# Homology self closeness number and cofibration sequence

Gopal Chandra Dutta  Department of Mathematics and Statistics
Indian Institute of Technology, Kanpur
Uttar Pradesh 208016
India<EMAIL_ADDRESS>
###### Abstract.

We study the homology self-closeness numbers of simply connected CW complex
and those of the homotopy cofiber. Self-maps of spaces in cofibrations which
appear in Homology decomposition are studied. We also consider Postnikov tower
and studied the relation of homology self-closeness numbers between homology
decomposition and homotopy decomposition.

###### Key words and phrases:

self-homotopy equivalence, cofibration, self-closeness number, homology
decomposition, homotopy decomposition

###### 2020 Mathematics Subject Classification:

Primary 55P10, 55Q05 ; Secondary: 55P05, 55P30, 55S45.

##  1\. Introduction

For a pointed space $X$, the set of homotopy classes of base point preserving
maps $X\to X$ is denoted by $[X,X]$. This is a monoid under composition of
maps. The subset of all invertible elements of this monoid is denoted by
$\operatorname{Aut}(X)$. It is the group of homotopy classes of self-
equivalences on $X$. For notational convenience, we do not distinguish between
a map $f\colon X\to X$ and that of its homotopy class in $[X,X]$. The group
$\operatorname{Aut}(X)$ has been studied by several authors (cf. [2, 19]).
Choi and Lee introduced the following sub-monoids in [7]:

$\mathcal{A}_{\\#}^{k}(X):=\big{\\{}f\in[X,X]:~{}f_{\\#}\colon\pi_{i}(X)\xrightarrow{\cong}\pi_{i}(X),\text{
for all }0\leq i\leq k\big{\\}}.$

For connected CW complex $X$, we have the chain:

$[X,X]=\mathcal{A}_{\\#}^{0}(X)\cdots\supseteq\mathcal{A}_{\\#}^{k}(X)\supseteq\mathcal{A}_{\\#}^{k+1}(X)\cdots\supseteq\mathcal{A}_{\\#}^{\infty}(X)=\operatorname{Aut}(X).$

They introduced a numeric homotopy invariant associated to the chain, called
homotopy self-closeness number $N\mathcal{A}_{\\#}(X)$ defined as

$N\mathcal{A}_{\\#}(X):=\min\big{\\{}k\geq
0:~{}\mathcal{A}_{\\#}^{k}(X)=\operatorname{Aut}(X)\big{\\}}.$

Similar kind of monoids $\mathcal{A}_{\ast}^{k}(X)$ and
$\mathcal{A}^{\ast}_{k}(X)$ for homology and cohomology were introduced in
[16].

$\mathcal{A}_{\ast}^{k}(X):=\big{\\{}f\in[X,X]:~{}f_{\ast}\colon
H_{i}(X)\xrightarrow{\cong}H_{i}(X),\text{ for all }0\leq i\leq k\big{\\}};$
$\mathcal{A}^{\ast}_{k}(X):=\big{\\{}f\in[X,X]:~{}f^{\ast}\colon
H^{i}(X)\xrightarrow{\cong}H^{i}(X),\text{ for all }0\leq i\leq k\big{\\}}.$

We know that for a simply connected CW-complex, a self-map is a homotopy
equivalence if and only if it is a homology equivalence (or a cohomology
equivalence). Therefore, we have the two types of chain for simply connected
CW complex

$[X,X]=\mathcal{A}_{\ast}^{0}(X)\cdots\supseteq\mathcal{A}_{\ast}^{k}(X)\supseteq\mathcal{A}_{\ast}^{k+1}(X)\cdots\supseteq\mathcal{A}_{\ast}^{\infty}(X)=\operatorname{Aut}(X),$

and

$[X,X]=\mathcal{A}^{\ast}_{0}(X)\cdots\supseteq\mathcal{A}^{\ast}_{k}(X)\supseteq\mathcal{A}^{\ast}_{k+1}(X)\cdots\supseteq\mathcal{A}^{\ast}_{\infty}(X)=\operatorname{Aut}(X).$

This motivates to define _homology and cohomology self-closeness numbers_ (cf.
[16])

$N\mathcal{A}_{\ast}(X):=\min\big{\\{}k\geq
0:~{}\mathcal{A}_{\ast}^{k}(X)=\operatorname{Aut}(X)\big{\\}},$
$N\mathcal{A}^{\ast}(X):=\min\big{\\{}k\geq
0:~{}\mathcal{A}^{\ast}_{k}(X)=\operatorname{Aut}(X)\big{\\}}.$

In [15], Oda and Yamaguchi studied the homotopy self-closeness number of the
following type of a _fibration sequence_ :

(1) $\cdots\to K(G,n+1)\to X\to Y\xrightarrow{\gamma}K(G,n+2),$

where $G$ is an abelian group and $K(G,n)$ is the Eilenberg-MacLane space of
type $(G,n)$. Note that $X$ is the homotopy fiber of the map $\gamma$.

In 2021, Li studied the cohomology self-closeness number of the following type
of _cofibration sequence_ :

(2) $A\xrightarrow{\gamma}X\xrightarrow{\iota}Y\to\Sigma{A}.$

He obtained the relations between cohomology self-closeness numbers of $X$ and
$Y$ (cf. [12]).

Recall that for an abelian group $G$ and integer $n\geq 2$, the Moore space
$M(G,n)$ of type $(G,n)$ is the simply connected CW-complex $M(G,n)$ unique
upto homotopy, such that

$\tilde{H}_{i}(M(G,n))=\begin{cases}G&\text{if}~{}i=n,\\\ 0&\text{if}~{}i\neq
n.\end{cases}$

If $G$ is free-abelian, then $M(G,n)$ is just the wedge of copies of $S^{n}$
and $M({\mathbb{Z}}_{m},n)=S^{n}\cup_{m}e^{n+1},$ where $(n+1)$-cell is
attached to $S^{n}$ by a degree $m$ map. Thus, for a finitely generated $G$,
Moore space $M(G,n)$ is a finite CW-complex of dimension $n$ if $G$ is free-
abelian and of dimension $n+1$ if $G$ has torsion. Note that
$\pi_{n}(X;G):=[M(G,n),X]$ is a group for $n\geq 2$ and abelian group for
$n\geq 3$. It is called the _$n$ -th homotopy group of $X$ with coefficients
in $G$_.

In this paper, we consider the dual _cofibration sequence_ of Equation 1:

(3) $M(G,n)\xrightarrow{\gamma}X\xrightarrow{\iota}Y\to M(G,n+1)\to\cdots.$

Here $n\geq 2$, $Y$ is the homotopy cofiber of the map $\gamma$ and $\iota$ is
a cofibration. This sequence is a particular case of the cofibration sequence
2. The group $\operatorname{Aut}(Y)$ was studied by Oka, Sawashita and
Sugawara in [13]. The homotopy self-closeness number corresponding to the
cofibration sequence 3 was studied in [14] by considering $G=\mathbb{Z}$.

In Section 2, we establish relations between $N\mathcal{A}_{\ast}(X)$ and
$N\mathcal{A}_{\ast}(Y)$. Indeed, we find $N\mathcal{A}_{\ast}(Y)$ by
utilizing the given map $\gamma$ and making certain assumptions about $X$.
First, we prove the inequality $N\mathcal{A}_{\ast}(X)\leq
N\mathcal{A}_{\ast}(Y)$, when the induced homomorphism
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ is an epimorphism (see
Theorem 2.2).

Moreover, the reverse inequality $N\mathcal{A}_{\ast}(Y)\leq
N\mathcal{A}_{\ast}(X)$ holds if the induced homomorphism
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ is a monomorphism and
$\gamma^{\\#}\big{(}\operatorname{Aut}(X)\big{)}\subseteq\gamma_{\\#}\big{(}\pi_{n}(M(G,n);G)\big{)}$
(see Theorem 2.4).

Further, we get the equality $N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(Y)$
if the induced homomorphism
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ is an isomorphism (see
Theorem 2.5).

As a special case, assume that
$\pi_{n}(M(G,n);G)\cong\mathbb{Z}/{q\mathbb{Z}}\langle\operatorname{Id}_{M}\rangle$
and $\pi_{n}(X,G)\cong\mathbb{Z}/{m\mathbb{Z}}\langle\gamma\rangle$, where
$q,m\geq 0$. Then

  1. (a)

$N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(Y),~{}~{}\text{if}~{}q=0\neq
m~{}\text{or}~{}q>m.$

  2. (b)

$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(Y)$ if $q=m$ and
$\operatorname{conn}(X)\geq 2~{}(\text{or}~{}G~{}\text{is free})$.

(see Proposition 2.6).

By taking $G=\mathbb{Z}/{p\mathbb{Z}}$ for a prime $p\neq 2$, we obtain the
relation $N\mathcal{A}_{\ast}(Y)\leq N\mathcal{A}_{\ast}(X)$ under some
conditions (see Proposition 2.11).

In Section 3, we consider the _homology decomposition_ $\\{X_{n}\\}$ of a
simply connected CW-complex $X$, where $X_{n}$ is the $n$-th _homology
section_ (cf. [5, Section 7.3]). In Theorem 3.3, we obtain the relations
between homology self-closeness numbers of consecutive homology sections.
Moreover, in Lemma 3.6, we prove the relationship between the group of self-
homotopy equivalences of $X$ and $X_{n}$. Using this Lemma, we obtain the
relation between the two numbers $N\mathcal{A}_{\ast}(X)$ and
$N\mathcal{A}_{\ast}(X_{n})$ under some conditions, see Theorem 3.9.

In Section 4, we consider the _homotopy decomposition (Postnikov tower)_
$\\{X^{(n)}\\}$ of a simply connected CW-complex $X$, where $X^{(n)}$ is the
$n$-th _homotopy section_. The main aim of this section is to obtain relations
between homology and homotopy self-closeness numbers and their corresponding
decomposition. In Lemma 4.3, we obtain a relation between the homology self-
closeness numbers of homotopy section and homology section. Theorem 4.4 gives
the analogous results of Theorem 3.3 for homotopy sections. Choi and Lee have
studied the relation of homotopy self-closeness numbers of $X$ and $X^{(n)}$
[8, Theorem 3.5]. We find the homological version of the result in Theorem
4.6. We prove the equality
$N\mathcal{A}_{\ast}(X^{(n)})=N\mathcal{A}_{\ast}(X_{n})$ under some
conditions in Corollary 4.7. Finally, we compute homology self-closeness
numbers of homotopy sections (Postnikov tower) for some CW-complexes by the
help of homology decomposition.

######  1.1 Acknowledgements.

The author would like to thank IIT Kanpur for Ph.D fellowship.

