Connectedness of Certain Graph Coloring Complexes

TL;DR

This paper investigates the connectedness properties of certain graph coloring complexes, specifically for bipartite graphs formed by the product of K2 and K_n, providing explicit formulas for their connectedness.

Contribution

It establishes exact connectedness values for Hom complexes of bipartite graphs with complete graphs, extending understanding of their topological structure.

Findings

Connectedness of Hom complexes is m-n-1 for m ≥ n

Connectedness is m-3 in other cases

For bipartite graphs K2 × K_n, connectedness is m - d - 2, where d is the maximum degree

Abstract

In this article, we consider the bipartite graphs $K_2 \times K_n$. We prove that the connectedness of the complex $\displaystyle \text{Hom}(K_2\times K_{n}, K_m) $ is $m-n-1$ if $m \geq n$ and $m-3$ in the other cases. Therefore, we show that for this class of graphs, $\text{Hom} (G, K_m)$ is exactly $m-d-2$ connected, $m \geq n$, where $d$ is the maximal degree of the graph $G$.

Equations

$$(v,w)\in E(G)⟹(\varphi (v),\varphi (w))\in E(H).$$

$$f(x_1)\succ x_1\prec f(x_2)\succ x_2\prec f(x_3)\succ x_3\dots f(x_t)\succ x_t\prec f(x_1),t\ge 2.$$

$$a=\left\{\begin{array}[]{@{}c@{\quad}l@{}}\text{min}\{x,z_{r}\}&\text{if}~{}~{}x\in X_{r}^{\tau}\setminus\bigcup\limits_{r\neq j}X_{j}^{\tau}\\ x&\text{if}~{}~{}x\in X_{i}^{\tau}\cap X_{j}^{\tau},i\neq j.\\ \end{array}\right.$$

$$w(c):=(-1{)}^t[\tau :\sigma ][\mu (y_t):\sigma ]\prod _{i=1}^t[\mu (y_i):y_i]\prod _{i=1}^{t-1}[\mu (y_i):y_{i+1}]$$

Full text

Keywords : Hom complexes, Exponential graphs, Discrete Morse theory.

2000 Mathematics Subject Classification primary 05C15, secondary 57M15

1 Introduction

A classical problem in graph theory is the determination of the chromatic number of a graph which finds application in several fields. In 1955, Kneser posed the Kneser conjecture which dealt with the computation of the chromatic number of a certain class of graphs [11]. This conjecture was proved in 1975 by Lovaśz [15]. Lovaśz constructed a simplicial complex called the neighborhood complex $\mathcal{N}(G)$ of a graph $G$ to prove the Kneser conjecture.

In [1], Lovaśz generalised the neighborhood complex by the prodsimplicial complex called the Hom complex, denoted by $\text{Hom}(G,H)$ for graphs $G$ and $H$. In particular, $\text{Hom}(K_{2},G)$ (where $K_{2}$ denotes a complete graph with $2$ vertices) and $\mathcal{N}(G)$ are homotopy equivalent. The idea was to be able to estimate the chromatic number of an arbitrary graph $G$ by understanding the connectivity of the Hom complex from some standard graph into $G$. If $H$ is the complete graph $K_{n}$, then the complexes $\text{Hom}(G,K_{n})$ are highly connected. These complexes $\text{Hom}(G,K_{n})$ can be considered to be the spaces of $n$- colorings of the graph $G$. The complexes $\text{Hom}(G,H)$ are not very well understood.

Babson and Kozlov made the following conjecture in [1].

Conjecture 1.1**.**

For a graph $G$ with maximal degree $d$, $\text{Hom}(G,K_{n})$ is at least $n-d-2$ connected.

Čukić and Kozlov proved this conjecture in [4]. They also proved that for those cases where $G$ is an odd cycle, $\text{Hom}(G,K_{n})$ is $n-4$ connected for all $n\geq 3.$ In [13], it is proved that for even cycles $C_{2m}$, $\text{Hom}(C_{2m},K_{n})$ is $n-4$ connected for all $n\geq 3.$ In [17], Malen proved a strong generalisation of [4], by showing that $\text{Hom}(G,K_{n})$ is at least $m-D-2$ connected, where $D=\max\limits_{H\subset G}\delta(H)$, $\delta(H)$ being the minimal degree of the graph $H$, and the summation is taken over all the induced subgraphs $H$ of $G$. Since $D\leq d$, this also implies Conjecture 1.1.

It is natural to ask whether it is possible to classify the class of graphs $G$ for which the Hom complexes $\text{Hom}(G,K_{n})$ are exactly $n-D-2$ connected. In this article, we consider the bipartite graphs $K_{2}\times K_{n}$, which are $n-1$ regular graphs and $D=d=n-1$. Since $\displaystyle\text{Hom}(K_{2}\times K_{n},K_{m})\simeq\text{Hom}(K_{2},K_{m}^{K_{n}})$, where $\displaystyle K_{m}^{K_{n}}$ is an exponential graph, it is sufficient to determine the connectedness of $\text{Hom}(K_{2},K_{m}^{K_{n}})$ which is the same as the connectedness of the neighborhood complex $\mathcal{N}(K_{m}^{K_{n}})$. The main result of this article is

Theorem 1.1**.**

Let $m,n\geq 2$. Then

\text{conn}(\text{Hom}(K_{2}\times K_{n},K_{m}))=\begin{cases}m-n-1&\text{if \hskip 8.5359ptm\geq n}\\ m-3&\text{otherwise}.\\ \end{cases}**

In the special case, where $m=n+1$, in [18], it is shown that

Theorem 1.2**.**

Let $n\geq 4$ and $p$ = $\frac{n!(n-1)n}{2}$. Then

H_{k}(\mathcal{N}(K_{n+1}^{K_{n}});\mathbb{Z}_{2})=\begin{cases}\ Z_{2}&\text{if \hskip 8.5359ptk=0,1$or$n-1}\\ \ Z_{2}^{p-n!+1}&\text{ if \hskip 8.5359ptk=2}\\ 0&\text{otherwise}.\\ \end{cases}**

2 Preliminaries

A graph $G$ is a pair $(V(G),E(G))$, where $V(G)$ is the set of vertices of $G$ and $E(G)\subset V(G)\times V(G)$ the set of edges. If $(x,y)\in E(G)$, it is also denoted by $x\sim y$. Here, $x$ is said to be adjacent to $y$. The degree of a vertex $v$ is defined by $\text{deg}(v)=|\{y\in V(G)\ |\ x\sim y\}|$. Here $|X|$ represents the cardinality of the set $X$.

A bipartite graph is a graph $G$ with subsets $X$ and $Y$ of $V(G)$ such that $V(G)=X\sqcup Y$ and $(v,w)\notin E(G)$ if $\{v,w\}\subseteq X$ or $\{v,w\}\subseteq Y$. Examples of bipartite graphs include the even cycles $C_{2n}$ where $V(C_{2n})=\{1,2,\dots,2n\}$ and $E(C_{2n})=\{(i,i+1)\ |\ 1\leq i\leq 2n-1\}\cup\{(1,2n)\}$. In this case $X=\{1,3,5,\ldots,2n-1\}$ and $Y=\{2,4,6,$ $\dots,2n\}$.

A graph homomorphism from a graph $G$ to graph $H$ is a function $\phi:V(G)\to V(H)$ such that,

[TABLE]

A finite abstract simplicial complex X is a collection of finite sets such that if $\tau\in X$ and $\sigma\subset\tau$, then $\sigma\in X$. The elements of $X$ are called simplices of $X$. If $\sigma\in X$ and $|\sigma|=k+1$, then $\sigma$ is said to be $k-dimensional$. A $k-1$ dimensional subset of a $k$ simplex $\sigma$ is called a facet of $\sigma$. A prodsimplicial complex is a polyhedral complex each of whose cells is a direct product of simplices ([14]).

Let $v$ be a vertex of the graph $G$. The *neighborhood of $v$ * is defined by $N(v)=\{w\in V(G)\ |\ (v,w)\in E(G)\}$. If $A\subset V(G)$, the neighborhood of $A$ is defined as $N(A)=\{x\in V(G)\ |\ (x,a)\in E(G)\,\,\forall\,\,a\in A\}$.

The neighborhood complex, $\mathcal{N}(G)$ of a graph $G$ is the abstract simplicial complex whose vertices are all the non isolated vertices of $G$ and whose simplices are those subsets of $V(G)$ which have a common neighbor.

Consider distinct vertices $u$ and $v$ in $G$ such that $N(u)\subset N(v)$. The subgraph $G\setminus\{u\}$ of $G$, where $V(G\setminus\{u\})=V(G)\setminus\{u\}$ and $E(G\setminus\{u\})=E(G)\setminus\{(v,w)\in E(G)\ |\ \text{at least one of}\,v\,\text{or}\,w\,\text{is}\,u\}$, is called a fold of $G$.

For any two graphs $G$ and $H$, $\text{Hom}(G,H)$ is the polyhedral complex whose cells are indexed by all functions $\eta:V(G)\to 2^{V(H)}\setminus\{\emptyset\}$, such that if $(v,w)\in V(G)$, then $\eta(v)\times\eta(w)\subset E(H)$.

Elements of $\text{Hom}(G,H)$ are called cells and are denoted by $(\eta(v_{1}),\dots,$ $\eta(v_{k}))$, where $V(G)=\{v_{1},\dots,v_{k}\}$. A cell $(A_{1},\dots,A_{k})$ is called a face of $B=(B_{1},\dots B_{k})$, if $A_{i}\subset B_{i}$ $\forall\,1\leq i\leq k$. The Hom complex is often referred to as a topological space. Here, we are referring to the geometric realisation of the order complex of the poset. The simplicial complex whose facets are the chains of the Poset $P$ is called the order complex of $P$.

