# The homotopy classification of four-dimensional toric orbifolds

Xin Fu  Department of Mathematics, Ajou University, Suwon 16499, Republic of
Korea<EMAIL_ADDRESS>,  Tseleung So  Department of Mathematics and
Statistics, Univeristy of Regina, Regina, SK S4S 0A2, Canada
<EMAIL_ADDRESS>and  Jongbaek Song  School of Mathematics, KIAS, Seoul
02455, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

Let $X$ be a $4$-dimensional toric orbifold. If $H^{3}(X)$ has a non-trivial
odd primary torsion, then we show that $X$ is homotopy equivalent to the wedge
of a Moore space and a CW-complex. As a corollary, given two 4-dimensional
toric orbifolds having no 2-torsion in the cohomology, we prove that they have
the same homotopy type if and only their integral cohomology rings are
isomorphic.

###### Key words and phrases:

cohomological rigidity, toric orbifold

###### 2010 Mathematics Subject Classification:

Primary 57R18, 55P15; Secondary 55P60

##  1\. Introduction

One of the central problems in topology is the rigidity question, namely when
a weaker equivalence between two spaces implies a stronger equivalence between
them. The Freedman’s work on the classification of closed oriented simply
connected topological $4$-manifolds via the intersection form is a nice
example of this type of question. In toric topology, a similar type of
question was posed in [MS08], which is now called the _cohomologial rigidity
problem_ , which asks if homeomorphism/diffeomorphism classes of quasitoric
manifolds can be classified by their integral cohomology rings.

Although the problem looks overambitious, it is a sensible question to ask on
the following basis. No counter-example has been found since it was
formulated. On the contrary, there is a piece of evidence supporting the
cohomological rigidity of quasitoric manifolds. Indeed, the classification
result in [OR70] together with the description of the cohomology ring of a
quasitoric manifold [DJ91, Theorem 4.14] implies the cohomological rigidity of
$4$-dimensional quasitoric manifolds. Besides, many affirmative answers have
been proved, for instance certain Bott manifolds [Cho15], generalized Bott
manifolds [CMS10], 6-dimensional quasitoric manifolds associated to
3-dimensional Pogorelov polytopes [BEM+17].

Being a generalized notion of quasitoric manifold, a toric orbifold [DJ91] is
a $2n$-dimensional compact orbifold equipped with a locally standard
$T^{n}$-action whose orbit space is a simple polytope. It is known that the
cohomology rings fail to classify toric orbifolds up to homeomorphism. For
instance, there are weighted projective spaces with isomorphic cohomology
rings that are not homeomorphic. Therefore, toric orbifolds do not satisfy
cohomological rigidity. However, in the above counter-examples, two weighted
projective spaces with isomorphic cohomology rings are homotopy equivalent
[BFNR13]. Hence we take a step back and ask a homotopical version of the
cohomogical rigidity:

######  Question 1.1.

Are two toric orbifolds homotopy equivalent if their integral cohomology rings
are isomorphic as graded rings?

This paper aims to answer this question for certain $4$-dimensional toric
orbifolds. We first study certain CW-complexes which model 4-dimensional toric
orbifolds and investigate their homotopy theory. In what follows,
$H^{\ast}(X)$ denotes the cohomology ring with integral coefficients unless
otherwise stated, and $P^{3}(k)$ denotes the $3$-dimensional mod-$k$ Moore
space for $k>1$. It is known that $H^{3}(X)$ is a finite cyclic group. We
refer to [Fis92, Jor98]. Let $H^{3}(X)\cong\mathbb{Z}_{m}$ with $m=2^{s}q$ for
$q$ odd and $s\geq 0$. When $q>1$, we show that $X$ decomposes into a wedge of
$P^{3}(q)$ and a recognizable space.

######  Theorem 1.2.

Let $X$ be a 4-dimensional toric orbifold such that
$H^{3}(X)\cong\mathbb{Z}_{m}$. If $m=2^{s}q$ for an odd integer $q>1$ and
$s\geq 0$, then $X$ is homotopy equivalent to $\hat{X}\vee P^{3}(q)$, where
$\hat{X}$ is a simply connected 4-dimensional CW-complex with
$H^{3}(\hat{X})=\mathbb{Z}_{2^{s}}$ and $H^{i}(\hat{X})\cong H^{i}(X)$ for
$i\neq 3$.

If $m$ is odd or equivalently $s=0$, then Theorem 1.2 implies
$X\simeq\hat{X}\vee P^{3}(m)$ where $H^{3}(\hat{X})=0$. As an application, we
can answer Question 1.1 for certain $4$-dimensional toric orbifolds in the
following theorem.

######  Theorem 1.3.

Let $X$ and $X^{\prime}$ be 4-dimensional toric orbifolds such that $H^{3}(X)$
and $H^{3}(X^{\prime})$ have no 2-torsion. Then $X$ is homotopy equivalent to
$X^{\prime}$ if and only if there is a ring isomorphism $H^{*}(X)\cong
H^{*}(X^{\prime})$.

This paper is organized as follows. In Section 2, we review the constructive
definition of a $4$-dimensional toric orbifold $X$. In particular, it is
important to see that $X$ is the mapping cone of a map from a lens space to a
wedge of $2$-spheres. This phenomenon is motivated by the work of [BSS17] and
can also be understood in terms of a $\mathbf{q}$-CW complex studied in
[BNSS19]. In Section 3, we define a category $\mathscr{C}_{n,m}$ of certain
CW-complexes which model 4-dimensional toric orbifolds and study the homotopy
theory of $\mathscr{C}_{n,m}$. Sections 4 aims to give a necessary and
sufficient condition for $X\in\mathscr{C}_{n,m}$ to decompose into a wedge of
$P^{3}(q)$ and a space in $\mathscr{C}_{n,2^{s}}$. In Section 5, we study the
$p$-local version of the discussion of Section 4 for some odd prime $p$ and
apply this to $4$-dimensional toric orbifolds. Combining the equivalent
condition (Proposition 4.8) and the $p$-local decomposition (Proposition 5.3),
we finally complete the proofs of Theorem 1.2 and Theorem 1.3 in Section 6.

## acknowledgment

We started this project during our participation to the Thematic Program on
Toric Topology and Polyhedral Products at Fields Institute. We gratefully
acknowledge the support of Fields Institute and the organizers of the program.
Furthermore, we thank Mikiya Masuda, Taras Panov, Dong Youp Suh and Donald
Stanley for discussing the topics, and thank Stephen Theriault for
proofreading our draft and giving helpful comments.

The first author was supported by Fields Institute and is supported by the
National Research Foundation of Korea funded by the Korean Government
(NRF-2019R1A2C2010989), the second author is supported by Pacific Institute
for the Mathematical Sciences (PIMS) Postdoctoral Fellowship and the third
author is supported by Basic Science Research Program through the National
Research Foundation of Korea (NRF) funded by the Ministry of Education
(NRF-2018R1D1A1B07048480) and a KIAS Individual Grant (MG076101) at Korea
Institute for Advanced Study.

##  2\. Toric Orbifolds of Dimension $4$

We begin with a summary of the constructive definition of a toric orbifold.
For our purpose, we focus on the $4$-dimensional case. For more details of
toric orbifolds see [DJ91, Section 7], [PS10, Section 2] and [CLS11, Chapter
3, Chapter 10].

Let $P$ be an $(n+2)$-gon on vertices $v_{1}\dots,v_{n+2}$ for some $n\geq 0$.
We denote by $E_{i}$ the edge connecting $v_{i}$ and $v_{i+1}$ for
$i=1,\dots,n+2$, where we take indices modulo $n+2$. To each edge $E_{i}$,
assign a primitive vector $\xi_{i}=(a_{i},b_{i})\in\mathbb{Z}^{2}$ such that
two adjacent vectors $\xi_{i}$ and $\xi_{i+1}$ are linearly independent. We
often describe this combinatorial data as in Figure 1.

$\cdots$$\ddots$$v_{i+2}$$v_{n+2}$$v_{1}$$v_{2}$$v_{i+1}$$v_{i}$$\xi_{n+2}$$\xi_{1}$$\xi_{i+1}$$\xi_{i}$
Figure 1. $(n+2)$-gon with primitive vectors on facets.

Identify $\mathbb{Z}^{2}$ with $\mathrm{Hom}(S^{1},T^{2})$. Each $\xi_{i}$
defines a one-parameter subgroup of $T^{2}$

$S^{1}_{\xi_{i}}=\\{(t^{a_{i}},t^{b_{i}})\in T^{2}\mid t\in S^{1}\\}.$

Now, define the following identification space

(2.1) $X=P\times T^{2}/_{\sim}$

where $(p,g)\sim(q,h)$ if and only if $p=q$ and

  1. (1)

$gh^{-1}\in S^{1}_{\xi_{i}}$ if $p$ is in the relative interior of $E_{i}$;

  2. (2)

$gh^{-1}\in T^{2}$ if $p=q$ is a vertex of $P$.

Here the torus $T^{2}$ acts on $X$ by the multiplication on the second factor,
which yields the orbit map $\pi\colon X\to P$ by the projection onto the first
factor.

We roughly describe the orbifold structure on $X$ following the identification
(2.1). First, there is a standard presentation of $\mathbb{C}^{2}$ given by a
homeomorphism $\mathbb{R}^{2}_{\geq}\times
T^{2}/_{\sim_{std}}\cong\mathbb{C}^{2}$ that maps $[(x,y),(t,s)]$ in
$\mathbb{R}^{2}_{\geq}\times T^{2}/_{\sim_{std}}$ to $(xt,ys)$ in
$\mathbb{C}^{2}$. Here the standard identification $\sim_{std}$ is defined by
$\big{(}(x_{1},y_{1}),g\big{)}\sim_{std}\big{(}(x_{2},y_{2}),h\big{)}$
whenever

$(x_{1},y_{1})=(x_{2},y_{2})~{}\text{in}~{}\mathbb{R}^{2}_{\geq}~{}\text{and}~{}\begin{cases}gh^{-1}\in
1\times S^{1}&~{}\text{if}~{}x_{1}=0,~{}y_{1}\neq 0;\\\ gh^{-1}\in S^{1}\times
1&~{}\text{if}~{}x_{1}\neq 0,~{}y_{1}=0;\\\ gh^{-1}\in
T^{2}&~{}\text{if}~{}x_{1}=y_{1}=0.\end{cases}$

Let $U_{i}$ be a neighborhood of $v_{i}$ in $P$, which is homeomorphic to
$\mathbb{R}^{2}_{\geq}$ as a manifold with corners. Let $\psi_{i}$ be a
homeomorphism $\mathbb{R}^{2}_{\geq}\cong U_{i}$ and let $\rho_{i}\colon
T^{2}\twoheadrightarrow T^{2}$ be an endomorphism of $T^{2}$ given by

(2.2) $\rho_{i}\colon T^{2}\to
T^{2},\quad(t_{1},t_{2})\overset{\rho_{i}}{\longmapsto}(t_{1}^{a_{i}}t_{2}^{a_{i+1}},t_{1}^{b_{i}}t_{2}^{b_{i+1}}).$

Since $\xi_{i}=(a_{i},b_{i})$ and $\xi_{i+1}=(a_{i+1},b_{i+1})$ are linearly
independent, the kernel $K_{i}=\ker\rho_{i}$ is a cyclic subgroup of $T^{2}$.
Then the map $\psi_{i}\times\rho_{i}$ induces a surjection

(2.3) $\mathbb{C}^{2}\cong\mathbb{R}^{2}_{\geq}\times
T^{2}/_{\sim_{std}}\xrightarrow{\psi_{i}\times\rho_{i}}U_{i}\times
T^{2}/_{\sim}.$

This shows that $U_{i}\times T^{2}/_{\sim}$ is homeomorphic to the quotient
$\mathbb{C}^{2}/K_{i}$, where $K_{i}$ acts on $\mathbb{C}^{2}$ as a subgroup
of $T^{2}$. Hence, the map (2.3) forms an orbifold chart around the point
$[v_{i},g]\in X$. The gluing maps among these orbifold charts are determined
by the underlying polygon.

