Aspherical completions and rationally inert elements

TL;DR

This paper investigates rationally inert elements in the homotopy groups of spaces, extending previous results to non-simply connected cases and characterizing their properties in Poincaré duality complexes and wedge spaces.

Contribution

It extends the theory of rationally inert elements to broader classes of spaces, including non-simply connected and Poincaré duality complexes, providing new characterizations and properties.

Findings

Rationally inert elements in Poincaré duality complexes require the algebra to have at least two generators.

In wedge of spheres, rationally non-trivial maps are rationally inert.

The homotopy fiber of the inclusion has a rational homotopy type as a completion of a free Lie algebra.

Abstract

Let $X$ be a connected space. An element $[f]\in \pi_n(X)$ is called rationally inert if $\pi_*(X)\otimes \mathbb Q \to \pi_*(X\cup_fD^{n+1})\otimes \mathbb Q$ is surjective. We extend the results obtained in the simply connected case, and prove in particular that if $X\cup_fD^{n+1}$ is a Poincar\'e duality complex and the algebra $H(X)$ requires at least two generators then $[f]\in \pi_n(X)$ is rationally inert. On the other hand, if $X$ is rationally a wedge of at least two spheres and $f$ is rationally non trivial, then $f$ is rationally inert. Finally if $f$ is rationally inert then the rational homotopy of the homotopy fibre of the injection $X \to X\cup_fD^{n+1}$ is the completion of a free Lie algebra.

Equations

$${\pi }_{\ast }(X)\otimes \mathbb{Q}\to {\pi }_{\ast }(X{\cup }_f{D}^{n+1})\otimes \mathbb{Q}$$

$$X\to S,$$

$$(-{)}^{\vee }:=\text{mbox}Hom(-,\mathbb{Q}),\text{mbox}andH(-):=H^{\ast }(-;\mathbb{Q}),$$

$${\pi }_{\ast }(i_{\mathbb{Q}}):{\pi }_{\ast }(X_{\mathbb{Q}})\to {\pi }_{\ast }((X{\cup }_f{D}^{n+1}{)}_{\mathbb{Q}}).$$

$$Y\stackrel{\simeq }{⟶}\underset{{\sigma }_1<\cdots <{\sigma }_r}{\underleftarrow{\lim }}(S^{n_{{\sigma }_1}}\vee \cdots \vee S^{n_{{\sigma }_r}}{)}_{\mathbb{Q}},$$

$$V(0)=V\cap \text{mbox}kerd\text{mbox}andV(n+1)=V\cap d^{-1}(\wedge V(n)).$$

$$X⟶X_{\mathbb{Q}}:=⟨\wedge W⟩,$$

$$f_{\mathbb{Q}}=⟨\phi ⟩:⟨\wedge W⟩\to ⟨\wedge V⟩.$$

$${\pi }_{\ast }(⟨\wedge W⟩)\cong W^{\vee }.$$

$${\pi }_{\ast }(⟨\phi ⟩)={\phi }^{\vee }:W^{\vee }\to V^{\vee }.$$

$$\wedge V\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$Z(0)=Z\cap d^{-1}(\wedge V)\text{mbox}andZ(n+1)=Z\cap d^{-1}(\wedge V\otimes \wedge Z(n)),$$

$$s(L_V{)}_p=\text{mbox}Hom(V^{p+1})$$

$$<v,s[x,y]>=(-1{)}^{1+degy}<d_{1}v,sx,sy>,v\in V,x,y\in L_V.$$

$$s(L_V)={\pi }_{\ast }⟨\wedge V⟩.$$

$$Z\cap \text{mbox}kerd\stackrel{\cong }{⟶}{H}^{\ge 1}(\wedge Z).$$

$$L_S=\underset{n}{\underleftarrow{\lim }}{L}_S/L_S^n$$

$$\wedge Z({\sigma }_1,\dots ,{\sigma }_k)\hookrightarrow \wedge Z$$

$$\wedge Z=\underset{{\sigma }_1<\cdots <{\sigma }_k}{\underrightarrow{\lim }}\wedge Z({\sigma }_1,\dots ,{\sigma }_k)$$

$$⟨\wedge Z⟩=\underset{{\sigma }_1<\cdots <{\sigma }_k}{\underleftarrow{\lim }}⟨\wedge Z({\sigma }_1,\dots ,{\sigma }_k)⟩=\underset{{\sigma }_1<\cdots <{\sigma }_k}{\underleftarrow{\lim }}(S^{{\sigma }_1}\vee \cdots \vee S^{{\sigma }_k}{)}_{\mathbb{Q}}.$$

$$F=\underset{{\sigma }_1<\cdots <{\sigma }_k}{\underleftarrow{\lim }}(S^{n_{{\sigma }_1}}\vee \cdots \vee S^{n_{{\sigma }_k}}{)}_{\mathbb{Q}}.$$

$$Z({\sigma }_1\dots {\sigma }_k;0)={\oplus }_i\mathbb{Q}{z}_{{\sigma }_i}$$

$$Z({\sigma }_1\dots {\sigma }_k;n+1)=Z\cap d^{-1}({\wedge }^{2}Z({\sigma }_1\dots {\sigma }_k;(n))).$$

$$⟨\wedge Z⟩=\underset{{\sigma }_1<\cdots <{\sigma }_k}{\underleftarrow{\lim }}⟨\wedge Z({\sigma }_i\dots {\sigma }_k)⟩=\underset{{\sigma }_1<\cdots <{\sigma }_k}{\underleftarrow{\lim }}{\left(S^{n_{{\sigma }_1}}\vee \cdots \vee S^{n_{{\sigma }_k}}\right)}_{\mathbb{Q}}=F.$$

$$i_{\mathbb{Q}}:(X\vee Y{)}_{\mathbb{Q}}\to (X\times Y{)}_{\mathbb{Q}}$$

$$\phi :\wedge T:=\wedge W\otimes \wedge Q\otimes \wedge R\stackrel{\simeq }{⟶}\wedge W{\times }_{\mathbb{Q}}\wedge Q.$$

$${\phi }_1:(\wedge T,d_1)\to (\wedge W,d_1){\times }_{\mathbb{Q}}(\wedge Q,d_1)$$

$$\wedge T{\otimes }_{\wedge W\otimes \wedge Q}\wedge W\otimes \wedge U_W\otimes \wedge Q\otimes \wedge U_Q\simeq A:=(\wedge W{\oplus }_{\mathbb{Q}}\wedge Q)\otimes \wedge U_W\otimes \wedge U_Q.$$

$$0\to {\wedge }^{\ge 1}W\otimes \wedge U_W\otimes \wedge U_Q\to A\to \wedge Q\otimes \wedge U_W\otimes \wedge U_Q\to 0.$$

$$0\to ({\wedge }^{\ge 1}W\otimes \wedge U_W\otimes \wedge U_Q)/J\to A/J\to \wedge Q\otimes \wedge U_W\otimes \wedge U_Q\to 0.$$

Full text

2010MSC: 55P62, 55P05

Key Words: rational homotopy, attaching cells, inertia

In [2] and [16] the authors define and establish the properties of rationally inert elements in the homotopy groups of simply connected CW complexes $X$ of finite type: $[f]\in\pi_{n}(X)$ is rationally inert if

[TABLE]

is surjective. Our objective here is to use Sullivan completions $X\to X_{\mathbb{Q}}$ to extend the definitions to $[f]\in\pi_{n}(X)$, $n\geq 1$, where $X$ is any connected CW complex, and then to extend the principal results of [16] to this more general setting and establish several applications. For details about Sullivan completions the reader is referred to [14]

Inverse homotopy equivalences between the homotopy categories of connected CW complexes, $X$, and connected simplicial sets, $S$, are provided by $X\mapsto$ Sing$\,X$, the singular simplices in $X$, and by $S\mapsto|S|$, its Milnor realization. These identify a map $X\to|S|$ with a morphism Sing$\,X\to S$. For simplicity we denote both by

[TABLE]

and refer to either a connected CW complex or a connected simplicial set simply as a connected space.

Additionally, for simplicity, we adopt the

Convention. Our base field is $\mathbb{Q}$. When the meaning is clear, we will suppress the differentials from the notation. For simplicity, we will also write

[TABLE]

for singular cohomology. Moreover, where there is no ambiguity we suppress the differential from the notation for a complex, and write $A$ instead of $(A,d)$.

As detailed in §1 below, a Sullivan completion $X_{\mathbb{Q}}$ appears naturally as a simplicial set. Sullivan models and Sullivan completions are reviewed in §1. In particular, if $X$ is simply connected and of finite type, then [8, Theorem 15.11] its Sullivan completion induces an isomorphism $\pi_{*}(X)\otimes\mathbb{Q}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\pi_{*}(X_{\mathbb{Q}})$. Thus we extend the definition of rationally inert elements as follows:

Definition. If $X$ is a connected space then $[f]\in\pi_{n}(X)$, some $n\geq 1$, is rationally inert if the inclusion $i:X\to X\cup_{f}D^{n+1}$ induces a surjection,

[TABLE]

This condition can be characterized in terms of the homotopy type of the fibre $F(f)$ of $i_{\mathbb{Q}}$ (Theorem 1). Applications are then provided in Theorems 2, 3 and 4. To state Theorem 1 we need the

Definition. A connected space $Y$ is rationally wedge-like if for some non-void linearly ordered set $S=\{\sigma\}$, and integers $n_{\sigma}>0$, there is a homotopy equivalence,

[TABLE]

where the inverse system is defined by the projections of $S^{n_{\sigma_{1}}}\vee\dots\vee S^{n_{\sigma_{r}}}$ on the sub wedges.

Remark: Note that in general $(X\vee Y)_{\mathbb{Q}}$ is different from $X_{\mathbb{Q}}\vee Y_{\mathbb{Q}}$ !

Applications are then provided in Theorems 2, 3, 4 and 5.

Theorem 3.3 in [16] is a special case of Theorem 1 since in that case the homotopy fibre of $X\to X\cup_{f}D^{n+1}$ is rationally a wedge of spheres if and only if its rationalization is rationally wedge-like.

An example of rationally inert elements is provided by the following theorem, established for simply connected spaces in ([16, Theorem 5.1]).