A topological space $X$ is said to be $n$-connected if $\pi_{\ast}(X)=0$ for all $\ast\leq n$. By convention, $\pi_{0}(X)=0$ means $X$ is connected. The connectivity of a topological space $X$ is denoted by $\text{conn}(X)$, i.e., $\text{conn}(X)$ is the largest integer $m$ such that $X$ is $m$-connected. If $X$ is a non empty, disconnected space, it is said to be $-1$ connected and if it is empty, it is said to be $-\infty$ connected.

We now review some of the constructions related to the existence of an internal hom which is related to the categorical product. All these details can be found in [8, 9, 16].

• •

The categorical product of two graphs $G$ and $H$, denoted by $G\times H$ is the graph where $V(G\times H)=V(G)\times V(H)$ and $(g,h)\sim(g^{\prime},h^{\prime})$ in $G\times H$ if $g\sim g^{\prime}$ and $h\sim h^{\prime}$ in $G$ and $H$ respectively.

• •

If $G$ and $H$ are two graphs, then the exponential graph $\displaystyle H^{G}$ is defined to be the graph where $\displaystyle V(H^{G})$ contains all the set maps from $V(G)$ to $V(H)$. Any two vertices $f$ and $f^{\prime}$ in $\displaystyle V(H^{G})$ are said to be adjacent, if $v\sim v^{\prime}$ in $G$ implies that $f(v)\sim f^{\prime}(v^{\prime})$ in $H$.

Using tools from poset topology ([3]), it can be shown that given a poset $P$ and a poset map $c:P\rightarrow P$ such that $c\circ c=c$ and $c(x)\geq x\,\forall\,x\in P$, there is a strong deformation retract induced by $c:P\rightarrow c(P)$ on the relevant spaces. Here, $c$ is called the closure map.

From [5, Proposition 3.5] we have a relationship between the exponential graph and the categorical product in the Hom-complex.

Proposition 2.1**.**

Let $G$, $H$ and $K$ be graphs. Then $\displaystyle\text{Hom}(G\times H,K)$ can be included in $\displaystyle\text{Hom}(G,K^{H})$ so that $\displaystyle\text{Hom}(G\times H,K)$ is the image of the closure map on $\displaystyle\text{Hom}(G,K^{H})$. In particular, there is a strong deformation retract $\displaystyle|\text{Hom}(G\times H,K)|\hookrightarrow|\text{Hom}(G,K^{H})|$.

From [1, Proposition 5.1] we have the following result which allows us to replace a graph by its fold in the Hom complex.

Proposition 2.2**.**

Let $G$ and $H$ be graphs such that $u,v$ are distinct vertices of $G$ and $N(u)\subset N(v)$. The inclusion $i:G\setminus\{u\}\hookrightarrow G$ respectively, the homomorphism $\phi:G\rightarrow G\setminus\{u\}$ which maps $u$ to $v$ and fixes all the other vertices, induces the homotopy equivalence $i_{H}:\text{Hom}(G,H)\rightarrow\text{Hom}(G\setminus\{u\},H)$,‪ respectively $\phi_{H}:\text{Hom}(G\setminus\{u\},H)\rightarrow\text{Hom}(G,H)$.

3 Tools from Discrete Morse Theory

We introduce some tools from Discrete Morse Theory which have been used in this article. R. Forman in [6] introduced what has now become a standard tool in Topological Combinatorics, Discrete Morse Theory. The principal idea of Discrete Morse Theory (simplicial) is to pair simplices in a complex in such a way that they can be cancelled by elementary collapses. This will reduce the original complex to a homotopy equivalent complex, which is not necessarily simplicial, but which has fewer cells. More details of discrete Morse theory can be found in [10] and [14].

Definition 3.1**.**

A partial matching in a Poset $P$ is a subset $\mathcal{M}$ of $P\times P$ such that

• •

$(a,b)\in\mathcal{M}$ implies $b\succ a$, i.e. $a<b$ and $\not\exists\,c$ such that $a<c<b$.

• •

Each element in $P$ belongs to at most one element in $\mathcal{M}$.

In other words, if $\mathcal{M}$ is a partial matching on a Poset $P$ then, there exists $A\subset P$ and an injective map $f:A\rightarrow P\setminus A$ such that $f(x)\succ x$ for all $x\in A$.

Definition 3.2**.**

An acyclic matching is a partial matching $\mathcal{M}$ on the Poset $P$ such that there does not exist a cycle

[TABLE]

If we have an acyclic partial matching on $P$, those elements of $P$ which do not belong to the matching are said to be *critical *. To obtain the desired homotopy equivalence, the following result is used.

Theorem 3.1**.**

*(Main theorem of Discrete Morse Theory)[6] *

Let $X$ be a simplicial complex and let $\mathcal{A}$ be an acyclic matching such that the empty set is not critical. Then, $X$ is homotopy equivalent to a cell complex which has a $d$ -dimensional cell for each $d$ -dimensional critical face of $X$ together with an additional [math]-cell.

4 Main Result

To prove Theorem 1.1, we first construct an acyclic matching on the face poset of $\mathcal{N}(K_{m}^{K_{n}})$. We then consider the corresponding Morse Complex to determine the connectedness of $\text{Hom}(K_{2}\times K_{n},K_{m})$. In this article $n\geq 3,m-n=p\geq 1$ and $[k]$ denotes the set $\{1,2,\dots,k\}$. Any vertex in the exponential graph $K_{m}^{K_{n}}$ is a set map $f:[n]\rightarrow[m]$.

Lemma 4.1**.**

The graph $K_{m}^{K_{n}}$ can be folded onto the graph $G$, where the vertices $f\in V(G)$ have images of cardinality either 1 or $n$.

Proof.

Consider the vertex $f$ such that $1<|$Im $f|$ $<n$. Since $f$ is not injective there exist distinct $i,j\in[n]$ such that $f(i)=f(j)=\alpha$. Consider $\tilde{f}\in V(K_{m}^{K_{n}})$ such that $\tilde{f}([n])=\alpha$. By the definition of the exponential graph, any neighbor $h$ of $f$ will not have $\alpha$ in its image and therefore $h$ will be a neighbor of $\tilde{f}$ thereby showing that $N(f)\subseteq N(\tilde{f})$. $K_{m}^{K_{n}}$ can be folded to the subgraph $K_{m}^{K_{n}}\setminus\{f\}$ of $K_{m}^{K_{n}}$. Repeating the above argument for all noninjective, non constant maps from $[n]$ to $[m]$, $K_{m}^{K_{n}}$ can be folded to the graph $G$ whose vertices are either constant maps or injective maps from $[n]$ to $[m]$. ∎

From Proposition 2.2, we observe that $\mathcal{N}(K_{m}^{K_{n}})\simeq\mathcal{N}(G)$. Hence, it is sufficient to study the homotopy type of $\mathcal{N}(G)$.

We now fix the following notations.

If $f\in V(G)$ and $f([n])=\{x\}$, then $f$ is denoted by $<x>$. In the other cases the string $a_{1}a_{2}\dots a_{n}$ denotes the vertex $f$ where $a_{i}=f(i)$, $1\leq i\leq n.$ Hence, if the notation $a_{1}a_{2}\dots a_{n}$ is used, it is understood that for $1\leq i<j\leq n$, $a_{i}\neq a_{j}$. Let $\sigma=\{f_{1},\ldots,f_{q}\}\subset V(G)$. Then $X_{i}^{\sigma}=\{f_{j}(i)\ |\ 1\leq j\leq q\}$, $A_{i}^{\sigma}=[m]\setminus\bigcup\limits_{j\neq i}X_{j}^{\sigma}$ and $X_{\sigma}=\bigcup\limits_{i=1}^{n}X_{i}^{\sigma}$.

4.1 Construction of an acyclic matching

Let $(P,\subset)$ be the face poset of $\mathcal{N}(G)$, $S_{1}=\{\sigma\in P\ |<1>\ \notin\sigma,\sigma\,\cup<1>\ \in P\}$ and the map $\mu_{1}:S_{1}\rightarrow P\setminus S_{1}$ be defined by $\mu_{1}(\sigma)=\sigma\,\cup<1>$.

Let $S_{1}^{\prime}=P\setminus\{S_{1},\mu_{1}(S_{1})\}$ and for $2\leq i\leq m,$ define $S_{i}=\{\sigma\in S_{i-1}^{\prime}\ |<i>\ \notin\sigma,\sigma\,\cup<i>\ \in S_{i-1}^{\prime}\}$ and $\mu_{i}:S_{i}\rightarrow S_{i-1}^{\prime}\setminus S_{i}$ by $\mu_{i}(\sigma)=\sigma\,\cup<i>,$ where $S_{i}^{\prime}=S_{i-1}^{\prime}\setminus\{S_{i},\mu_{i}(S_{i})\}.$ Since $S_{i}\cap S_{j}=\emptyset\ \forall\ 1\leq i\neq j\leq m$, if $\sigma\in S=\bigcup\limits_{i=1}^{m}S_{i}$, then $\sigma$ belongs to exactly one $S_{i}.$ Define $\mu:S\rightarrow P\setminus S$ by $\mu(\sigma)=\mu_{i}(\sigma)$ where $\sigma\in S_{i}$. By the construction, $\mu$ is an injective map and hence $\mu$ is a well defined partial matching on the poset $P$.

Proposition 4.2**.**

$\mu$* is an acyclic matching on $P$.*

Proof.

Assume that $\mu$ is not an acyclic matching, i.e., there exist distinct simplices $\sigma_{1},\ldots,$ $\sigma_{t}$ in $S$ such that $\mu(\sigma_{i})\succ\sigma_{i+1(\text{mod}\ t)},1\leq i\leq t$.