A certain cofibration construction of $X$ is studied in [BSS17] based on the
orbifold structure on $X$. Pick a vertex $v_{i}$ of $P$ and $U_{i}$ is its
neighborhood as above. Consider a line segment $\ell_{i}$ in $P$ connecting
two points lying in the relative interior of $E_{i}$ and $E_{i+1}$,
respectively. The restriction of identification (2.1) to $\ell_{i}$ gives rise
to a subspace of $X$

$L_{i}=\ell_{i}\times T^{2}/_{\sim}.$

By assuming that the homeomorphism $\psi_{i}\colon\mathbb{R}^{2}\to U_{i}$
sends the arc $S^{1}_{\geq}=S^{1}\cap\mathbb{R}^{2}_{\geq}$ to $\ell_{i}$, the
restriction of (2.3) to $S^{1}_{\geq}\times T^{2}/_{\sim_{std}}$ induces a
homeomorphism $\big{(}S^{1}_{\geq}\times T^{2}/_{\sim_{std}}\big{)}/K_{i}\cong
L_{i}$. Here, we notice that $K_{i}$ is isomorphic to
$\mathbb{Z}_{m_{i,i+1}}$, where

(2.4) $m_{i,j}=|\det\begin{bmatrix}\xi_{i}^{t}&\xi_{j}^{t}\end{bmatrix}|.$

As $S^{1}_{\geq}\times T^{2}/_{\sim_{std}}$ is homeomorphic to $S^{3}$, we
conclude that $L_{i}$ is homeomorphic to $S^{3}/\mathbb{Z}_{m_{i,i+1}}$ which
is $S^{3}$ if $m_{i,i+1}=1$ and is a lens space otherwise. This description
can be found in [SS18, Proposition 2.3] including higher dimensional cases.

Moreover, the subspace $U_{i}\times T^{2}/_{\sim}$ is homeomorphic to a
tubular neighborhood of the cone on $L_{i}$. Let $B$ be the union of all edges
$E_{j}$ where $j\neq i,i+1$. The subspace $B\times T^{2}/_{\sim}$ is homotopic
to a wedge of $n$ copies of $2$-spheres and the subspace
$(P-\\{v_{i}\\})\times T^{2}/_{\sim}$ retracts to $B\times T^{2}/_{\sim}$. As
$X$ is a union of $(P-\\{v_{i}\\})\times T^{2}/_{\sim}$ and $U_{i}\times
T^{2}/_{\sim}$, it implies a homotopy cofibration

(2.5) $L_{i}\xrightarrow{f_{i}}\bigvee^{n}_{j=1}S^{2}\to X$

where the map $f_{i}$ is induced by the composition of the inclusion $\iota$
and the retraction $r$

$\ell_{i}\times
T^{2}/_{\sim}\overset{\iota}{\hookrightarrow}(P-\\{v_{i}\\})\times
T^{2}/_{\sim}\xrightarrow{r}B\times T^{2}/_{\sim}.$

See Figure 2 for a pictorial illustration of (2.5).

$v_{i}$$\ell_{i}$$=$$\cup$$v_{i}$$\ell_{i}$ Figure 2.
$X=\bigvee_{i=1}^{3}S^{2}\cup_{f_{i}}CL_{i}$.

Applying the cohomology functor to the cofibre sequence (2.5) and referring to
[BNSS19, Theorem 1.1], we can compute the free part of $H^{\ast}(X)$. The
cohomology of $X$ has been discussed using various tools in the works [Jor98,
Theorem 2.5.5], [Fis92, Theorem 2.3] and [KMZ17, Corollary 5.1] which can be
summarised as follows.

######  Proposition 2.1.

Let $X$ be a toric orbifold of dimension $4$. Then we have

(2.6) $\begin{array}[]{c|c c c c c c}i&0&1&2&3&4&\geq 5\\\ \hline\cr
H^{i}(X)&\mathbb{Z}&0&\mathbb{Z}^{n}&\mathbb{Z}_{m}&\mathbb{Z}&0\end{array}$

where $m$ is the greatest common divisor of $\\{m_{i,j}\mid 1\leq i<j\leq
n+2\\}$ for $m_{i,j}$’s defined in (2.4). We set $\mathbb{Z}_{m}=0$ if $m=1$.

######  Remark 2.2.

The way of realising $X$ as a cofibre in (2.5) can be understood in a more
general framework of a _$\mathbf{q}$ -CW complex_. A $\mathbf{q}$-CW complex
is defined inductively starting from a discrete set $X_{0}$ of points. Then,
$X_{i}$ is defined by the pushout

${\bigsqcup_{\alpha}S^{i-1}/K_{\alpha}}$${\bigsqcup_{\alpha}e^{i}/K_{\alpha}}$${X_{i-1}}$${X_{i},}$$\scriptstyle{\\{\phi_{\alpha}\\}}$${\ulcorner}$

where $e^{i}$ and $S^{i-1}$ are $i$-dimensional cell and its boundary,
respectively, and $K_{\alpha}$ is a finite group acting linearly on $e_{i}$.
Every toric orbifold is a $\mathbf{q}$-CW complex. We refer to [BNSS19] for
more details.

##  3\. Cohomology of 4-dimensional CW-complexes

###  3.1. A category of 4-dimensional CW-complexes

Suppose that $X$ is a simply connected CW-complex satisfying (2.6). By [Hat02,
Proposition 4H.3] it is homotopy equivalent to a CW-complex

(3.1) $\left(\underset{i=1}{\overset{n}{\bigvee}}S^{2}\vee
P^{3}(m)\right)\cup_{f}e^{4}$

where $f\colon S^{3}\to\bigvee_{i=1}^{n}S^{2}\vee P^{3}(m)$ is the attaching
map of the 4-cell. In this section we study the homotopy theory of CW-
complexes in this form.

Define $\mathscr{C}_{n,m}$ to be the full subcategory of $\text{Top}_{*}$
consisting of mapping cones as in (3.1). Here the orientation of the 4-cell
$e^{4}$ is the induced orientation of the upper hemisphere in $S^{5}$. We
label the $i^{\text{th}}$ copy of 2-spheres in $\bigvee_{i=1}^{n}S^{2}$ by
$S^{2}_{i}$ for $1\leq i\leq n$ and write

$Y=\underset{i=1}{\overset{n}{\bigvee}}S^{2}_{i}\vee P^{3}(m)$

for short. Let $\mu_{i},\nu\in H_{2}(Y)$ be homology classes representing
$S^{2}_{i}$ and the 2-cell of $P^{3}(m)$ respectively. Then, we have

(3.2)
$H_{2}(Y)\cong\mathbb{Z}\langle{\mu_{1},\ldots,\mu_{n}}\rangle\oplus\mathbb{Z}_{m}\langle{\nu}\rangle.$

Let $g\colon Y\to Y$ be a map. Then the induced homology map $g_{\ast}\colon
H_{2}(Y)\to H_{2}(Y)$ is given by
$g_{*}(\mu_{i})=\sum^{n}_{j=1}x_{ij}\mu_{j}+y_{i}\nu$ and $g_{*}(\nu)=z\nu$
for some integers $x_{ij}$ and mod-$m$ congruence classes $y_{i}$ and $z$.
Conversely, we have the following lemma.

######  Lemma 3.1.

Given a vector $(y_{1},\ldots,y_{n},z)\in(\mathbb{Z}_{m})^{n+1}$ and an
$(n\times n)$-integral matrix

$\left(\begin{array}[]{c c c}x_{11}&\ldots&x_{1n}\\\ \vdots&\ddots&\vdots\\\
x_{n1}&\cdots&x_{nn}\end{array}\right)\in\text{Mat}_{n}(\mathbb{Z}),$

there exists a map $g\colon Y\to Y$ such that
$g_{*}(\mu_{i})=\sum^{n}_{j=1}x_{ij}\mu_{j}+y_{i}\nu$ and $g_{*}(\nu)=z\nu$.

###### Proof.

First, consider the string of isomorphisms

$\displaystyle\big{[}\underset{i=1}{\overset{n}{\bigvee}}S^{2}_{i},\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(m)\big{]}$ $\displaystyle\cong$
$\displaystyle\bigoplus^{n}_{i=1}\big{[}S^{2}_{i},\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(m)\big{]}$ $\displaystyle\cong$
$\displaystyle\bigoplus^{n}_{i=1}\pi_{2}\big{(}\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(m)\big{)}$ $\displaystyle\cong$
$\displaystyle\bigoplus^{n}_{i=1}H_{2}\big{(}\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(m)\big{)}$ $\displaystyle\cong$
$\displaystyle\bigoplus^{n}_{i=1}\left(\bigoplus^{n}_{j=1}\mathbb{Z}\oplus\mathbb{Z}_{m}\right)$

where the third isomorphism is due to Hurewicz Theorem. Under these
isomorphisms, take $g^{\prime}\colon\bigvee_{i=1}^{n}S^{2}_{i}\to Y$ to be the
map corresponding to

$\left(\begin{array}[]{c c c}x_{11}&\cdots&x_{1n}\\\ \vdots&\ddots&\vdots\\\
x_{n1}&\cdots&x_{nn}\end{array}\right)\oplus(y_{1},\dots,y_{n})\in\left(\bigoplus^{n}_{i=1}\bigoplus^{n}_{j=1}\mathbb{Z}\right)\oplus\left(\bigoplus^{n}_{i=1}\mathbb{Z}_{m}\right).$

Then $g^{\prime}_{*}(\mu_{i})=\sum^{n}_{j=1}x_{ij}\mu_{j}+y_{i}\nu$.

Next, for $z\in\mathbb{Z}_{m}$ let $g^{\prime\prime}\colon P^{3}(m)\to Y$ be
the composition

$g^{\prime\prime}\colon
P^{3}(m)\xrightarrow{\underline{z}}P^{3}(m)\hookrightarrow Y,$

where $\underline{z}\colon P^{3}(m)\to P^{3}(m)$ is the degree-$z$ map. Let
$g:Y\to Y$ be the wedge sum $g=g^{\prime}\vee g^{\prime\prime}$. Then
$g_{*}(\mu_{i})=\sum^{n}_{j=1}x_{ij}\mu_{j}+y_{i}\nu$ and $g_{*}(\nu)=z\nu$. ∎

###  3.2. Cellular cup product representation

Let $C_{f}\in\mathscr{C}_{n,m}$ be the mapping cone of a map $f\colon S^{3}\to
Y$. As the inclusion $Y\hookrightarrow C_{f}$ induces an isomorphism
$H_{2}(Y)\to H_{2}(C_{f})$, we do not distinguish $\mu_{i},\nu\in H_{2}(Y)$
and their images in $H_{2}(C_{f})$. Let $u_{i}\in H^{2}(C_{f})\text{ and }e\in
H^{4}(C_{f})$ be cohomology classes dual to $\mu_{i}$ and the homology class
represented by the 4-cell in $C_{f}$ respectively. Let $v\in H^{3}(C_{f})$ be
the Ext image of $\nu$. Then

(3.3) $\begin{array}[]{c c
c}H^{2}(C_{f})\cong\mathbb{Z}\langle{u_{1},\ldots,u_{n}}\rangle,&H^{3}(C_{f})\cong\mathbb{Z}_{m}\langle{v}\rangle,&H^{4}(C_{f})\cong\mathbb{Z}\langle{e}\rangle.\end{array}$

We call the set $\\{u_{1},\ldots,u_{n},v,e\\}$ the _cellular basis of
$H^{*}(C_{f})$_.