As described above, and in detail in ([14, §1]) the Sullivan completion $X_{\mathbb{Q}}$ of a space $X$ is a simplicial set $\langle\land W\rangle$ constructed from a minimal Sullivan model for $X$. This is used in ([14, §4]) to construct a completion, $\overline{H(\Omega X)}$, of the rational loop space homology of $X$. The homotopy fibre $F(f)$ of Theorem 1 also has the form $\langle\land Z\rangle$ for some minimal Sullivan algebra $\land Z$ (§3), although $\land Z$ may not be the Sullivan model of a space. Nevertheless, ([14, §4]), for any minimal Sullivan algebra, $\land Z$, $\pi_{*}\Omega\langle\land Z\rangle$ is naturally a graded Lie algebra, complete with respect to a natural filtration. Its Lie bracket is given explicitly in terms of the Whitehead products in $\pi_{*}\langle\land Z\rangle$. We generalize ([16, Theorem 3.3 (I)]) in

A general question asks what conditions on a group $G$ imply that $(BG)_{\mathbb{Q}}$ is aspherical; i.e., a $K(\pi,1)$. This is true when $G$ is a finitely generated free group, when $G$ is the fundamental group of a Riemann surface or when $G$ is a right-angled Artin group ([19], [7]). We consider here the one-relator groups, $\pi_{1}(X\cup_{f}D^{2})$, obtained by adding a 2-cell to a wedge of circles along a continuous map $f:S^{1}\to X$. The well known Lyndon theorem ([18], [20],[6]) states that if $f$ is not a proper power, then $X\cup_{f}D^{2}$ is aspherical. In general it may happen that a connected space $X$ is aspherical, but $X_{\mathbb{Q}}$ is not. However, the spaces considered by Lyndon remain aspherical when rationalized:

Remark. Note that even if $f$ is a proper power, where Lyndon’s theorem does not apply, it is true that $(X\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical.

Finally recall a famous unsolved problem of JHC Whitehead [21]: is a subcomplex of an aspherical two-dimensional CW complex aspherical ? As observed by Anick [1] it is sufficient to consider the case that both subcomplexes share the same $1$-skeleton and base point. The problem then reduces to the question: If $X$ is a finite 2-dimensional connected CW complex and $X\cup\left(\amalg_{k=1}^{p}D_{k}^{2}\right)$ is aspherical, is $X$ aspherical ?

In [1] Anick provides a positive answer to an analogous question for simply connected rational spaces. Here we have a positive answer for Sullivan completions of connected spaces.

1 Sullivan models and Sullivan completions

We review briefly the basic facts and notation from Sullivan’s theory. For details the reader is referred to [14]. A $\Lambda$-algebra is a commutative differential graded algebra (cdga) of the form $(\land V,d)$, where $V=V^{\geq 0}$ is a graded vector space and $\land V$ is the free graded commutative algebra generated by $V$. Moreover the differential is required to satisfy the Sullivan condition: $V=\cup_{n\geq 0}V(n)$, where

[TABLE]

Here $V$ is a generating vector space for $\land V$. If $V=V^{\geq 1}$ then $V$ is a Sullivan algebra.

Moreover, $\land V=\oplus_{p\geq 0}\land^{p}V$, where $\land^{p}V$ denotes the linear span of the monomials in $V$ of length $p$; $p$ is called the wedge degree. In particular, a $\Lambda$-algebra is minimal if $d:V\to\land^{\geq 2}V$ and quadratic if $d:V\to\land^{2}V$. Thus associated with a minimal $\Lambda$-algebra $(\land V,d)$ is the quadratic $\Lambda$-algebra $(\land V,d_{1})$ defined by: $d_{1}v$ is the component of $dv$ in $\land^{2}V$.

Note that if $V=V^{\geq 1}$, then the inclusion of a subspace $W\subset\land^{\geq 1}V$ extends to an isomorphism $\land W\stackrel{{\scriptstyle\cong}}{{\to}}\land V$ if and only if $W\oplus\land^{\geq 2}V=\land V$. In this case $\land W$ satisfies the same condition as $\land V$: the definition of a Sullivan algebra does not depend on the choice of generating vector space. Observe as well that if $V=V^{\geq 1}$ then the natural map

[TABLE]

is an isomorphism.

With each connected space $Y$ is associated a cdga $A_{PL}(Y)$ and a unique isomorphism class of minimal Sullivan algebras $(\land V,d)$ characterized by the existence of a quasi-isomorphism $(\land V,d)\stackrel{{\scriptstyle\simeq}}{{\to}}A_{PL}(Y)$. By definition $(\land V,d)$ is the minimal Sullivan model of $Y$. Among their properties are the natural isomorphisms $H(\land V,d)\cong H(Y)$ of graded algebras. Moreover, any map, $f:X\to Y$ determines a ”homotopy class” of morphisms, $\varphi:\land V\to\land W$, from the minimal Sullivan model of $Y$ to that of $X$; $\varphi$ is a Sullivan representative of $f$.

On the other hand, the construction of Sullivan completions is accomplished by a functor associating to a $\Lambda$-algebra, $\land W$, a simplicial set $\langle\land W\rangle$, with the property that $<\,\,>$ converts direct limits to inverse limits. In particular, if $\land W$ is a minimal Sullivan model of a connected space $X$ then this determines a based homotopy class of maps

[TABLE]

the Sullivan completion of $X$. In particular, if $\varphi:\land V\to\land W$ is a Sullivan representative of $f:X\to Y$ then

[TABLE]

Moreover, ([9, Theorem 1.3]) for any minimal Sullivan algebra, $\land W$, there is a natural bijection $\pi_{*}(\langle\land W\rangle)\cong\mbox{Hom}(\land^{\geq 1}W/\land^{\geq 2}W)$, and the isomorphism $W\stackrel{{\scriptstyle\cong}}{{\to}}\land^{\geq 1}W/\land^{\geq 2}W$ then induces a bijection

[TABLE]

Therefore, for any morphism $\varphi:\land V\to\land W$ of minimal Sullivan algebras, it follows that $\pi_{*}(\langle\varphi\rangle)$ is surjective if and only if $\varphi:\land^{\geq 1}V/\land^{\geq 2}V\to\land^{\geq 1}W/\land^{\geq 2}W$ is injective, or equivalently, if the generating vector space $W\subset\land W$ can be chosen so that $\varphi:V\to W$ is the inclusion of a subspace. In this case

[TABLE]

Now a general morphism $\varphi:\land V\to\land W$ of Sullivan algebras factors ([9, Theorem 3.1]) as

[TABLE]

in which (i) $\eta(v)=v\otimes 1$, (ii) $\gamma$ is a quasi-isomorphism, (iii) $Z=Z^{\geq 0}$, (iv) $Z=\cup_{n}Z(n)$ satisfying

[TABLE]

and (v) the quotient $(\land Z,\overline{d})=\mathbb{Q}\otimes_{\land V}(\land V\otimes\land Z)$ is a minimal $\Lambda$-algebra. Here $\land V\otimes\land Z$ is a minimal $\Lambda$-extension of $\land V$.

Remark. If $\pi_{*}\langle\varphi\rangle$ is surjective we take $\eta=\varphi$ to be an inclusion $V\to W$ and $\land W=\land V\otimes\land Z$.

In particular, with each minimal Sullivan algebra $(\land V,d)$ is associated a unique isomorphism class of $\Lambda$-extensions, $(\land V\otimes\land U,d)$, its acyclic closures. These are characterized by the following two properties: (i) the augmentation $\land V\to\mathbb{Q}$ extends to a quasi-isomorphism $\land V\otimes\land U\stackrel{{\scriptstyle\simeq}}{{\to}}\mathbb{Q}$ with $U\to 0$, and (ii) the quotient differential in $\land U=\mathbb{Q}\otimes_{\land V}(\land V\otimes\land U)$ is zero.

Finally, a minimal Sullivan algebra $\land V$ determines the graded homotopy Lie algebra $L_{V}=(L_{V})_{\geq 0}$ given by

[TABLE]

and

[TABLE]

(Here $s$ is the degree 1 suspension isomorphism.) Thus

[TABLE]

2 Rationally wedge-like spaces

Lemma 1. The following two conditions on a minimal Sullivan algebra, $\land Z$, are equivalent:

1. (i)

The generating vector space $Z\subset\land Z$ can be chosen so that

[TABLE] 2. (ii)

$\land Z$ is the minimal Sullivan model of a cdga $A=\mathbb{Q}\oplus A^{\geq 1}$ in which the differential and products in $A^{\geq 1}$ are zero.

If these hold then $Z$ can be chosen so that $Z\cap\mbox{ker}\,d\stackrel{{\scriptstyle\cong}}{{\to}}H^{\geq 1}(\land Z)$ and $(\land Z,d)$ is quadratic.

proof: If (i) holds let $A$ be the quotient of $\land Z$ by $\land^{\geq 2}Z$ and by a direct summand of the image of ker$\,d$ in $Z$. If (ii) holds set $V_{0}=A^{\geq 1}$ and define a quadratic Sullivan algebra $\land V$ by setting $V(k)=\oplus_{j\leq k}V_{k}$, with $d:V_{k+1}\stackrel{{\scriptstyle}}{{\to}}\land^{2}V(k)\cap\mbox{ker}\,d$ inducing an isomorphism in homology. Then $(\land V,d)$ has zero homology in wedge degree $2$, and it follows that $\land V$ has zero homology in wedge degrees $\geq 2$. Hence $\land V$ is a quadratic Sullivan model for $A$. Thus $\land V\cong\land Z$, and so $Z$ can be chosen so that $d:Z\to\land^{2}Z$. Thus the final assertion is part of ([14, Proposition 6]). $\square$

Example: Finite wedges of spheres: $S=S^{\sigma_{1}}\vee\dots\vee S^{\sigma_{k}}$.

The quasi-isomorphism $A_{PL}(S)\to\oplus_{\mathbb{Q}}A_{PL}(S^{\sigma_{i}})\simeq\oplus_{\mathbb{Q}}H(S^{\sigma_{i}})$ identifies the minimal Sullivan model of $S$ as a minimal Sullivan algebra $\land Z$ satisfying the conditions of Lemma 1. Here $Z\cap\mbox{ker}\,d$ has a basis $z_{1},\dots,z_{k}$ representing orientation classes of $S^{\sigma_{1}},\dots,S^{\sigma_{k}}$.

