_2(\widetilde{B'};\mathbb Z)$ is the group of the cycles of $C_2(\widetilde{B'})$. By TOP4 there are maps $\varphi_j:S^2\to\widetilde{B'}$ such that the image of the fundamental class of $S^2$ is the cycle $\alpha_j$ of Lemma 2.1. Attach to $B'$ one cell $e_j$ of dimension three along $p\varphi_j:S^2\to B'$, for $j=1,\dots,r$. This is $X$.

Property 1. By TOP1 the cells of dimension three do not change the fundamental group, and $\pi_1(B)\to\pi_1(B')$ is $P\to G$.

The universal cover of $X$. Since $\pi_1(B')\to\pi_1(X)$ is an isomorphism, the preimage of $B'$ in $\widetilde X$ is the universal cover $\widetilde{B'}$. By TOP2 the cells of dimension three of $\widetilde X$ are the lifts of the $e_j$; let $\tilde e_j$ be the lift that is attached along the lift $\varphi_j$ of $p\varphi_j$. By TOP3, $\partial\tilde e_j=\alpha_j .$ So $C_3(\widetilde X)=R^r$ with the basis $\tilde e_1,\dots,\tilde e_r$, $C_2(\widetilde X)=C_2(\overline B)\oplus R^r$, and $C_j(\widetilde X)=C_j(\overline B)$ for $j\le1$. With the projection $q$ and the identification of the two modules $R^r$ by $\tilde e_j\mapsto\tilde d_j$, Lemma 2.1 says $q\,\partial_3=\mathrm{id}.$ We refer to this equation as (3.1).

Property 2. By (3.1), $\partial_3$ is injective, so $H_3(\widetilde X;\mathbb Z)=0$; there are no cells of a dimension above three. Let $M$ be a left $R$-module and $\psi:C_3(\widetilde X)\to M$ an $R$-linear map. Then $\psi q:C_2(\widetilde X)\to M$ is $R$-linear, and $(\psi q)\partial_3=\psi$ by (3.1). So every element of $C^3(X;M)$ is a coboundary, and $H^3(X;M)=0$.

Property 3. Let $N$ be a right $R$-module. Write $C=C_*(\widetilde X)$ and $C'=C_*(\overline B)$, a subcomplex. By the definition, $C_*(B;N)=N\otimes_{\mathbb Z[P]}C_*(\widetilde B)$. By the formula for covers of the first lesson, $C_*(\overline B)=\mathbb Z[G]\otimes_{\mathbb Z[P]}C_*(\widetilde B)$, because $\overline B$ is the cover $B_K$ of $B$ (Lemma 1.1). So $\begin{gathered} C_*(B;N)=N\otimes_RC',\\ C_*(X;N)=N\otimes_RC , \end{gathered}$ and we must compare the homology of these two complexes. An element of $N\otimes_RC_2$ is a pair $(a,\nu)$ with $a\in N\otimes_RC'_2$ and $\nu\in N^r$, and an element of $N\otimes_RC_3$ is a $\nu\in N^r$. Write $c(\nu)$ for the image of $\nu$ under the map $N^r\to N\otimes_RC'_2$ that the chains $c_j$ define. Then $\begin{gathered} \partial_3\nu=(-c(\nu),\ \nu),\\ \partial_2(a,\nu)=\partial a+\partial c(\nu), \end{gathered}$ the second because $\partial\tilde d_j=\partial c_j$.

Degree $3$: $\partial_3\nu=0$ forces $\nu=0$. So $H_3(N\otimes C)=0=H_3(N\otimes C')$.
Degree $2$: let $(a,\nu)$ be a cycle. Then $(a,\nu)-\partial_3\nu=(a+c(\nu),0)$ is a cycle of $N\otimes C'$. So every class of $H_2(N\otimes C)$ comes from a cycle of $N\otimes C'$. If a cycle $(a,0)$ is a boundary $\partial_3\nu$, then $\nu=0$ and $a=0$. Since $N\otimes C'$ has no chains of degree three, $H_2(N\otimes C')\to H_2(N\otimes C)$ is an isomorphism.
Degree $1$: the two complexes have the same chains and the same cycles of degree one. The boundaries are the same too, because $\partial_2(a,\nu)=\partial(a+c(\nu))$.
Degree $0$: the two complexes agree in the degrees $0$ and $1$. $\blacksquare$

Remarks. The equation (3.1) says more than $H_3(\widetilde X)=0$: the boundary $\partial_3$ is an isomorphism onto a direct summand of $C_2(\widetilde X)$. This is why the cohomology vanishes for every module $M$, and why no condition on $M$ or $N$ is needed.