With coefficient $\mathbb{Z}_{m}$, let $\bar{u}_{i}\in
H^{2}(C_{f};\mathbb{Z}_{m})$ and $\bar{e}\in H^{4}(C_{f};\mathbb{Z}_{m})$ be
the mod-$m$ images of $u_{i}$ and $e$, and let $\bar{v}\in
H^{2}(C_{f};\mathbb{Z}_{m})$ be the cohomology class dual to $\nu$. Then

$\begin{array}[]{c c
c}H^{2}(C_{f};\mathbb{Z}_{m})\cong\mathbb{Z}_{m}\langle{\bar{u}_{1},\ldots,\bar{u}_{n},\bar{v}}\rangle,&H^{3}(C_{f};\mathbb{Z}_{m})\cong\mathbb{Z}_{m}\langle{\beta(\bar{v})}\rangle,&H^{4}(C_{f};\mathbb{Z}_{m})\cong\mathbb{Z}_{m}\langle{\bar{e}}\rangle,\end{array}$

where $\beta$ is the Bockstein homomorphism. We call the set
$\\{\bar{u}_{1},\ldots,\bar{u}_{n},\bar{v};\bar{e}\\}$ the _mod- $m$ cellular
basis_ of $H^{*}(C_{f};\mathbb{Z}_{m})$.

######  Definition 3.2.

Let $C_{f}$ be a mapping cone in $\mathscr{C}_{n,m}$. Then the cellular cup
product representation $M_{cup}(C_{f})$ of $C_{f}$ is
$A\in\text{Mat}_{n}(\mathbb{Z})$ if $m=1$, and is a triple
$(A,\textbf{b},c)\in\text{Mat}_{n}(\mathbb{Z})\oplus(\mathbb{Z}_{m})^{n}\oplus\mathbb{Z}_{m}$
if $m>1$, where

$A=\left(\begin{array}[]{c c c}a_{11}&\cdots&a_{1n}\\\ \vdots&\ddots&\vdots\\\
a_{n1}&\cdots&a_{nn}\end{array}\right)~{}\text{and}~{}\mathbf{b}=(b_{1},\dots,b_{n})$

are given by $u_{i}\cup u_{j}=a_{ij}e,~{}\bar{u}_{i}\cup\bar{v}=b_{i}\bar{e}$
and $\bar{v}\cup\bar{v}=c\bar{e}$.

Here $A$ is a symmetric matrix since it is the matrix representation of the
bilinear form

$(-\cup-)_{\mathbb{Z}}\colon H^{2}(C_{f};\mathbb{Z})\otimes
H^{2}(C_{f};\mathbb{Z})\to H^{4}(C_{f};\mathbb{Z})$

with respect to the cellular basis $\\{u_{1},\ldots,u_{n};e\\}$. Furthermore,
Universal Coefficient Theorem implies
$\bar{u}_{i}\cup\bar{u}_{j}=a_{ij}\bar{e}\pmod{m}$. So
$(-\cup-)_{\mathbb{Z}_{m}}\colon H^{2}(C_{f};\mathbb{Z}_{m})\otimes
H^{2}(C_{f};\mathbb{Z}_{m})\to H^{4}(C_{f};\mathbb{Z}_{m})$ can be recovered
from $M_{cup}(C_{f})$ as well.

######  Remark 3.3.

When $X$ is an oriented compact smooth $4$-manifold, the intersection form
$I(X)$ is the bilinear form given by cup products of degree $2$ cohomology
classes modulo torsion

$I(X)\colon H^{2}(X)/\text{Tor}\otimes H^{2}(X)/\text{Tor}\to\mathbb{Z},\quad
x\otimes y\mapsto\langle{x\cup y,[X]}\rangle,$

where $[X]\in H_{4}(X)$ is the fundamental class. Although defined in a
similar fashion, $I(X)$ and $M_{cup}(X)$ are different. First, $I(X)$ only
concerns cup products of free elements in $H^{2}(X)$ and its matrix
representation is a symmetric matrix, while $M_{cup}(X)$ concerns cup products
of cohomology with integral and $\mathbb{Z}_{m}$-coefficients and is a triple
consisting of a matrix, a mod-$m$ vector and a mod-$m$ congruence class that
record all data. Second, a matrix representation of $I(X)$ depends on the
choice of generators of $H^{2}(X)$, whereas we define $M_{cup}(X)$ using a
fixed CW-complex structure of $X$. In the following section, we will discuss
the transformation between cellular map representations of two CW-complex
structures of the same $X$. It is similar to matrix congruence but is slightly
more complicated, as cup products of cohomology with $\mathbb{Z}_{m}$
coefficient are involved.

Let $g\colon S^{3}\to Y$ be another map and let $C_{g}\in\mathscr{C}_{n,m}$ be
its mapping cone. Recall that $f+g$ is the composition

$f+g\colon S^{3}\xrightarrow{\text{comult}}S^{3}\vee S^{3}\xrightarrow{f\vee
g}Y\vee Y\xrightarrow{\text{fold}}Y.$

Denote its mapping cone by $C_{f+g}$.

######  Lemma 3.4.

Let $Y$ be (1) $S^{2}_{1}\vee S^{2}_{2}$ or (2) $S^{2}_{1}\vee P^{3}(m)$ and
let $f,g:S^{3}\to Y$ be two maps. Then
$M_{cup}(C_{f+g})=M_{cup}(C_{f})+M_{cup}(C_{g})$.

###### Proof.

In the following we only prove Case (2). The proof also works for Case (1) but
is simpler. Let

  * •

$\\{u,v;e\\}$, $\\{u_{1},v_{1};e_{1}\\}$ and $\\{u_{2},v_{2};e_{2}\\}$ be the
cellular bases of $H^{*}(C_{f+g}),H^{*}(C_{f})$ and $H^{*}(C_{g})$,
respectively;

  * •

$\\{\bar{u},\bar{v};\bar{e}\\}$, $\\{\bar{u}_{1},\bar{v}_{1};\bar{e}_{1}\\}$
and $\\{\bar{u}_{2},\bar{v}_{2};\bar{e}_{2}\\}$ be the mod-$m$ cellular bases
of $H^{*}(C_{f+g};\mathbb{Z}_{m})$, $H^{*}(C_{f};\mathbb{Z}_{m})$ and
$H^{*}(C_{g};\mathbb{Z}_{m})$, respectively;

  * •

the cellular cup product representations of $C_{f+g},C_{f}$ and $C_{g}$ be
$M_{cup}(C_{f+g})=(A,\textbf{b},c)$,
$M_{cup}(C_{f})=(A_{1},\textbf{b}_{1},c_{1})$ and
$M_{cup}(C_{g})=(A_{2},\textbf{b}_{2},c_{2})$, respectively.

Here $A,A_{1},A_{2}$ are integers and
$\textbf{b},\textbf{b}_{1},\textbf{b}_{2}$ are mod-$m$ congruence classes. We
claim that

$A=A_{1}+A_{2},~{}\textbf{b}=\textbf{b}_{1}+\textbf{b}_{2}\text{ and
}c=c_{1}+c_{2}.$

Consider the mapping cone $C^{\prime}=Y\cup_{f\vee g}(e^{4}_{1}\vee
e^{4}_{2})$ of $g\vee h\colon S^{3}\vee S^{3}\to Y$. Let

  * •

$u^{\prime}\in H^{2}(C^{\prime})$, $e^{\prime}_{1},e^{\prime}_{2}\in
H^{4}(C^{\prime})$ be cohomology classes dual to $S^{2}$, $e^{4}_{1}$ and
$e^{4}_{2}$;

  * •

$\bar{u}^{\prime}\in H^{2}(C^{\prime};\mathbb{Z}_{m})$,
$\bar{e}^{\prime}_{1},\bar{e}^{\prime}_{2}\in
H^{4}(C^{\prime};\mathbb{Z}_{m})$ be the mod-$m$ images of $u,e^{\prime}_{1}$
and $e^{\prime}_{2}$;

  * •

$\bar{v}^{\prime}\in H^{2}(C^{\prime};\mathbb{Z}_{m})$ be the cohomology class
dual to the 2-cell of $P^{3}(m)$.

Observe that $C_{f}$ and $C_{g}$ are subcomplexes of $C^{\prime}$. Let
$\imath_{1}\colon C_{f}\to C^{\prime}$ and $\imath_{2}\colon C_{g}\to
C^{\prime}$ be natural inclusions and let $q\colon C_{f+g}\to C^{\prime}$ be
the map collapsing the equatorial disk of the 4-cell in $C_{f+g}$ to a point.
Then

$\begin{array}[]{l l
l}q^{*}(u^{\prime})=u,&q^{*}(\bar{v}^{\prime})=\bar{v},&q^{*}(e^{\prime}_{1})=q^{*}(e^{\prime}_{2})=e,\\\\[5.0pt]
\imath^{*}_{j}(u^{\prime})=u_{j},&\imath^{*}_{j}(\bar{v}^{\prime})=\bar{v}_{j},&\imath^{*}_{j}(e^{\prime}_{k})=\delta_{jk}e_{j}\end{array}$

for $j,k\in\\{1,2\\}$, where $\delta_{jk}$ is the Kronecker symbol. On the one
hand, $u^{\prime}\cup
u^{\prime}=\alpha_{1}e^{\prime}_{1}+\alpha_{2}e^{\prime}_{2}$ for some
integers $\alpha_{1}$ and $\alpha_{2}$. Now the naturality of cup products
implies

$\displaystyle\imath^{*}_{j}(u^{\prime}\cup u^{\prime})$ $\displaystyle=$
$\displaystyle\imath^{*}_{j}(\alpha_{1}e^{\prime}_{1}+\alpha_{2}e^{\prime}_{2})$
$\displaystyle\imath^{*}_{j}(u^{\prime})\cup\imath^{*}_{j}(u^{\prime})$
$\displaystyle=$
$\displaystyle\alpha_{1}\imath^{*}_{j}(e^{\prime}_{1})+\alpha_{2}\imath^{*}_{j}(e^{\prime}_{2})$
$\displaystyle u_{j}\cup u_{j}$ $\displaystyle=$
$\displaystyle\alpha_{j}e_{j}$

for $j\in\\{1,2\\}$. So $\alpha_{j}=A_{j}$. On the other hand,

$\displaystyle u\cup u$ $\displaystyle=$ $\displaystyle q^{*}(u^{\prime})\cup
q^{*}(u^{\prime})$ $\displaystyle=$ $\displaystyle q^{*}(u^{\prime}\cup
u^{\prime})$ $\displaystyle=$ $\displaystyle
q^{*}(A_{1}e^{\prime}_{1}+A_{2}e^{\prime}_{2})$ $\displaystyle=$
$\displaystyle A_{1}q^{*}(e^{\prime}_{1})+A_{2}q^{*}(e^{\prime}_{2})$
$\displaystyle=$ $\displaystyle(A_{1}+A_{2})e.$

So $A=A_{1}+A_{2}$. Similarly we can show
$\textbf{b}=\textbf{b}_{1}+\textbf{b}_{2}$ and $c=c_{1}+c_{2}$. Therefore we
have

$M_{cup}(C_{f+g})=M_{cup}(C_{f})+M_{cup}(C_{g}).$

∎

###  3.3. Cellular map representations

Let $f,f^{\prime}\colon S^{3}\to Y$ be two maps and
$C_{f},C_{f^{\prime}}\in\mathscr{C}_{n,m}$ be their mapping cones. Let

  * •

$\\{u_{1},\ldots,u_{n},v,e\\}$ and
$\\{u^{\prime}_{1},\ldots,u^{\prime}_{n},v^{\prime},e^{\prime}\\}$ be the
cellular bases of $H^{*}(C_{f})$ and $H^{*}(C_{f^{\prime}})$,

  * •

$\\{\bar{u}_{1},\ldots,\bar{u}_{n},\bar{v},\bar{e}\\}$ and
$\\{\bar{u}^{\prime}_{1},\ldots,\bar{u}^{\prime}_{n},\bar{v}^{\prime},\bar{e}^{\prime}\\}$
be the mod-$m$ cellular bases of $H^{*}(C_{f};\mathbb{Z}_{m})$ and
$H^{*}(C_{f^{\prime}};\mathbb{Z}_{m})$.