Now choose elements $x_{i}$ in the homotopy Lie algebra $L_{S}$ of $S$ so that $\langle z_{i},sx_{j}\rangle=\delta_{ij}$. The $x_{j}$ then freely generate a free sub Lie algebra $E\subset L_{S}$. In fact, the rescaling argument in ([9, p.230]) generalizes to reduce to the case $S=S^{\geq 2}$, in which case the result is established in [8, §23, Example 2]. Moreover, it follows from [9, Chap. 2] that

[TABLE]

where $L_{S}^{n}$ is the ideal spanned by the iterated commutators in $L_{S}$ of length $n$. According to [9, Chapter 2], the $x_{\sigma_{i}}$ map to a basis of $L_{S}/L_{S}^{2}$ and hence the inclusion $E\hookrightarrow L_{S}$ induces isomorphisms $E/E^{n}\stackrel{{\scriptstyle\cong}}{{\to}}L_{S}/L_{S}^{n}$.

Proposition 1. A connected space $F$ is rationally wedge-like if and only il it has the form $F=\langle\land Z\rangle$, where $\land Z$ satisfies the equivalent conditions of Lemma 1.

proof: Suppose first that $F=\langle\land Z\rangle$, where $\land Z$ satisfies the conditions of Lemma 1, and pick a linearly ordered basis of $Z\cap\mbox{ker}\,d$. Then each finite subset $z_{\sigma_{1}}<\dots<z_{\sigma_{k}}$ determines an inclusion

[TABLE]

of quadratic Sullivan algebras with $Z(\sigma_{1},\dots,\sigma_{k})\subset Z$, and for which $\{z_{\sigma_{i}}\}$ is a basis of $H^{\geq 1}(\land Z(\sigma_{1},\dots,\sigma_{p}))$, and $\land Z(\sigma_{1},\dots,\sigma_{k})$ is a Sullivan model for $S^{\sigma_{1}}\vee\dots\vee S^{\sigma_{k}}$. Moreover, the inclusions $\land Z(\sigma_{i_{1}},\dots\sigma_{i_{r}})\to\land Z(\sigma_{1},\dots,\sigma_{k})$ are Sullivan representatives for the projections $S^{\sigma_{1}}\vee\dots\vee S^{\sigma_{k}}\to S^{\sigma_{i_{1}}}\vee\dots\vee S^{\sigma_{i_{r}}}.$

Now

[TABLE]

and so

[TABLE]

In the reverse direction, suppose $F$ is rationally wedge like, so that

[TABLE]

Then let $\land Z$ be a Sullivan algebra satisfying the conditions of Lemma 1 in which $Z\cap\mbox{ker}\,d$ has a basis $\{z_{\sigma}\}$ of degrees $n_{\sigma}$. Thus any subset $\sigma_{1}<\dots<\sigma_{k}$ determines a sub Sullivan algebra $Z(\sigma_{1}\dots\sigma_{k})$ by the requirement that $Z(\sigma_{1}\dots\sigma_{k})=\cup_{n}\,Z(\sigma_{1}\dots\sigma_{k};n)$ in which

[TABLE]

and

[TABLE]

This gives as above that

[TABLE]

$\square$

Corollary 1. If $X=\vee_{\sigma}S^{n_{\sigma}}$ is a wedge of spheres, then $X_{\mathbb{Q}}$ is rationally wedge-like. If all the spheres are circles then $X_{\mathbb{Q}}$ is aspherical.

Corollary 2. If $\langle\land Z\rangle$ is rationally wedge-like and dim$\,H^{\geq 1}(\land Z)>1$, then the sum of the solvable ideals in $L_{Z}$ is zero.

proof: It follows from Lemma 1 that cat$(\land Z)=1$, and so from [11], Sdepth$\,L_{Z}<\infty$. Now [12, Theorem 1] asserts that the sum, rad$\,L_{Z}$, of the solvable ideals in $L_{Z}$ is finite dimensional, and that $L_{Z}$ acts nilpotently in rad$\,L_{Z}$. In particular, if rad$\,L_{Z}\neq 0$ then the center of $L_{Z}$ is non-zero. Let $x\in L_{Z}$ be an element in the center.

Since $\langle\land Z\rangle$ is rationally wedge-like, $\land Z=\varinjlim_{\mathcal{S}}\land Z(\sigma_{1},\dots,\sigma_{k})$ where $\land Z(\sigma_{1},\dots,\sigma_{k})$ is the minimal Sullivan model of a wedge of $k$ spheres, and ${\mathcal{S}}$ has by hypothesis at least two elements. Then $L_{Z}=\varprojlim_{\mathcal{S}}L_{Z(\sigma_{1},\dots,\sigma_{k})}$, and the maps $L_{Z}\to L_{Z(\sigma_{1},\dots,\sigma_{k})}$ are surjective. Thus if $x\neq 0$ it maps to a non-zero element in some $L_{Z(\sigma_{1},\dots,\sigma_{k})}$ with $k>1$. This would contradict the Example above. $\square$

Remark. Rationally wedge-like spaces provide examples of minimal Sullivan algebras $\land Z$ for which $\langle\land Z\rangle$ is not the Sullivan completion of a space. For example, suppose $Z=Z^{3}$ has a countably infinite basis, so that $\pi_{*}\langle\land Z\rangle=\pi_{3}\langle\land Z\rangle=(Z^{3})^{\vee}$.

Thus for any minimal Sullivan algebra $\land V$, the condition $\langle\land V\rangle=\langle\land Z\rangle$ would imply that $V=V^{3}$ and $(V^{3})^{\vee}\cong(Z^{3})^{\vee}$. But if $\land V$ were the minimal model of a space $X$ then we would have $V^{3}\cong H^{3}(X)=H_{3}(X)^{\vee}$ and so either dim$\,V^{3}<\infty$ or card$\,(V^{3})\geq$ card$\,\mathbb{R}$. In the second case, card$\,((V^{3})^{\vee})>$ card$\,\mathbb{R}$ and so $(V^{3})^{\vee}$ and $(Z^{3})^{\vee}$ are not isomorphic.

Proposition 2. Suppose $X$ and $Y$ are connected spaces, one of which has rational homology of finite type. Then

1. (i)

The homotopy fibre, $F$, of the natural map

[TABLE]

is rationally wedge-like. 2. (ii)

If $X_{\mathbb{Q}}$ and $Y_{\mathbb{Q}}$ are aspherical then so are $F$ and $(X\vee Y)_{\mathbb{Q}}$.

This result is analogous to the fact that the usual fibre of the injection $X\vee Y\to X\times Y$ is the join of $\Omega X$ and $\Omega Y$ and thus a suspension. (But note that $(X\vee Y)_{\mathbb{Q}}$ may be different from $X_{\mathbb{Q}}\vee Y_{\mathbb{Q}}$.)

Proposition 2 follows easily from a result about Sullivan algebras (Proposition 3, below). For this, consider minimal Sullivan algebras, $\land W$ and $\land Q$. The natural surjection $\land W\otimes\land Q\to\land W\times_{\mathbb{Q}}\land Q$ is surjective in homology, and so extends to a minimal Sullivan model

[TABLE]

Filtering by wedge degree then yields a morphism

[TABLE]

between the associated bigraded cdga’s. (Here $(\land-,d_{1})$ is the associated quadratic Sullivan algebra.)

Proposition 3. With the hypotheses and notation above,

1. (i)

$\langle\land R\rangle$ is rationally wedge-like. 2. (ii)

$\varphi_{1}$ is a quasi-isomorphism.

proof: (i) Let $\land W\otimes\land U_{W}$ and $\land Q\otimes\land U_{Q}$ denote the respective acyclic closures. Then $\land R$ is quasi-isomorphic to

[TABLE]

Dividing $A$ by the ideal generated by $W$ yields the short exact sequence

[TABLE]

Decompose the differential in $\land W\otimes\land U_{W}$ in the form $d=d_{1}+d^{\prime}$ with $d_{1}(W)\subset\land^{2}W$, $d_{1}(U_{W})\subset W\otimes\land U_{W}$, $d^{\prime}(W)\subset\land^{\geq 3}W$ and $d^{\prime}(U_{W})\subset\land^{\geq 2}W\otimes\land U_{W}$. Then $d_{1}$ is a differential and $(\land W\otimes\land U_{W},d_{1})$ is the acyclic closure of $(\land W,d_{1})$. Choose a direct summand, $S$, of $d_{1}(\land^{\geq 1}U_{W})$ in $W\otimes\land U_{W}$. Then $I=(\land^{\geq 2}W\otimes\land U_{W})\oplus S$ is acyclic for the differential $d_{1}$ and therefore also for the differential $d$. Thus $J=I\otimes\land U_{Q}$ is an acyclic ideal in $A$ and $A\stackrel{{\scriptstyle\cong}}{{\to}}A/J$.

Now consider the short exact sequence

[TABLE]

The inclusion of $\land U_{W}$ in the right hand term is a quasi-isomorphism. This yields a quasi-isomorphism

[TABLE]

Since the differential and the multiplication in $d_{1}(\land^{\geq 1}U_{W})\otimes\land^{\geq 1}U_{Q}$ are zero, it follows from Proposition 1 that $\langle\land R\rangle$ is rationally wedge-like.

(ii) The surjection $(\land W\otimes\land Q,d_{1})\to(\land W\times_{\mathbb{Q}}\land Q,d_{1})$ extends to a quasi-isomorphism

[TABLE]

from a minimal Sullivan algebra. We first show that $\widehat{R}$ can be chosen so that $(\land\widehat{T},\delta)$ is quadratic. Then we extend $\delta$ to a differential $\widehat{d}=\sum_{i\geq 1}\widehat{d}_{i}$ in which $\widehat{d}_{1}=\delta$ and

[TABLE]

It is automatic that $(\land\widehat{T},\widehat{d})$ will be a minimal Sullivan algebra. Moreover, filtering by wedge degree shows that $\widehat{\varphi}$ is a quasi-isomorphism and so $\land\widehat{T}$ is a minimal Sullivan model for $\land W\times_{\mathbb{Q}}\land Q$. In particular this identifies $\widehat{T}$ with $T$, $R$ with $\widehat{R}$ and $\widehat{\varphi}$ with $\varphi$, thereby establishing (ii).