##  2\. Homology self-closeness number

In this section, we study the relation between the homology self-closeness
numbers of $X,Y$ for the cofibration sequence of Equation 3. Recall that the
_connectivity_ of a space is defined as

$\operatorname{conn}(X):=\min\big{\\{}k\geq 0~{}:~{}\pi_{k+1}(X)\neq
0\big{\\}}.$

Thus, $\pi_{k}(X)=0$ for all $k\leq\operatorname{conn}(X)$. Moreover, we
denote the _homological dimension_ by

$H_{\ast}\text{-}\dim(X):=\max\big{\\{}k\geq 0~{}:~{}H_{k}(X)\neq 0\big{\\}}.$

Therefore, $H_{k}(X)=0$ for all $k>H_{\ast}\text{-}\dim(X)$. For any simply
connected CW-complex $X$, we have

$2=\operatorname{conn}(X)+1\leq N\mathcal{A}_{\ast}(X)\leq
H_{\ast}\text{-}\dim(X).$

Recall that, a map $f\colon X\to X^{\prime}$ is called _homotopical
$n$-equivalence_ if $f_{\\#}\colon\pi_{k}(X)\to\pi_{k}(X^{\prime})$ is
isomorphism for all $k<n$ and an epimorphism for $k=n$. A map $f\colon X\to
X^{\prime}$ is called _homological $n$-equivalence_ if $f_{\ast}\colon
H_{k}(X)\to H_{k}(X^{\prime})$ is isomorphism for all $k<n$ and an epimorphism
for $k=n$.

If $X^{\prime}$ is simply connected then the mapping cone $C_{f}$ of a map
$f\colon X\to X^{\prime}$ is simply connected (cf [5, Proposition B.4 ]).

######  2.1 Lemma.

Let $f\colon X\to X^{\prime}$ be a map between simply connected spaces. Then
the following are equivalent:

  1. (a)

$f$ is homotopical $n$-equivalence.

  2. (b)

$f$ is homological $n$-equivalence.

  3. (c)

$C_{f}$ is $n$-connected.

###### Proof.

Follows from [22, Theorem 4] and [5, Lemma 6.4.12].

∎

From now onwards, $X$ will denote a simply connected CW-complex and $G$ a
finitely generated abelian group.

######  2.2 Theorem.

$M(G,n)\xrightarrow{\gamma}X\xrightarrow{\iota}Y$ be a cofibration sequence
such that the induced homomorphism
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ is an epimorphism. Then

  1. (a)

$2\leq N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(Y)\leq n+1$ if
$H_{\ast}\text{-}\dim(X)<n$.

  2. (b)

$n\leq N\mathcal{A}_{\ast}(Y)$ if $H_{n}(X)=H_{n+1}(X)=0$ and $2\leq n\leq
N\mathcal{A}_{\ast}(X)$.

###### Proof.

  1. (a)

Note that $N\mathcal{A}_{\ast}(X)\leq H_{\ast}\text{-}\dim(X)<n.$ Moreover,
$H_{k}(Y)=0$ for all $k\geq n+2$ using the long exact sequence of homology
groups of the cofibration sequence. Therefore, $N\mathcal{A}_{\ast}(Y)\leq
n+1$. Assume that $N\mathcal{A}_{\ast}(Y)=m$ and
$g\in\mathcal{A}_{\ast}^{m}(X)$. Let $m<n$, otherwise we get the desired
result.

Since $\gamma_{\\#}$ is epimorphism, so there exists $h\colon M(G,n)\to
M(G,n)$ such that $g\circ\gamma=\gamma\circ h$. Therefore, there is a map
$f\in[Y,Y]$ such that $f\circ\iota=\iota\circ g$. Hence, we get a homotopy
commutative diagram

(10)

Note that $\iota_{\ast}\colon H_{k}(X)\to H_{k}(Y)$ is isomorphism for all
$k\leq n-1$. Therefore, $f_{\ast}\colon H_{k}(Y)\to H_{k}(Y)$ is isomorphism
for all $k\leq m$. So $f\in\mathcal{A}_{\ast}^{m}(Y)=\operatorname{Aut}(Y)$.
Hence, using the commutativity of the right side diagram, we have
$g\in\mathcal{A}_{\ast}^{n-1}(X)=\operatorname{Aut}(X)$. Consequently,
$N\mathcal{A}_{\ast}(X)\leq m=N\mathcal{A}_{\ast}(Y)$ and we get the desired
result.

  2. (b)

Let $N\mathcal{A}_{\ast}(Y)=m$ and $g\in\mathcal{A}_{\ast}^{m}(X)$. As in the
proof of part (a), we have the diagram 10. If possible, assume that $m<n$.
Therefore, $f\in\mathcal{A}_{\ast}^{m}(Y)=\operatorname{Aut}(Y)$. Observe the
following commutative diagram:

(17)

Thus $h\in\operatorname{Aut}(M(G,n))$. Therefore, using the five Lemma on the
long exact sequence of homology groups of the diagram 10, we have
$g\in\operatorname{Aut}(X)$. This contradicts the fact that
$N\mathcal{A}_{\ast}(X)\geq n$. Hence, $N\mathcal{A}_{\ast}(Y)=m\geq n$.

∎

######  2.3 Remark.

If $\gamma=0$, in that case $Y=X\vee M(G,n+1)$ for $n\geq 2$. This follows
that the difference between $N\mathcal{A}_{\ast}(X)$ and
$N\mathcal{A}_{\ast}(Y)$ only depends on $X$. Let $X=S^{2}$, then
$N\mathcal{A}_{\ast}(X)=2<n+1=N\mathcal{A}_{\ast}(Y).$

######  2.4 Theorem.

$M(G,n)\xrightarrow{\gamma}X\xrightarrow{\iota}Y$ be a cofibration sequence
such that the induced homomorphism
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ is a monomorphism and
$\gamma^{\\#}\big{(}\operatorname{Aut}(X)\big{)}$
$\subseteq\gamma_{\\#}\big{(}\pi_{n}(M(G,n);G)\big{)}$. Then

  1. (a)

$3\leq N\mathcal{A}_{\ast}(Y)\leq N\mathcal{A}_{\ast}(X)\leq n-1$ if $X$ is
$2$-connected, $H_{\ast}\text{-}\dim(X)<n$ and $G$ has torsion.

  2. (b)

$2\leq N\mathcal{A}_{\ast}(Y)\leq N\mathcal{A}_{\ast}(X)\leq n-1$ if
$H_{\ast}\text{-}\dim(X)<n$ and $G$ is free.

###### Proof.

  1. (a)

Observe that $\iota\colon X\to Y$ is homological $n$-equivalence. So
homotopical $n$-equivalence by Lemma 2.1. Since $H_{\ast}\text{-}\dim(X)<n$,
therefore using [10, Proposition 4C.1] there exits a CW-complex say $Z$ which
is homotopy equivalence to $X$ and $\dim(Z)\leq n$. Thus the induced map
$\iota_{\\#}\colon[Z,X]\to[Z,Y]$ is onto. Moreover, let $p\colon Z\to X$ is a
homotpy equivalence map with homotopy inverse $q\colon X\to Z$. Therefore,
$p^{\\#}\colon[X,X]\to[Z,X]$ is a bijection map defined as $f\mapsto f\circ
p$. Similarly, $q^{\\#}\colon[Z,Y]\to[X,Y]$ is bijection. Consequently, the
induced map $\iota_{\\#}\colon[X,X]\to[X,Y]$ is surjective.

Let $N\mathcal{A}_{\ast}(X)=l<n$ and $f\in\mathcal{A}_{\ast}^{l}(Y)$. By
surjectivity of $\iota_{\\#}\colon[X,X]\to[X,Y]$ implies that there exist some
$g\in[X,X]$ such that $\iota\circ g=f\circ\iota$. Note that
$\operatorname{conn}(X)=2$ implies that $\operatorname{conn}(Y)=2$. Therefore,
the homotopical connectivity of the map $\gamma\colon M(G,n)\to X$ is 2 by
Lemma 2.1. Therefore, using [18, Proposition 4.4] there exists a map $h\colon
M(G,n)\to M(G,n)$ such that we get the homotopy commutative diagram 10. Since
$\iota_{\ast}\colon H_{k}(X)\xrightarrow{\cong}H_{k}(Y)$ for all $k\leq l$,
therefore $g\in\mathcal{A}_{\ast}^{l}(X)=\operatorname{Aut}(X)$. Let
$\bar{g}\in\operatorname{Aut}(X)$ such that $g\circ\bar{g}=\bar{g}\circ
g=\operatorname{Id}_{X}$. It is sufficient to show that
$h\in\operatorname{Aut}(M(G,n))$.

From the assumption (ii),
$\gamma^{\\#}(\bar{g})\in\gamma_{\\#}\big{(}[M(G,n),M(G,n)]\big{)}$. Then
there exits $\bar{h}\in[M(G,n),M(G,n)]$ such that
$\gamma_{\\#}(\bar{h})=\gamma^{\\#}(\bar{g})$. So
$\gamma\circ\bar{h}=\bar{g}\circ\gamma$. Therefore, we have

$\gamma\circ\bar{h}\circ h=\bar{g}\circ\gamma\circ h=\bar{g}\circ
g\circ\gamma=\gamma,$

and similarly, $\gamma\circ h\circ\bar{h}=\gamma$. Injectivity of
$\gamma_{\\#}$ implies that $h\circ\bar{h}=\bar{h}\circ
h=\operatorname{Id}_{M(G,n)}$. So
$h\in\operatorname{Aut}\big{(}M(G,n)\big{)}$. Using five Lemma on the long
exact sequence of homology groups of the Diagram 10, we have
$f\in\operatorname{Aut}(Y)$. Consequently,

$N\mathcal{A}_{\ast}(Y)\leq l=N\mathcal{A}_{\ast}(X).$

  2. (b)

Note that $\operatorname{conn}(X)=1$ implies that $\operatorname{conn}(Y)=1$.
Moreover, if $G$ is free abelian group, then $\dim(M(G,n))=n$. Therefore, we
can use [18, Proposition 4.4] to obtain the diagram 10. Rest of the proof
follows similarly.

∎

######  2.5 Theorem.

$M(G,n)\xrightarrow{\gamma}X\xrightarrow{\iota}Y$ be a cofibration sequence
such that the induced homomorphism
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ is an isomorphism. Then

  1. (a)

$N\mathcal{A}_{\ast}(Y)=N\mathcal{A}_{\ast}(X)$ if $X$ is $2$-connected,
$H_{\ast}\text{-}\dim(X)<n$ and $G$ has torsion.

  2. (b)

$N\mathcal{A}_{\ast}(Y)=N\mathcal{A}_{\ast}(X)$ if $H_{\ast}\text{-}\dim(X)<n$
and $G$ is free.

###### Proof.

From Theorem 2.2, we have $N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(Y)$.

For the converse part, surjectivity of the map
$\gamma_{\\#}\colon\pi_{n}(M(G,n);G)\to\pi_{n}(X;G)$ implies that

$\gamma^{\\#}\big{(}\operatorname{Aut}(X)\big{)}\subseteq\pi_{n}(X;G)=\gamma_{\\#}\big{(}\pi_{n}(M(G,n);G)\big{)}.$

Therefore, using Theorem 2.4, we have $N\mathcal{A}_{\ast}(Y)\leq
N\mathcal{A}_{\ast}(X)$. Hence, we get the desired result.