Given a map $\psi\colon C_{f^{\prime}}\to C_{f}$ and a coefficient ring $R$,
let

$\psi^{\ast}_{R}\colon H^{2}(C_{f};R)\to H^{2}(C_{f^{\prime}};R)$

be the induced morphism on the second cohomology with coefficient $R$.

######  Definition 3.5.

Let $\psi\colon C_{f^{\prime}}\to C_{f}$ be a map. Then the _cellular map
representation_ $M(\psi)$ of $\psi$ is $W\in\text{Mat}_{n}(\mathbb{Z})$ if
$m=1$, and is the triple
$(W,\textbf{y},z)\in\text{Mat}_{n}(\mathbb{Z})\oplus(\mathbb{Z}_{m})^{n}\oplus\mathbb{Z}_{m}$
if $m>1$, where

(3.4) $W=\left(\begin{array}[]{c c c}x_{11}&\cdots&x_{1n}\\\
\vdots&\ddots&\vdots\\\
x_{n1}&\cdots&x_{nn}\end{array}\right)~{}\text{and}~{}\mathbf{y}=(y_{1},\dots,y_{n})$

are given by
${\psi^{\ast}_{\mathbb{Z}}}(u_{j})=\sum^{n}_{i=1}x_{ij}u^{\prime}_{i}$ and
${\psi^{\ast}_{\mathbb{Z}_{m}}}(\bar{v})=\sum^{n}_{i=1}y_{i}\bar{u}^{\prime}_{i}+z\bar{v}^{\prime}$.

######  Lemma 3.6.

For $R=\mathbb{Z}$ or $\mathbb{Z}_{m}$, consider $\psi$, $\psi^{\ast}_{R}$ and
$M(\psi)$ as above. For $1\leq j\leq n$, we have
$\psi^{\ast}_{\mathbb{Z}_{m}}(\bar{u}_{j})=\sum^{n}_{i=1}x_{ij}\bar{u}^{\prime}_{i}$.
Furthermore, if $\psi$ is a homotopy equivalence, then $W$ is an invertible
matrix and $z$ is a unit in $\mathbb{Z}_{m}$.

###### Proof.

Since $C_{f}$ and $C_{f^{\prime}}$ are simply connected, Universal Coefficient
Theorem implies that

$H^{2}(C_{f}{;R})\cong\text{Hom}(H_{2}(C_{f}),R),~{}H^{2}(C_{f^{\prime}}{;R})\cong\text{Hom}(H_{2}(C_{f^{\prime}}),R)\text{
and }\psi^{\ast}_{R}=\text{Hom}(\psi_{\ast},R)$

is dual to $\psi_{\ast}\colon{H_{2}(C_{f^{\prime}})\to H_{2}(C_{f})}$. So
$\psi^{\ast}_{\mathbb{Z}_{m}}(\bar{u}_{j})$ is the mod-$m$ image of
$\psi^{\ast}_{\mathbb{Z}}(u_{j})$ and the first part of the lemma follows.

If $\psi$ is a homotopy equivalence, then $W\in\text{Mat}_{n}(\mathbb{Z})$ and

$\left(\begin{array}[]{c c c c}\bar{x}_{11}&\cdots&\bar{x}_{1n}&y_{1}\\\
\vdots&\ddots&\vdots&\vdots\\\ \bar{x}_{n1}&\cdots&\bar{x}_{nn}&y_{n}\\\
0&\cdots&0&z\end{array}\right)\in\text{Mat}_{n+1}(\mathbb{Z}_{m}),\quad\text{where
}\bar{x}_{ij}\equiv x_{ij}\pmod{m}$

are invertible matrices. So the second part follows. ∎

The cellular map representation records the data of $\psi_{\mathbb{Z}}^{\ast}$
and $\psi_{\mathbb{Z}_{m}}^{\ast}$. The square matrix $W$ in (3.4) is the map
representation of $\psi^{\ast}_{\mathbb{Z}}$ with respect to bases
$\\{u_{1},\ldots,u_{n}\\}$ and $\\{u^{\prime}_{1},\ldots,u^{\prime}_{n}\\}$.
Lemma 3.6 implies that

$\left(\begin{array}[]{c c}\overline{W}&\textbf{y}^{t}\\\
\textbf{0}&z\end{array}\right)\in\text{Mat}_{n+1}(\mathbb{Z}_{m}),$

where $\overline{W}$ is the mod-$m$ image of $W$ and
$\textbf{0}=(0,\ldots,0)$, is the matrix representation of
$\psi^{\ast}_{\mathbb{Z}_{m}}$ with respect to bases
$\\{\bar{u}_{1},\ldots,\bar{u}_{n},\bar{v}\\}$ and
$\\{\bar{u}^{\prime}_{1},\ldots,\bar{u}^{\prime}_{n},\bar{v}^{\prime}\\}$.

Recall that in linear algebra, matrix representations of a bilinear form
$V\otimes V\to\mathbb{Z}$ with respect to different bases of $V$ are congruent
to each other. So we have the following lemma.

######  Lemma 3.7.

For $C_{f},C_{f^{\prime}}\in\mathscr{C}_{n,m}$, let
$M_{cup}(C_{f})=(A,\textbf{b},c)$ and
$M_{cup}(C_{f^{\prime}})=(A^{\prime},\textbf{b}^{\prime},c^{\prime})$ be their
cellular cup product representations. If there is a homotopy equivalence
$\psi\colon C_{f^{\prime}}\to C_{f}$ with
$M(\psi)=(W,\textbf{y},z)\in\text{GL}_{n}(\mathbb{Z})\oplus(\mathbb{Z}_{m})^{n}\oplus\mathbb{Z}^{*}_{m}$,
then

$\begin{array}[]{c c
c}A^{\prime}=W^{t}AW,&\textbf{b}^{\prime}=\textbf{y}AW+z\textbf{b}W,&c^{\prime}=\textbf{y}A\textbf{y}^{t}+2z\textbf{y}\textbf{b}^{t}+z^{2}c.\end{array}$

∎

In particular, if two maps $f$ and $f^{\prime}\colon S^{3}\to Y$ are
homotopic, then the matrix cup product representations of their mapping cones
are the same.

######  Lemma 3.8.

If $f$ is homotopic to $f^{\prime}$, then
$M_{cup}(C_{f})=M_{cup}(C_{f^{\prime}})$.

###### Proof.

Take a homotopy $\phi\colon S^{3}\times I\to Y$ between $f$ and $f^{\prime}$.
It induces a homotopy equivalence $\Phi\colon C_{f}\to C_{f^{\prime}}$ such
that its restriction to $Y$ is the identity map. So
$M(\Phi)=(I_{n},\textbf{0},1)$. Then the lemma follows from Lemma 3.7. ∎

######  Lemma 3.9.

Let $C_{f}\in\mathscr{C}_{n,m}$ and let $(W,\textbf{y},z)$ be a triple in
$\text{GL}_{n}(\mathbb{Z})\oplus(\mathbb{Z}_{m})^{n}\oplus\mathbb{Z}_{m}^{*}$.
Then there exist a CW-complex $C_{f^{\prime}}\in\mathscr{C}_{n,m}$ and a
homotopy equivalence $\psi\colon C_{f}\to C_{f^{\prime}}$ such that the
cellular map representation $M(\psi)$ is $(W,\textbf{y},z)$.

###### Proof.

Let

$W=\left(\begin{array}[]{c c c}x_{11}&\cdots&x_{1n}\\\ \vdots&\ddots&\vdots\\\
x_{n1}&\cdots&x_{nn}\end{array}\right)~{}\text{and}~{}\mathbf{y}=(y_{1},\dots,y_{n}).$

By Lemma 3.1, there exists a map $\tilde{\psi}\colon Y\to Y$ such that
$\tilde{\psi}_{\ast}(\mu_{i})=\sum^{n}_{j=1}x_{ij}\mu_{j}+y_{i}\nu$ and
$\tilde{\psi}_{\ast}(\nu)=z\nu$, where $\mu_{1},\dots,\mu_{n}$ and $\nu$ are
elements in $H_{2}(Y)$ as in (3.2). Thus, we have
$\tilde{\psi}^{\ast}_{\mathbb{Z}}(u_{i})=\sum^{n}_{j=1}x_{ji}u_{j}$ and
$\tilde{\psi}^{\ast}_{\mathbb{Z}_{m}}(\bar{v})=\sum^{n}_{i=1}y_{i}\bar{u}_{i}+z\bar{v}$.

Let $f^{\prime}=\tilde{\psi}\circ f$ and let $C_{f^{\prime}}$ be its mapping
cone. Then there is a diagram of cofibration sequences

$\textstyle{S^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{\psi}}$$\textstyle{C_{f}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\textstyle{S^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C_{f^{\prime}},}$

where $\psi$ is an induced map. Since $W\in GL_{n}(\mathbb{Z})$ and
$z\in\mathbb{Z}^{*}_{m}$, the middle vertical arrow $\tilde{\psi}$ induces an
isomorphism in homology. By five lemma, $\psi_{\ast}$ is an isomorphism, which
implies that $\psi$ is a homotopy equivalence. Finally, we have
$M(\psi)=(W,\textbf{y},z)$ by the construction. ∎

##  4\. The Homotopy theory of complexes in $\mathscr{C}_{n,m}$

###  4.1. The $\mathscr{C}_{n,1}$ case

When $m=1$, the mapping cone $C_{f}\in\mathscr{C}_{n,1}$ is in the form
$\bigvee_{i=1}^{n}S^{2}_{i}\cup_{f}e^{4}$ where $f\colon
S^{3}\to\bigvee_{i=1}^{n}S_{i}^{2}$ is the attaching map of the 4-cell.
Hilton–Milnor Theorem (see for instance [Sel97, Theorem 7.9.4]) implies that
$f$ is homotopic to a wedge sum

$\sum^{n}_{i=1}a_{i}\eta_{i}+\sum_{1\leq j<k\leq n}a_{jk}\omega_{jk},$

for some integers $a_{i}$’s and $a_{jk}$’s. Here $\eta_{i}$’s and
$\omega_{jk}$’s are compositions

$\begin{array}[]{l}\eta_{i}\colon
S^{3}\overset{\eta}{\to}S^{2}_{i}\hookrightarrow\underset{\ell=1}{\overset{n}{\bigvee}}S^{2}_{\ell}\\\\[8.0pt]
\omega_{jk}\colon S^{3}\xrightarrow{[\imath_{1},\imath_{2}]}S^{2}_{j}\vee
S^{2}_{k}\hookrightarrow\underset{l=1}{\overset{n}{\bigvee}}S^{2}_{\ell}\end{array}$

of Hopf map $\eta$, Whitehead product $[\imath_{1},\imath_{2}]$ and canonical
inclusions of $S^{2}_{i}$ and $S^{2}_{j}\vee S^{2}_{k}$ into
${\bigvee}_{\ell=1}^{n}S^{2}_{\ell}$. The lemma below shows that the
coefficients $a_{i}$ and $a_{jk}$ are determined by $M_{cup}(C_{f})$.