Claim 1**.**

$\sigma_{i+1(\text{mod}\ t)}=\mu(\sigma_{i})\setminus\{f\}$, where $|\text{Im}\,f|=1$.

$\mu(\sigma_{i})\succ\sigma_{i+1(\text{mod}\ t)}$ implies that $\sigma_{i+1}=\mu(\sigma_{i})\setminus\{f\}$, for some $f\in V(G)$. If $|\text{Im}\,f|=n$, then $f\notin\sigma_{i+1(\text{mod}\ t)}\Rightarrow f\notin\mu(\sigma_{i+1(\text{mod}\ t)})\Rightarrow f\notin\sigma_{i+2(\text{mod}\ t)}$. Hence, $f\notin\mu(\sigma_{i})$, a contradiction. So $f=\ <j>$ for some $j\in\{1,\ldots,m\}$.

Let $k\in\{1,\ldots,m\}$ be the least integer such that $\{\sigma_{1},\ldots,\sigma_{t}\}\cap S_{k}\neq\emptyset$. Without loss of generality assume that $\sigma_{1}\in S_{k}$, i.e. $<k>\ \notin\sigma_{1}$ and $\mu(\sigma_{1})=\sigma_{1}\cup<k>$. Since $\sigma_{1}\neq\sigma_{2}$, from Claim 1, $\sigma_{2}=\mu(\sigma_{1})\setminus\{<i>\}$, $i\neq k$. Therefore, $<k>\in\sigma_{2}$. If $<k>\ \in\sigma_{i},2\leq i\leq t,$ then $<k>\ \in\mu(\sigma_{t})$. $\mu(\sigma_{1})\setminus\{<k>\}=\sigma_{1}=\mu(\sigma_{t})\setminus\{<k>\}\Rightarrow\mu(\sigma_{1})=\mu(\sigma_{t})$, a contradiction (since $\sigma_{1}\neq\sigma_{t}$ and $\mu$ is injective).

Let $l\in[m]$ be such that $<k>\ \notin\sigma_{l}$ and $<k>\ \in\sigma_{j},2\leq j<l$. Here, $<k>\ \in\mu(\sigma_{l-1})$ and therefore, $\sigma_{l}=\mu(\sigma_{l-1})\setminus\{<k>\}$. Since $\sigma_{l},\mu(\sigma_{l-1})\notin S_{j},\mu_{j}(S_{j})$ $\forall\,j<k$, we see that $\sigma_{l},\mu(\sigma_{l-1})\in S_{k-1}^{\prime}$, which implies that $\sigma_{l}\in S_{k}$ (since $\mu(\sigma_{l-1})=\sigma_{l}\cup<k>$). Here, $\sigma_{l}\neq\sigma_{l-1}$ and $\mu(\sigma_{l})=\mu(\sigma_{l-1})$, a contradiction. Thus our assumption that $\mu(\sigma_{i})\succ\sigma_{i+1(\text{mod}\ t)},1\leq i\leq t$ is incorrect i.e. there exists no such cycle in $P$. Hence, $\mu$ is an acyclic matching.

∎

4.2 Critical Cells

A subset $\sigma$ of $V(G)$ is a simplex in $\mathcal{N}(G)$ if and only if $A_{i}^{\sigma}\neq\emptyset\ \forall\ i\in[n]$. In this section, we characterize the critical cells of $\mathcal{N}(G)$. If there exists $z_{1}\ldots z_{n}\in N(\sigma)$ such that $1\notin\{z_{1},\ldots,z_{n}\}$, then $\sigma\,\cup<1>\ \in\mathcal{N}(G)$, which implies that $\sigma$ is not a critical cell. Further, if $\sigma\in N(<x>)$ for some $x\neq 1$, then $<1>\ \in N(\sigma)$ and $\sigma\in S_{1}$. Therefore, for $\sigma$ to be a critical cell, either $N(\sigma)=\ <1>$ or $\sigma\subset N(z_{1}\ldots z_{n})$ with $1\in\{z_{1},z_{2},\dots z_{n}\}$.

Case 1. $\sigma\subset N(z_{1}\ldots z_{n})$, where $z_{k}=1$ for some $k\in\{1,2,\dots,n\}$.

In this case, clearly, $\sigma=\{<x_{1}>,\ldots,<x_{l}>\}\ \cup\tau$, where $\tau=\{f_{1},\ldots,f_{q}\}$ with $|\text{Im}\,f_{i}|=n$ $\forall\ i\in\{1,\dots,q\}$ and $x_{1},\dots x_{l}\in[m]\setminus\{z_{1},\ldots,z_{n}\}$. Since $\sigma\sim z_{1}\ldots z_{n}$ and $<z_{i}>$ is not a neighbor of $z_{1}\ldots z_{n}$ for all $i\in[n]$ we see that $<z_{i}>\ \notin\sigma$. If $z_{r}\in X_{\sigma}$ for some $r\in[n]$, then $z_{r}\in X_{\tau}$, which implies that there exists $f\in\tau$ such that $z_{r}\in\text{Im}\,f$. Since $f\sim z_{1}\ldots z_{n}$, $f(i)\neq z_{r}\ \forall\ i\neq r$, which implies that $f(r)=z_{r}$.

Proposition 4.3**.**

For $x\in[m]$ with $x\neq z_{i}$, $\forall~{}i\in[n]$ and $<x>\ \notin\sigma$, define

[TABLE]

Let $\eta$ be the set $\{\sigma\,\cup<a_{1}>,\ldots,<a_{s}>\}\setminus\{<b_{1}>,\ldots,<b_{t}>\},s,t\geq 0,a_{i},b_{j}<a,1\leq i\leq s,1\leq j\leq t$. Then

• (i)

$\eta\in\mathcal{N}(G)\Longleftrightarrow\eta\,\cup<a>\ \in\mathcal{N}(G).$**

• (ii)

$\eta\in S_{1}\Longleftrightarrow\eta\,\cup<a>\ \in S_{1}.$**

Proof.

If $\eta\,\cup<a>\ \in\mathcal{N}(G)\ \text{or}\ S_{1}$, then $\eta\in\mathcal{N}(G)\ \text{or}\ S_{1}$, respectively.

Case (i) $a=\text{min}\{x,z_{r}\}$.

Since $a\leq x,z_{r}\in X_{r}^{\tau}\setminus\bigcup\limits_{r\neq j}X_{j}^{\tau}$ and $<x>,<z_{r}>\ \notin\sigma$, we observe that $x,z_{r}\in X_{r}^{\eta}\setminus\bigcup\limits_{r\neq j}X_{j}^{\eta}$. Any neighbor $f$ of $\eta$ has the property that $f(i)\neq x,z_{r}$ (i.e. $f(i)\neq a$) if $i\neq r$. If $<y>\ \sim\eta,$ i.e. $\eta\in\mathcal{N}(G)$, then $y\neq a$ and $<y>\ \sim\ <a>$ implying that $<y>\ \sim\eta\,\cup<a>$ and thus $\eta\,\cup<a>\ \in\mathcal{N}(G)$.

If $\eta\in S_{1}$ and $<y>\ \sim\eta\,\cup<1>$, then $<y>\ \sim\ <a>$ and $<1>$. Therefore $\eta\,\cup<a>\cup<1>\ \in\mathcal{N}(G)$ and $\eta\,\cup<a>\ \in S_{1}.$

Let $x_{r}=\text{max}\{x,z_{r}\}$. Clearly, $a,x_{r}\in X_{r}^{\tau}\setminus\bigcup\limits_{i\neq r}X_{i}^{\tau}$. For any neighbor $y=y_{1}\ldots y_{n}$ of $\eta$, $y_{i}\neq x_{r},a\ \forall\ i\neq r$ and so $y_{1}\ldots y_{r-1}x_{r}y_{r+1}\ldots y_{n}\sim\eta$ and $<a>$. Therefore $\eta\,\cup<a>\ \in\mathcal{N}(G)$ if $\eta\in\mathcal{N}(G)$.

If $\eta\in S_{1}$ and $y\sim\eta$, then $y_{i}\neq 1\ \forall\ i\in[n]$. Since $x_{r}\neq 1,y_{1}\ldots y_{r-1}x_{r}y_{r+1}$ $\ldots y_{n}\sim\eta,<a>$ and $<1>$. Thus, $\eta\,\cup<a>\ \in S_{1}$.

Case (ii) $x=a\in X_{i}^{\tau}\cap X_{j}^{\tau},i\neq j$.

In this case, $a\neq z_{t}$ and $a\notin A_{t}^{\eta}\ \forall\ t\in[n]$. Since $A_{t}^{\eta\cup<a>}=A_{t}^{\eta}\setminus\{a\}$ and $a\notin A_{t}^{\eta}$, we get $A_{t}^{\eta\cup<a>}=A_{t}^{\eta}$. If $f\in N(\eta)$, then $\forall\ t\in[n]$, $f(t)\in A_{t}^{\eta}=A_{t}^{\eta\cup<a>}$ implying that $f\in N(\eta\,\cup<a>)$. $N(\eta\,\cup<a>)\subset N(\eta)$ always, and therefore, $N(\eta)=N(\eta\,\cup<a>)$. $(i)$ and $(ii)$ now follow. ∎

Proposition 4.4**.**

Let $a$ be as defined in Proposition 4.3, $\sigma\notin S_{t}\cup\mu_{t}(S_{t})$ $\forall\ t<a$ and $\sigma\,\cup<a>\ \in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$, $i_{1}<a$. Then, there exists $s\in\{2,\dots,m\}$ and a cell

$\xi_{s}=\sigma\cup\begin{cases}\{<a>,<b_{1}>,\ldots,<b_{r}>\}\setminus\{<l_{1}>,\ldots,<l_{t}>\}&\text{if}~{}\text{s}~{}\text{is}~{}\text{even}\\ \{<b_{1}>,\ldots,<b_{r}>\}\setminus\{<l_{1}>,\ldots,<l_{t}>\}&\text{if}~{}\text{s}~{}\text{is}~{}\text{odd},\end{cases}$**

where $r,t\geq 0$ and $b_{1},\ldots,b_{r},l_{1},\ldots,l_{t}<a$ such that $\xi_{s}\in S_{1}\cup\mu_{1}(S_{1})$ but $\xi_{s}\setminus\{<a>\}$ (if $<a>\ \in\xi_{s}$) or $\xi_{s}\,\cup<a>$ ( if $<a>\ \notin\xi_{s}$) $\notin S_{1}\cup\mu_{1}(S_{1})$.