∎

######  2.6 Proposition.

$M(G,n)\xrightarrow{\gamma}X\xrightarrow{\iota}Y$ be a cofibration sequence
such that $H_{\ast}\text{-}\dim(X)<n$. Assume that there exist non-negative
integers $q,m$ such that

$\pi_{n}(M(G,n);G)\cong\mathbb{Z}/{q\mathbb{Z}}\langle\operatorname{Id}_{M}\rangle,~{}~{}\pi_{n}(X,G)\cong\mathbb{Z}/{m\mathbb{Z}}\langle\gamma\rangle.$

Then

  1. (a)

$N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(Y)$ if $q=0\neq m\text{ or
}~{}q>m$.

  2. (b)

$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(Y)$ if $q=m$, $X$ is $2$-connected
and $G$ has torsion.

  3. (c)

$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(Y)$ if $q=m$ and $G$ is a free
abelian group.

###### Proof.

  1. (a)

Note that $\gamma_{\\#}(f_{1}+f_{2})=\gamma\circ(f_{1}+f_{2})\cong\gamma\circ
f_{1}+\gamma\circ f_{2}$. So $\gamma_{\\#}$ is a homomorphism. It is
sufficient to show that $\gamma_{\\#}$ is surjective. Let $g\in\pi_{n}(X;G)$.
Then there exists an integer $s$ such that $g=s\gamma$. Observe that

$g=s\gamma=\gamma\circ
s\operatorname{Id}_{M}=\gamma_{\\#}(s\operatorname{Id}_{M}).$

Thus $\gamma_{\\#}$ is surjective. Therefore using Theorem 2.2(a), we get the
desired result.

  2. (b)

Assume that $q=m$. As in the proof of part(a), we have surjectivity of the
homomorphism $\gamma_{\\#}$. Moreover, $\gamma_{\\#}$ is injective as $q=m$.
Therefore, we get the equality using Theorem 2.5(a).

  3. (c)

Similar as part (b).

∎

The following Corollary is a homological version of [14, Theorem 4].

######  2.7 Corollary.

Let $X$ be a simply connected CW-complex such that $\dim(X)\leq n-1$ and
$\pi_{n}(X)\cong\mathbb{Z}$, where $n\geq 2$. If $\gamma\colon S^{n}\to X$ is
a generator of $\pi_{n}(X)$, then

$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X\cup_{\gamma}e^{n+1}).$

###### Proof.

It is enough to show that $\gamma_{\\#}\colon[S^{n},S^{n}]\to[S^{n},X]$ is a
bijection. For surjectivity, let $g\in[S^{n},X].$ Therefore there exists an
integer $s$ such that $g=s\gamma$. We know that $s\gamma=\gamma\circ
s\operatorname{Id}_{S^{n}}$, where $\operatorname{Id}_{S^{n}}$ is a generator
of $\pi_{n}(S^{n})$. Thus $g=\gamma_{\\#}(s\operatorname{Id}_{S^{n}})$. Hence
$\gamma_{\\#}$ is surjective.

For injectivity, let $f_{1},f_{2}\in[S^{n},S^{n}]$ such that
$\gamma_{\\#}(f_{1})=\gamma_{\\#}(f_{2})$. This implies $\gamma\circ
f_{1}=\gamma\circ f_{2}$. Moreover observe that
$f_{1}=k_{1}\operatorname{Id}_{S^{n}},~{}~{}f_{2}=k_{2}\operatorname{Id}_{S^{n}}$
for some $k_{1},k_{2}\in\mathbb{Z}$. Therefore $\gamma\circ
k_{1}\operatorname{Id}_{S^{n}}=\gamma\circ k_{2}\operatorname{Id}_{S^{n}}$.
This implies $k_{1}\gamma=k_{2}\gamma$. So that $k_{1}=k_{2}$. Then using
Theorem 2.5 we get the desired result.

∎

######  2.8 Remark.

Corollary 2.7 can also be deduced directly from Proposition 2.6.

Recall the generalised Fruedenthal suspension theorem: Let $Y$ be an
$n$-connected CW-complex and $X$ is finite dimensional CW-complex. Then the
suspension map $\Sigma\colon[X,Y]\to[\Sigma X,\Sigma Y]$ is isomorphism if
$\dim(X)<2n+1$ and epimorphism if $\dim(X)=2n+1$ ([9, Theorem 1.21]).

######  2.9 Proposition.

Consider the Hopf map $S^{3}\xrightarrow{\eta_{2}}S^{2}$ and all its
suspension map $S^{n+1}\xrightarrow{\eta_{n}}S^{n}$, where $n\geq 3$. Then

$N\mathcal{A}_{\ast}(S^{n}\cup_{\eta_{n}}e^{n+2})=\begin{cases}2&\text{if}~{}n=2,\\\
n+2&\text{if}~{}n\geq 3.\end{cases}$

###### Proof.

For $n=2$, the Hopf map is $S^{3}\xrightarrow{\eta_{2}}S^{2}$ generates the
group $\pi_{3}(S^{2})\cong\mathbb{Z}$. Therefore, using Corollary 2.7, we have
$N\mathcal{A}_{\ast}(S^{2})=N\mathcal{A}_{\ast}(S^{2}\cup_{\eta_{2}}e^{4})$.
Hence $N\mathcal{A}_{\ast}(S^{2}\cup_{\eta_{2}}e^{4})=2$.

For $n\geq 3,$ the suspension of $\eta_{2}$ is
$S^{n+1}\xrightarrow{\eta_{n}}S^{n}$. Using [10, Corollary 4J.4], we have
$\pi_{n+1}(S^{n})\cong\mathbb{Z}_{2}\langle\eta_{n}\rangle$. Therefore, we
have $N\mathcal{A}_{\ast}(S^{n})\leq
N\mathcal{A}_{\ast}(S^{n}\cup_{\eta_{n}}e^{n+2})$ by Proposition 2.6(a).
Hence,

$n\leq N\mathcal{A}_{\ast}(S^{n}\cup_{\eta_{n}}e^{n+2})\leq n+2,~{}\text{ for
all }n\geq 3.$

It is sufficient to show that
$N\mathcal{A}_{\ast}(S^{n}\cup_{\eta_{n}}e^{n+2})\neq n$. Let
$P=S^{n}\cup_{\eta_{n}}e^{n+2}$. By [1, Section 8], we have

$[P,P]\cong\mathbb{Z}\langle\operatorname{Id}_{P}\rangle\oplus\mathbb{Z}\langle\iota\circ\bar{\xi}\rangle,$

where $\operatorname{Id}_{P}\colon P\to P$ is the identity map, $\iota\colon
S^{n}\to P$ is inclusion map, $q\colon P\to S^{n+2}$ is quotient map and
$\bar{\xi}\in[P,S^{n}],~{}~{}\widetilde{\xi}\in[S^{n+2},P]$ satisfying the
following relations

(18)
$\bar{\xi}\circ\iota=2\cdot\operatorname{Id}_{S^{n}},~{}~{}q\circ\widetilde{\xi}=2\cdot\operatorname{Id}_{S^{n+2}},~{}~{}~{}\iota\circ\bar{\xi}+\widetilde{\xi}\circ
q=2\cdot\operatorname{Id}_{p}.$

Observe that the homomorphisms $\iota_{\ast}\colon
H_{n}(S^{n})\xrightarrow{\cong}H_{n}(P)$ and $q_{\ast}\colon
H_{n+2}(P)\xrightarrow{\cong}H_{n+2}(S^{n+2})$ are isomorphisms. Let
$H_{0}(S^{0})\cong{\mathbb{Z}}\langle a_{0}\rangle.$ Consider the suspension
map $H_{0}(S^{0})\xrightarrow{\Sigma^{k}}H_{k}(S^{k})$ defined as
$a_{0}\mapsto a_{k}=\Sigma^{k}(a_{0})$, which is an isomorphism for each
$k\geq 1$. Observe that

$H_{k}(P)=\begin{cases}\mathbb{Z}\langle b\rangle&\text{if}~{}k=n,\\\
\mathbb{Z}\langle c\rangle&\text{if}~{}~{}k=n+2,\\\ 0&\text{ otherwise
},\end{cases}$

where $b=\iota_{\ast}(a_{n})$, and $c=(q_{\ast})^{-1}(a_{n+2})$. From the
relation 18, we have

$(\iota\circ\bar{\xi})_{\ast}(b)=(\iota\circ\bar{\xi}\circ\iota)_{\ast}(a_{n})=(2\cdot\iota\circ\operatorname{Id}_{S^{n}})_{\ast}(a_{n})=2\cdot\iota_{\ast}(a_{n})=2\cdot
b.$

Moreover, $\widetilde{\xi}_{\ast}(a_{n+2})=(q_{\ast})^{-1}\circ
q_{\ast}\circ\widetilde{\xi}_{\ast}(a_{n+2})=2\cdot(q_{\ast})^{-1}(a_{n+2})=2\cdot
c.$

Therefore,

$(\iota\circ\bar{\xi})_{\ast}(c)=(2\cdot\operatorname{Id}_{P}-\widetilde{\xi}\circ
q)_{\ast}(c)=2\cdot c-\widetilde{\xi}_{\ast}\circ q_{\ast}(c)=2\cdot
c-\widetilde{\xi}_{\ast}(a_{n+2})=0.$

So for $f\in\mathcal{A}_{\ast}^{n}(P)$, we have $f_{\ast}\colon
H_{k}(P)\xrightarrow{\cong}H_{k}(P)$ for all $k\leq n$. Note that

$f=m\operatorname{Id}_{P}+r(\iota\circ\bar{\xi}),\text{ where
}m,r\in\mathbb{Z}.$

Thus $f_{\ast}(b)=mb+2rb=(m+2r)b$. This implies that $m+2r=+1\text{ or }-1$.
Therefore,
$f=3\operatorname{Id}_{P}-(\iota\circ\bar{\xi})\in\mathcal{A}_{\ast}^{n}(P)$.
But $f\notin\operatorname{Aut}(P)$ as $f_{\ast}(c)=3c.$ Hence
$N\mathcal{A}_{\ast}(P)\neq n.$

∎

######  2.10 Remark.

Note that $\Sigma(C_{\eta_{n}})=C_{\Sigma\eta_{n}}=C_{\eta_{n+1}}$ for all
$n\geq 2$. Therefore,
$\Sigma^{n-2}\mathbb{C}P^{2}=S^{n}\cup_{\eta_{n}}e^{n+2}$ for all $n\geq 2$.
Hence,

$N\mathcal{A}_{\ast}(\Sigma^{n-2}\mathbb{C}P^{2})=\begin{cases}2&\text{if}~{}n=2,\\\
n+2&\text{if}~{}n\geq 3.\end{cases}$

######  2.11 Proposition.

$M(G,n)\xrightarrow{\gamma}X\xrightarrow{\iota}Y$ be a cofibration sequence
such that $X$ is $2$-connected and $H_{\ast}\text{-}\dim(X)<n$. Assume that
any one of the following conditions hold:

  1. (i)

For $q\geq
0,~{}\pi_{n}(M(G,n);G)\cong\mathbb{Z}/{q\mathbb{Z}}\langle\operatorname{Id}_{M}\rangle$
and $\gamma$ is a generator of any $\mathbb{Z}$ direct summand of the abelian
group $\pi_{n}(X;G)$.