######  Lemma 4.1.

Let $C_{f}\in\mathscr{C}_{n,1}$ be the mapping cone of
$f\simeq\sum^{n}_{i=1}a_{i}\eta_{i}+\sum_{1\leq j<k\leq n}a_{jk}\omega_{jk}$.
If

$M_{cup}(C_{f})=\begin{pmatrix}a^{\prime}_{1}&a^{\prime}_{12}&\cdots&a^{\prime}_{1n}\\\
a^{\prime}_{12}&a^{\prime}_{2}&\cdots&a^{\prime}_{2n}\\\
\vdots&\vdots&\ddots&\vdots\\\
a^{\prime}_{1n}&a^{\prime}_{2n}&\cdots&a^{\prime}_{n}\end{pmatrix},$

then $a_{i}=a^{\prime}_{i}$ and $a_{jk}=a^{\prime}_{jk}$ for all $i,j$ and
$k$.

###### Proof.

By Lemma 3.8, we may assume $f=\sum^{n}_{i=1}a_{i}\eta_{i}+\sum_{1\leq j<k\leq
n}a_{jk}\omega_{jk}$. For $n=2$, let $C_{1},C_{2}$ and $C_{12}$ be the mapping
cones of $a_{1}\eta_{1},a_{2}\eta_{2}$ and $a_{12}\omega_{12}$. Then their
cellular cup product representations are

$M_{cup}(C_{1})=\left(\begin{array}[]{c c}a_{1}&0\\\
0&0\end{array}\right),\quad M_{cup}(C_{2})=\left(\begin{array}[]{c c}0&0\\\
0&a_{2}\end{array}\right),\quad M_{cup}(C_{12})=\left(\begin{array}[]{c
c}0&a_{12}\\\ a_{12}&0\end{array}\right).$

By Lemma 3.4, we have

$M_{cup}(C_{f})=\left(\begin{array}[]{c c}a_{1}&a_{12}\\\
a_{12}&a_{2}\end{array}\right).$

So the lemma holds.

For $n\geq 3$, let $\\{u_{1},\ldots,u_{n},e\\}$ be the cellular basis of
$H^{*}(C_{f})$. We claim that

$u_{i}\cup u_{i}=a_{i}e\quad\text{and}\quad u_{j}\cup u_{k}=a_{jk}e,$

for each $1\leq i\leq n$ and $1\leq j<k\leq n$. The composition

$f_{jk}\colon
S^{3}\overset{f}{\to}\underset{l=1}{\overset{n}{\bigvee}}S^{2}_{l}\xrightarrow{\text{pinch}}S^{2}_{j}\vee
S^{2}_{k}$

is homotopic to
$a_{j}\eta^{\prime}_{1}+a_{k}\eta^{\prime}_{2}+a_{jk}\omega^{\prime}_{12}$,
where $\eta^{\prime}_{1}\colon
S^{3}\overset{\eta}{\to}S^{2}_{j}\hookrightarrow S^{2}_{j}\vee S^{2}_{k}$ and
$\eta^{\prime}_{2}\colon S^{3}\overset{\eta}{\to}S^{2}_{k}\hookrightarrow
S^{2}_{j}\vee S^{2}_{k}$ are compositions of Hopf map $\eta$ and canonical
inclusions and $\omega^{\prime}_{12}\colon S^{3}\to S^{2}_{j}\vee S^{2}_{k}$
is the Whitehead product. Let $C_{jk}$ be the mapping cone of $f_{jk}$. By
Lemma 3.8 and the above argument, we have

$M_{cup}(C_{jk})=\left(\begin{array}[]{c c}a_{j}&a_{jk}\\\
a_{jk}&a_{k}\end{array}\right).$

Let $\\{u^{\prime}_{j},u^{\prime}_{k};e^{\prime}\\}$ be the cellular basis of
$H^{*}(C_{jk})$ and let $\alpha\colon C_{f}\to C_{jk}$ be the map which
pinches all 2-spheres in $C_{f}$ to the basepoint except for $S^{2}_{j}$ and
$S^{2}_{k}$. Then

$\alpha^{*}(u^{\prime}_{j})=u_{j},~{}\alpha^{*}(u^{\prime}_{k})=u_{k}\text{
and }\alpha^{*}(e^{\prime})=e.$

By the naturality of cup products, we have

$\displaystyle\alpha^{*}(u^{\prime}_{j})\cup\alpha^{*}(u^{\prime}_{k})$
$\displaystyle=$ $\displaystyle\alpha^{*}(u^{\prime}_{j}\cup u^{\prime}_{k})$
$\displaystyle u_{j}\cup u_{k}$ $\displaystyle=$
$\displaystyle\alpha^{*}(a_{jk}e^{\prime})$ $\displaystyle=$ $\displaystyle
a_{jk}e.$

So $a^{\prime}_{jk}=a_{jk}$. Similarly we can show $a^{\prime}_{i}=a_{i}$.
Hence, the lemma follows. ∎

Now we classify the homotopy types of CW-complexes in $\mathscr{C}_{n,1}$ by
their integral cohomology rings in the next statement.

######  Proposition 4.2.

Let $f,f^{\prime}\colon S^{3}\to\bigvee_{i=1}^{n}S^{2}_{i}$ be two maps and
let $C_{f},C_{f^{\prime}}\in\mathscr{C}_{n,1}$ be their mapping cones. Then
$C_{f}\simeq C_{f^{\prime}}$ if and only if there is a ring isomorphism
$H^{*}(C_{f})\cong H^{*}(C_{f^{\prime}})$.

###### Proof.

The “only if” part is trivial. Assume that $H^{*}(C_{f})\cong
H^{*}(C_{f^{\prime}})$. Then there is an invertible matrix
$W\in\text{GL}_{n}(\mathbb{Z})$ such that

$W^{t}\cdot M_{cup}(C_{f})\cdot W=M_{cup}(C_{f^{\prime}}).$

By Lemma 3.9, there are a CW-complex $\tilde{C}\in\mathscr{C}_{n,1}$ and a
homotopy equivalence $\psi\colon\tilde{C}\to C_{f}$ such that $M(\psi)=W$. We
claim that $\tilde{C}\simeq C_{f^{\prime}}$. By Lemma 3.7, we have

$M_{cup}(\tilde{C})=W^{t}\cdot M_{cup}(C_{f})\cdot W=M_{cup}(C_{f^{\prime}}).$

Let $\tilde{f}$ be the attaching map of the 4-cell in $\tilde{C}$ and let

$M_{cup}(\tilde{C})=M_{cup}(C_{f^{\prime}})=\left(\begin{array}[]{c c c
c}a_{1}&a_{12}&\cdots&a_{1n}\\\ a_{12}&a_{2}&\cdots&a_{2n}\\\
\vdots&\vdots&\ddots&\vdots\\\ a_{1n}&a_{2n}&\cdots&a_{n}\end{array}\right).$

Then Lemma 4.1 implies that $f^{\prime}$ and $\tilde{f}$ are homotopic to the
wedge sum

$\sum^{n}_{i=1}a_{i}\eta_{i}+\sum_{1\leq i<j\leq n}a_{ij}\omega_{ij},$

which means $C_{f^{\prime}}\simeq\tilde{C}$. Hence, we have established the
claim. ∎

######  Corollary 4.3.

Two $4$-dimensional toric orbifolds without torsion in (co)homology are
homotopy equivalent if and only if their integral cohomology rings are
isomorphic.∎

As 4-dimensional quasitoric manifolds always have torsion-free (co)homology,
Corollary 4.3 implies that the homotopy types of 4-dimensional quasitoric
manifolds are classified by their cohomology rings. As we mentioned in
Introduction, the homeomorphism types of 4-dimensional toric manifolds are
cohomologically rigid. One can deduce the conclusion from the topological
classification of 4-dimensional smooth manifolds with $T^{2}$-action studied
in [OR70] together with the cohomology formula [DJ91, Theorem 4.14].

We note that the method in Proposition 4.2 applies to CW-complexes in
$\mathscr{C}_{n,1}$ which are not necessarily manifolds. We also refer to
[DKS18, Section 5] for the computation of the cohomology ring of toric
orbifolds considered in Corollary 4.3.

###  4.2. The $\mathscr{C}_{n,m}$ case

From now on, we assume $m=2^{s}q$, where $q>1$ is odd and $s\geq 0$. Recall
from (3.3) that $H^{3}(C_{f})\cong\mathbb{Z}_{m}$ for
$C_{f}\in\mathscr{C}_{n,m}$. In this subsection, we discuss the homotopy type
of $C_{f}$. To be more precise, we study a necessary and sufficient condition
for a wedge decomposition

(4.1) $C_{f}\simeq\hat{C}\vee P^{3}(q)$

where $\hat{C}$ is a complex in $\mathscr{C}_{n,2^{s}}$ so that
$H^{i}(C_{f})\cong H^{i}(\hat{C})$ for $i\neq 3$ and
$H^{3}(\hat{C})\cong\mathbb{Z}_{2^{s}}$.

######  Lemma 4.4.

Let $q$ be odd and greater than $1$. Consider

  1. (i)

a map $g_{1}\colon S^{3}\to P^{3}(q)$ and its mapping cone $C_{1}$,

  2. (ii)

a map $g_{2}\colon S^{3}\to P^{4}(q)$ and the mapping cone $C_{2}$ of the
composition

$S^{3}\xrightarrow{g_{2}}P^{4}(q)\xrightarrow{[\kappa_{1},\kappa_{2}]}S^{2}\vee
P^{3}(q),$

where $[\kappa_{1},\kappa_{2}]$ is the Whitehead product of inclusions

$\kappa_{1}\colon S^{2}\to S^{2}\vee P^{3}(q)~{}\text{and}~{}\kappa_{2}\colon
P^{3}(m)\to S^{2}\vee P^{3}(q).$

For $i=1$ or $2$, if $H^{*}(C_{i};\mathbb{Z}_{m})$ has trivial cup products,
then $g_{i}$ is null homotopic.

###### Proof.

Let $q=p_{1}^{r_{1}}\ldots p_{\ell}^{r_{\ell}}$ be a primary factorization of
$q$ such that $p_{j}$’s are different odd primes and all $r_{j}$’s are at
least 1. By Hurewicz Theorem, $\pi_{3}(P^{4}(q))\cong\mathbb{Z}_{q}$. By
[Por66, Theorem 4] and [ST19, Lemma 2.1],

$\pi_{3}(P^{3}(q))\cong\bigoplus^{\ell}_{j=1}\pi_{3}(P^{3}(p_{j}^{r_{j}}))\cong\bigoplus^{\ell}_{j=1}\mathbb{Z}_{p_{j}^{r_{j}}}\cong\mathbb{Z}_{q}.$

It suffices to prove the two cases after localization at $p_{j}$.