Proof.

Clearly, $i_{1}>1$ as $\sigma\,\cup<a>\ \in S_{1}\cup\mu_{1}(S_{1})$ implies that $\sigma\in S_{1}\cup\mu_{1}(S_{1})$, a contradiction. Define the cell

$\xi_{1}=\begin{cases}\sigma\setminus\{<i_{1}>\}&\text{if}~{}~{}<i_{1}>\ \in\sigma\\ \sigma\,\cup<i_{1}>&\text{if}~{}~{}<i_{1}>\ \notin\sigma.\\ \end{cases}$

Since, $\sigma\ \cup<a>\ \in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$, $\xi_{1}\ \cup<a>$ is always a simplex, thereby showing that $\xi_{1}\in\mathcal{N}(G)$. If $\xi_{1}\in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$, i.e. $\xi_{1}\notin S_{j}\cup\mu_{j}(S_{j})$ $\forall\ j<i_{1}$, then $\sigma$ will belong to $S_{i_{1}}\,\cup\mu_{i_{1}}(S_{i_{1}})$, which contradicts the hypothesis. Therefore, $\exists\ i_{2}<i_{1}$ such that $\xi_{1}\in S_{i_{2}}\cup\mu_{i_{2}}(S_{i_{2}})$. By Proposition 4.3(i), $\xi_{1}\,\cup<a>$, which is $\{\sigma\,\cup<a>\}\setminus\{<i_{1}>\}$ or $\sigma\,\cup<a>\cup<i_{1}>$, belongs to $S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}}).$ So, if $i_{2}=1$ the result holds. Let $i_{2}>1$. Define the cell

$\xi_{2}=\begin{cases}\{\xi_{1}\,\cup<a>\}\setminus\{<i_{2}>\}&\text{if}~{}~{}<i_{2}>\ \in\sigma\\ \xi_{1}\,\cup<a>\cup<i_{2}>&\text{if}~{}~{}<i_{2}>\ \notin\sigma.\end{cases}$

By Proposition 4.3, $\xi_{2}\in\mathcal{N}(G)$. Since $\xi_{1}\,\cup<a>\ \notin S_{i_{2}}\cup\mu_{i_{2}}(S_{i_{2}})$, there exists $i_{3}<i_{2}$ such that $\xi_{2}\in S_{i_{3}}\cup\mu_{i_{3}}(S_{i_{3}})$. Here, $\xi_{2}\setminus\{<a>\}$ which is $\xi_{1}\setminus\{<i_{2}>\}$ or $\xi_{1}\,\cup<i_{2}>\ \in S_{i_{2}}\cup\mu_{i_{2}}(S_{i_{2}})$, which implies that the result holds if $i_{3}=1$. Inductively, assume that there exists $l,1<l<m$, where $i_{l+1}<i_{l}<\ldots<i_{1}<a$ and $\xi_{t}\in S_{i_{t+1}}\cup\mu_{i_{t+1}}(S_{i_{t+1}})$, $1\leq t\leq l$ such that

$\xi_{t}=\begin{cases}\{\xi_{t-1}\,\cup<a>\}\setminus\{<i_{t}>\}&\text{if}~{}<i_{t}>\ \in\xi_{t-1},~{}t~{}\text{is}~{}\text{even}\\ \xi_{t-1}\,\cup<a>\cup<i_{t}>&\text{if}~{}~{}<i_{t}>\ \notin\xi_{t-1},~{}t~{}\text{is}~{}\text{even}\\ \xi_{t-1}\setminus\{<a>,<i_{t}>\}&\text{if}~{}~{}<i_{t}>\ \in\xi_{t-1},~{}t~{}\text{is}~{}\text{odd}\\ \{\xi_{t-1}\,\cup<i_{t}>\}\setminus\{<a>\}&\text{if}~{}~{}<i_{t}>\ \notin\xi_{t-1},~{}t~{}\text{is}~{}\text{odd}.\end{cases}$

Since $\xi_{l}\setminus\{<a>\}(\text{or}~{}~{}\xi_{l}\cup<a>)=\xi_{l-1}\setminus\{<i_{l}>\}$ or $\xi_{l-1}\,\cup<i_{l}>\ \in S_{i_{l}}\cup\mu_{i_{l}}(S_{i_{l}})$ and $i_{l}>i_{l+1}$, the result holds if $i_{l+1}=1$. If $i_{l+1}>1$, by induction the result follows.

∎

We first get some necessary conditions for $\sigma$ to be a critical cell.

Lemma 4.5**.**

Let $\sigma$ be a critical cell. Then

• (i)

$X_{\sigma}=[m]\ \text{or}\ [m]\setminus\{1\}.$**

• (ii)

$x\in[m]\setminus\{z_{1},\ldots,z_{n}\}\Rightarrow\ <x>\ \in\sigma$.

Proof.

A critical cell $\sigma$ does not belong to $S_{i}\cup\mu_{i}(S_{i})$ for all $i\in\{1,\ldots,m\}.$

• (i)

If $x\in[m]\setminus\{1\}$ such that $x\notin X_{\sigma}$, then $\sigma,<1>\ \sim\ <x>$, thereby implying that $\sigma\in S_{1}$. Hence $[m]\setminus\{1\}\subset X_{\sigma}$.

• (ii)

Let $x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$ such that $<x>\ \notin\sigma$. Since $z_{1}\ldots z_{n}$ is a neighbor of both $<x>$ and $\sigma$, we see that $\sigma\,\cup<x>\ \in\mathcal{N}(G)$. $x\neq 1\in X_{\sigma},<x>\ \notin\sigma$ implies that there exists $i\in[n]$ such that $x\in X_{i}^{\tau}$. If $x\in X_{k}^{\tau}\setminus\bigcup\limits_{j\neq k}X_{j}^{\tau}$, then $\sigma\sim z_{1}\ldots z_{k-1}xz_{k+1}\ldots z_{n}$, which implies that $\sigma\in S_{1}$. Thus $x\notin X_{k}^{\tau}\setminus\bigcup\limits_{j\neq k}X_{j}^{\tau}$. If $a$ is as defined in Proposition 4.3, then $\sigma\,\cup<a>\ \in\mathcal{N}(G)$ and $\xi_{s}\in S_{1}\Rightarrow\xi_{s}\,\cup<a>$ or $\xi_{s}\setminus\{<a>\}\in S_{1}$. This contradicts Proposition 4.4 and thus, $<x>\ \in\sigma$ if $x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$.

∎

Lemma 4.6**.**

Let $\sigma$ be a critical cell. Then

• (i)

$X_{r}^{\tau}\cap X_{s}^{\tau}=\emptyset$* if $r\neq s$.*

• (ii)

$x\in X_{i}^{\tau}\Rightarrow x\geq z_{i},i\in\{1,\ldots,n\}$.

• (iii)

$\exists\ a\in X_{k}^{\tau}$* such that $a<\text{min}\{z_{1},\ldots,\widehat{z_{k}},\ldots,z_{n}\}$.*

Proof.

• (i)

Assume that $x\in X_{r}^{\tau}\cap X_{s}^{\tau}.$ For any $f\in\tau,f(i)\neq z_{j}\ \forall\ i\neq j,x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$ and $x\neq 1$. Therefore, $<x>\ \in\sigma$ by Lemma 4.5. Since $<x>\ \notin\sigma\setminus\{<x>\}$, Proposition 4.3 holds, when $\sigma$ is replaced by $\sigma\setminus\{<x>\}$ in the definition of $\eta$.

Since $\sigma\notin S_{i}\cup\mu_{i}(S_{i})$ $\forall\ i$, we see that $\sigma\setminus\{<x>\}\notin S_{x}\cup\mu_{x}(S_{x})$. Therefore there exists $i<x$ such that $\sigma\setminus\{<x>\}\in S_{i}\cup\mu_{i}(S_{i})$. Considering $\sigma\setminus\{<x>\}$ instead of $\sigma\ \cup<a>$ in Proposition 4.4 and by an argument similar to that in the proof of Proposition 4.4, we see that there exists $s\in\{1,\dots,m\}$ and a cell $\xi_{s}$ such that

$\xi_{s}=\sigma\cup\begin{cases}\{<b_{1}>,\ldots,<b_{r}>\}\setminus\{<x>,<l_{1}>,\ldots,<l_{t}>\}&\text{if}~{}s~{}\text{is}~{}\text{even}\\ \{<b_{1}>,\ldots,<b_{r}>\}\setminus\{<l_{1}>,\ldots,<l_{t}>\}&\text{if}~{}s~{}\text{is}~{}\text{odd},\end{cases}$

where $r,t\geq 0$ and $b_{1},\ldots,b_{r},l_{1},\ldots,l_{t}<x$ such that, $\xi_{s}\in S_{1}\cup\mu_{1}(S_{1})$ but $\xi_{s}\setminus\{<x>\}$ (if $<x>\ \in\xi_{s}$) or $\xi_{s}\,\cup<x>$ (if $<x>\ \notin\xi_{s}$) $\notin S_{1}\cup\mu_{1}(S_{1})$. But, this is not possible by Proposition 4.3. Hence $X_{r}^{\tau}\cap X_{s}^{\tau}=\emptyset$.