  2. (ii)

For a prime $p>2$, if $G=\mathbb{Z}/{p\mathbb{Z}}$ and $\gamma$ is a generator
of any $\mathbb{Z}$ direct summand of the abelian group $\pi_{n}(X;G)$.

Then $N\mathcal{A}_{\ast}(Y)\leq N\mathcal{A}_{\ast}(X)$.

###### Proof.

Let $N\mathcal{A}_{\ast}(X)=l<n$ and $f\in\mathcal{A}_{\ast}^{l}(Y)$. As in
the proof of Theorem 2.4, we have $g\colon X\to X$ and $h\colon M(G,n)\to
M(G,n)$, which satisfies the commutative diagram 10. Moreover,
$g\in\mathcal{A}_{\ast}^{l}(X)=\operatorname{Aut}(X)$. Let $\bar{g}$ be the
homotopy inverse of $g$. It is sufficient to show that
$h\in\operatorname{Aut}(M(G,n))$.

  1. (i)

Let $\pi_{n}(X,G)\cong\mathbb{Z}\langle\gamma\rangle\oplus U$ for some
subgroup $U$. Then we have $h=s\operatorname{Id}_{M}$ for some
$s\in\mathbb{Z}$. Moreover $\bar{g}\circ\gamma=t\gamma+u$ for some
$t\in\mathbb{Z}$ and some $u\in U$. It follows that

$\gamma=\bar{g}\circ g\circ\gamma=\bar{g}\circ\gamma\circ h=(t\gamma+u)\circ
s\operatorname{Id}_{M}=ts\gamma+su.$

This implies that $ts=1$ and $su=0$. So $s=+1\text{ or}-1$, hence
$h\in\operatorname{Aut}(M(G,n)).$

  2. (ii)

From universal coefficient theorem for homotopy, we have

$0\to\operatorname{Ext}\big{(}G,\pi_{n+1}(Z)\big{)}\to\pi_{n}(Z,G)\to\operatorname{Hom}\big{(}G,\pi_{n}(Z)\big{)}\to
0.$

Take $Z=M(\mathbb{Z}/{p\mathbb{Z}},n)$. From [4, Section 1], we have

$\pi_{n+1}(Z)=\pi_{n+1}(M(\mathbb{Z}/{p\mathbb{Z}},n))=\mathbb{Z}/{p\mathbb{Z}}\otimes\mathbb{Z}/{2\mathbb{Z}}=0.$

Therefore,
$\pi_{n}(M(\mathbb{Z}/{p\mathbb{Z}},n);\mathbb{Z}/{p\mathbb{Z}})\cong\operatorname{Hom}(\mathbb{Z}/{p\mathbb{Z}},\mathbb{Z}/{p\mathbb{Z}})\cong\mathbb{Z}/{p\mathbb{Z}}\langle\operatorname{Id}_{M}\rangle$.
Hence, using part (i), we get the desired result.

∎

######  2.12 Remark.

If $G$ is a free abelian group, then simply connectedness is sufficient for
Proposition 2.11.

The following Corollary is a homological version of [16, Theorem 5].

######  2.13 Corollary.

Let $X$ be a simply connected CW-complex such that $\dim(X)\leq n-1$, where
$n\geq 2$. If $\gamma\colon S^{n}\to X$ is a generator of a direct summand
$\mathbb{Z}\subset\pi_{n}(X)$, then

$N\mathcal{A}_{\ast}(X\cup_{\gamma}e^{n+1})\leq N\mathcal{A}_{\ast}(X).$

###### Proof.

Follows from Proposition 2.11(i), by taking $G=\mathbb{Z}$.

∎

######  2.14 Example.

Finiteness theorem of Serre [21, Section 6]

$\pi_{4n-1}(S^{2n})\cong\mathbb{Z}\oplus F_{n},\text{ for all }n\geq 1,$

where $F_{n}$ is a finite abelian group. Let $\gamma$ be a generator of the
$\mathbb{Z}$ direct summand of the abelian group $\pi_{4n-1}(S^{2n})$.
Therefore, $N\mathcal{A}_{\ast}(S^{2n}\cup_{\gamma}e^{4n})\leq
N\mathcal{A}_{\ast}(S^{2n})$ by using Corollary 2.13. Hence,

$N\mathcal{A}_{\ast}(S^{2n}\cup_{\gamma}e^{4n})\leq 2n.$

Observe that
$\mathcal{A}_{\ast}^{2n-1}(S^{2n}\cup_{\gamma}e^{4n})(X)=[S^{2n}\cup_{\gamma}e^{4n}~{},~{}S^{2n}\cup_{\gamma}e^{4n}]$.
Let $f\colon S^{2n}\cup_{\gamma}e^{4n}\to S^{2n}\cup_{\gamma}e^{4n}$ be a
constant map at base point. Therefore,
$f\in[S^{2n}\cup_{\gamma}e^{4n}~{},~{}S^{2n}\cup_{\gamma}e^{4n}]$ but
$f\notin\operatorname{Aut}(S^{2n}\cup_{\gamma}e^{4n})$. Hence,

$N\mathcal{A}_{\ast}(S^{2n}\cup_{\gamma}e^{4n})=N\mathcal{A}_{\ast}(S^{2n})=2n.$

##  3\. Homology decomposition

In this section, we consider homology decomposition of a space and its
relation with homology self-closeness numbers. First, we recall the
definition.

######  3.1 Definition.

Let $X$ be a simply connected CW-complex. A _homology decomposition_ of $X$
consists of a sequence of simply connected CW-complexes $\\{X_{n}\\}$ and
structure maps

$j_{n}\colon X_{n}\to X,\iota_{n}\colon X_{n}\to
X_{n+1},~{}k_{n+1}:M\big{(}H_{n+1}(X),n\big{)}\to X_{n},$

that satisfy the following conditions:

  1. (i)

$j_{n\ast}\colon H_{k}(X_{n})\to H_{k}(X)$ is an isomorphism for all $k\leq n$
and $H_{k}(X_{n})=0$ for all $k>n$.

  2. (ii)

$M\big{(}H_{n+1}(X),n\big{)}\xrightarrow{k_{n+1}}X_{n}\xrightarrow{\iota_{n}}X_{n+1}$
is a cofibration sequence of the cellular map $k_{n+1}$ such that the induced
map $k_{n+1\ast}\colon H_{n}(M(H_{n+1}(X),n))\to H_{n}(X_{n})$ is trivial.

  3. (iii)

The following diagram commutes.

|
---|---
$\textstyle{X_{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j_{n+1}}$$\textstyle{X}$$\textstyle{X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j_{n}}$$\scriptstyle{\iota_{n}}$

The collection $\big{\\{}X_{n},j_{n},\iota_{n},k_{n}\big{\\}}$ is called a
homology decomposition of $X$. The spaces $X_{n}$ are called the $n$-th
homology sections of $X$ and the maps $k_{n+1}\colon
M\big{(}H_{n+1}(X),n\big{)}\to X_{n}$ are called homological $k$-invariants.
It is well known that every simply connected CW-complex admits a homology
decomposition (see [5, 10]). Note that if all homology groups are free then
the cellular skeletons serve as a homology decompositions, with the cell
attaching maps as $k$-invariants.

######  3.2 Remark.

It follows from the definition that $N\mathcal{A}_{\ast}(X_{n})\leq n.$

The following Theorem deduce relations between the homology self-closeness
numbers of two consecutive homology section.

######  3.3 Theorem.

Let $X$ be a simply connected $CW$-complex. For a homology decomposition
$\big{\\{}X_{n},j_{n},\iota_{n},k_{n}\big{\\}}$ of $X$, we have the following:

  1. (a)

If the induced homomorphism

$k_{m+1\\#}\colon\pi_{m}\big{(}M(H_{m+1}(X),m);H_{m+1}(X)\big{)}\to\pi_{m}\big{(}X_{m};H_{m+1}(X)\big{)}$

is an epimorphism for some $m\in\mathbb{N}$, then
$N\mathcal{A}_{\ast}(X_{m})\leq N\mathcal{A}_{\ast}(X_{m+1})$.

  2. (b)

If the induced homomorphism

$k_{m+1\\#}\colon\pi_{m}\big{(}M(H_{m+1}(X),m);H_{m+1}(X)\big{)}\to\pi_{m}\big{(}X_{m};H_{m+1}(X)\big{)}$

is a monomorphism such that

$k^{\\#}_{m+1}(\operatorname{Aut}(X_{m}))\subseteq
k_{m+1\\#}\Big{(}\pi_{m}\big{(}M(H_{m+1}(X),m);H_{m+1}(X)\big{)}\Big{)},$

where $H_{k}(X)$ is free-abelian group for $k=m,m+1$, then

$N\mathcal{A}_{\ast}(X_{m+1})\leq N\mathcal{A}_{\ast}(X_{m}).$

  3. (c)

If $H_{k}(X)$ is free-abelian group for $k=m,m+1$ and the induced homomorphism
$k_{m+1\\#}$ is an isomorphism, then
$N\mathcal{A}_{\ast}(X_{m})=N\mathcal{A}_{\ast}(X_{m+1})$.

###### Proof.

  1. (a)

Let $N\mathcal{A}_{\ast}(X_{m+1})=l\leq m+1$ and
$g\in\mathcal{A}_{\ast}^{l}(X_{m})$. If $l\geq m$, then
$\mathcal{A}_{\ast}^{l}(X_{m})=\operatorname{Aut}(X_{m})$ and we get the
desired result.

Let us assume that $l<m$. Surjectivity of $k_{m+1\\#}$ implies that there
exists $h\colon M\big{(}H_{m+1}(X),m\big{)}\to M\big{(}H_{m+1}(X),m\big{)}$
such that $k_{m+1}\circ h=g\circ k_{m+1}$. Therefore, we get a homotopy
commutative diagram

(25)

Observe that $\iota_{m\ast}\colon
H_{k}(X_{m})\xrightarrow{\cong}H_{k}(X_{m+1})$ for all $k\leq m$. Therefore,
$f\in\mathcal{A}_{\ast}^{l}(X_{m+1})=\operatorname{Aut}(X_{m+1})$. Using the
commutativity of the right side diagram, we have
$g\in\mathcal{A}_{\ast}^{m}(X_{m})=\operatorname{Aut}(X_{m})$. Consequently,

$N\mathcal{A}_{\ast}(X_{m})\leq l=N\mathcal{A}_{\ast}(X_{m+1}).$

  2. (b)

Observe that $H_{\ast}\text{- }\dim(X_{m})\leq m$ and the map $\iota_{m}\colon
X_{m}\to X_{m+1}$ is homotopical $m$-equivalence. So the induced map
$\iota_{\\#}\colon[X_{m},X_{m}]\to[X_{m},X_{m+1}]$ is surjective.