For the $i=1$ case, the lemma is a special case of [ST19, Proposition 4.4].
For the $i=2$ case, it can be proved by the argument of [ST19, Proposition
3.2] and replacing $P^{3}(p^{t})$ by $S^{2}$ and the index $t$ by $\infty$,
respectively. ∎

######  Lemma 4.5.

Let $m=2^{s}q$, where $q$ is odd and greater than 1. Let $f\colon
S^{3}\to\bigvee_{i=1}^{n}S^{2}_{i}\vee P^{3}(m)$ be the attaching map of the
4-cell in $C_{f}$ and let $M_{cup}(C_{f})=(A,\textbf{b},c)$. If
$\textbf{b}\equiv(0,\ldots,0)\pmod{q}$ and $c\equiv 0\pmod{q}$, then there is
a CW-complex $\hat{C}\in\mathscr{C}_{n,2^{s}}$ such that
$C_{f}\simeq\hat{C}\vee P^{3}(q)$.

###### Proof.

Since $2^{s}$ and $q$ are coprime, we have $P^{3}(m)\simeq P^{3}(2^{s})\vee
P^{3}(q)$. By Hilton–Milnor Theorem, $f$ is homotopic to a wedge sum

$f\simeq\sum^{n}_{i=1}a_{i}\eta_{i}+\sum_{1\leq j<k\leq
n}a_{jk}\omega_{jk}+\eta^{\prime}+\sum^{n}_{i=1}\omega^{\prime}_{i}+\eta_{q}+\sum_{i=1}^{n}\omega_{iq}$

for some integers $a_{i}$’s and $a_{jk}$’s. Here $\eta^{\prime}$,
$\omega^{\prime}_{i}$, $\eta_{P}$ and $\omega_{iP}$ are compositions

$\begin{array}[]{l}\eta^{\prime}\colon
S^{3}\xrightarrow{b^{\prime}}P^{3}(2^{s})\hookrightarrow\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(2^{s})\vee P^{3}(q)\\\\[8.0pt] \omega^{\prime}_{i}\colon
S^{3}\xrightarrow{b^{\prime}_{i}}P^{4}(2^{s})\xrightarrow{[\kappa^{\prime}_{1},\kappa^{\prime}_{2}]}S^{2}_{i}\vee
P^{3}(2^{s})\hookrightarrow\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(2^{s})\vee P^{3}(q)\\\\[8.0pt] \eta_{q}\colon
S^{3}\xrightarrow{b_{q}}P^{3}(q)\hookrightarrow\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(2^{s})\vee P^{3}(q)\\\\[8.0pt] \omega_{iq}\colon
S^{3}\xrightarrow{b_{iq}}P^{4}(q)\xrightarrow{[\kappa_{1},\kappa_{2}]}S^{2}_{i}\vee
P^{3}(q)\hookrightarrow\underset{j=1}{\overset{n}{\bigvee}}S^{2}_{j}\vee
P^{3}(2^{s})\vee P^{3}(q)\\\\[8.0pt] \end{array}$

for some maps $b^{\prime}$, $b^{\prime}_{i}$ and $b_{q}$, $b_{iq}$. Here

$[\kappa^{\prime}_{1},\kappa^{\prime}_{2}]:P^{4}(2^{s})\to S^{2}_{i}\vee
P^{3}(2^{s})$

is the Whitehead product of inclusions $\kappa^{\prime}_{1}:S^{2}_{i}\to
S^{2}_{i}\vee P^{3}(2^{s})$ and $\kappa^{\prime}_{2}:P^{3}(2^{s})\to
S^{2}_{i}\vee P^{3}(2^{s})$, and

$[\kappa_{1},\kappa_{2}]:P^{4}(q)\to S^{2}_{i}\vee P^{3}(q)$

is the Whitehead product of inclusions $\kappa_{1}:S^{2}_{i}\to S^{2}_{i}\vee
P^{3}(q)$ and $\kappa_{2}:P^{3}(q)\to S^{2}_{i}\vee P^{3}(q)$. If $\eta_{q}$
and $\omega_{iP}$’s are null homotopic, then $f$ factors through a map
$\hat{f}\colon S^{3}\to\bigvee_{i=1}^{n}S^{2}_{i}\vee P^{3}(2^{s})$. Let
$\hat{C}$ be the mapping cone of $\hat{f}$. Then $C_{f}\simeq\hat{C}\vee
P^{3}(q)$.

Hence it suffices to show that $\eta_{q}$ and $\omega_{\ell q}$ are null
homotopic for any $\ell$ with $1\leq\ell\leq n$. After localization away from
2, $P^{3}(2^{s})$ becomes contractible and the composition

$f_{\ell q}\colon S^{3}\overset{f}{\to}\bigvee^{n}_{j=1}S^{2}_{j}\vee
P^{3}(q)\xrightarrow{\text{pinch}}S^{2}_{\ell}\vee P^{3}(q)$

is homotopic to the wedge sum $a_{\ell}\tilde{\eta}_{\ell}+\jmath\circ
b_{q}+[\kappa_{1},\kappa_{2}]\circ b_{\ell q}$, where $\tilde{\eta}$ is the
composition of Hopf map $\eta$ and inclusion $S^{2}_{\ell}\to S^{2}_{\ell}\vee
P^{3}(q)$, and $\jmath\colon P^{3}(q)\to S^{2}_{\ell}\vee P^{3}(q)$ is the
inclusion.

Consider the diagram of homotopy cofibration sequences

${\ast}$${\underset{i\neq\ell}{\overset{n}{\bigvee}}S^{2}_{i}}$${\underset{i\neq\ell}{\overset{n}{\bigvee}}S^{2}_{i}}$${S^{3}}$${\underset{i=1}{\overset{n}{\bigvee}}S^{2}_{i}\vee
P^{3}(q)}$${C_{f}}$${S^{3}}$${S^{2}_{\ell}\vee P^{3}(q)}$${C_{\ell
q}}$$\scriptstyle{f}$pinch$\scriptstyle{\pi}$$\scriptstyle{f_{\ell q}}$

where $C_{\ell q}$ is the mapping cone of $f_{\ell q}$ and $\pi$ is an induced
map. Let $\\{\bar{u}_{1},\ldots,\bar{u}_{n},\bar{v};\bar{e}\\}$ and
$\\{\bar{u}^{\prime},\bar{v}^{\prime};\bar{e}^{\prime}\\}$ be the mod-$q$
cellular bases of $H^{*}(C_{f};\mathbb{Z}_{q})$ and $H^{*}(C_{\ell
q};\mathbb{Z}_{q})$. Then

$\begin{array}[]{c c
c}\pi^{*}(\bar{u}^{\prime})=\bar{u}_{\ell},&\pi^{*}(\bar{v}^{\prime})=\bar{v},&\pi^{*}(\bar{e}^{\prime})=\bar{e}.\end{array}$

By the hypothesis,
$\pi^{*}(\bar{u}^{\prime}\cup\bar{v}^{\prime})=\bar{u}_{\ell}\cup\bar{v}=0$.
Since $\pi^{*}\colon H^{4}(C_{\ell q};\mathbb{Z}_{q})\to
H^{4}(C_{f};\mathbb{Z}_{q})$ is isomorphic, we have
$\bar{u}^{\prime}\cup\bar{v}^{\prime}=0$. Similarly we have
$\bar{v}^{\prime}\cup\bar{v}^{\prime}=0$ and
$\bar{u}^{\prime}\cup\bar{u}^{\prime}=a_{\ell}\bar{e}$ so that
$M_{cup}(C_{\ell q})=(a_{\ell},0,0)$. Let $C_{1}$ and $C_{2}$ be the mapping
cones of $[\kappa_{1},\kappa_{2}]\circ b_{\ell q}$ and $\jmath\circ b_{q}$. By
Lemma 3.4,

$M_{cup}(C_{1})=M_{cup}(C_{2})=(0,0,0)$

so $H^{*}(C_{1};\mathbb{Z}_{q})$ and $H^{*}(C_{2};\mathbb{Z}_{q})$ have
trivial cup products. By Lemma 4.4, $b_{\ell q}$ is null homotopic and so is
$\omega_{\ell q}$. Also, notice that $C_{2}\simeq S^{2}_{\ell}\vee C^{\prime}$
where $C^{\prime}$ is the mapping cone of $b_{q}$. So
$H^{*}(C^{\prime};\mathbb{Z}_{q})$ has trivial cup products. By Lemma 4.4,
$b_{q}$ is null homotopic and so is $\eta_{q}$. ∎

######  Remark 4.6.

In general $\hat{C}$ cannot be further decomposed into a wedge of non-
contractible spaces, for example $\hat{C}=\Sigma\mathbb{R}\mathbb{P}^{3}$.

Notice that $\hat{C}\vee P^{3}(q)$ is not contained in $\mathscr{C}_{n,m}$,
but it is homotopic to a mapping cone in $\mathscr{C}_{n,m}$ as follows. Since
$2^{s}$ and $q$ are coprime to each other, there exist integers $\alpha$ and
$\beta$ such that $2^{s}\alpha+q\beta=1$, where the mod-$q$ congruence class
of $\alpha$ and the mod-$2^{s}$ congruence class of $\beta$ are unique.
Identify $\mathbb{Z}_{2^{s}}\oplus\mathbb{Z}_{q}$ with $\mathbb{Z}_{m}$ via
the isomorphism

(4.2)
$\rho\colon\mathbb{Z}_{2^{s}}\oplus\mathbb{Z}_{q}\to\mathbb{Z}_{m},\quad(x,y)\mapsto
q\beta x+2^{s}\alpha y.$

It induces a homotopy equivalence $\rho:P^{3}(2^{s})\vee P^{3}(q)\to
P^{3}(m)$. Consider the diagram of homotopy cofibrations

(4.3) ${S^{3}}$${\underset{i=1}{\overset{n}{\bigvee}}S^{2}_{i}\vee
P^{3}(2^{s})\vee P^{3}(q)}$${\hat{C}\vee
P^{3}(q)}$${S^{3}}$${\underset{i=1}{\overset{n}{\bigvee}}S^{2}_{i}\vee
P^{3}(m)}$${C^{\prime}}$$\scriptstyle{\hat{f}\vee\ast}$$\scriptstyle{id\vee\rho}$$\scriptstyle{\tilde{\rho}}$$\scriptstyle{(id\vee\rho)\circ(\hat{f}\vee\ast)}$

where $C^{\prime}$ is the mapping cone of $(id\vee\rho)\circ(\hat{f}\vee\ast)$
and $\tilde{\rho}$ is an induced homotopy equivalence. Then
$C^{\prime}\simeq\hat{C}\vee P^{3}(q)$ via $\tilde{\rho}$.