• (ii)

Let $x\in X_{i}^{\tau}$ such that $x<z_{i}$. As in the case $(i)$, $x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$ and hence $<x>\ \in\sigma$. By exactly the same proof as the one above, we get a contradiction and thus $x\geq z_{i}$.

• (iii)

Let $z=\text{min}\{z_{1},\ldots,\widehat{z_{k}},\ldots,z_{n}\}$ and $a=\text{min}\{x\ |\ x\in X_{k}^{\tau}\}$. If $X_{\sigma}=[m],$ then $a=1$ and the result follows. Let $X_{\sigma}=[m]\setminus\{1\}$ and $a\geq z$. $\sigma,<z>\ \sim\ <1>$ implies that $\sigma\,\cup<z>\ \in\mathcal{N}(G)$ and $\sigma$ is a critical cell implies that $\sigma\,\cup<z>\ \notin S_{1}$. If $z=2$, then $\sigma,\sigma\,\cup<2>\ \notin S_{1}$ and hence $\sigma\in S_{2}$, which is not possible. Therefore, $z>2$. Choose $t$, $1<t<z$. Here, $t\neq z_{i}\ \forall\ i$ and therefore $t\in[m]\setminus\{z_{1},\ldots,z_{n}\}$ and $t\notin X_{i}^{\tau}\ \forall\ i\in[n]$ (by $(ii)$). Since $t\notin X_{\sigma\setminus<t>}$ and $t\neq 1$, $\{\sigma\,\cup<z>\}\setminus\{<t>\}\in S_{1}$, which implies $\sigma\,\cup<z>\ \notin\mu_{t}(S_{t})\ \forall\ t<z$. Therefore, $\sigma\,\cup<z>\ \in\mu_{z}(S_{z})$, which implies that $\sigma$ is not a critical cell. Thus, $a<z$.

∎

We now consider those cells $\sigma$ for which $X_{\sigma}=[m]$ or $[m]\setminus\{1\}$ and $<x>\ \in\sigma$ if $x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$. We now get sufficient conditions for $\sigma$ to be a critical cell.

Lemma 4.7**.**

$\sigma$* is a critical cell if it satisfies the following conditions*

• (i)

$X_{r}^{\tau}\cap X_{s}^{\tau}=\emptyset$* if $r\neq s$.*

• (ii)

$x\in X_{i}^{\tau}$* implies $x\geq z_{i},i\in\{1,\ldots,n\}$.*

• (iii)

$\exists\ a\in X_{k}^{\tau}$, where $a<\text{min}\{z_{1},\ldots,\widehat{z_{k}},\ldots,z_{n}\}$.

Proof.

If $x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$ such that $x\notin X_{\sigma\setminus<x>}$, then $x\neq 1$ and $\sigma\setminus\{<x>\}\sim\ <x>$. Thus, $\sigma\setminus\{<x>\}\in S_{1}$. In this case, $\sigma\notin S_{x}\cup\mu_{x}(S_{x})$. We consider the following two cases.

Case (i) : $y\in\{z_{1},\ldots,z_{n}\}$.

(a) $X_{\sigma}=[m]$.

In this case, $A_{i}^{\sigma}=\{z_{i}\}\ \forall\ i\in\{1,\ldots,n\}$. Therefore, for each $i\in[n],A_{i}^{\sigma\cup<z_{i}>}$ $=\emptyset$. Thus, $\sigma\,\cup<z_{i}>\ \notin\mathcal{N}(G)$ which implies that $\sigma\notin S_{z_{i}}\cup\mu_{z_{i}}(S_{z_{i}})$.

(b) $X_{\sigma}=[m]\setminus\{1\}$.

Here, $A_{i}^{\sigma}=\{1,z_{i}\}\ \forall\ i\neq k$ and $A_{k}^{\sigma}=\{1\}$. Since $A_{k}^{\sigma\cup<1>}=\emptyset$, $\sigma\notin S_{1}$. Let $a=\text{min}\{x\ |\ x\in X_{k}^{\tau}\}$. Since $X_{\sigma}=[m]\setminus\{1\},a\neq 1$ and by $(iii),a<z_{i}\ \forall\ i\neq k$. Hence, $<a>\ \in\sigma$.

If $l\neq k$ then, $\sigma\,\cup<z_{l}>\ \sim\ <1>$ and so $\sigma\,\cup<z_{l}>\ \in\mathcal{N}(G)$. $A_{l}^{\sigma\cup<z_{l}>\setminus<a>}=A_{l}^{\sigma\cup<z_{l}>}=\{1\}$ and therefore both $\{\sigma\,\cup<z_{l}>\}\setminus\{<a>\}$ and $\sigma\,\cup<z_{l}>\ \notin S_{1}$. If $a=2$, then $\sigma\,\cup<z_{l}>\ \in S_{2}\cup\mu_{2}(S_{2})$ (since $\sigma\cup<z_{l}>,\{\sigma\,\cup<z_{l}>\}\setminus\{<2>\}\notin S_{1}$), and therefore $\sigma\ \notin S_{z_{l}}\cup\mu_{z_{l}}(S_{z_{l}})$.

Assume $a>2$ and choose $t$, such that $1<t<a$. Clearly $t\neq z_{i}\ \forall\ i\neq k$ and $<t>\ \in\sigma$. Since $t<z_{i}\ \forall\ i\neq k$, from $(ii)\ t\notin X_{i}^{\tau}\ \forall\ i\neq k$. Since $a$ is the least element in $X_{k}^{\tau}$ and $t<a,t\notin X_{k}^{\tau}$ and thus $t\notin X_{\tau}$. Therefore, $\{\sigma\,\cup<z_{l}>\}\setminus\{<t>\},\{\sigma\,\cup<z_{l}>\}\setminus\{<a>,<t>\}$ will both be neighbor of $<t>$ and hence, they both belong to $S_{1}$. Neither $\sigma\,\cup<z_{l}>$ nor $\{\sigma\,\cup<z_{l}>\}\setminus\{<a>\}$ belong to $S_{t}\cup\mu_{t}(S_{t})\ \forall\ t<a$. Therefore $\sigma\,\cup<z_{l}>\ \in S_{a}\cup\mu_{a}(S_{a})$. We can now conclude that $\sigma\notin S_{z_{i}}\cup\mu_{z_{i}}(S_{z_{i}})\ \forall\ i\in[n]$.

Case (ii): $y\in[m]\setminus\{z_{1},\ldots,z_{n}\},y\in X_{\sigma\setminus<y>}$.

Since $y\in X_{\sigma\setminus<y>},y\in X_{i}^{\tau}$ for some $i\in[n]$. From $(i)$ and $(ii)$, $y\in X_{i}^{\tau}\setminus\bigcup\limits_{j\neq i}X_{j}^{\tau}$ and $y>z_{i}$. We first assume that $\sigma\in S_{y}\cup\mu_{y}(S_{y})$, i.e., $\sigma\setminus\{<y>\}\notin S_{j}\cup\mu_{j}(S_{j}),1\leq j<y$. The set $\{\sigma\,\cup<z_{i}>\}\setminus\{<y>\}$ has a neighbor $z_{1}\ldots z_{i-1}yz_{i+1}\ldots z_{n}$ and is therefore a simplex in $\mathcal{N}(G)$.

Claim 2**.**

$\{\sigma\,\cup<z_{i}>\}\setminus\{<y>\}\notin S_{t}\cup\mu_{t}(S_{t}),1\leq t<z_{i}$.

Suppose $\exists\ i_{1}<z_{i}$ such that $\{\sigma\,\cup<z_{i}>\}\setminus\{<y>\}\in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$. Since $<y>\ \notin\sigma\setminus\{<y>\}$, Proposition 4.3 holds when $\sigma$ is replaced by $\sigma\setminus\{<y>\}$ in the definition of $\eta$. Now, using Proposition 4.4, we get a simplex $\xi_{s},1\leq s<i_{1}\leq m$, where $\xi_{s}=\{\sigma\,\cup\{<j_{1}>,\ldots,<j_{l}>\}\}\setminus\{<y>,<k_{1}>,\ldots,<k_{t}>\}$ if $s$ is odd and $\xi_{s}=\{\sigma\,\cup\{<z_{i}>,<j_{1}>,\ldots,<j_{l}>\}\}\setminus\{<y>,<k_{1}>,\ldots,<k_{t}>\}$ if $s$ is even, $j_{1},\ldots,j_{l},k_{1},\ldots,k_{t}<z_{i}$ such that exactly one element in $\{\xi_{s},\xi_{s}\setminus\{<z_{i}>\}\}\in S_{1}\cup\mu_{1}(S_{1})$ (or $\xi_{s},\xi_{s}\,\cup<z_{i}>\in S_{1}\cup\mu_{1}(S_{1})$). From Proposition 4.3, this is not possible and therefore $\{\sigma\,\cup<z_{i}>\}\setminus\{<y>\}\notin S_{t}\cup\mu_{t}(S_{t})\ \forall\ 1\leq t<z_{i}$. Thus the claim is true.

Since, $\sigma\setminus\{<y>\}\notin S_{t}\cup\mu_{t}(S_{t})\ \forall\ 1\leq t<z_{i}$ (by the assumption), $\sigma\setminus\{<y>\}\in S_{z_{i}}$, a contradiction. Thus $\sigma\notin S_{y}\cup\mu_{y}(S_{y})$. Therefore, $\sigma$ is a critical cell.