To accomplish the first step, define $d_{1}:U_{W}\to W\otimes\land U_{W}$ and $d_{1}:U_{Q}\to Q\otimes\land U_{Q}$ as in (i). Assign $\land W$ and $\land Q$ wedge degree as a second degree and assign $U_{W}$ and $U_{Q}$ second degree [math]. Then $(\land W\otimes\land U_{W},d_{1})$ and $(\land Q\otimes\land U_{Q},d_{1})$ are the respective acyclic closures of $(\land W,d_{1})$ and $(\land Q,d_{1})$, and $d_{1}$ increases the second degree by 1. Now $\widehat{\varphi}$ and $\widehat{T}$ may be constructed so that $\widehat{R}$ is equipped with a second gradation for which $\delta$ increases the second degree by one and $\widehat{\varphi}$ is bihomogeneous of degree zero.

The argument in the proof of (i) now yields a sequence of bihomogeneous quasi-isomorphisms connecting

[TABLE]

Thus $H^{\geq 1}(\land\widehat{R})$ is concentrated in second degree 1. Therefore $\land\widehat{R}$ satisfies condition (i) of Proposition 1, and it follows that we may choose $\widehat{R}$ so that the quotient cdga $\land\widehat{R}$ is quadratic and $H^{\geq 1}(\land\widehat{R})$ embeds in $\widehat{R}$. This implies that $\widehat{R}$ is concentrated in second degree 1 and that

[TABLE]

In particular, $(\land\widehat{T},\delta)$ is a quadratic Sullivan algebra.

The construction of $\widehat{d}$ proceeds as follows. Write the differential in $\land W\times_{\mathbb{Q}}\land Q$ as $d=\sum_{r\geq 1}d_{r}$ in which $d_{r}$ is a derivation raising wedge degree by $r+1$. Thus for each $r$, $\sum_{i+j=r}d_{i}d_{j}=0$. Now we construct by induction a sequence of derivations $\widehat{d}_{1}=\delta,\dots,\widehat{d}_{r}\dots$, in $\land\widehat{T}$ in which $\widehat{d}_{r}$ increases the wedge degree by $r+1$, and

[TABLE]

Thus, in view of (i), $\widehat{d}:=\sum\widehat{d}_{i}$ will define a differential in $\widehat{T}$, $(\land\widehat{T},\widehat{d})$ will be a Sullivan algebra, and

[TABLE]

will be a cdga morphism. Filtering by wedge degree shows that $\widehat{\varphi}$ is a quasi-isomorphism.

It remains to construct the $\widehat{d}_{i}$, $i\geq 2$. For this, set $\widehat{T}(k)=W\oplus Q\oplus\widehat{R}^{\leq k}$. Since $(\land\widehat{T},\delta)$ is a Sullivan algebra it follows that each $\widehat{R}^{k}$ is the union of an increasing family of subspaces $F^{p}(\widehat{R}^{k})$ such that

[TABLE]

Set $\widehat{d}_{1}=\delta$ and assume by induction that $\widehat{d}_{1},\dots,\widehat{d}_{r}$ have been constructed, and that $\widehat{d}_{r+1}$ has been constructed in $\widehat{R}^{<k}\oplus F^{p}(\widehat{R}^{k})$.

Let $y_{i}$ be a basis for a direct summand of $F^{p}(\widehat{R}^{k})$ in $F^{p+1}(\widehat{R}^{k})$. Then

[TABLE]

It follows that

[TABLE]

Since $\widehat{\varphi}$ is a surjective quasi-isomorphism with respect to $\widehat{d}_{1}$ and $d_{1}$, this implies that

[TABLE]

with $\widehat{\varphi}\Phi_{i}=-\widehat{d}_{r+1}\widehat{\varphi}y_{i}$. Extend $\widehat{d}_{r+1}$ to $F^{p+1}(\widehat{R}^{k})$ by setting $\widehat{d}_{r+1}y_{i}=-\Phi_{i}$.

$\square$

proof of Proposition 2: (i) Let $\land W$ and $\land Q$ be the minimal Sullivan models of $X$ and $Y$. A Sullivan representative of the inclusion $i:X\vee Y\to X\times Y$ is then the inclusion

[TABLE]

It follows that $i_{\mathbb{Q}}$ is the surjection

[TABLE]

But this surjection is a fibration ([8, Proposition 17.9]) with fibre $\langle\land R\rangle$, which is a rationally wedge-like by Proposition 3.

(ii) When $X_{\mathbb{Q}}$ and $Y_{\mathbb{Q}}$ are aspherical, then $U_{W}$ and $U_{Q}$ are concentrated in degree [math] and $W$ is concentrated in degree 1. This shows that $F$ is aspherical. Since one of $X,Y$ has rational homology of finite type, $(X\times Y)_{\mathbb{Q}}=X_{\mathbb{Q}}\times Y_{\mathbb{Q}}$ is aspherical. We deduce then from the homotopy sequence of the fibration $F\to(X\vee Y)_{\mathbb{Q}}\to(X\times Y)_{\mathbb{Q}}$ that $(X\vee Y)_{\mathbb{Q}}$ is also aspherical.

$\square$

3 Cell attachments and Theorem 1

Before undertaking the proof of Theorem 1 we set up the basic framework that translates the topology of a cell attachment to Sullivan’s theory, and establish two preliminary Propositions.

Suppose $f:S^{n}\to X$ is the map of Theorem 1, and denote by $(\land W,d)$ the Sullivan minimal model of $X$. A Sullivan representative of $f$ is a morphism from $\land W$ to the minimal model of $S^{n}$. Composing with the quasi-isomorphism from that model to $H(S^{n})$ gives a morphism $\psi:\land W\to H(S^{n})$. Now define a linear map of degree $-n$,

[TABLE]

by setting $\varepsilon(1)=0$ and $\psi(\Phi)=\varepsilon(\Phi)\cdot[S^{n}]$, $\Phi\in\land^{\geq 1}W$, where $[S^{n}]$ denotes an orientation class in $S^{n}$. In particular, $\varepsilon\circ d=0$ and $\varepsilon(\land^{\geq 2}W)=0$.

Now define a cdga $(\land W\oplus\mathbb{Q}a,D)$ as follows: deg$\,a=n+1$, $a^{2}=a\cdot\land^{+}W=0$, and

[TABLE]

By [8, (13)b and (13)d], division by $a$ yields the commutative diagram,

[TABLE]

in which $(\land V,d)$ is a minimal Sullivan model for $X\cup_{f}D^{n+1}$, and $\lambda$ is a Sullivan representative for the inclusion $i:X\to X\cup_{f}D^{n+1}$. In particular, $i_{\mathbb{Q}}:X_{\mathbb{Q}}\to(X\cup_{f}D^{n+1})_{\mathbb{Q}}$ is identified with $\langle\lambda\rangle:\langle\land W\rangle\to\langle\land V\rangle$.

As described in $\S 1$, $\lambda$ factors as

[TABLE]

in which $\land V\otimes\land Z$ is a $\Lambda$-extension of $\land V$, $\gamma$ is a quasi-isomorphism, and the quotient

[TABLE]

is a minimal $\Lambda$-algebra. Since $H^{1}(i)$ is injective, it follows that $\lambda:V^{1}\to W^{1}$ is injective. Therefore $Z=Z^{\geq 1}$ and $\land Z$ is a minimal Sullivan algebra.

Further, because $\gamma$ is a quasi-isomorphism of Sullivan algebras, $\langle\gamma\rangle$ is a homotopy equivalence, which (up to homotopy) identifies $\langle\eta\rangle$ with $\langle\lambda\rangle$. But ([8, Proposition 17.9]) $\langle\eta\rangle$ is the projection of a Serre fibration with fibre $\langle\land Z\rangle$. Thus $\langle\land Z\rangle$, the homotopy fibre of $\langle\lambda\rangle$, and the homotopy fibre $F(f)$ of $i_{\mathbb{Q}}$, all have the same homotopy type:

[TABLE]

On the other hand, we have

Proposition 4. With the hypotheses and notation of (3), let $\land V\otimes\land U$ be the acyclic closure of $\land V$. Then there is a degree 1 isomorphism,

[TABLE]

and $H^{\geq 1}(\land Z,\overline{d})\cdot H^{\geq 1}(\land Z,\overline{d})=0.$

proof: First observe that in diagram (3), $\tau\Phi=\lambda\Phi+\alpha(\Phi)a$. Thus $\tau$ must coincide with $\lambda$ in $\land^{\geq 2}V$, and that also $D\circ\tau=D\circ\lambda$. Thus for $\Phi\in\land V$,

[TABLE]

Hence

[TABLE]

Now let $\land V\otimes\land U$ be the acyclic closure of $\land V$. Apply $-\otimes_{\land V}\land V\otimes\land U$ to diagram (9) to obtain a short exact sequence of complexes,

[TABLE]

in which the differential in $\mathbb{Q}a\otimes\land U$ is zero and the homology of the central complex is $\mathbb{Q}\,1$ in positive degrees. It follows that $H^{0}(\land W\otimes\land U)=\mathbb{Q}1$ and that the connecting homomorphism is an isomorphism of degree 1. By (11), $\varepsilon$ vanishes on $\land V$, and hence $(\varepsilon\otimes id)\circ(\lambda\otimes id)=0$ in $\land V\otimes\land U$. Now a straightforward calculation shows that the connecting homomorphism is given explicitly by

[TABLE]

On the other hand, applying $-\otimes_{\land V}\land V\otimes\land U$ to the quasi-isomorphism $\gamma$ yields quasi-isomorphisms \textstyle{(\land Z,\overline{d})}$$\textstyle{\land V\otimes\land U\otimes\land Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\gamma}$$\textstyle{\land W\otimes\land U} , so that we have a degree 1 isomorphism $H^{\geq 1}(\land Z,\overline{d})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\mathbb{Q}a\otimes\land U$. It is immediate that $H(\varepsilon\otimes id)$ vanishes on products, which gives the second assertion.