######  Lemma 4.7.

Let $M_{cup}(C^{\prime})=(A,\textbf{b},c)$. Then
$\textbf{b}\equiv(0,\cdots,0)$ and $c\equiv 0\pmod{q}.$

###### Proof.

We prove $b_{i}\equiv 0\pmod{q}$. Let

  * •

$\bar{u}_{i}\in H^{2}(\hat{C};\mathbb{Z}_{m})$ and $\bar{e}\in
H^{4}(\hat{C};\mathbb{Z}_{m})$ be the mod-$m$ cohomology classes dual to
homology classes representing $S^{2}_{i}$ and the 4-cell in $\hat{C}$,
respectively;

  * •

$\mu_{i},\,\omega_{2^{s}},\,\omega_{q}\in H_{2}(\hat{C}\vee
P^{3}(q);\mathbb{Z})$ be the homology classes representing $S_{i}^{2}$, the
bottom cells of $P^{3}(2^{s})$ and $P^{3}(q)$;

  * •

$\bar{w}_{2^{s}},\,\bar{w}_{q}\in H^{2}(\hat{C}\vee P^{3}(q);\mathbb{Z}_{m})$
be the cohomology classes such that

$\begin{array}[]{c c c}\bar{w}_{2^{s}}(\omega_{2^{s}})\equiv
q\beta&\bar{w}_{2^{s}}(\omega_{q})\equiv 0&\bar{w}_{2^{s}}(\mu_{i})\equiv
0\\\\[8.0pt] \bar{w}_{q}(\omega_{2^{s}})\equiv 0&\bar{w}_{q}(\omega_{q})\equiv
2^{s}\alpha&\bar{w}_{q}(\mu_{i})\equiv 0\end{array}\pmod{m}.$

Denote $\bar{v}=\bar{w}_{2^{s}}+\bar{w}_{q}$. Then
$\bar{u}_{1},\cdots,\bar{u}_{n}$ and $\bar{v}$ form a basis of
$H^{2}(\hat{C}\vee P^{3}(q);\mathbb{Z}_{m})$.

The right square of (4.3) implies

$\tilde{\rho}^{*}(\bar{u}^{\prime}_{i})=\bar{u}_{i},\quad\tilde{\rho}^{*}(\bar{v}^{\prime})=\bar{v}=\bar{w}_{2^{s}}+\bar{w}_{q}.$

By the naturality of cup products, we have

$\displaystyle\tilde{\rho}^{*}(\bar{u}^{\prime}_{i}\cup\bar{v}^{\prime})$
$\displaystyle=$ $\displaystyle\tilde{\rho}^{*}(b_{i}\bar{e}^{\prime})$
$\displaystyle\bar{u}_{i}\cup(\bar{w}_{2^{s}}+\bar{w}_{q})$ $\displaystyle=$
$\displaystyle b_{i}\bar{e}$ $\displaystyle\bar{u}\cup\bar{w}_{2^{s}}$
$\displaystyle=$ $\displaystyle b_{i}\bar{e}.$

Since $\bar{w}_{2^{s}}$ is a multiple of $q$ and $\bar{e}$ is a generator,
$b_{i}\equiv 0\pmod{q}$. Similarly we can show that $c\equiv 0\pmod{q}$. ∎

######  Proposition 4.8.

Let $m=2^{s}q$ as before. For $C_{f}\in\mathscr{C}_{n,m}$, let
$M_{cup}(C_{f})=(A,\textbf{b},c)$ where

$A=\left(\begin{array}[]{c c c}a_{11}&\cdots&a_{1n}\\\ \vdots&\ddots&\vdots\\\
a_{n1}&\cdots&a_{nn}\end{array}\right)\text{ and
}\mathbf{b}=(b_{1},\dots,b_{n}).$

Then $C_{f}\simeq\hat{C}\vee P^{3}(q)$ for some
$\hat{C}\in\mathscr{C}_{n,2^{s}}$ if and only if the system of mod-$q$ linear
equations

(4.4) $\left\\{\begin{array}[]{c c
l}a_{11}y_{1}+\cdots+a_{1n}y_{n}&\equiv&-b_{1}\\\ \vdots&&\\\
a_{n1}y_{1}+\cdots+a_{nn}y_{n}&\equiv&-b_{n}\\\
b_{1}y_{1}+\cdots+b_{n}y_{n}&\equiv&-c\end{array}\right.\quad{\pmod{q}}$

has a solution $(y_{1},\ldots,y_{n})\in{(\mathbb{Z}_{q})^{n}}$.

###### Proof.

Suppose $g\colon C^{\prime}\simeq\hat{C}\vee P^{3}(q)\to C_{f}$ is a homotopy
equivalence. Let $M(g)=(W,\textbf{y},z)$ and let
$M_{cup}(C^{\prime})=(A^{\prime},\mathbf{b}^{\prime},c^{\prime})$. By Lemma
3.7, we have

(4.5)
$A^{\prime}=W^{t}AW,~{}\textbf{b}^{\prime}=\textbf{y}AW+z\textbf{b}W~{}\text{
and
}~{}c^{\prime}=\textbf{y}A\textbf{y}^{t}+2z\textbf{y}\textbf{b}^{t}+z^{2}c.$

Lemma 4.7 implies $\mathbf{b}^{\prime}\equiv(0,\cdots,0)$ and
$c^{\prime}\equiv 0$ modulo $q$. Since $W$ and $z$ are invertible in
$\mathbb{Z}_{q}$, we can rewrite equations in (4.5) as

$~{}z^{-1}\textbf{y}A\equiv-\textbf{b}~{}\text{ and
}~{}z^{-1}\textbf{y}\textbf{b}^{t}\equiv-c\pmod{q}.$

Therefore, $z^{-1}\textbf{y}$ is a solution of (4.4).

Conversely, suppose there is a solution
$\mathbf{y}=(y_{1},\ldots,y_{n})\in(\mathbb{Z}_{q})^{n}$ of (4.4). By Lemma
3.9, there exist $C^{\prime\prime}\in\mathscr{C}_{n,m}$ and a homotopy
equivalence $g\colon C^{\prime\prime}\to C_{f}$ such that
$M(g)=(I,\textbf{y}^{\prime},1)$ and

$M_{cup}(C^{\prime\prime})=\Big{(}A,~{}\mathbf{y}^{\prime}\bar{A}+\mathbf{b},~{}\mathbf{y}^{\prime}\bar{A}(\mathbf{y}^{\prime})^{t}+2\mathbf{b}(\mathbf{y}^{\prime})^{t}+c\Big{)},$

where
$\textbf{y}^{\prime}=(\rho(y_{1},0),\cdots,\rho(y_{n},0))\in(\mathbb{Z}_{m})^{n}$
for $\rho$ defined in (4.2) and $\bar{A}$ is the mod-$m$ image of $A$. Then

$\mathbf{y}^{\prime}\bar{A}+\mathbf{b}\equiv 0~{}\text{ and
}~{}\mathbf{y}^{\prime}\bar{A}(\mathbf{y}^{\prime})^{t}+2\mathbf{b}(\mathbf{y}^{\prime})^{t}+c\equiv
0\pmod{q}.$

Note that Lemma 4.5 implies $C^{\prime\prime}\simeq\hat{C}\vee P^{3}(q)$ for
some $\hat{C}\in\mathscr{C}_{n,2^{s}}$. Consequently, we have
$C_{f}\simeq\hat{C}\vee P^{3}(q)$. ∎

##  5\. Odd primary local decomposition of toric orbifolds

Let $X=P\times T^{2}/_{\sim}$ be a 4-dimensional toric orbifold associated
with the combinatorial data described in Section 2. Since $X$ is simply
connected and $H^{*}(X)$ satisfies (2.6), [Hat02, Proposition 4H.3] implies
that $X$ is in $\mathscr{C}_{n,m}$ up to homotopy. Let $m=2^{s}q$, where $q$
is odd and $s\geq 0$. In this section, we show that for any odd prime $p$,
there is a $p$-local equivalence

(5.1) $X\simeq_{(p)}\hat{X}\vee P^{3}(q)$

for a CW-complex $\hat{X}$ in $\mathscr{C}_{n,2^{s}}$ and $P^{3}(q)$ denotes a
point if $q=1$.

The q-CW complex structure of $X$ with respect to a vertex $v_{i}$ (see Remark
2.2) implies that $X$ is homotoppy equivalent to the mapping cone of a map

$f\colon L_{i}\to\bigvee^{n}_{j=1}S^{2}$

where $L_{i}$ is the quotient $S^{3}/\mathbb{Z}_{m_{i,i+1}}$ and
$m_{i,i+1}=|\det\begin{bmatrix}\xi^{t}_{i},\xi^{t}_{i+1}\end{bmatrix}|$.
Recall that $\mathbb{Z}_{m_{i,i+1}}$ is isomorphic to a subgroup
$\ker\rho_{i}$ of $T^{2}$, where $\rho_{i}$ is defined in (2.2). The
$\mathbb{Z}_{m_{i,i+1}}$-action on $S^{3}$ is given by the inclusion
$\ker\rho_{i}\hookrightarrow T^{2}$ and the standard $T^{2}$-action on
$S^{3}$. If $m_{i,i+1}=1$, then $L_{i}\cong S^{3}$ and $X$ is in
$\mathscr{C}_{n,1}$. So the equivalence (5.1) holds. If $m_{i,i+1}>1$, then
$L_{i}$ is a lens space $L(m_{i,i+1};k_{i})$ for some $k_{i}$ coprime to
$m_{i,i+1}$.

In the following, the _$p$ -component_ $\nu_{p}(t)$ of a number $t$ is defined
to be the $p$-power $p^{r}$ such that $p^{r}$ divides $t$ but $p^{r+1}$ does
not.

######  Lemma 5.1.

For $p$ odd prime, let $\nu_{p}(m_{i,i+1})=p^{r}$ and let
$L_{i}=L(m_{i,i+1};k_{i})$ be a lens space. Then there is a map
$\alpha_{p}\colon\Sigma L_{i}\to S^{4}\vee P^{3}(p^{r})$ that is a $p$-local
equivalence.

###### Proof.

Let $m_{i,i+1}=p^{r}{t}$ where $p$ and $t$ are coprime. Then
$P^{3}(m_{i,i+1})\simeq P^{3}(p^{r})\vee P^{3}(t)$. Consider the diagram of
homotopy cofibration sequences

${\ast}$${P^{3}(t)}$${P^{3}(t)}$${S^{3}}$${P^{3}(m_{i,i+1})}$${\Sigma
L_{i}}$${S^{3}}$${P^{3}(p^{r})}$${C}$$\scriptstyle{\phi}$pinch$\scriptstyle{\alpha_{p}}$$\scriptstyle{\phi^{\prime}}$

where $\phi$ is the attaching map of the 4-cell in $\Sigma L_{i}$,
$\phi^{\prime}$ is the composition of $\phi$ and the pinch map, $C$ is the
mapping cone of $\phi^{\prime}$ and $\alpha_{p}$ is an induced map. The right
column induces an exact sequence

$\cdots\to\tilde{H}^{i-1}(P^{3}(t);\mathbb{Z}_{p^{r}})\to\tilde{H}^{i}(C;\mathbb{Z}_{p^{r}})\xrightarrow{\alpha_{p}^{*}}\tilde{H}^{i}(\Sigma
L_{i};\mathbb{Z}_{p^{r}})\to\tilde{H}^{i}(P^{3}(t);\mathbb{Z}_{p^{r}})\to\cdots$

Since $\tilde{H}^{*}(P^{3}(t);\mathbb{Z}_{p^{r}})=0$, the map
$\alpha_{p}^{*}\colon H^{*}(C;\mathbb{Z}_{p^{r}})\to H^{*}(\Sigma
L_{i};\mathbb{Z}_{p^{r}})$ is an isomorphism. Moreover,
$H^{*}(C;\mathbb{Z}_{p^{r}})$ has trivial cup products because $\Sigma L_{i}$
is a suspension. Now, Lemma 4.4 shows that $\phi^{\prime}$ is null homotopic,
which means $C\simeq S^{4}\vee P^{3}(p^{r})$. Therefore, we consider
$\alpha_{p}$ as a map from $\Sigma L_{i}$ to $S^{4}\vee P^{3}(p^{r})$. Since
$P^{3}(t)$ is contractible after $p$-localization, the right column implies
that $\alpha_{p}$ is a $p$-local equivalence. ∎

######  Lemma 5.2.

Let $p$ be a prime and let $H^{3}(X)\cong\mathbb{Z}_{m}$. Then there exists
$i\in\\{1,\ldots,n+2\\}$ such that $\nu_{p}(m_{i,i+1})=\nu_{p}(m)$.