∎

Combining Lemmas 4.5, 4.6 and 4.7, we have the following necessary and sufficient conditions for $\sigma$ to be a critical cell.

Theorem 4.8**.**

$\sigma$* is a critical cell if and only if*

• (i)

$X_{\sigma}=[m]\ \text{or}\ [m]\setminus\{1\}.$**

• (ii)

$<x>\ \in\sigma$* for each $x\in[m]\setminus\{z_{1},\ldots,z_{n}\}$.*

• (iii)

$x\in X_{i}^{\tau}\Rightarrow x\geq z_{i}\ \forall\ 1\leq i\leq n$.

• (iv)

$X_{r}^{\tau}\cap X_{s}^{\tau}=\emptyset\ \forall\ r\neq s$.

• (v)

there exists $z<\text{min}\{z_{1},\ldots,z_{k-1},\widehat{z_{k}},z_{k+1},\ldots,z_{n}\}$ where $z\in X_{k}^{\tau}$.

Case 2. $N(\sigma)=\ <1>$.

In this case, for all critical cells $\sigma$, $X_{\sigma}=[m]\setminus\{1\}$, which implies that $1\in A_{i}^{\sigma}\ \forall\ i\in[n]$. Further, since $A_{i}^{\sigma}\cap A_{j}^{\sigma}=\{1\}$ for $i\neq j$, $A_{i_{1}}^{\sigma}=A_{i_{2}}^{\sigma}=\{1\}$ for at least two distinct elements $i_{1},i_{2}$ in $[n]$ ($A_{i}^{\sigma}=\{1\}$ for only one $i\Rightarrow z_{1}\ldots z_{i-1}1z_{i+1}\ldots z_{n}\in N(\sigma),z_{j}\in A_{j}^{\sigma},j\neq i$ and $A_{i}^{\sigma}\neq\{1\}\ \forall\ i\Rightarrow\sigma\in S_{1}$).

Lemma 4.9**.**

If $\sigma$ is critical, then $\sigma=\{<2>,\ldots,<m>\}$.

Proof.

If $<i>\ \notin\sigma$ for $i\neq 1$, then $\sigma\,\cup<i>\ \subset N(<1>)$ thereby implying that $\sigma\,\cup<i>\ \in\mathcal{N}(G)$.

The proof is by induction on $i$. If $<2>\ \notin\sigma$, then $\sigma,\sigma\,\cup<2>\ \notin S_{1}$ implies that $\sigma\in S_{2}$, a contradiction.

If $2\in X_{\sigma\setminus<2>}$, then $2\in X_{l}^{\sigma\setminus<2>}$ for at least one $l\in[n]$. If $2\in X_{l}^{\sigma\setminus<2>}$ for exactly one $l\in[n]$, then at least one of $A_{i_{1}}^{\sigma\setminus<2>}$ or $A_{i_{2}}^{\sigma\setminus<2>}$ is $\{1\}$ and if $2\in X_{l}^{\sigma\setminus<2>}\cap X_{k}^{\sigma\setminus<2>}$, then $A_{i_{1}}^{\sigma\setminus<2>}=A_{i_{2}}^{\sigma\setminus<2>}=\{1\}$. In both these cases $\sigma\setminus<2>\ \notin S_{1}$ and therefore, $\sigma\in S_{2}$, a contradiction. Therefore, $<2>\ \in\sigma$ and $2\notin X_{\sigma\setminus<2>}$.

Assume that $<t>\ \in\sigma$ and $t\notin X_{\sigma\setminus<t>}\ \forall\ 1<t<m$. Suppose $<m>\ \notin\sigma$. Since $t\notin X_{\sigma\setminus<t>}\ \forall\ 1<t<m$, $<t>\ \in N(\{\sigma\,\cup<m>\}\setminus\{<t>\})$, which implies that $\{\sigma\,\cup<m>\}\setminus\{<t>\}\in S_{1}$. Therefore, $\sigma\,\cup<m>\ \notin S_{t}\cup\mu_{t}(S_{t})\ \forall\ 1\leq t<m$, which implies that $\sigma\in S_{m}$, a contradiction. Hence $<m>\ \in\sigma$.

If $m\in X_{\sigma\setminus<m>}$, then $m\in\text{Im}\,f$, where $f\in\sigma$ and $|\text{Im}\,f|=n$. By the induction hypothesis, $1,\ldots,m-1\notin\text{Im}\,f$. Further, since $n>1$, this case is not possible. Hence, $m\notin X_{\sigma\setminus<m>}$.

∎

We now describe the Morse Complex $\mathcal{M}=(\mathcal{M}_{i},\partial)$ corresponding to this acyclic matching on the poset $P$. If $c_{i}$ denotes the number of critical $i$ cells, then the free abelian group generated by these critical cells is denoted by $\mathcal{M}_{i}$. We use the following version of Theorem 3.1, from which we explicitly compute the boundary maps in the Morse Complex $\mathcal{M}$.

Proposition 4.10**.**

(Theorem 11.13 [14])

Let $X$ be a simplicial complex and $\mu$ be an acyclic matching on the face poset of $X\setminus\{\emptyset\}$. Let $c_{i}$ denote the number of critical $i$ cells of $X$. Then

• (a)

$X$* is homotopy equivalent to $X_{c}$, where $X_{c}$ is a $CW$ complex with $c_{i}$ cells in dimension $i$.*

• (b)

There is a natural indexing of cells of $X_{c}$ with the critical cells of $X$ such that for any two cells $\tau$ and $\sigma$ of $X_{c}$ satisfying dim $\tau$ = dim $\sigma+1$, the incidence number $[\tau:\sigma]$ is given by

$[\tau:\sigma]$* = $\sum\limits_{c}w(c).$*

The sum is taken over all (alternating) paths $c$ connecting $\tau$ with $\sigma$ i.e., over all sequences $c$ = $\{\tau,y_{1},\mu(y_{1}),\ldots,y_{t},\mu(y_{t}),\sigma\}$ such that $\tau\succ y_{1}$, $\mu(y_{t})\succ\sigma$, and $\mu(y_{i})\succ y_{i+1}$ for $i=1,\ldots,t-1$. The quantity $w(c)$ associated to this alternating path is defined by

[TABLE]

where all the incidence numbers are taken in the complex $X$.

4.3 Alternating paths

Since we have a description of the critical cells in $\mathcal{N}(G)$, we now study the alternating paths between them. We first consider critical $p+1$ cells, which are of the type $\sigma=\{<x>|\ x\in T\}\cup\{f_{1},f_{2}\}\subset N(z_{1}\ldots z_{n})$, where $T=[m]\setminus\{z_{1},\ldots,z_{n}\}$, $|\text{Im}\,f_{1}|=|\text{Im}\,f_{2}|=n$ and $z_{k}=1$.

Let $\tau=\{f_{1},f_{2}\}$ and the non-empty sets $X_{i}^{\tau}\cap\,T,1\leq i\leq n$ be labelled $Y_{i_{1}},\ldots,Y_{i_{q}}$, where $i_{1}\leq i_{2}\leq\ldots\leq i_{q}$. By Theorem 4.8 $(iv)$, $Y_{i_{j}}\cap Y_{i_{k}}=\emptyset\ \forall\ j\neq k.$ For any $l,1\leq l\leq q$, let $z=\text{min}\{z_{i_{1}},\ldots,z_{i_{l}}\},D$ any subset of $\{z_{i_{1}},\ldots,z_{i_{l}}\}$ and $C\subset\bigcup\limits_{j=1}^{l}Y_{i_{j}}$ be such that $C\cap Y_{i_{j}}\neq\emptyset\ \forall\ 1\leq j\leq l$. We first prove the following:

Lemma 4.11**.**

The cell $\alpha=\{\sigma\cup\{<y>|\ y\in D\}\}\setminus\{<x>|\ x\in C\}\in S_{z}\cup\mu_{z}(S_{z})$.

Proof.

Any element of $C$ is different from $1$ and therefore belongs to $X_{\sigma}.$ For $j\in\{1,\ldots,l\}$, there exists at least one element $y\in C$ such that $y\in A_{i_{j}}^{\alpha}$ and therefore in $X_{i_{j}}^{\alpha}$. If $t\in[n]\setminus\{i_{1},\ldots,i_{l}\}$, then $z_{t}\in A_{t}^{\alpha}$. Hence, $A_{t}^{\alpha}\neq\emptyset\ \forall\ t\in[n]$ and therefore $\alpha\in\mathcal{N}(G).$

• (i)

Let $z=1$. If $t\in[n]\setminus\{i_{1},\ldots,i_{l}\}$, then $z_{t}\neq 1$ and $z_{t}\in A_{t}^{\alpha\cup<1>}$, which implies that $A_{t}^{\alpha\cup<1>}\neq\emptyset$. Since $T\cap\{1\}=\emptyset$, $y\in A_{i_{j}}^{\alpha}$ implies that $y\in A_{i_{j}}^{\alpha\cup<1>}\ \forall\ j\in\{1,\ldots,l\}.$ Thus, $\alpha\,\cup<1>\ \in\mathcal{N}(G)$ and therefore $\alpha\in S_{1}\cup\mu_{1}(S_{1})$.