Let $N\mathcal{A}_{\ast}(X_{m})=r\leq m$ and
$f\in\mathcal{A}_{\ast}^{r}(X_{m+1})$. By surjectivity of $\iota_{\\#}$ there
exists $g\in[X_{m},X_{m}]$ such that $\iota_{m}\circ g=f\circ\iota_{m}$.
Therefore, using [18, Proposition 4.4] there is a map $h\colon
M\big{(}H_{m+1}(X),m\big{)}\to M\big{(}H_{m+1}(X),m\big{)}$ such that we have
the homotopy commutative Diagram 25. Since $\iota_{m\ast}\colon
H_{k}(X_{m})\xrightarrow{\cong}H_{k}(X_{m+1})$ for all $k\leq m$, therefore
$g\in\mathcal{A}_{\ast}^{r}(X_{m})=\operatorname{Aut}(X_{m})$. Thus, there is
a $\bar{g}\in\operatorname{Aut}(X_{m})$ such that
$g\circ\bar{g}=\operatorname{Id}_{X_{m}}=\bar{g}\circ g$. From the given
assumption

$k^{\\#}_{m+1}(\bar{g})\in
k_{m+1\\#}\big{(}\pi_{m}\big{(}M(H_{m+1}(X),m);H_{m+1}(X)\big{)}\big{)}.$

So there exists $\bar{h}\colon M\big{(}H_{m+1}(X),m\big{)}\to
M\big{(}H_{m+1}(X),m\big{)}$ such that $k_{m+1\\#}(h)=k^{\\#}_{m+1}(\bar{g})$,
i.e. $k_{m+1}\circ\bar{h}=\bar{g}\circ k_{m+1}$. Observe that

$k_{m+1}\circ\bar{h}\circ h=\bar{g}\circ k_{m+1}\circ h=\bar{g}\circ g\circ
k_{m+1}=k_{m+1}.$

Similarly, $k_{m+1}\circ h\circ\bar{h}=k_{m+1}.$ Therefore, injectivity of
$k_{m+1\\#}$ implies that
$h\in\operatorname{Aut}\big{(}M\big{(}H_{m+1}(X),m\big{)}\big{)}$. Using five
Lemma on the long exact sequence of homology groups we have
$f\in\operatorname{Aut}(X_{m+1})$. Consequently,

$N\mathcal{A}_{\ast}(X_{m+1})\leq N\mathcal{A}_{\ast}(X_{m}).$

  3. (c)

Note that
$k^{\\#}_{m+1}(\operatorname{Aut}(X_{m}))\subseteq\pi_{m}\big{(}X_{m};H_{m+1}(X)\big{)}$.
Surjectivity of $k_{m+1\\#}$ implies that
$k_{m+1\\#}\Big{(}\pi_{m}\big{(}M(H_{m+1}(X),m);H_{m+1}(X)\big{)}\Big{)}=\pi_{m}\big{(}X_{m};H_{m+1}(X)\big{)}$.
Therefore using (a) and (b), we get the desired result.

∎

The following Proposition gives a relation between two consecutive homology
sections which is independent of $k_{\\#}$ and $k^{\\#}$ maps.

######  3.4 Proposition.

Let $X$ be a simply connected CW-complex. If $H_{m}(X)$ is finitely generated
free abelian group for some $m\in\mathbb{N}$ then either
$N\mathcal{A}_{\ast}(X_{m+1})=m+1$ or $N\mathcal{A}_{\ast}(X_{m+1})\leq
N\mathcal{A}_{\ast}(X_{m})$.

###### Proof.

Observe that the map $\iota_{m}\colon X_{m}\to X_{m+1}$ is a homotopical
$m$-equivalence. Then $\iota_{m\\#}\colon[X_{m},X_{m}]\to[X_{m},X_{m+1}]$ is a
surjective map. Assume that $N\mathcal{A}_{\ast}(X_{m+1})<m+1.$ Let
$N\mathcal{A}_{\ast}(X_{m})=l\leq m$ and
$f\in\mathcal{A}_{\ast}^{l}(X_{m+1})$. Then there exits $g\colon X_{m}\to
X_{m}$ which make the following diagram homotopy commutative:

(32)

Thus, $g\in\mathcal{A}_{\ast}^{l}(X_{m})=\operatorname{Aut}(X_{m})$.
Therefore, $f\in\mathcal{A}_{\ast}^{m}(X_{m+1})=\operatorname{Aut}(X_{m+1})$.
Hence,

$N\mathcal{A}_{\ast}(X_{m+1})\leq l=N\mathcal{A}_{\ast}(X_{m}).$

∎

######  3.5 Example.

Consider $X=S^{n}\vee S^{n+1}$ for $n\geq 2$. Observe that $X_{n}=S^{n}$ and
$X_{n+1}=X$. From Proposition 3.4, we have
$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X_{n+1})=n+1$ or
$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X_{n+1})\leq
N\mathcal{A}_{\ast}(X_{n})=n$. Let $f\colon X\to X$ be a map defined as

$f(x)=\begin{cases}x&\text{if}~{}x\in S^{n},\\\ \ast&\text{if}~{}~{}x\in
S^{n+1}.\end{cases}$

Then clearly $f\in\mathcal{A}_{\ast}^{n}(X)$ but
$f\notin\mathcal{A}_{\ast}^{n+1}(X)$. Hence
$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X_{n+1})=n+1$.

Moreover consider $Y=\displaystyle\bigvee^{\infty}_{i=2}S^{k}$. For any $m\geq
2$ we have $Y_{m}=\displaystyle\bigvee^{m}_{i=2}S^{k}$. Now we proceed by
induction. From the above we have $N\mathcal{A}_{\ast}(Y_{3})=3$ and let
$N\mathcal{A}_{\ast}(Y_{l})=l$. Then $Y_{l+1}=Y_{l}\bigvee S^{l+1}$ and
consider a map $g\colon Y_{l+1}\to Y_{l+1}$ defined as

$g(y)=\begin{cases}y&\text{if}~{}y\in Y_{l},\\\ \ast&\text{if}~{}y\in
S^{l+1}.\end{cases}$

Thus, clearly $g\in\mathcal{A}_{\ast}^{l}(Y_{l+1})$ but
$g\notin\mathcal{A}_{\ast}^{l+1}(Y_{l+1})$. Hence
$N\mathcal{A}_{\ast}(Y_{l+1})=l+1$ using Proposition 3.4. Therefore by
induction we have $N\mathcal{A}_{\ast}(Y_{m})=m$ for $m\geq 2$ and
$N\mathcal{A}_{\ast}(Y)=\infty$.

Recall that the homotopical dimension of $X$ is defined by

$\pi_{\ast}\text{-}\dim(X):=\max\big{\\{}k:~{}\pi_{k}(X)\neq 0\big{\\}}.$

Given any $f\colon X\to Y$, the induced maps of homology sections $f_{n}\colon
X_{n}\to Y_{n}$ do not necessarily exist. In [3], the author gives some
sufficient conditions to have induced maps in homology decomposition.

######  3.6 Lemma.

Let $X$ be a simply connected CW-complex such that $H_{k}(X)$ is finitely
generated free abelian group for each $k$. If $\pi_{\ast}\text{-}\dim(X)=m$,
then for each $n\geq m+1$

$\operatorname{Aut}(X)\cong\operatorname{Aut}(X_{n}).$

###### Proof.

Since $H_{k}(X)$ is free abelian group for all $k$, therefore homology
decomposition of $X$ gives the CW-decomposition of $X$. Therefore for any map
$f\colon X\to X$, we have $f^{\prime}_{n}\colon X_{n}\to X_{n}$ such that
$f\circ j_{n}\simeq j_{n}\circ f^{\prime}_{n}$ by cellular approximations,
where $f^{\prime}_{n}\colon X_{n}\to X_{n}$ is the restriction of the cellular
map $f^{\prime}\colon X\to X$ on $X_{n}$ such that $f\simeq f^{\prime}$. Note
that if $f\colon X\to X$ is self-homotopy equivalence then
$f^{\prime}_{n}\colon X_{n}\to X_{n}$ is self-homotopy equivalence. Consider
the map

(33) $\displaystyle\phi\colon[X,X]\to[X_{n},X_{n}]~{}~{}\text{ defined as
}\phi(f)=f^{\prime}_{n}.$

Let $n\geq m+1$. To show the surjectivity of $\phi$, let $r\in[X_{n},X_{n}]$
then $r\colon X_{n}\to X_{n}\subset X.$ Since $\pi_{n}(X)=0$ for all $n\geq
m+1$, so $H^{k+1}\big{(}X,X_{n},\pi_{k}(X)\big{)}=0$ for all $k\geq n$.
Therefore by Obstruction we have a map $\widetilde{r}\colon X\to X$ such that
$\phi(\widetilde{r})=r$. For injectivity, let $f,g\in[X,X]$ such that
$\phi(f)=\phi(g)$. This implies $f^{\prime}_{n}\simeq g^{\prime}_{n}\colon
X_{n}\to X_{n}\subset X$. Let $F\colon X_{n}\times I\to X$ be the homotopy map
between $f^{\prime}_{n}$ and $g^{\prime}_{n}$. Consider a map $G\colon
X\times\partial{I}\cup X_{n}\times I\to X$ defined as

$G(x,t)=\begin{cases}f^{\prime}(x)&\forall(x,t)\in X\times\\{0\\},\\\
g^{\prime}(x)&\forall(x,t)\in X\times\\{1\\},\\\ F(x,t)&\forall(x,t)\in
X_{n}\times I.\end{cases}$

Note that $H^{k+1}\big{(}X\times I,X\times\partial{I}\cup X_{n}\times
I;\pi_{k}(X)\big{)}=0$ for all $k\geq n+1$. Therefore there exists an
extension map $\widetilde{G}\colon X\times I\to X$ of $G$ by obstruction
theory. Hence $\widetilde{G}$ be the homotopy between $f^{\prime}$ and
$g^{\prime}$. Thus $f\simeq g$, so $\phi$ is injective. Consequently for
$n\geq m+1$ we get the bijective map

$\phi\colon[X,X]\to[X_{n},X_{n}].$

Moreover Observe that $\phi(f\circ g)=\phi(f^{\prime}\circ
g^{\prime})=(f^{\prime}\circ g^{\prime})_{n}=f^{\prime}_{n}\circ
g^{\prime}_{n}=\phi(f)\circ\phi(g).$ Hence

$\phi\colon\operatorname{Aut}(X)\xrightarrow{\cong}\operatorname{Aut}(X_{n}).$

∎

######  3.7 Remark.

For a CW-decomposition of X, a self-homotopy equivalence f: X → X does not
necessarily induce a self-homotopy equivalence on the skeleton [20, Remark].
Further, if the CW-decomposition of $X$ comes from the homology decomposition
(when all the homology groups of X are free abelian), then the induced map of
f on each skeleton gives a self-homotopy equivalence.

Recall the theorem of Serre ([21]): For a simply connected CW-complex $X$, the
homology group $H_{k}(X)$ is finitely generated abelian group for all $k$ if
and only if $\pi_{k}(X)$ is finitely generated abelian group for all $k$.

######  3.8 Lemma.

Let $X$ be a simply connected $CW$ complex such that $H_{k}(X)$ is finitely
generated for each $k\in\mathbb{N}$. Then
$N\mathcal{A}_{\\#}(X_{n})=N\mathcal{A}_{\ast}(X_{n})\leq n$.