###### Proof.

By [KMZ17, Corollary 5.1] and [Fis92, Lemma 3.1],
$m=\text{gcd}\\{m_{i,j}|1\leq i<j\leq n+2\\}$. If $n=1$, the lemma is trivial.
So we prove the lemma for $n\geq 2$.

Without loss of generality, suppose $\nu_{p}(m_{1,j})=\nu_{p}(m)=p^{r}$ for
some $j\in\\{3,\cdots,n+1\\}$. Let $\xi_{1}=(a,b)$. Since $a$ and $b$ are
coprime, there exist $u$ and $v$ such that

$\left(\begin{array}[]{c c}u&-b\\\ v&a\end{array}\right)\in
SL_{2}(\mathbb{Z}).$

Changing the basis of $\mathbb{Z}^{2}$ if necessary, we may assume
$\xi_{1}=(1,0)$.

Let $\xi_{2}=(x,y)$ and $\xi_{j}=(z,w)$. Then

$m_{1,2}=\left|\det\begin{pmatrix}1&x\\\ 0&y\end{pmatrix}\right|=|y|,\quad
m_{1,j}=\left|\det\begin{pmatrix}1&z\\\ 0&w\end{pmatrix}\right|=|w|.$

Write $w=cp^{r}$ and $y=c^{\prime}p^{s}$, where $c$ and $c^{\prime}$ are
integers coprime to $p$ and $s\geq r$. If $s=r$, then
$v_{p}(m_{1,2})=\nu_{p}(m)$ and consequently the lemma holds. If $s>r$, then

$\displaystyle m_{2,j}=|xw-
yz|=|cxp^{r}-c^{\prime}yp^{s}|=p^{r}|cx-c^{\prime}yp^{s-r}|.$

Since $x$ is coprime to $y$, $x$ is coprime to $p$. So $cx-c^{\prime}yp^{s-r}$
is coprime to $p$ and $\nu_{p}(m_{2,j})=p^{r}$. If $j=3$, then we are done. If
not, iterate this argument to $m_{2,j}$, $\xi_{3}$ and $\xi_{j}$. Then we can
conclude that $\nu_{p}(m_{j-1,j})=p^{r}$. ∎

For any odd prime $p$, let $\nu_{p}(m)=p^{r}$. By Lemma 5.2, there is an
$i\in\\{1,\ldots,n+2\\}$ such that $\nu_{p}(m_{i,i+1})=\nu_{p}(m)=p^{r}$. Pick
the vertex $v_{i}$ and construct the q-CW-complex structure with respect to
$v_{i}$. Then there is a homotopy cofibration sequence

$L_{i}\overset{f}{\to}\bigvee^{n}_{j=1}S^{2}\to X$

with a coaction $c:X\to X\vee\Sigma L_{i}$. Furthermore, the 3-skeleton of $X$
is $\bigvee^{n}_{j=1}S^{2}\vee P^{3}(m)$ for $m=2^{s}q$. Let $\hat{X}$ be the
quotient $X/P^{3}(q)$ and let $\phi_{p}$ be the composition

(5.2) $\phi_{p}\colon X\overset{c}{\to}X\vee\Sigma
L\xrightarrow{\jmath\vee\alpha}\hat{X}\vee P^{3}(p^{r})\vee
S^{4}\xrightarrow{\text{pinch}}\hat{X}\vee P^{3}(p^{r})$

where $\alpha$ is the map in Lemma 5.1 and $\jmath\colon X\to\hat{X}$ is the
quotient map.

######  Proposition 5.3 ($p$-local version of Main Theorem).

Let $p$ be an odd prime. If $\nu_{p}(m)=p^{r}$ for some $r\geq 1$, then
$\phi_{p}\colon X\to\hat{X}\vee P^{3}(p^{r})$ is a $p$-local equivalence.

###### Proof.

We claim that the map $\phi_{p}$ in (5.2) induces an isomorphism on
$\mathbb{Z}_{(p)}$-cohomology

(5.3) $\phi^{*}_{p}\colon H^{*}(\hat{X}\vee P^{3}(p^{r});\mathbb{Z}_{(p)})\to
H^{*}(X;\mathbb{Z}_{(p)})$

where $\mathbb{Z}_{(p)}$ is the ring of $p$-local integers.

The cofibration sequence $P^{3}(q)\hookrightarrow
X\overset{\jmath}{\to}\hat{X}$ induces an exact sequence

$\cdots\to\tilde{H}^{i-1}(P^{3}(q))\to\tilde{H}^{i}(\hat{X})\overset{\jmath^{*}}{\to}\tilde{H}^{i}(X)\to\tilde{H}^{i}(P^{3}(q))\to\tilde{H}^{i+1}(\hat{X})\to\cdots$

For $i\neq 3$, since $\tilde{H}^{i}(P^{3}(p^{r}))=0$ and $\jmath^{\ast}\colon
H^{i}(\hat{X})\to H^{i}(X)$ is an isomorphism, the map (5.3) is an
isomorphism.

Next, consider the cofibration sequence
$L_{i}\overset{f}{\to}\bigvee_{i=1}^{n}S^{2}\overset{\imath}{\to}X\overset{\delta}{\to}\Sigma
L_{i}$, where $\imath$ is the inclusion and $\delta$ is the coboundary map. It
induces an exact sequence

$\cdots\to\tilde{H}^{i}(\Sigma
L_{i};\mathbb{Z}_{(p)})\xrightarrow{\delta^{*}}\tilde{H}^{i}(X;\mathbb{Z}_{(p)})\to\tilde{H}^{i}(\underset{i=1}{\overset{n}{\bigvee}}S^{2};\mathbb{Z}_{(p)})\to\tilde{H}^{i+1}(\Sigma
L;\mathbb{Z}_{(p)})\to\cdots$

Since $\imath^{*}\colon H^{2}(X;\mathbb{Z}_{(p)})\to
H^{2}(\bigvee^{n}_{i=1}S^{2}_{i};\mathbb{Z}_{(p)})$ is an isomorphism,
$\delta^{*}\colon H^{3}(\Sigma L;\mathbb{Z}_{(p)})\to
H^{3}(X;\mathbb{Z}_{(p)})$ is an isomorphism. Consider the following
commutative diagram

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{c}$$\scriptstyle{=}$$\textstyle{X\vee\Sigma
L_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\jmath}\vee\alpha_{p}}$$\textstyle{\hat{X}\vee
P^{3}(p^{r})\vee
S^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\hat{X}\vee
P^{3}(p^{r})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\textstyle{\Sigma
L_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{p}}$$\textstyle{P^{3}(p^{r})\vee
S^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P^{3}(p^{r})}$

where the composite in the upper row is $\phi_{p}$ in (5.2) and the unnamed
arrows are pinch maps. The left square commutes due to the property of the
coaction map. By Lemma 5.1, the map $\alpha_{p}^{*}\colon
H^{3}(P^{3}(p^{r})\vee S^{4};\mathbb{Z}_{(p)})\to H^{3}(\Sigma
L;\mathbb{Z}_{(p)})$ is isomorphic, so the composite in the lower row induces
an isomorphism $H^{3}(P^{3}(p^{r});\mathbb{Z}_{(p)})\to
H^{3}(X;\mathbb{Z}_{(p)})$. Since $H^{3}(\hat{X};\mathbb{Z}_{(p)})=0$, the map
(5.3) is an isomorphism for $i=3$.

Therefore $\phi^{*}_{p}\colon H^{*}(\hat{X}\vee
P^{3}(p^{r});\mathbb{Z}_{(p)})\to H^{*}(X;\mathbb{Z}_{(p)})$ is an isomorphism
and $\phi_{p}$ is a $p$-local equivalence. ∎

######  Lemma 5.4.

Let $X$ be a 4-dimensional toric orbifold with $H^{3}(X)\cong\mathbb{Z}_{m}$,
and let $\nu_{p}(m)=p^{r}$ for some odd prime $p$ and $r\geq 1$. If
$M_{cup}(X)=(A,\textbf{b},c)$ where

$A=\left(\begin{array}[]{c c c}a_{11}&\cdots&a_{1n}\\\ \vdots&\ddots&\vdots\\\
a_{n1}&\ldots&a_{nn}\end{array}\right)\text{ and
}\mathbf{b}=(b_{1},\dots,b_{n}),$

then the system of mod-$p^{r}$ linear equations

$\left\\{\begin{array}[]{c c l}a_{11}y_{1}+\ldots+a_{1n}y_{n}&\equiv&-b_{1}\\\
\vdots&&\\\ a_{n1}y_{1}+\ldots+a_{nn}y_{n}&\equiv&-b_{n}\\\
b_{1}y_{1}+\ldots+b_{n}y_{n}&\equiv&-c\end{array}\right.\quad\pmod{p^{r}}$

has a solution in $(\mathbb{Z}_{p^{r}})^{n}$.

###### Proof.

By Proposition 5.3 there is a map $\phi_{p}\colon X\to\hat{X}\vee
P^{3}(p^{r})$ that becomes a homotopy equivalence after localized at $p$,
where $\hat{X}\in\mathscr{C}_{n,2^{s}}$ is the quotient $X/P^{3}(q)$. Let

$M(\phi_{p})=(W,\textbf{y},z)\in\text{Mat}_{n}(\mathbb{Z})\oplus(\mathbb{Z}_{m})^{n}\oplus\mathbb{Z}_{m}$

be the cellular map representation of $\phi_{p}$. After $p$-localization, $W$
is an invertible matrix and $z$ is a unit. The lemma follows from Proposition
4.8. ∎

##  6\. Proof of the Main Theorems

###### Proof of Theorem 1.2.

Let $q=p_{1}^{r_{1}}\ldots p_{k}^{r_{k}}$ be the primary factorization where
$p_{i}$’s are different odd primes and $r_{i}\geq 1$. For each prime $p_{i}$,
Lemma 5.4 implies that the mod-$p_{i}^{r_{i}}$ version of (4.4) has a
solution. By Chinese Remainder Theorem, they give a mod-$q$ solution for
(4.4). By Proposition 4.8, $X$ is homotopy equivalent to $\hat{X}\vee
P^{3}(q)$. ∎

###### Proof of Theorem 1.3.

The “only if” part is trivial. To prove the “if” part, let $X$ and
$X^{\prime}$ be 4-dimensional toric orbifolds such that
$H^{3}(X)\cong\mathbb{Z}_{m}$ and
$H^{3}(X^{\prime})\cong\mathbb{Z}_{m^{\prime}}$ for $m$ and $m^{\prime}$ odd.
The hypothesis implies that $H^{3}(X)\cong H^{3}(X^{\prime})$, hence we have
$m=m^{\prime}$. By Theorem 1.2, we have $X\simeq\hat{X}\vee P^{3}(m)$ and
$X^{\prime}\simeq\hat{X}^{\prime}\vee P^{3}(m)$ for some
$\hat{X},\hat{X}^{\prime}\in\mathscr{C}_{n,1}$. Since $H^{i}(X)\cong
H^{i}(\hat{X})$ and $H^{i}(X^{\prime})\cong H^{i}(\hat{X}^{\prime})$ for
$i\neq 3$, we have $H^{*}(\hat{X})\cong H^{*}(\hat{X}^{\prime})$. Then
Proposition 4.2 implies that $\hat{X}\simeq\hat{X}^{\prime}$, which yields
$X\simeq X^{\prime}$. ∎

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