• (ii)

Let $z>1.$ Consider the cell $\eta=\{\sigma\cup\{<a_{i}>,1\leq i\leq s\}\}\setminus\{<b_{j}>,1\leq j\leq r\}$, $s,r\geq 0,a_{i},b_{j}<z$. Let $D^{\prime}=\{<x>|\ x\in D\}$, $C^{\prime}=\{<y>|\ y\in C\}$ and

$\gamma=\begin{cases}\{\eta\cup D^{\prime}\}\setminus C^{\prime}&\text{if}~{}<z>\ \notin\ D^{\prime}\\ \{\eta\cup D^{\prime}\}\setminus\{<z>,C^{\prime}\}&\text{if}~{}<z>\ \in D^{\prime},\end{cases}$

Proposition 4.12**.**

$\eta\in\mathcal{N}(G)\Leftrightarrow\gamma,\gamma\,\cup<z>\ \in\mathcal{N}(G)$. In particular, $\eta\in S_{1}\Leftrightarrow\gamma,\gamma\,\cup<z>\ \in S_{1}$.

Since $\sigma$ is critical, by Theorem 4.8 $(iii)$ $y\in C$ has the property $y>z$,. If $j\in\{1,\ldots,l\},$ then there exists $y\neq 1,z$ in $C$ such that $y$ belongs to $A_{i_{j}}^{\gamma}$ and therefore also to $A_{i_{j}}^{\gamma\cup<1>},A_{i_{j}}^{\gamma\cup<z>}$ and $A_{i_{j}}^{\gamma\cup<z>\cup<1>}$. If $1\leq j\leq l$, $z_{i_{j}}\geq z>1$ and $z_{i_{j}}\in A_{i_{j}}^{\eta}$ implies that $z_{i_{j}}\in A_{i_{j}}^{\eta\cup<1>}.$ Let $t\in[n]\setminus\{i_{1},\ldots,i_{l}\}$. $X_{\sigma}=[m]$ or $[m]\setminus\{1\}$, $j\in\{1,\ldots,l\}$ implies that $z_{i_{j}}\in X_{i_{j}}^{\sigma}$. Therefore, $z_{i_{j}}\notin A_{t}^{\eta}\ \forall\ t\neq i_{j}$ and $D\cap A_{t}^{\eta}=\emptyset.$ By the definition of $\gamma,A_{t}^{\eta}=A_{t}^{\gamma}\ \forall\ t\in[n]\setminus\{i_{1},\ldots,i_{l}\}$. $A_{t}^{\gamma\cup<z>}=A_{t}^{\gamma}$ or $A_{t}^{\gamma}\setminus\{z\}$. Now, we only have to consider $t\in[n]\setminus\{i_{1},\ldots,i_{l}\}.$

If $\eta\in\mathcal{N}(G),$ then $A_{t}^{\eta}\neq\emptyset$ and $z\notin A_{t}^{\eta}$. Therefore $A_{t}^{\eta}=A_{t}^{\gamma}=A_{t}^{\gamma\cup<z>}$ thereby showing that $\gamma,\gamma\,\cup<z>\ \in\mathcal{N}(G)$.

If $\eta\notin\mathcal{N}(G),$ then for some $t\in[n]\setminus\{i_{1},\ldots,i_{l}\},A_{t}^{\eta}=\emptyset$ and hence so are $A_{t}^{\gamma}$ and $A_{t}^{\gamma\cup<z>}.$ Thus $\gamma$ and $\gamma\,\cup<z>\ \notin\mathcal{N}(G).$

If $\eta\in S_{1},A_{t}^{\eta\cup<1>}\neq\emptyset$ and therefore $A_{t}^{\gamma\cup<1>}$, $A_{t}^{\gamma\cup<z>\cup<1>}\neq\emptyset$.

If $\eta\notin S_{1}$, then for some $t\in[n]\setminus\{i_{1},\ldots,i_{l}\},A_{t}^{\eta\cup<1>}=\emptyset$, thereby implying that $A_{t}^{\eta}=\{1\}$. Thus, $A_{t}^{\gamma}=A_{t}^{\gamma\cup<z>}=\{1\}$ and the proof of Proposition 4.12 follows.

Suppose there exists $i_{1}<z$ such that $\alpha\in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$.

Clearly, $i_{1}>1$ as $\alpha\ \in S_{1}\cup\mu_{1}(S_{1})$ implies that $\sigma\in S_{1}\cup\mu_{1}(S_{1})$, by Proposition 4.12, which is a contradiction. Define the cell

$\xi_{1}=\begin{cases}\sigma\setminus\{<i_{1}>\}&\text{if}~{}~{}<i_{1}>\ \in\sigma\\ \sigma\,\cup<i_{1}>&\text{if}~{}~{}<i_{1}>\ \notin\sigma.\\ \end{cases}$

Since $\alpha\in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$, $\{\xi_{1}\cup D^{\prime}\}\setminus C^{\prime}$ is always a simplex, which implies that $\xi_{1}\in\mathcal{N}(G)$, by Proposition 4.12. If $\xi_{1}\in S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}})$, i.e. $\xi_{1}\notin S_{j}\cup\mu_{j}(S_{j})$ $\forall\ j<i_{1}$, then $\sigma$ will belong to $S_{i_{1}}\,\cup\mu_{i_{1}}(S_{i_{1}})$, a contradiction. Therefore, $\exists\ i_{2}<i_{1}$ such that $\xi_{1}\in S_{i_{2}}\cup\mu_{i_{2}}(S_{i_{2}})$. Now, $\{\xi_{1}\,\cup D^{\prime}\}\setminus C^{\prime}$, which is $\alpha\setminus\{<i_{1}>\}$ or $\alpha\ \cup<i_{1}>$, belongs to $S_{i_{1}}\cup\mu_{i_{1}}(S_{i_{1}}).$ If $i_{2}=1$, then $\xi_{1}\in S_{1}\cup\mu_{1}(S_{1})$ and $\{\xi_{1}\,\cup D^{\prime}\}\setminus C^{\prime}\notin S_{1}\cup\mu_{1}(S_{1})$, which is a contradiction from Proposition 4.12. Let $i_{2}>1$. Define the cell

$\xi_{2}=\begin{cases}\{\xi_{1}\,\cup D^{\prime}\}\setminus\{C^{\prime},<i_{2}>\}&\text{if}~{}~{}<i_{2}>\ \in\sigma\\ \xi_{1}\,\cup D^{\prime}\ \cup<i_{2}>&\text{if}~{}~{}<i_{2}>\ \notin\sigma.\end{cases}$

By Proposition 4.12, $\xi_{2}\in\mathcal{N}(G)$. Since $\{\xi_{1}\,\cup D^{\prime}\}\setminus C^{\prime}\ \notin S_{i_{2}}\cup\mu_{i_{2}}(S_{i_{2}})$, there exists $i_{3}<i_{2}$ such that $\xi_{2}\in S_{i_{3}}\cup\mu_{i_{3}}(S_{i_{3}})$. Here, $\{\xi_{2}\cup C^{\prime}\}\setminus D^{\prime}$ which is $\xi_{1}\setminus\{<i_{2}>\}$ or $\xi_{1}\,\cup<i_{2}>\ \in S_{i_{2}}\cup\mu_{i_{2}}(S_{i_{2}})$. If $i_{3}=1$, then we get a contradiction by Proposition 4.12. Inductively, assume that there exists $l,1<l<m$, where $i_{l+1}<i_{l}<\ldots<i_{1}<z$ and $\xi_{t}\in S_{i_{t+1}}\cup\mu_{i_{t+1}}(S_{i_{t+1}})$, $1\leq t\leq l$ such that

$\xi_{t}=\begin{cases}\{\xi_{t-1}\,\cup D^{\prime}\}\setminus\{C^{\prime},<i_{t}>\}&\text{if}~{}<i_{t}>\ \in\xi_{t-1},~{}t~{}\text{is}~{}\text{even}\\ \{\xi_{t-1}\,\cup D^{\prime}\cup<i_{t}>\}\setminus C^{\prime}&\text{if}~{}~{}<i_{t}>\ \notin\xi_{t-1},~{}t~{}\text{is}~{}\text{even}\\ \{\xi_{t-1}\cup C^{\prime}\}\setminus\{D^{\prime},<i_{t}>\}&\text{if}~{}~{}<i_{t}>\ \in\xi_{t-1},~{}t~{}\text{is}~{}\text{odd}\\ \{\xi_{t-1}\,\cup C^{\prime}\cup<i_{t}>\}\setminus D^{\prime}&\text{if}~{}~{}<i_{t}>\ \notin\xi_{t-1},~{}t~{}\text{is}~{}\text{odd}.\end{cases}$

Since $\{\xi_{l}\cup C^{\prime}\}\setminus D^{\prime}(\text{or}~{}~{}\{\xi_{l}\cup D^{\prime}\}\setminus C^{\prime})=\xi_{l-1}\setminus\{<i_{l}>\}$ or $\xi_{l-1}\,\cup<i_{l}>\ \in S_{i_{l}}\cup\mu_{i_{l}}(S_{i_{l}})$ and $i_{l}>i_{l+1}$, if $i_{l+1}=1$, using Proposition 4.12 we arrive at a contradiction.

By induction, there exists $1<s\leq m$, and a cell

$\xi_{s}=\sigma\cup\begin{cases}\{D^{\prime},\langle b_{1}\rangle,\ldots,\langle b_{r}\rangle\}\setminus\{C^{\prime},\langle l_{1}\rangle,\ldots,\langle l_{t}\rangle\}&\text{if}~{}\text{s}~{}\text{is}~{}\text{even}\\ \{<b_{1}>,\ldots,<b_{r}>\}\setminus\{\langle l_{1}\rangle,\ldots,\langle l_{t}\rangle\}&\text{if}~{}\text{s}~{}\text{is}~{}\text{odd},\end{cases}$

where $r,t\geq 0$ and $b_{1},\ldots,b_{r},l_{1},\ldots,l_{t}<z$ such that, $\xi_{s}\in S_{1}\cup\mu_{1}(S_{1})$ but $\{\xi_{s}\cup C^{\prime}\}\setminus D^{\prime}$ (if $s$ is even) or $\{\xi_{s}\cup D^{\prime}\}\setminus C^{\prime}$ (if $s$ is odd) $\notin S_{1}\cup\mu_{1}(S_{1})$. But, this is not possible by Proposition 4.12.