###### Proof.

Let $H_{k}(X)$ is finitely generated abelian group for each $k$. Therefore
from the definition of homology decomposition we have $H_{k}(X_{n})$ is
finitely generated abelian group for each $k$. Hence we get the desired result
using [16, Corollary 42].

∎

The following Theorem deduce the relation between homology self-closeness
number of $X$ with its homology section.

######  3.9 Theorem.

Let $X$ be a simply connected CW-complex such that
$\pi_{\ast}\text{-}\dim(X)=m$. Then following holds:

  1. (a)

$2\leq N\mathcal{A}_{\ast}(X)\leq m+1$.

  2. (b)

$2\leq N\mathcal{A}_{\ast}(X)\leq m$, if the group $\pi_{m}(X)$ is finitely
generated.

  3. (c)

$N\mathcal{A}_{\ast}(X_{n})=N\mathcal{A}_{\ast}(X)$, if $n\geq m+1$ and
$H_{k}(X)$ is finitely generated free abelian group for each $k$.

  4. (d)

$N\mathcal{A}_{\ast}(X_{n})\leq N\mathcal{A}_{\ast}(X)$, if $n\geq m+1$ and
$H_{k}(X)$ is finite generated for each $k$.

  5. (e)

$N\mathcal{A}_{\ast}(X_{n})\leq m$ for $n\leq m$.

  6. (f)

$N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(X_{n})$, if $\dim(X)\leq
n<\infty$.

###### Proof.

  1. (a)

Since $X$ is a simply connected space, thus $N\mathcal{A}_{\ast}(X)\geq 2$.
Let $f\in\mathcal{A}_{\ast}^{m+1}(X)$ then we have
$f\in\mathcal{A}_{\\#}^{m}(X)=\operatorname{Aut}(X)$. Hence $2\leq
N\mathcal{A}_{\ast}(X)\leq m+1$.

  2. (b)

Let $f\in\mathcal{A}_{\ast}^{m}(X)$. Then $f_{\ast}\colon H_{k}(X)\to
H_{k}(X)$ is isomorphism for all $k\leq m$. Thus,
$f_{\\#}\colon\pi_{k}(X)\to\pi_{k}(X)$ is isomorphism for all $k\leq m-1$ and
$f_{\\#}\colon\pi_{m}(X)\to\pi_{m}(X)$ is surjective. Since $\pi_{m}(X)$ is
finitely generated therefore $f_{\\#}\colon\pi_{m}(X)\to\pi_{m}(X)$ is also an
isomorphism. So, $f\in\mathcal{A}_{\\#}^{m}(X)=\operatorname{Aut}(X)$ and we
get the desired result.

  3. (c)

Note that $\pi_{k}(X)$ is finitely generated abelian group for all $k$ as
$H_{k}(X)$ is finitely generated abelian group for each $k$ by Serre. Assume
that $n\geq m+1$. Let $N\mathcal{A}_{\ast}(X)=l\leq m$ and
$g\in\mathcal{A}_{\ast}^{l}(X_{n})$. Then
$\phi^{-1}(g)\in\mathcal{A}_{\ast}^{l}(X)=\operatorname{Aut}(X)$ from Lemma
3.6. Thus $g\in\operatorname{Aut}(X_{n})$. So $N\mathcal{A}_{\ast}(X_{n})\leq
l=N\mathcal{A}_{\ast}(X)$. If possible, assume that
$N\mathcal{A}_{\ast}(X_{n})<l$. Let $h\in\mathcal{A}_{\ast}^{l-1}(X)$ then
$\phi(h)\in\mathcal{A}_{\ast}^{l-1}(X_{n})=\operatorname{Aut}(X_{n})$.
Therefore, $h\in\operatorname{Aut}(X)$, which contradict the fact that
$N\mathcal{A}_{\ast}(X)=l$. Therefore, for $n\geq m+1$, we have

$N\mathcal{A}_{\ast}(X_{n})=N\mathcal{A}_{\ast}(X).$

  4. (d)

Let $N\mathcal{A}_{\ast}(X)=r$ and $h\in\mathcal{A}_{\\#}^{r}(X_{n})$. Since
$n\geq m+1$, as in the proof of Lemma 3.6, we see that there exists a map
$\bar{h}\colon X\to X$ such that the following diagram is homotopy
commutative.

(40)

Note that the map $j_{n\ast}\colon H_{i}(X_{n})\xrightarrow{\cong}H_{i}(X)$
for all $i\leq n$ and $r\leq m<n$. Thus,
$\bar{h}\in\mathcal{A}_{\\#}^{r}(X)=\operatorname{Aut}(X).$ Therefore using
the commutativity of the diagram we have
$h\in\mathcal{A}_{\\#}^{n}(X_{n})=\operatorname{Aut}(X_{n})$. Hence
$N\mathcal{A}_{\ast}(X_{n})\leq r=N\mathcal{A}_{\ast}(X).$

  5. (e)

Observe that $N\mathcal{A}_{\ast}(X_{n})\leq n$. Therefore we get the desired
result.

  6. (f)

Let $\dim(X)<\infty$ and $n\geq\dim(X)$. Thus
$H^{n+1}\big{(}X_{n},H_{n+1}(X)\big{)}=0$. Therefore, for any map $f\colon
X\to X$ there exists a map $f_{n}\colon X_{n}\to X_{n}$ such that we have a
diagram as of Equation 40, which is homotopy commutative [3, Theorem 3.1]. Let
$N\mathcal{A}_{\ast}(X_{n})=k$ and $h\in\mathcal{A}_{\\#}^{k}(X)$. Using the
commutativity of the diagram, we have
$h_{n}\in\mathcal{A}_{\\#}^{k}(X_{n})=\operatorname{Aut}(X_{n})$. Further,
using the diagram we get $h\in\mathcal{A}_{\\#}^{n}(X)=\operatorname{Aut}(X)$
since $N\mathcal{A}_{\ast}(X)\leq\dim(X).$ Hence, $N\mathcal{A}_{\ast}(X)\leq
k=N\mathcal{A}_{\ast}(X_{n}).$

∎

######  3.10 Example.

Let $X=\mathbb{C}P^{\infty}.$ Then $\pi_{\ast}\text{-}\dim(X)=2$. Note that

$H_{k}(\mathbb{C}P^{\infty})=\begin{cases}\mathbb{Z}&\text{if}~{}k\text{ is
even,}\\\ 0&\text{if}~{}k\text{ is odd}.\end{cases}$

From Theorem 3.9, we have $N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X_{n})$
for all $n\geq 3$, where $X_{n}$ is the $n$-th homology decomposition of $X$.
Observe that $X_{2n+1}=X_{2n}=\mathbb{C}P^{n}$ for all $n\geq 1$. Therefore,

$N\mathcal{A}_{\ast}(\mathbb{C}P^{\infty})=N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X_{3})=N\mathcal{A}_{\ast}(X_{2})=N\mathcal{A}_{\ast}(\mathbb{C}P^{1})=N\mathcal{A}_{\ast}(S^{2})=2.$

Moreover, $N\mathcal{A}_{\ast}(\mathbb{C}P^{n})=N\mathcal{A}_{\ast}(X)=2$ for
all $n\geq 2$ by Theorem 3.9. Consequently,

$N\mathcal{A}_{\ast}(\mathbb{C}P^{n})=N\mathcal{A}_{\ast}(X_{2n})=2,~{}~{}\text{for}~{}n\in\mathbb{N}\cup\\{\infty\\}.$

######  3.11 Remark.

Consider $X=\mathbb{C}P^{\infty}$. Then $X_{2n+1}=X_{2n}=\mathbb{C}P^{n}$ for
$n\geq 1$. Therefore,

$N\mathcal{A}_{\ast}(X_{n})=\begin{cases}0&\text{if}~{}~{}n<2,\\\
2&\text{if}~{}~{}n\geq 2.\end{cases}$

######  3.12 Example.

Consider $X=K(G,n)$ for $n\geq 2$. Note that the Hurewicz map
$h_{\ast}\colon\pi_{n+1}(X)\to H_{n+1}(X)$ is epimorphism, thus
$H_{n+1}(X)=0$. For homology sections $\\{X_{m}\\}$, we have

$X_{m}=\begin{cases}\ast&\text{if}~{}~{}m<n,\\\
M(G,n)&\text{if}~{}~{}m=n,n+1.\end{cases}$

By definition, the map $j_{n+2}\colon X_{n+2}\to X$ is a homologial
$(n+2)$-equivalence, and hence a homotopical $(n+2)$-equivalence (cf. Lemma
2.1). So $j_{n+2\\#}\colon\pi_{k}(X_{n+2})\xrightarrow{\cong}\pi_{k}(X)$ is an
isomorphism for all $k\leq n+1$. Hence, $X_{n+2}$ is $(n-1)$-connected CW-
complex. Therefore,

$\mathcal{A}_{\\#}^{n-1}(X_{n+2})=\cdots=\mathcal{A}_{\\#}^{1}(X_{n+2})=[X_{n+2},X_{n+2}].$

Thus, $N\mathcal{A}_{\\#}(X_{n+2})\geq n$. Using Lemma 3.8, we get
$N\mathcal{A}_{\ast}(X_{n+2})=N\mathcal{A}_{\\#}(X_{n+2})\geq n$.

Moreover, from Theorem 3.9(d), we have $N\mathcal{A}_{\ast}(X_{n+2})\leq
N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\\#}(X)=n$. Hence,
$N\mathcal{A}_{\ast}(X_{n+2})=n$. Similarly, we can prove that
$N\mathcal{A}_{\ast}(X_{m})=n$ for $m>n+2$. Consequently,

$N\mathcal{A}_{\ast}(X_{m})=\begin{cases}0&\text{if}~{}~{}m<n,\\\
n&\text{if}~{}~{}m\geq n.\end{cases}$

##  4\. Homotopy decomposition

In this section, we consider homotopy decomposition (Postnikov tower) of a
space $X$ which was studied in previous (cf. [6]). This can be thought as a
dual construction to homology decomposition. Here, we prove some results
related to homotopy and homology decomposition and their intermediate
relations. These results help us to compute homology self-closeness numbers of
Postnikov tower, some of them can be thought as generalization of those
results in [8]. First, recall the definition

######  4.1 Definition.

Let $X$ be a simply connected CW-complex. A homotopy decomposition (Postnikov
decomposition) of $X$ consists of a sequence of simply connected CW-complexes
$\\{X^{(n)}\\}$ and maps $g_{n}\colon X\to X^{(n)}$ such that

  1. (i)

Each inclusion map $g_{n}\colon X\to X^{(n)}$ induces isomorphisms
$g_{n\\#}\colon\pi_{k}(X)\to\pi_{k}(X^{(n)})$ for all $k\leq n$ and
$\pi_{k}(X^{(n)})=0$ for $k>n$.

  2. (ii)

There exist maps $p_{n+1}\colon X^{(n+1)}\to X^{(n)}$ such that

$K\big{(}\pi_{n+1}(X),n+1\big{)}\to X^{(n+1)}\xrightarrow{p_{n+1}}X^{(n)}$

is a fiber sequence.