$\square$

Theorem 1 is now contained in

Theorem 1’. Suppose $X$ is a connected CW complex, and $[f]\in\pi_{n}(X)$, some $n\geq 1$. Then in the factorization (3)

[TABLE]

the following conditions are equivalent:

1. (i)

$[f]$ is rationally inert. 2. (ii)

The generating space $Z$ can be chosen so that

[TABLE] 3. (iii)

The homotopy fibre of $F(f)$ of $i_{\mathbb{Q}}:X_{\mathbb{Q}}\to(X\cup_{f}D^{n+1})_{\mathbb{Q}}$ is rationally wedge-like.

proof: (i) $\Longrightarrow$ (ii): Since $\langle\lambda\rangle$ is identified with $i_{\mathbb{Q}}$, $[f]\in\pi_{n}(X)$ is rationally inert if and only if the generating space $W$ can be chosen so that $\lambda$ restricts to an inclusion $V\to W$. In this case, $\land W$ decomposes as a Sullivan extension $\land V\to\land V\otimes\land Z=\land W$. Thus we may take $\eta=\lambda$ and $\gamma=id_{\land W}$. Note that if $\land V\otimes\land U$ is the acyclic closure of $\land V$, then the augmentation $\land V\otimes\land U\stackrel{{\scriptstyle\simeq}}{{\to}}\mathbb{Q}$ defines a quasi-isomorphism $\land W\otimes\land U=\land V\otimes\land Z\otimes\land U\stackrel{{\scriptstyle\simeq}}{{\to}}\land Z$.

If dim$\,H^{\geq 1}(\land Z)=1$, then necessarily $\land Z$ is the minimal Sullivan model of a sphere $S^{k}$ and $\langle\land Z\rangle=S^{k}_{\mathbb{Q}}$. If dim$\,H^{\geq 1}(\land Z)\geq 2$, let $\sigma:\land Z\to\land W\otimes\land U$ be a right inverse to the quasi-isomorphism $\land W\otimes\land U\simeq\land Z$ above. Since $\land V\otimes\land Z$ is a minimal Sullivan algebra, it will follow that

[TABLE]

But this will imply that $\sigma:\land^{\geq 2}Z\to\land^{\geq 2}W\otimes\land U$. Now a simple calculation shows that the connecting homomorphism vanishes on any $\overline{d}$-cycle in $\land^{\geq 2}Z$. Since the connecting homomorphism is an isomorphism it follows that division by $\land^{\geq 2}Z$ induces an injection $H^{\geq 1}(\land Z)\to Z$, and (ii) follows from Lemma 1.

To complete this direction of the proof we need to establish (15). For this write $Z=\cup_{k}Z(k)$ in which $Z(0)=Z\cap\mbox{ker}\,\overline{d}$ and $Z(k+1)=Z\cap\overline{d}^{-1}(\land Z(k))$. Assuming by induction that $\sigma:Z(k)\to\land^{\geq 1}W\otimes\land U$ we obtain that for $z\in Z(k+1)$, $d\sigma(z)=\sigma(\overline{d}z)\in\land^{\geq 2}W\otimes\land U$. Now let $\Phi$ be the component of $\sigma(z)$ in $1\otimes\land U$. Since $\land W$ is minimal it follows that $d:\land^{\geq 1}W\otimes\land U\to\land^{\geq 2}W\otimes\land U$. But if $\Phi\neq 0$ then $d(1\otimes\Phi)$ has a non-zero component in $V\otimes\land U$. Therefore $\Phi=0$ and (15) follows by induction on $k$.

(ii) $\Rightarrow$ (iii): Since $\langle\land Z\rangle\simeq F(f)$ it follows from Proposition 1 that $F(f)$ is rationally wedge-like.

(iii) $\Rightarrow$ (i): First suppose that $F(f)$ is a rational sphere $S^{k}_{\mathbb{Q}}$. Then $\land Z$ is the minimal Sullivan model of a sphere, and so dim$\,Z\cap\mbox{ker}\,\overline{d}=1$. Thus it follows from Proposition 4 that $U=0=V$. Since $\land V$ is the minimal Sullivan model for $X\cup_{f}D^{n+1}$ this implies that $\pi_{*}(X\cup_{f}D^{n+1})_{\mathbb{Q}}=0$ and $[f]$ is rationally inert.

Otherwise $F(f)$ is the inverse limit of rational wedges of at least two spheres. If $[f]$ is not inert then in the sequence

[TABLE]

the image of $\pi_{*}(\Omega(X\cup_{f}D^{n+1})_{\mathbb{Q}}$ contains a non-zero class $\omega\in\pi_{*+1}(F(f))$. Because $\Omega(X\cup_{f}D^{n+1})_{\mathbb{Q}}$ acts on $F(f)$, it follows that the Whitehead product $\omega\bullet\beta$ of $\omega$ and any $\beta\in\pi_{*}(F(f))$ is zero.

Then, because $\pi_{*}(F(f))=\varprojlim\pi_{*}(S^{\sigma_{1}}\vee\dots\vee S^{\sigma_{k}})$ it follows that for some $r\geq 2$, the image $\overline{\omega}$ of $\omega$ in some $\pi_{*}(S^{\sigma_{1}}\vee\dots\vee S^{\sigma_{k}})_{\mathbb{Q}}$ is non-zero, and that

[TABLE]

As observed in (4), $\pi_{*}(S^{\sigma_{1}}\vee\dots\vee S^{\sigma_{r}})_{\mathbb{Q}}$ is the suspension of its homotopy Lie algebra $L$, and it follows from [9, Chapter 2] that $\overline{\omega}$ determines a non-zero element in the center of $L$. But the center of $L$ is zero, and therefore $[f]$ is rationally inert.

$\square$

4 Poincaré duality complexes

We say a CW complex $Y=X\cup_{f}D^{n+1}$ is a rational Poincaré duality complex if $H(Y)$ is a Poincaré duality algebra and the top class is in the image of $H(Y,X)$. In this case it follows that $H^{\leq n}(X)\stackrel{{\scriptstyle\cong}}{{\to}}H(X)$. Poincaré duality complexes are rational Poincaré duality complexes, and so Theorem 2 follows from

Theorem 2’. If $Y=X\cup_{f}D^{n+1}$ is a rational Poincaré duality complex and the algebra $H(Y)$ requires at least two generators, then $[f]\in\pi_{n}(X)$ is rationally inert.

Before undertaking the proof we establish some notation. Let $\land V$ be the minimal Sullivan model of $Y$, and let $S$ be a direct summand in $(\land V)^{n+1}$ of $(\land V)^{n+1}\cap\mbox{ker}\,d$. Then division by $S$ and by $(\land V)^{>n+1}$ defines a surjective quasi-isomorphism $\land V\stackrel{{\scriptstyle\simeq}}{{\to}}A$, and

[TABLE]

where $\omega$ is a cycle representing the top cohomology class of $Y$. As shown in ([16, §5]), a cdga model of the inclusion $X\hookrightarrow Y$ is then provided by the inclusion

[TABLE]

where deg$\,t=n$, $t\cdot A^{+}=0$, and $dt=\omega$.

Thus if $A\otimes\land U$ is the acyclic closure of $A$, then a cdga model for the homotopy fibre of $j$ is given by

[TABLE]

Thus from the short exact sequence

[TABLE]

we deduce that

[TABLE]

is an isomorphism of graded vector spaces.

For the proof of Theorem 2 we first eliminate two special cases. First if $V^{1}=0$ the argument of ([16, §5]) shows that $(A\oplus\mathbb{Q}t)\otimes\land U$ is a cdga model of a wedge of spheres, and so $[f]$ is rationally inert. (Note that in [16] it is assumed that $X$ is simply connected; however the proof of this assertion relies only on the fact that $V^{1}=0$.) Secondly, if $n=1$ then $X\simeq_{\mathbb{Q}}S^{1}_{1}\vee\dots\vee S^{1}_{2q}$ and so $Y$ is rationally equivalent to an oriented Riemann surface. In this case Theorem 2’ is established in [13].

Thus to prove Theorem 2’ we may assume that $n\geq 2$ and that $A^{1}$ contains a non-zero cycle $x$. Since $H(A)$ is a Poincaré duality algebra there is a cycle $w\in A^{n}$ such that $wx=\omega$. The first step for the proof is then

Lemma 2. With the hypotheses and notation above, $A^{n+1}\otimes\land U\subset d(A^{n}\otimes\land U)$.

proof: Choose $\overline{x}\in U^{0}$ so that $d\overline{x}=x$. Since $\land V$ is a minimal Sullivan algebra, $V$ is the union of an increasing sequence of subspaces $V(0)\subset\dots\subset V(q)\subset\dots$ in which $V(0)=\mathbb{Q}x$ and $d:V(q+1)\to\land V(q)$. It follows that $U$ is the union of an increasing sequence of subspaces $U(0)\subset\dots\subset U(q)\subset\dots$ in which $U(0)=\mathbb{Q}\overline{x}$ and

[TABLE]

We show by induction on $q$ that

[TABLE]

First note that any $z\in A^{n+1}$ has the form $z=dy+\lambda wx$, some $\lambda\in\mathbb{Q}$. Thus

[TABLE]

Then for $r\geq 1$,

[TABLE]

It follows by induction on $r$ that $A^{n+1}\otimes\land U(0)\subset d(A^{n}\otimes\land U(0)).$

Now fix a direct summand, $T$, of $U(q)$ in $U(q+1)$, and assume by induction that for some $s$,

[TABLE]

Then write $\Phi\in A^{n+1}\otimes\land U(q)\otimes\land^{\leq s+1}T$ as $\Phi=\sum\Phi_{i}\otimes\Psi_{i}$ with $\Phi_{i}\in A^{n+1}\otimes\land U(q)$ and $\Psi_{i}\in\land^{\leq s+1}T$. By the hypothesis $\Phi_{i}=d\Omega_{i}$ with $\Omega_{i}\in A^{n}\otimes\land U(q)$. Therefore

[TABLE]

The first term is in $d(A^{n}\otimes\land U(q)\otimes\land^{\leq s+1}T)$. On the other hand, $d\Psi_{i}\in A^{\geq 1}\otimes\land U(q)\land^{\leq s}T$ and so the second term is in $A^{n+1}\otimes\land U(q)\otimes\land^{\leq s}T$. By hypothesis, the second term is contained in $d(A^{n}\otimes\land U(q)\otimes\land^{\leq s}T)$. This closes the induction. $\square$

proof of Theorem 2’: Let $\Phi\in\land U$. Then

[TABLE]

is a cycle, and

[TABLE]

By Lemma 2, $w\overline{x}\,d\Phi=d\Psi$ for some $\Psi\in A^{n}\otimes\land U$. Thus $(t-(-1)^{n})w\overline{x})\Phi+\Psi$ is a cycle projecting to $t\otimes\Phi$ in $\mathbb{Q}t\otimes\land U$. Then such cycles map to a basis of $\mathbb{Q}t\otimes\land U$. But because $n\geq 2$, $2n>n+1$ and so the product of any two of those cycles is zero. Therefore this defines a cdga quasi-isomorphism from the cohomology of a wedge of spheres to $(A\oplus\mathbb{Q}t)\otimes\land U$. Lemma 1 and Theorem 1’ together then imply that $[f]$ is rationally inert. $\square$

5 The structure of $L_{Z}$ and Theorem 3

Any minimal Sullivan algebra $\land V$ equips $L_{V}$ with a natural additional structure ([14, §3]), defined as follows. Associated with $\land V$ is the set, directed by inclusion, of the finite dimensional subspaces $V_{\alpha}\subset V$ for which $\land V_{\alpha}$ is preserved by $d$. For convenience we denote this set by ${\mathcal{J}}_{V}=\{\alpha\}$. In particular,

[TABLE]

That structure permits the explicit description of the Whitehead products in $\pi_{*}\langle\land V\rangle$ in terms of the Lie brackets in $L_{V}$ ([14, Formula (11)]).