The complex $X$ depends on choices: the chains $c_j$ and the maps $\varphi_j$. The proposition holds for every choice.

For a character $\rho$ of $G$, property 3 with $N=\mathbb C_\rho$ gives $H_i(B;\rho\pi)\cong H_i(X;\rho),\qquad\pi:P\to G .$ So the twisted homology of $X$ is computed on the complex $B$ of dimension two, by Fox derivatives. But $X$ has a different fundamental group than $B$: elements of $P$ of infinite order may have finite order in $G$. The obstruction of the lesson on the tower applies to complexes of dimension two with the fundamental group $G$, and $B$ is not one of them.

4. Exercises

4.1. Let $B$ be a single vertex with one loop, $P=\mathbb Z$ with the generator $x$, and $t_1=x$. Is $K$ perfect? What goes wrong in Lemma 2.1?

4.2. Show that a normal subgroup $K$ of $P$ acts trivially on $K_{\mathrm{ab}}$ by conjugation, and that $K_{\mathrm{ab}}$ is a left module over $\mathbb Z[P/K]$, generated by the classes of $t_1,\dots,t_r$ if $K$ is their normal closure.

4.3. Let $X$ be as in the proposition. Compute $\chi(X)$ from $\chi(B)$, where the Euler characteristic of a complex of dimension three is $c_0-c_1+c_2-c_3$.

4.4. Where does the proof of property 3 use that the chain groups are tensored over $R$, and where would it fail for a module over $\mathbb Z[P]$ on which $K$ acts nontrivially?

4.5. Derive from (3.1) that $C_2(\widetilde X)=\ker q\oplus\operatorname{im}\partial_3$ and that $\ker q=C_2(\overline B)$.

5. Solutions

4.1. $K=P=\mathbb Z$ is not perfect. $B'$ is a disc, $\overline B=B$ is the circle with $H_1=\mathbb Z$, and $\partial\tilde d_1$ is a generator of $H_1(\overline B)$: it is not the boundary of a chain of $\overline B$. Indeed $B'$ is contractible and has no cycle of dimension two.

4.2. For $k,k'\in K$ the class of $kk'k^{-1}$ in $K_{\mathrm{ab}}$ is the class of $k'$ times the class of the commutator $kk'k^{-1}k'^{-1}$, which is trivial. So the action of $P$ on $K_{\mathrm{ab}}$ by conjugation factors through $P/K$. Every element of $K$ is a product of conjugates $gt_j^{\pm1}g^{-1}$, whose classes are $\pm g\cdot[t_j]$.

4.3. $X$ has $r$ more cells of dimension two and $r$ cells of dimension three: $\chi(X)=\chi(B)+r-r=\chi(B)$.

4.4. The identification $C_*(B;N)=N\otimes_RC_*(\overline B)$ needs that $K$ acts trivially on $N$: it uses $\begin{aligned} &N\otimes_{\mathbb Z[P]}C_*(\widetilde B)\\ &\quad=N\otimes_R\bigl(R\otimes_{\mathbb Z[P]}C_*(\widetilde B)\bigr). \end{aligned}$ The complex $X$ has the fundamental group $G$, so modules over $\mathbb Z[P]$ that do not come from $G$ are not coefficients on $X$ at all.

4.5. For $c\in C_2(\widetilde X)$: $c=(c-\partial_3q(c))+\partial_3q(c)$, and $q(c-\partial_3q(c))=q(c)-q(c)=0$. If $c=\partial_3\nu$ and $q(c)=0$, then $\nu=q\partial_3\nu=0$. The kernel of the projection $q$ is the first summand $C_2(\overline B)$.

References

[OpenAI-W] OpenAI, A Counterexample to Wall's D(2) Problem, OpenAI Math Release preprint, 6 October 2026, Section 3. https://github.com/openai/math/tree/main/preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026
[Hatcher] A. Hatcher, Algebraic Topology, Cambridge University Press, 2002; free on the author's page. https://pi.math.cornell.edu/~hatcher/AT/AT.pdf