  3. (iii)

The following diagram is commutative

|
---|---
$\textstyle{X^{(n+1)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{n+1}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{n+1}}$$\scriptstyle{g_{n}}$$\textstyle{X^{(n)}}$

  4. (iv)

The fibration sequence in (ii) is equivalent to the principal fibration

$K\big{(}\pi_{n+1}(X),n+1\big{)}\to X^{(n+1)}\to
X^{(n)}\xrightarrow{k^{n+1}}K\big{(}\pi_{n+1}(X),n+2\big{)},$

determined by a map $k^{n+1}$, where $[k^{n+1}]\in
H^{n+2}\big{(}X^{(n)};\pi_{n+1}(X)\big{)}$.

We write this as $\\{X^{(n)},g_{n},p_{n},k^{n}\\}$ and call it a homotopy
decomposition (Postnikov decomposition) of $X$. The maps $k^{n+1}\colon
X^{(n)}\to K\big{(}\pi_{n+1}(X),n+2\big{)}$ are called $k$-invariants. The
space $X^{(n)}$ is called the $n$-th homotopy section and is obtained as the
homotopy fiber of $k^{n}.$

######  4.2 Remark.

From the definition, we have $N\mathcal{A}_{\\#}(X^{(n)})\leq n.$

The following Lemma establish a relation between the homology self-closeness
numbers of homotopy and homology sections.

######  4.3 Lemma.

Let $X$ be a simply connected CW-complex such that $H_{k}(X)$ is finitely
generated abelian group for each $k$. Then the following results hold.

  1. (i)

$N\mathcal{A}_{\ast}(X^{(n)})=N\mathcal{A}_{\\#}(X^{(n)})\leq n.$

  2. (ii)

$N\mathcal{A}_{\ast}(X^{(n)})\leq N\mathcal{A}_{\ast}(X_{n}),$ if $H_{n}(X)$
is free-abelian group.

###### Proof.

  1. (i)

From the definition of homotopy decomposition, we have $\pi_{k}(X^{(n)})$ is
finitely generated for all $k$. Therefore using [16, Corollary 42], we get the
desired result.

  2. (ii)

Let $j_{n}\colon X_{n}\to X$ and $g_{n}\colon X\to X^{(n)}$ be two maps over
the homology decomposition and homotopy decomposition of $X$ respectively.
Observe that $g_{n}\colon X\to X^{(n)}$ is homotopical $(n+1)$-equivalence.
From Lemma 2.1, we say that $g_{n}$ is homological $(n+1)$-equivalence.
Therefore, $(g_{n}\circ j_{n})_{\ast}\colon
H_{i}(X_{n})\xrightarrow{\cong}H_{i}(X^{(n)})$ for all $i\leq n$. Moreover,
for any $h\colon X^{(n)}\to X^{(n)}$ there exists a $\bar{h}\colon X_{n}\to
X_{n}$ such that the following diagram is homotopy commutative by Compression
Lemma.

(47)

Let $N\mathcal{A}_{\ast}(X_{n})=k$ and $f\in\mathcal{A}_{\ast}^{k}(X^{(n)})$.
Thus, $\bar{f}\in\mathcal{A}_{\ast}^{k}(X_{n})=\operatorname{Aut}(X_{n})$.
Therefore, $f\in\mathcal{A}_{\ast}^{n}(X^{(n)})=\operatorname{Aut}(X^{(n)}).$
Hence,

$N\mathcal{A}_{\ast}(X^{(n)})\leq N\mathcal{A}_{\ast}(X_{n}).$

∎

Next we deduce relations between the homology self-closeness numbers of two
consecutive homotopy sections in Postnikov tower.

######  4.4 Theorem.

Let $X$ be a simply connected CW-complex such that $H_{k}(X)$ is finitely
generated for all $k$. For homotopy decomposition
$\\{X^{(n)},g_{n},p_{n},k^{n}\\}$ of $X$, we have the following:

  1. (a)

If the induced map

$k^{m+1\\#}\colon\big{[}K\big{(}\pi_{m+1}(X),m+2\big{)},K\big{(}\pi_{m+1}(X),m+2\big{)}\big{]}\to\big{[}X^{(m)},K\big{(}\pi_{m+1}(X),m+2\big{)}\big{]}$

is surjective for some $m\in\mathbb{N}$, then
$N\mathcal{A}_{\ast}(X^{(m)})\leq N\mathcal{A}_{\ast}(X^{(m+1)})$.

  2. (b)

If the induced map

$k^{m+1\\#}\colon\big{[}K\big{(}\pi_{m+1}(X),m+2\big{)},K\big{(}\pi_{m+1}(X),m+2\big{)}\big{]}\to\big{[}X^{(m)},K\big{(}\pi_{m+1}(X),m+2\big{)}\big{]}$

is injective such that

$k^{m+1}_{\\#}(\operatorname{Aut}(X^{(m)}))\subseteq
k^{m+1\\#}\big{(}\big{[}K\big{(}\pi_{m+1}(X),m+2\big{)},K\big{(}\pi_{m+1}(X),m+2\big{)}\big{]}\big{)},$

then $N\mathcal{A}_{\ast}(X^{(m+1)})\leq N\mathcal{A}_{\ast}(X^{(m)})$.

  3. (c)

If the induced map $k^{m+1\\#}$ is bijective, then
$N\mathcal{A}_{\ast}(X^{(m)})=N\mathcal{A}_{\ast}(X^{(m+1)})$.

###### Proof.

Note that $N\mathcal{A}_{\ast}(X^{n})=N\mathcal{A}_{\\#}(X^{n})$ for homotopy
sections $\\{X^{n}\\}$ by Lemma 4.3. Rest of the proof directly follows from
[15, Theorem 3.3 & Theorem 3.9].

∎

######  4.5 Proposition.

Let $X$ be a simply connected CW-complex such that $H_{k}(X)$ is finitely
generated for all $k$. Then either $N\mathcal{A}_{\ast}(X^{(n+1)})=n+1$ or
$N\mathcal{A}_{\ast}(X^{(n+1)})\leq N\mathcal{A}_{\ast}(X^{(n)})$.

###### Proof.

Observe that the map
$p^{\\#}_{n+1}\colon[X^{(n)},X^{(n)}]\to[X^{(n+1)},X^{(n)}]$ is surjective,
see [5, Proposition 8.2.2]. As in the proof of Proposition 3.4, we have either
$N\mathcal{A}_{\\#}(X^{(n+1)})=n+1$ or $N\mathcal{A}_{\\#}(X^{(n+1)})\leq
N\mathcal{A}_{\\#}(X^{(n)})$. Moreover, using Lemma 4.3, we get the desired
result.

∎

The following Theorem is a homological version of [8, Theorem 3.5].

######  4.6 Theorem.

Let $X$ be a simply connected CW-complex of dimension $m$ such that $H_{k}(X)$
is finitely generated for each $k$. Then

  1. (a)

$N\mathcal{A}_{\ast}(X^{(n)})=N\mathcal{A}_{\ast}(X)$, if the conditions hold
$N\mathcal{A}_{\ast}(X)\leq n$ and $m-1\leq n$.

  2. (b)

$N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(X^{(n)})$, if the condition
holds $N\mathcal{A}_{\ast}(X)\leq n<m-1$.

  3. (c)

$N\mathcal{A}_{\ast}(X^{(n)})<N\mathcal{A}_{\ast}(X)$, if the condition holds
$n<N\mathcal{A}_{\ast}(X)$.

###### Proof.

  1. (a)

We want to prove the inequality first $N\mathcal{A}_{\ast}(X^{(n)})\leq
N\mathcal{A}_{\ast}(X)$. Assume that $N\mathcal{A}_{\ast}(X)=k$. Consider the
case $k<n$, otherwise we get the inequality trivially. Given any map
$\phi\colon X^{(n)}\to X^{(n)}$ there exists a map $\bar{\phi}\colon X\to X$
such that the following diagram is homotopy commutative by Compression Lemma
[10, Lemma 4.6].

(54)

Let $f\in\mathcal{A}_{\ast}^{k}(X^{(n)})$. Then there exists a
$\bar{f}\in\mathcal{A}_{\ast}^{k}(X)$ by Compression Lemma and the fact that
$g_{n\ast}\colon H_{i}(X)\xrightarrow{\cong}H_{i}(X^{(n)})$ for all $i\leq n$.
Since $\bar{f}\in\mathcal{A}_{\ast}^{k}(X)=\operatorname{Aut}(X)$, therefore
using the commutativity of the diagram, we have
$f\in\mathcal{A}_{\ast}^{n}(X^{(n)})=\operatorname{Aut}(X^{(n)}).$ Hence
$N\mathcal{A}_{\ast}(X^{(n)})\leq N\mathcal{A}_{\ast}(X).$

Conversely, let $N\mathcal{A}_{\ast}(X^{(n)})=l$ and
$h\in\mathcal{A}_{\ast}^{l}(X)$. Then, there exists $h^{n}\colon X^{(n)}\to
X^{(n)}$ which makes the above Diagram 54 homotopy commutative (cf. [11]).
Thus, $h^{n}\in\mathcal{A}_{\ast}^{l}(X^{(n)})=\operatorname{Aut}(X^{(n)}).$
Therefore, $h\in\mathcal{A}_{\ast}^{n}(X)=\operatorname{Aut}(X)$. Hence
$N\mathcal{A}_{\ast}(X)\leq N\mathcal{A}_{\ast}(X^{(n)})$ and we get the
desired result.

  2. (b)

Follows from the converse part of (a).

  3. (c)

Observe that $N\mathcal{A}_{\ast}(X^{(n)})\leq n$. Thus, we get the desired
result.

∎

######  4.7 Corollary.

Let $X$ be a simply connected CW-complex such that $H_{k}(X)$ is finitely
generated for all $k$. Assume that any one of the following conditions hold:

  1. (a)

$\pi_{\ast}\text{-}\dim(X)<n<\infty$ and $H_{k}(X)$ is free for each $k$.

  2. (b)

$\dim(X)\leq n<\infty$.

Then, $N\mathcal{A}_{\ast}(X^{(n)})=N\mathcal{A}_{\ast}(X_{n})$.

###### Proof.

  1. (a)

Observe that the structure map of the Postnikov tower $g_{n}\colon X\to
X^{(n)}$ is a homotopy equivalence if $n\geq\pi_{\ast}\text{-}\dim(X)$. Thus
for $n\geq\pi_{\ast}\text{-}\dim(X)$, we have
$N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(X^{(n)})$. Therefore from Theorem
3.9 we have $N\mathcal{A}_{\ast}(X^{(n)})=N\mathcal{A}_{\ast}(X_{n})$ for
$n\geq\pi_{\ast}\text{-}\dim(X)+1$.

  2. (b)

Note that for $n\geq\dim(X)$, we have $X_{n}=X$. Hence using Theorem 4.6, we
get the desired result.