Moreover, for any augmented graded algebra, $A$, the classical completion is defined by $\widehat{A}=\varprojlim_{n}A/I^{n}$, $I^{n}$ denoting the $n^{th}$ power of the augmentation ideal. The Sullivan completion of $UL_{V}$ is then the inverse limit,

[TABLE]

Further, by ([10, Proposition 3.3]), there are natural isomorphisms $\widehat{H}_{*}(\Omega\langle\land V_{\alpha}\rangle;\mathbb{Q})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\widehat{UL_{\alpha}}$. Passing to inverse limits then yields the isomorphism of the Sullivan completions,

[TABLE]

Similarly, the Sullivan central series is the filtration of $L_{V}$ given by

[TABLE]

where $L_{\alpha}^{r}$ is the ideal spanned by iterated commutators of length $r$. It satisfies ([14, §6])

[TABLE]

In the case that $\langle\land V\rangle$ is the homotopy fibre of $i_{\mathbb{Q}}:X_{\mathbb{Q}}\to(X\cup_{f}D^{n+1})_{\mathbb{Q}}$ when $[f]\in\pi_{n}(X)$ is rationally inert, this additional structure has the striking properties provided in Theorem 3’ below.

Suppose next that $\land W=\land V\otimes\land Z$ is the decomposition of a minimal Sullivan algebra determined by an inclusion $\land V\to\land W$ with $V\subset W$, and denote $\mathbb{Q}\otimes_{\land V}\land W=(\land Z,\overline{d})$. Then the short exact sequence $V\to W\to Z$ dualizes to the short exact sequence

[TABLE]

of Lie algebra morphisms, which identifies $L_{Z}$ as an ideal in $L_{W}$. The holonomy representation $\overline{\theta}$ of $L_{V}$ in $H(\land Z)$, ([9, Chapter 4]), then extends ([14, §7]) to a holonomy representation of $\overline{UL_{V}}$ in $H(\land Z)$.

On the other hand, the right adjoint representation of $L_{W}$ in $L_{Z}$ extends to the right adjoint representation of $\overline{UL_{W}}$ in $L_{Z}$, which further factors to give a right representation of $\overline{UL_{V}}$ in $L_{Z}/L_{Z}^{(2)}$ ([14, Proposition 7]).

Now suppose $(\land Z,\overline{d}$ is a quadratic Sullivan algebra. The surjection $\land^{\geq 1}Z\to Z$ with kernel $\land^{\geq 2}Z$ induces a surjection $H^{\geq 1}(\land Z)\to Z\cap\mbox{ker}\,\overline{d}$ of $\overline{UL_{V}}$-modules. This in turn dualizes to an inclusion

[TABLE]

of right $\overline{UL_{V}}$-modules. Moreover, according to ([14, Propositions 6 and 7]) the pairing $Z\times sL_{Z}\to\mathbb{Q}$ induces an isomorphism

[TABLE]

of right $\overline{UL_{V}}$-modules.

For the rest of this section we fix a map to a connected CW complex,

[TABLE]

some $n\geq 1$, for which $[f]$ is rationally inert.

As observed in the Remark in $\S 1$, a Sullivan representative $\land V\to\land W$ for the inclusion $X\to X\cup_{f}D^{n+1}$ has the form

[TABLE]

and as above we denote the quotient differential in $\land Z$ by $(\land Z,\overline{d})$. It follows from Theorem 1’ that $(\land Z,\overline{d})$ is a quadratic Sullivan algebra and that $H^{\geq 1}(\land Z,\overline{d})=Z\cap\mbox{ker}\,\overline{d}$.

Now recall from $\S 2$ the linear map

[TABLE]

of degree $-n$. Since $\varepsilon(V)=0$, $\varepsilon$ factors to give

[TABLE]

Thus, in view of (17), Theorem 3 is contained in

Theorem 3’. With the hypotheses and notation above, let $\overline{\varepsilon}\in L_{Z}/L_{Z}^{(2)}$ denote the image of $\widehat{\varepsilon}$. Then

1. (i)

Both $L_{Z}/L_{Z}^{(2)}$ and $H^{\geq 1}(\land Z)^{\vee}$ are free $\overline{UL_{V}}$-modules, respectively generated by $\overline{\varepsilon}$ and $\widehat{\varepsilon}$. 2. (ii)

The map $\Phi\mapsto\varepsilon\cdot\Phi,\Phi\in\overline{UL_{W}}$, is a surjection

[TABLE]

of $\overline{UL_{W}}$-modules. 3. (iii)

Any subspace $S\subset L_{Z}$ with $S\stackrel{{\scriptstyle\cong}}{{\to}}L_{Z}/L_{Z}^{(2)}$ freely generates a free sub Lie algebra, $E\subset L_{Z}$, and

[TABLE]

Remark. When $X$ is simply connected with finite Betti numbers and $n\geq 2$, then Theorem 3’ is established in ([16, Theorem 3.3]).

Before undertaking the proof of Theorem 3’ we establish a preliminary Proposition. For this, denote by $\varepsilon_{W}:\land V\otimes\land U\stackrel{{\scriptstyle\simeq}}{{\to}}\mathbb{Q}$ the augmentation in the acyclic closure of $\land V$ defined by $\varepsilon_{W}(U)=0$. Since the quotient differential in $\land U$ is zero, the holonomy representation of $\overline{UL_{V}}$ is a representation in $\land U$. On the other hand, the holonomy representation of $\overline{UL_{V}}$ in $H^{\geq 1}(\land Z)$ is a representation in $Z\cap\mbox{Ker}\,\overline{d}$. Now we strengthen Proposition 4 with

Proposition 5. With the hypotheses and notation above, there is a commutative diagram

[TABLE]

in which $\psi$ is an isomorphism of $\overline{UL_{V}}$-modules of degree $n+1$.

proof. Implicit in the isomorphism $\land W=\land V\otimes\land Z$ is the choice of a left inverse, $\land Z\to\land W$, of graded algebras for the surjection $\land W\to\land Z=\mathbb{Q}\otimes_{\land V}\land W$. This, with $id_{\land V}$, defines an isomorphism $\land V\otimes\land Z\stackrel{{\scriptstyle\cong}}{{\to}}\land W$, and identifies $id\otimes\widehat{\varepsilon}$ with $\varepsilon$. A simple and standard argument using Proposition 1 shows that this left inverse can be chosen so that the image of $\land V\otimes\left((Z\cap\mbox{Ker}\,\overline{d})\oplus\mathbb{Q}\right)$ is preserved by $d$. It is then immediate that the inclusion of this subcomplex in $(\land V\otimes\land Z)$ is a quasi-isomorphism. Thus from the commutative diagram (3) we obtain the row exact sequence

[TABLE]

Since $\varepsilon(\land V)=0$, $\land V$ is a subcomplex. Division by this subcomplex yields the row exact sequence of complexes,

[TABLE]

in which the middle complex has zero homology. It is immediate that the connecting quasi-isomorphism $\delta$, is then given by

[TABLE]

With a shift of degrees, regard $\varepsilon_{W}$ as a quasi-isomorphism $\land V\otimes\land U\stackrel{{\scriptstyle\simeq}}{{\to}}\mathbb{Q}a$, sending $1\mapsto a$. Then, since $\land V\otimes\land U$ is $\land V$-semifree, in the diagram,

[TABLE]

we may lift $\varepsilon_{W}$ through $\delta$ to obtain the quasi-isomorphism, $\chi$, of $\land V$-modules. But $\land V\otimes(Z\cap\mbox{Ker}\,\overline{d})$ is also $\land V$-semifree. Therefore applying $\mathbb{Q}\otimes_{\land V}-$ yields a quasi-isomorphism $\psi:\land U\stackrel{{\scriptstyle\simeq}}{{\to}}Z\cap\mbox{Ker}\,\overline{d}$.

Now the differentials in $\land U$ and in $Z\cap\mbox{Ker}\,\overline{d}$ are zero, and so $\psi$ is an isomorphism. Moreover, $\mathbb{Q}\otimes_{\land V}-$ converts morphisms between $\land V$-semifree modules to morphisms of $L_{V}$-modules. In this case $\psi$ is then automatically a morphism of $\overline{UL_{V}}$-modules. Finally, it is also immediate that the diagram of the Proposition commutes. $\square$

proof of Theorem 2 (i). Here we rely consistently on the notation and conventions of $\S 2$.

First, observe that the dual of a $\overline{UL_{V}}$-module inherits a right $\overline{UL_{V}}$-module structure in the standard way. Thus replacing $\psi$ by $\psi^{-1}$ in the diagram of Proposition 5 and then dualizing yields the commutative diagram

[TABLE]

in which $1\in\mathbb{Q}$ maps to $\varepsilon_{W}\in(\land U)^{\vee}$ and to $\widehat{\varepsilon}\in(Z\cap\mbox{ker}\,\overline{d})^{\vee}$. By ([14, Proposition 8]) $(\land U)^{\vee}$ is a free right $\overline{UL_{V}}$-module, freely generated by $\varepsilon_{W}$. Since $H^{\geq 1}(\land Z)=(Z\cap\mbox{ker}\,\overline{d})$, it follows from (23) that $H^{\geq 1}(\land Z)^{\vee}$ is a free right $\overline{UL_{V}}$-module freely generated by $\widehat{\varepsilon}$.

(ii) To establish that the map

[TABLE]

is surjective, note that if $\beta\geq\alpha\in{\mathcal{J}}$ and $s\geq r$, then since $Z_{\beta}\supset Z_{\alpha}$,

[TABLE]

is a surjection of finite dimensional spaces. Thus it is sufficient to show that the composites

[TABLE]

are all surjective.

When $r=1$, this is immediate from part (i) of the Theorem. Moreover, it follows from the construction of $\tau$ that its image is an ideal in $L_{Z}$. This, together with the surjectivity of (24) when $r=1$ implies via the obvious induction that (24) is surjective for all $r$.

(iii) To show that $E$ is free it is sufficient to show that any linearly independent elements $x_{1},\dots,x_{k}\in S$ generate a free sub Lie algebra $F$. But by (ii) the restriction of $S$ to $Z\cap\mbox{ker}\,\overline{d}$ is an isomorphism $sS\stackrel{{\scriptstyle\cong}}{{\to}}(Z\cap\mbox{ker}\,\overline{d})^{\vee}$. It follows that there are $z_{1},\dots,z_{k}\in Z\cap\mbox{ker}\,\overline{d}$ such that

[TABLE]

Let $T$ be the linear span of the $z_{i}$, so that $\mathbb{Q}\oplus T\subset\mathbb{Q}\oplus(Z\cap\mbox{ker}\,\overline{d})$ is a sub cdga, with minimal Sullivan model $\land Z_{T}\subset\land Z$ satisfying $T=Z_{T}\cap\mbox{ker}\,\overline{d}$, and with homotopy Lie algebra $L_{T}$. The surjection $L_{Z}\to L_{T}$ maps the generating set $\{x_{i}\}$ of $F$ bijectively to a dual basis for $T$. As shown in the Example in §2, it follows that $F$ is free.

Finally, let $S_{\alpha}$ be the image of $S$ in $L_{Z_{\alpha}}$. Since $L_{Z}^{(2)}\to L_{Z_{\alpha}}^{(2)}$ is surjective, it follows that $S_{\alpha}+L_{Z_{\alpha}}^{2}=L_{Z_{\alpha}}$. Therefore, because $L_{Z_{\alpha}}$ is nilpotent, the induced maps $E\to L_{Z_{\alpha}}$ are surjective. Hence, these induce surjections $E/E\cap L_{Z}^{(r)}\to L_{Z_{\alpha}}/L_{Z_{\alpha}}^{r}$.

Since each $L_{Z_{\alpha}}/L^{r}_{Z_{\alpha}}$ is finite dimensional, it follows that passing to inverse limits yields surjections

[TABLE]

It is immediate from this that $\varprojlim_{r}E/E\cap L_{Z}^{(r)}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}L_{Z}.$

$\square$

6 One-relator groups

Our objective here is the proof of

Theorem 4 If $X$ is a wedge of at least two circles then any non-zero $[f]\in\pi_{1}(X)$ is rationally inert or, equivalently, $(X\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical.

proof: First observe that in fact

[TABLE]

In fact, the same argument as in the Example in §2 shows that the minimal Sullivan model of $X$ is cdga equivalent to $\mathbb{Q}\oplus H^{1}(X_{\mathbb{Q}})$. It follows that the homotopy Lie algebra, $L$, is concentrated in degree [math] and since $\pi_{*}(X_{\mathbb{Q}})=sL$, $X_{\mathbb{Q}}$ is aspherical. Thus if $[f]$ is rationally inert then $(X\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical. On the other hand, a Sullivan representative for the inclusion $i:X\to X\cup_{f}D^{2}$ is a morphism $\gamma:\land V\to\land W$ of minimal Sullivan algebras. Since $\pi_{1}(i)$ is injective, $H^{1}(i)$ is surjective and it follows that $\gamma:V^{1}\to W^{1}$ is injective. But if $(X\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical, then $V=V^{1}$, $\gamma$ is injective, and by definition $[f]$ is rationally inert.

Next note that it is sufficient to prove the Theorem when $X$ is a finite wedge of circles. Simply write $X=Y\vee Y^{\prime}$ in which $Y$ is a finite wedge of circles, $Y^{\prime}$ is a wedge of circles, and $f:S^{1}\to Y$. Then, as just observed, $Y^{\prime}_{\mathbb{Q}}$ is aspherical. It follows from Proposition 2 that if $(Y\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical, then so is $(X\cup_{f}D^{2})_{\mathbb{Q}}=\left[(Y\cup_{f}D^{2})\vee Y^{\prime}\right]_{\mathbb{Q}}$. Thus by (25), $[f]\in\pi_{1}(Y\cup_{f}D^{2})_{\mathbb{Q}}$ is rationally inert if and only if $[f]\in\pi_{1}(X_{\mathbb{Q}})$ is rationally inert.

In summary, we may and do assume henceforth that

[TABLE]

On the other hand, we observe that

[TABLE]

In fact, denote $G=\pi_{1}(X)$, so that $G_{\mathbb{Q}}=\pi_{1}(X_{\mathbb{Q}})$. According to [9, Theorem 7.5], $G^{n}/G^{n+1}\otimes\mathbb{Q}\stackrel{{\scriptstyle\cong}}{{\to}}G^{n}_{\mathbb{Q}}/G^{n+1}_{\mathbb{Q}}$. But by [15], $G^{n}/G^{n+1}$ is a free abelian group, and hence $G^{n}/G^{n+1}\to G^{n}_{\mathbb{Q}}/G^{n+1}_{\mathbb{Q}}$ is injective. Since $G$ is a free group, $G\to\varprojlim_{n}(G/G^{n})_{\mathbb{Q}}$ is injective and the image of $[f]$ in $G_{\mathbb{Q}}$ is non-zero. In particular, a Sullivan representative of $f$ is non-zero.

Next recall from the Example in §2 and Lemma 1 that $S^{1}\vee\dots\vee S^{1}\vee S^{2}$ has a quadratic minimal Sullivan model, $(\land W,d_{1})$ in which $W\cap\mbox{ker}\,d_{1}=H^{\geq 1}(S^{1}\vee\dots\vee S^{1}\vee S^{2})$. In particular, $W^{1}\cap\mbox{ker}\,d_{1}=H^{1}(S^{1}\vee\dots\vee S^{1})$. Moreover, $W^{>1}\cap\mbox{ker}\,d_{1}=W^{2}\cap\mbox{ker}\,d_{1}=\mathbb{Q}a$, where $a$ represents the orientation class of $S^{2}$. It follows that

[TABLE]

and that the identity in $\land W^{1}$ extends to a quasi-isomorphism

[TABLE]

with $\varphi(a)=a$ and $\varphi(R)=0$.

Note: In comparing with the general situation described in §3, observe that the $\land W^{1}$ here corresponds to the $\land W$ in §3, and that the $\land W$ here has no analogue in §3.

In particular $\varphi$ preserves wedge degrees when $a$ is assigned wedge degree 1. Thus not only is $H(\mbox{ker}\,\varphi)=0$, but in fact for cycles $\Phi\in\land W$,

[TABLE]

The proof of Theorem 3 is now accomplished in the following steps:

*Step One: Construction of a linear map of degree 1, $d_{0}:W\to W$, whose extension, also denoted $d_{0}$, to a derivation in $\land W$ provides a cdga $(\land W,d_{1}+d_{0})$ connected by cdga quasi-isomorphisms to $A_{PL}(X\cup_{f}D^{2})$. *

Step Two: $(\land W,d_{0}+d_{1})$ is a Sullivan algebra, and hence a Sullivan model for $X\cup_{f}D^{2}$.

Step Three: The minimal Sullivan model of $(\land W,d_{1}+d_{0})$ has the form $(\land V^{1},D)$, and so $(X\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical, and $[f]$ is rationally inert.

Step One: Construction of $d_{0}:W\to W$ whose extension to a derivation (also denoted by $d_{0}$) provides a cdga $(\land W,d_{0}+d_{1})$ connected by cdga quasi-isomorphisms to $A_{PL}(X\cup_{f}D^{2})$.

For this, fix a Sullivan representative $\psi:(\land W^{1},d)\to(\land v,0)$ for $f$ and, as at the start of §3, define $\varepsilon:\land W^{1}\to\mathbb{Q}$ by

[TABLE]

Then define a derivation $\delta$ in $\land W^{1}\oplus\mathbb{Q}a$ by setting

[TABLE]

Then $d_{1}\delta=0=\delta d_{1}$ and $\delta^{2}=0$, so that $(\land W^{1}\oplus\mathbb{Q}a,d_{1}+\delta)$ is a cdga. As observed at the start of §3, this cdga is connected by cdga quasi-isomorphisms to $A_{PL}(X\cup_{f}D^{2})$.

Next, we construct a linear map $d_{0}:W\to W$ of degree $1$ such that $d_{0}d_{1}+d_{1}d_{0}=0$ and $\varphi\circ d_{0}=\delta\circ\varphi$.

For this, recall that $W=\cup_{n}W(n)$ with $W(0)=W\cap\mbox{ker}\,d_{1}$ and $W(n+1)=W\cap d_{1}^{-1}(\land W(n))$. By convention, $W(-1)=0$. We assume by induction that $d_{0}$ is constructed in $W(n-1)$, and write $W(n)=W(n-1)\oplus S$. If $w\in S$, then

[TABLE]

and so $d_{0}d_{1}w$ is a cycle in $(\land^{2}W,d_{1})$.