∎

######  4.8 Example.

For $n\in\mathbb{N}\cup\\{\infty\\}$, consider $X=\mathbb{C}P^{n}$. From Lemma
4.3(ii), we have

$N\mathcal{A}_{\ast}(X^{(k)})\leq N\mathcal{A}_{\ast}(X_{k})~{}\text{ for all
}k.$

For $k\geq 2$, we know that $X^{(k)}$ are simply connected, thus
$N\mathcal{A}_{\ast}(X^{(k)})\geq 2$. Therefore,

$2\leq N\mathcal{A}_{\ast}(X^{(k)})\leq N\mathcal{A}_{\ast}(X_{k})=2~{}\text{
for }k\geq 2.$

Hence,

$N\mathcal{A}_{\ast}(X^{(k)})=\begin{cases}0&\text{if}~{}k<2,\\\
2&\text{if}~{}k\geq 2.\end{cases}$

######  4.9 Example.

Let $G$ and $H$ be finitely generated abelian groups. Consider
$X=M(G,n)\bigvee M(H,m)$, where $m>n>1$. Observe that

$X^{(k)}=\begin{cases}\ast&\text{if}~{}k<n,\\\
K(G,n)&\text{if}~{}k=n.\end{cases}$

Moreover, $H_{k}(X)=0$ for all $n<k<m$. Therefore,

$X_{k}=\begin{cases}\ast&\text{if}~{}k<n,\\\ M(G,n)&\text{if}~{}n\leq k<m,\\\
X&\text{if}~{}k=m.\end{cases}$

Note that, for $k\geq n$, we have $n\leq N\mathcal{A}_{\ast}(X^{(k)})$ since
$X^{(k)}$ is $n-1$ connected. From Lemma 4.3(ii), we have
$N\mathcal{A}_{\ast}(X^{(k)})\leq N\mathcal{A}_{\ast}(X_{k})$ for $n<k<m$.
Thus,

$n\leq N\mathcal{A}_{\ast}(X^{(k)})\leq
N\mathcal{A}_{\ast}(X_{k})=N\mathcal{A}_{\ast}(M(G,n))=n,~{}~{}\text{for}~{}n<k<m.$

Note that $\mathcal{A}_{\ast}^{n}(X)=\cdots=\mathcal{A}_{\ast}^{m-1}(X)$ and
$\operatorname{Aut}(X)=\mathcal{A}_{\ast}^{m}(X)\subsetneq\mathcal{A}_{\ast}^{m-1}(X)$,
see Example 3.5. Therefore, $N\mathcal{A}_{\ast}(X)=m$

From Theorem 4.6(a), we have
$N\mathcal{A}_{\ast}(X^{(k)})=N\mathcal{A}_{\ast}(X)$ for all $k\geq m$.
Hence, $N\mathcal{A}_{\ast}(X^{(k)})=N\mathcal{A}_{\ast}(X)=m$ for $k\geq m$.
Consequently,

$N\mathcal{A}_{\ast}(X^{(k)})=\begin{cases}0&\text{if}~{}k<n,\\\
n&\text{if}~{}n\leq k<m,\\\ m&\text{if}~{}k\geq m.\par\end{cases}$

In general we have the following result.

######  4.10 Proposition.

Let $G_{1},\cdots,G_{m}$ be finitely generated abelian groups. Consider
$X=\displaystyle\bigvee^{m}_{i=1}M(G_{i},n_{i})$, where $n_{i+1}>n_{i}>1$.
Then

$N\mathcal{A}_{\ast}(X^{(k)})=\begin{cases}0&\text{if}~{}k<n_{1},\\\
n_{i}&\text{if}~{}n_{i}\leq k<n_{i+1},\\\ n_{m}&\text{if}~{}k\geq
n_{m}.\end{cases}$

###### Proof.

Note that
$\mathcal{A}_{\ast}^{n_{i+1}}(X)\subsetneq\mathcal{A}_{\ast}^{n_{i}}(X)$ for
$i=1,\ldots,m-1$, see Example 3.5. We have already proved for the case $m=2$.
It is sufficient to prove the result for the case $m=3$ and then, we can say
that the result holds inductively. Let
$X=\bigvee\limits_{i=1}^{3}M(G_{i},n_{i}).$ Observe that

$X^{(k)}=\begin{cases}\ast&\text{ if }k<n_{1},\\\ K(G_{1},n_{1})&\text{ if
}k=n_{1}.\end{cases}$

Further,

$X_{k}=\begin{cases}\ast&\text{ if }k<n_{1},\\\
\bigvee\limits_{j=1}^{i}M(G_{j},n_{j})&\text{ if }n_{i}\leq k<n_{i+1},\\\
X&\text{ if }k=n_{m}.\end{cases}$

Moreover, $X^{(k)}$ is $(n_{1}-1)$ connected for $k\geq n_{1}$. Thus,
$n_{1}\leq N\mathcal{A}_{\ast}(X^{(k)})$ for $k\geq n_{1}$. From Lemma
4.3(ii), we have

$n_{1}\leq N\mathcal{A}_{\ast}(X^{(k)})\leq
N\mathcal{A}_{\ast}(X_{k})=N\mathcal{A}_{\ast}(X_{n_{1}})=n_{1},~{}\text{for}~{}n_{1}<k<n_{2}.$

Using Proposition 4.5, we have

$N\mathcal{A}_{\ast}(X^{(n_{2})})=n_{2}~{}\text{or}~{}N\mathcal{A}_{\ast}(X^{(n_{2})})\leq
N\mathcal{A}_{\ast}(X^{(n_{2}-1)})=n_{1}.$

If possible, assume that
$\mathcal{A}_{\ast}^{n_{1}}(X^{(n_{2})})=\operatorname{Aut}(X^{(n_{2})})$. Let
$f\in\mathcal{A}_{\ast}^{n_{1}}(X)$ but
$f\notin\mathcal{A}_{\ast}^{n_{2}}(X)$. By a functorial construction of
Postnikov tower, there exist $f^{(n_{2})}$ which makes the following diagram
homotopy commutative ( cf. [5, Proposition 7.2.11] ).

(61)

Thus,
$f^{(n_{2})}\in\mathcal{A}_{\ast}^{n_{1}}(X^{(n_{2})})=\operatorname{Aut}(X^{(n_{2})})$.
This contradicts the fact that $f\in\mathcal{A}_{\ast}^{n_{2}}(X)$. Hence
$N\mathcal{A}_{\ast}(X^{(n_{2})})=n_{2}$. Further, using Lemma 4.3(ii), we
have

$N\mathcal{A}_{\ast}(X^{(k)})\leq
N\mathcal{A}_{\ast}(X_{k})=N\mathcal{A}_{\ast}(X_{n_{2}})=n_{2},~{}\text{for}~{}n_{2}<k<n_{3}.$

If possible, assume that $N\mathcal{A}_{\ast}(X^{(k)})<n_{2}$ for some
$k\in(n_{2},n_{3})$. Then

$\mathcal{A}_{\ast}^{n_{1}}(X^{(k)})=\cdots=\mathcal{A}_{\ast}^{n_{2}-1}(X^{(k)})=\operatorname{Aut}(X^{(k)})~{}(\text{since}~{}H_{j}(X^{(k)})\cong
H_{j}(X)~{}\forall j\leq k).$

Let $h\in\mathcal{A}_{\ast}^{n_{1}}(X)$ but
$h\notin\mathcal{A}_{\ast}^{n_{2}}(X)$. Then by the similar argument as above,
we arrive at contradiction. Hence $N\mathcal{A}_{\ast}(X^{(k)})=n_{2}$ for
$n_{2}<k<n_{3}$.

From Theorem 4.6(a), we have
$N\mathcal{A}_{\ast}(X^{(k)})=N\mathcal{A}_{\ast}(X)=n_{3}$ for all $k\geq
n_{3}$. Consequently

$N\mathcal{A}_{\ast}(X^{(k)})=\begin{cases}0&\text{if}~{}k<n_{1},\\\
n_{1}&\text{if}~{}n_{1}\leq k<n_{2},\\\ n_{2}&\text{if}~{}n_{2}\leq
k<n_{3},\\\ n_{3}&\text{if}~{}k\geq n_{3}.\end{cases}$

∎

######  4.11 Example.

Consider $X=S^{m}\times S^{n}$, where $m>n>1$. Observe that

$X^{(k)}=\begin{cases}\ast&\text{if}~{}k<n,\\\
K(\mathbb{Z},n)&\text{if}~{}k=n.\end{cases}$

Therefore $N\mathcal{A}_{\ast}(X^{(k)})=0$ for all $k<n$. Note that $X^{(k)}$
is $n-1$ connected for $k\geq n$. So $n\leq N\mathcal{A}_{\ast}(X^{(k)})$ for
all $k\geq n$. From Lemma 4.3(ii), we have

$n\leq N\mathcal{A}_{\ast}(X^{(k)})\leq
N\mathcal{A}_{\ast}(X_{k})=N\mathcal{A}_{\ast}(X_{n})=N\mathcal{A}_{\ast}(S^{n})=n~{}\text{for
all}~{}n\leq k<m.$

Observe that $N\mathcal{A}_{\ast}(X)=N\mathcal{A}_{\ast}(S^{m}\times
S^{n})=N\mathcal{A}_{\\#}(S^{m}\times S^{n})=m$ (cf. [14, Proposition 5]).

Moreover, using Theorem 4.6(a), we have
$N\mathcal{A}_{\ast}(X^{(k)})=N\mathcal{A}_{\ast}(X)$ for all $k\geq m+n-1$.
Therefore,

$N\mathcal{A}_{\ast}(X^{(k)})=m~{}\text{for all}~{}k\geq m+n-1.$

Further, for $m\leq k<m+n-1$, we have $m\leq N\mathcal{A}_{\ast}(X^{(k)})$
from Theorem 4.6(b) and using Lemma 4.3(ii), we have

$N\mathcal{A}_{\ast}(X^{(k)})\leq
N\mathcal{A}_{\ast}(X_{k})=N\mathcal{A}_{\ast}(X_{m})=N\mathcal{A}_{\ast}(S^{m}\vee
S^{n})=m.$

Hence, $N\mathcal{A}_{\ast}(X^{(k)})=m$ for all $m\leq k<m+n-1$. Consequently,

$N\mathcal{A}_{\ast}(X^{(k)})=\begin{cases}0&\text{if}~{}k<n,\\\
n&\text{if}~{}n\leq k<m,\\\ m&\text{if}~{}k\geq m.\end{cases}$

######  4.12 Example.

For $l>0$, consider $X=\Sigma^{l}(S^{m}\times S^{n})$, where $m>n>1$. Note
that

$\Sigma(S^{m}\times S^{n})\simeq S^{n+1}\bigvee S^{m+1}\bigvee
S^{m+n+1}~{}~{}\text{(cf. \cite[cite]{[\@@bibref{}{AHAT}{}{}, Proposition
4I.1]})}.$

Therefore, $X=S^{n+l}\bigvee S^{m+l}\bigvee S^{m+n+l}$. From Proposition 4.10,
we have

$N\mathcal{A}_{\ast}(X^{(k)})=\begin{cases}0&\text{if}~{}k<n+l,\\\
n+l&\text{if}~{}n+l\leq k<m+l,\\\ m+l&\text{if}~{}m+l\leq k<m+n+l,\\\
m+n+l&\text{if}~{}k\geq m+n+l.\end{cases}$

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