Suppose first that $w\in W^{1}$. Then $d_{1}w\in\land^{2}W^{1}(n-1)$ and

[TABLE]

Thus by (27), for some $u\in\mbox{ker}\,\varphi\cap W^{2}$,

[TABLE]

Moreover, $\delta:W^{1}\to\mathbb{Q}a$, and so we may regard $\delta w$ as an element of $W^{2}$ for which $d_{1}\delta w=0$ in $\land W$. Set $d_{0}w=\delta w-u$. Then

[TABLE]

and, since $\varphi u=0$,

[TABLE]

On the other hand suppose $w\in W^{k}$, some $k\geq 2$. Then $d_{0}d_{1}w\in(\land^{2}W)^{k+2}$ and so $d_{0}d_{1}w\in R\land\land W\oplus\mathbb{Q}a^{2}$. Thus $\varphi(d_{0}d_{1}w)=0$ and again by (27) $d_{0}d_{1}w=d_{1}u$ for some $u\in W^{\geq 3}\subset R$. Set $d_{0}w=-u$, so that again

[TABLE]

Then, since $u\in R$, $\varphi u=0$ while $\varphi w\in\mathbb{Q}a$ and so $\delta\varphi w=0$ as well. This completes the construction of $d_{0}$. By construction,

[TABLE]

Finally we show that $d_{0}^{2}=0$ so that $d_{1}+d_{0}$ is a differential, and that

[TABLE]

In fact $d_{1}d_{0}^{2}=d_{0}^{2}d_{1}$. Assume by induction that $d_{0}^{2}=0$ in $W(n-1)$. Then for $w\in S$, $d_{0}^{2}w$ is a $d_{1}$-cycle and $\varphi(d_{0}^{2}w)=\delta^{2}\varphi w=0$. Thus by (27), $d_{0}^{2}w$ is a $d_{1}$-boundary, and hence $d_{0}^{2}w=0$. Thus $(\land W,d_{1}+d_{0})$ is a cdga and $\varphi$ is a morphism of cdga’s with respect to $d_{1}+d_{0}$ and $d_{1}+\delta$. Filter both sides by the difference between degree and wedge degree. The map induced by $\varphi$ in the $0^{th}$ term of the spectral sequence is the quasi-isomorphism $\varphi:(\land W,d_{1})\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}(\land W^{1}\oplus\mathbb{Q}a,d_{1})$. This establishes (28)

Note that by (16), the Sullivan representative $\psi$ is non-zero, and so for some $w\in W^{1}$, $\delta w=a$, and $d_{0}w\neq 0$.

Step Two: $(\land W,d_{1}+d_{0})$ is a Sullivan algebra, and hence is a Sullivan model for $X\cup_{f}D^{2}$.

Here we prove a more general result: if $(\land V,d)$ is any minimal Sullivan algebra and $d_{0}:V\to V$ is a linear map of degree 1 such that $d_{0}^{2}=dd_{0}+d_{0}d=0$, then $(\land V,d+d_{0})$ is a Sullivan algebra.

For this, fix an increasing filtration $0=V(0)\subset\dots\subset V(n)\subset\dots$ such that $V=\cup_{n}V(n)$ and $d:V(n+1)\to\land^{\geq 2}V(n)$. Then, as follows, define by induction a sequence of subspaces of $V$ of the form

[TABLE]

so that

[TABLE]

[TABLE]

and

[TABLE]

First, we set $Q(0)=P(0)=0$. Then suppose $Q(k)$, and $P(k)$ are constructed for $k\leq n$. Write

[TABLE]

and set

[TABLE]

It is immediate that

[TABLE]

Moreover, if $x\in S(n+1)$ then

[TABLE]

In particular, $d:Q(n+1)\to\land^{\geq 2}P(n)$. Further $d_{0}^{2}(S(n+1))=0$ and so $d_{0}(Q(n+1))=d_{0}(P(n))\subset P(n)$.

On the other hand, if $x\in S(n+1)$ then $d_{0}x\in Q(n+1)$ by construction, while $dx\in\land^{\geq 2}V(n)\subset\land P(n)$. This closes the induction and exhibits $(\land V,d+d_{0})$ as a Sullivan algebra.

Step Three: The minimal Sullivan model of $(\land W,d_{1}+d_{0})$ has the form $(\land V^{1},D)$, and so $(X\cup_{f}D^{2})_{\mathbb{Q}}$ is aspherical.

Recall from the Example in §2 that the homotopy Lie algebra of $(\land W,d_{1})$ is the completion, $\widehat{\mathbb{L}}$ of the free Lie algebra $\mathbb{L}(x_{1},\dots,x_{r},y)$ generated by vectors $x_{i}$ dual to the orientation classes of the circles, and by $y$ dual to the orientation class of $S^{2}$. By construction, $W^{\geq 2}=\mathbb{Q}a\oplus R$, and we may choose $y$ so that

[TABLE]

Now dualize $d_{0}:W\to W$ to $d:\widehat{\mathbb{L}}\to\widehat{\mathbb{L}}$. Since deg$\,d=-1$ it follows that $d:\widehat{\mathbb{L}}(x_{1},\dots,x_{r})\to 0$ and $dy\in\widehat{\mathbb{L}}(x_{1},\dots,x_{r})$. Moreover, because $d_{0}$ is a derivation satisfying $d_{0}d_{1}+d_{1}d_{0}=0=d_{0}^{2}$, it follows that $d$ is a derivation in the Lie algebra $\widehat{\mathbb{L}}$ and that $d^{2}=0$.

Moreover, if $(\land V,D)$ is the minimal Sullivan model of $(\land W,d_{1}+d_{0})$ then $V\cong H(W,d_{0})$. Therefore $H(\widehat{\mathbb{L}},d)=(H(W,d_{0}))^{\vee}$, and so it is sufficient to prove that

[TABLE]

Recall also from Step One that a Sullivan representative for $f$ determines a linear map $\varepsilon:W^{1}\to\mathbb{Q}$. Thus $\varepsilon$ desuspends to $\alpha\in{L}_{W^{1}}=\widehat{\mathbb{L}}(x_{1},\dots,x_{r})$. We show now that

[TABLE]

so that $dy\neq 0$.

For this, recall from Step One that if $w\in W^{1}$ then $d_{0}w=\varepsilon(w)a-u$, where $u\in W^{2}\cap\mbox{ker}\,\varphi=Z$. It follows that

[TABLE]

which establishes (29).

Denote by $\mathbb{L}_{q}(x_{i})$ the linear span of the commutators of length $q$ in the $x_{i}$. Write $dy$ as a series

[TABLE]

where $\alpha_{q}\in\mathbb{L}_{q}(x_{i})$ and $\alpha_{n}\neq 0$. Then form the differential graded Lie algebra $(\mathbb{L}(x_{i},y),\partial)$ with $\partial(x_{i})=0$ and $\partial(y)=\alpha_{n}$. Since $\alpha_{n}$ belongs to $\mathbb{L}_{n}(x_{i})$ we can modify the degrees in $\mathbb{L}(x_{i})$ by assigning deg$\,2$ to the $x_{i}$, without changing the homology with respect to $\partial$. Thus it follows from [16, Theorem 3.12] that $H_{q}(\mathbb{L}(x_{i},y),\partial)=0$ for $q>0$.

Now let $\omega=\sum_{q\geq p}\omega_{q}$ be a $d$-cycle in degree $r>0$ in $\widehat{\mathbb{L}}(x_{i},y)$, with $\omega_{q}\in\mathbb{L}_{q}(x_{i},y)$. Then $\omega_{p}$ is a $\partial$-cycle, and so a $\partial$-boundary. Choose $\beta_{p-n+1}\in\mathbb{L}_{p-n+1}(x_{i},y)$ with $\partial(\beta_{p-n+1})=\omega_{p}$. Write $\omega(1)=\omega-d(\beta_{p-n+1})$, then $\omega(1)$ is a sum $\sum_{s\geq p+1}\omega(1)_{s}$. One again $\omega(1)_{p+1}$ is a $\partial$-cycle. This determines $\beta_{p-n+2}$. Continue in this way to obtain at the and an element

[TABLE]

with $d\beta=\omega$.

$\square$

Corollary. With the notation of Theorem 3, set $V=W^{1}\cap\mbox{ker}\,d_{0}$. Then $(\land V,d_{1})\to(\land W^{1},d_{1}+d_{0})$ is the minimal Sullivan model of $X\cup_{f}D^{2}$.

proof: First note that any element in $\land^{2}W^{1}$ can be written as $\Phi=\sum_{i=1}^{n}w_{i}\land w_{i}^{\prime}$ in which $w_{1},\dots,w_{n},w_{1}^{\prime},\dots,w_{n}^{\prime}$ are all linearly independent. Thus if $d_{0}\Phi=0$ then each $d_{0}w_{i}=d_{0}w_{i}^{\prime}=0$. But $d_{1}:V\to\land^{2}W^{1}\cap\mbox{ker}\,d_{0}$, and so $\land V$ is preserved by $d_{1}$. It is immediate from Step Three that $V\stackrel{{\scriptstyle\cong}}{{\to}}H(W,d_{0})$, and it follows that $(\land V,d_{1})$ is the minimal Sullivan model of $X\cup_{f}D^{2}$.

$\square$

7 Whitehead’s problem and Theorem 5

Theorem 5. If $X$ is a connected CW complex and $(X\cup\amalg_{k=1}^{p}D^{2})_{\mathbb{Q}}$ is aspherical then $X_{\mathbb{Q}}$ is aspherical.

proof: The obvious induction reduces the statement to the case $p=1$. Then, since $\pi_{*}((X\cup_{f}D^{2})_{\mathbb{Q}})\cong V^{\vee}$ as sets where $\land V$ is the minimal Sullivan model of $X\cup_{f}D^{2}$, our hypothesis simply implies that $V=V^{1}$. Let $\varphi:(\land V,d)\to(\land W,d)$ be a Sullivan representative for the inclusion $i:X\to X\cup_{f}D^{2}$. Since $H^{1}(X\cup_{f}D^{2})\to H^{1}(X)$ is injective, it follows that $\varphi$ is injective and so $\land W$ decomposes as $\land V\otimes\land Z$, with $Z=Z^{\geq 1}$. In particular $|f|$ is rationally inert. Moreover, it follows from Proposition 1 that

[TABLE]

where deg$\,b=1$ and $\land V\otimes\land U$ is the acyclic closure of $\land V$. Since $V=V^{1}$, $U=U^{0}$ and $H^{\geq 1}(\land Z,\overline{d})=H^{1}(\land Z,\overline{d})$. This in turn implies $Z=Z^{1}$ and $X_{\mathbb{Q}}$ is aspherical.

$\square$