Hence $\alpha\notin S_{i}\cup\mu_{i}(S_{i})\ \forall\ i<z$. Replacing $\alpha$ by $\alpha\,\cup<z>$ (if $<z>\ \notin\alpha$) or by $\alpha\setminus\{<z>\}$ (if $<z>\ \in\alpha$) and applying an argument similar to the one above, we see that $\alpha\,\cup<z>$ (if $<z>\ \notin\alpha$) or $\alpha\setminus\{<z>\}$ (if $<z>\ \in\alpha$) $\notin S_{i}\cup\mu_{i}(S_{i})\ \forall\ i<z.$ Hence $\alpha\in S_{z}\cup\mu_{z}(S_{z})$.

∎

Lemma 4.13**.**

Let $\eta=\{<x_{1}>,\ldots,<x_{p}>,f_{1},f_{2}\}\subset N(w_{1}\ldots w_{n})$ be a simplex, where $w_{k}=1,w_{i_{0}}=\text{min}\{w_{1},\ldots,\widehat{w_{k}},\ldots,w_{n}\}$, $f_{2}(i)=w_{i}\ \forall\ i\neq k$ and $[m]\setminus\{w_{1},\ldots,w_{n}\}=\{x_{1},\ldots,x_{p}\}$. If $f_{2}(k)>w_{i_{0}}$, then $\eta\setminus\{f_{1}\}\in S_{w_{i_{0}}}$.

Proof.

$f_{2}(k)>w_{i_{0}}>1$ implies that $f_{2}(k)\in\{x_{1},\ldots,x_{p}\}$ and $1\notin\text{Im}\,f_{2}$. Therefore, $\{\eta\,\cup<w_{i_{0}}>\}\setminus\{f_{1}\}\sim\ <1>$. If $y_{1}\ldots y_{n}\sim\eta\setminus\{f_{1}\}$, then $y_{k}\notin\{w_{1},\ldots,\widehat{w_{k}},\ldots,w_{n},x_{1},\ldots,x_{p}\}$. Thus $y_{k}=1$ and $\eta\setminus\{f_{1}\}\notin S_{1}$. Consider $j$ such that $1<j<w_{i_{0}}.$ Since $j\in\{x_{1},\ldots,x_{p}\}$ and $j\notin X_{\eta\setminus\{f_{1}\}}$, $\eta\setminus\{f_{1},<j>\}$ and $\{\eta\,\cup<w_{i_{0}}>\}\setminus\{f_{1},<j>\}\sim\ <j>$, thereby showing that $\eta\setminus\{f_{1}\}$ and $\{\eta\,\cup<w_{i_{0}}>\}\setminus\{f_{1}\}\notin S_{j}\cup\mu_{j}(S_{j})\ \forall\ j<w_{i_{0}}$. Thus $\eta\setminus\{f_{1}\}\in S_{w_{i_{0}}}$.

∎

Lemma 4.14**.**

No element of $S_{1}\cup\mu_{1}(S_{1})$ belongs to any alternating path between two critical cells.

Proof.

Let $c$ = $\{\tau,y_{1},\mu(y_{1}),\ldots,y_{t},\mu(y_{t}),\alpha\}$ be an alternating path between the critical cells $\tau$ and $\alpha$. Let $S=\bigcup\limits_{i=1}^{m}S_{i}$ and $\mu:S\rightarrow P\setminus S$ be the map such that $\mu|_{S_{i}}=\mu_{i}$. Here, $y_{i}\in S$, $\mu(y_{i})\in\mu(S)$ and $S\cap\mu(S)=\emptyset.$ If $\gamma\in c$ and $\gamma\in\mu_{1}(S_{1})$, then $\gamma=\mu(y_{i})$ for some $i\in[t]$. Since $[\mu(y_{i}):y_{i}]=\pm 1$, $y_{i}=\mu(y_{i})\setminus\{<1>\}$. In the alternating path, $y_{i+1}\prec\mu(y_{i})$ and $y_{i+1}\neq y_{i}$ implies that $<1>\ \in\mu(y_{i})\setminus\{<1>\}$. This implies that $y_{i+1}\in\mu(S)$, a contradiction. Therefore $\gamma\notin c.$ If $\gamma\in S_{1}$, then $\mu(\gamma)\in c$, a contradiction. ∎

We are now ready to prove the main result of this section.

Theorem 4.15**.**

There exist two critical $p$-cells $\beta$ and $\gamma$ such that there exists exactly one alternating path from $\sigma$ to each of $\beta$ and $\gamma$. Further, there exists no alternating path from $\sigma$ to any other critical $p$-cell.

Proof.

Define the sets $A_{i},B_{i},i\in\{1,2\}$ to be $A_{i}=T\cap\{\text{Im}f_{i}\setminus\{f_{i}(k)\}\}$ and $B_{i}=\{z_{1},\ldots,\widehat{z_{k}},\ldots,z_{n}\}\setminus\text{Im}\,f_{i}$. Since $A_{i}\subset T$ and $B_{i}\subset\{z_{1},\ldots,z_{n}\},A_{i}\cap B_{i}=\emptyset\ \text{for}\ 1\leq i\leq 2.$ If $z_{i}\notin\{f_{1}(i),f_{2}(i)\}$ for $i\neq k$, then $z_{i}\notin\text{Im}\,f_{1},\text{Im}\,f_{2},T$ and so $z_{i}\notin X_{\sigma}$. Since $z_{i}\neq 1,<z_{i}>\ \sim\sigma,<1>$ and thus $\sigma\in S_{1}$. Therefore, $z_{i}\in\{f_{1}(i),f_{2}(i)\}\ \forall\ i\in[n]\setminus\{k\}$, thus implying that $B_{1}\cap B_{2}=\emptyset,B_{1}\subset\text{Im}\,f_{2}$ and $B_{2}\subset\text{Im}\,f_{2}$.

Let $x\in A_{1}\cap A_{2}$. For $i\in[n]\setminus\{k\},z_{i}\in\{f_{1}(i),f_{2}(i)\}$ and therefore, $\exists\ i\neq j\in[n]$ such that $f_{1}(i)=f_{2}(j)=x$. Here, $X_{i}^{\tau}\cap X_{j}^{\tau}\neq\emptyset$, a contradiction to 4.8 $(iv)$ and hence $A_{1}\cap A_{2}=\emptyset.$

Let $W=\{z\ |\ z=f_{1}(i)=f_{2}(i)\}$ and $T_{1}=T\setminus\{A_{1}\cup A_{2}\}$. Given a set $B\subset B_{1}\cup B_{2}$, define a corresponding set $A\subset A_{1}\cup A_{2}$ as follows. If $z_{i}\in B\cap B_{1}$, then $z_{i}\neq f_{1}(i)$ and hence, $f_{1}(i)\in A$. For $z_{i}\in B\cap B_{2}$, $f_{2}(i)\in A$. The number of elements in $A$ and $B$ are the same.

Let the facet of $\sigma$ in the alternating path $c$ be $y_{1}$. We consider the following two cases.

Case 1. $A_{1},A_{2}=\emptyset$.

In this case, $f_{1}(i)=f_{2}(i)=z_{i}\ \forall\ i\neq k$ and $\{f_{1}(k),f_{2}(k)\}\cap T\neq\emptyset.$ If an element $x$ of $T$ does not belong to $X_{\tau}$, then $\sigma\setminus\{<x>\}\sim\ <x>$ and so $\sigma\setminus\{<x>\}\in S_{1}.$ If $x\in X_{\tau}$, then $x\in\{f_{1}(k),f_{2}(k)\}$ and $\sigma\setminus\{<x>\}\subset N(z_{1}\ldots z_{k-1}xz_{k+1}\ldots z_{n})$ which implies $\sigma\setminus\{<x>\}\in S_{1}$ and from 4.14, $\sigma\setminus\{<x>\}\notin c$. Therefore, $y_{1}$ is either $\sigma\setminus\{f_{1}\}$ or $\sigma\setminus\{f_{2}\}$. Each of $X_{\sigma\setminus\{f_{1}\}}$ and $X_{\sigma\setminus\{f_{2}\}}$ will be either $[m]$ or $[m]\setminus\{1\}.$ Further, $\sigma\setminus\{f_{1}\}$ and $\sigma\setminus\{f_{2}\}$ satisfy the properties $(ii),(iii)$ and $(iv)$ of Theorem 4.8. By Theorem 4.8 $(v)$, at least one of $f_{1}(k)\ \text{or}\ f_{2}(k)<z_{i_{0}}$, where $z_{i_{0}}=\text{min}\{z_{1},\ldots,\widehat{z_{k}},\ldots,z_{n}\}$.

If $f_{1}(k)<z_{i_{0}}$, then $\sigma\setminus\{f_{2}\}$ is a critical $p$-cell as it satisfies all the five conditions of Theorem 4.8 and $c=\{\sigma,\sigma\setminus\{f_{2}\}\}$.

Similarly, $f_{2}(k)<z_{i_{0}}$ implies $\sigma\setminus\{f_{1}\}$ is critical and $c=\{\sigma,\sigma\setminus\{f_{1}\}\}$.

Now, assume that $f_{1}(k)<z_{i_{0}},f_{2}(k)>z_{i_{0}}$ and $y_{1}=\sigma\setminus\{f_{1}\}.$ Since $f_{2}(i)=z_{i}\ \forall\ i\neq k$ and $f_{2}(k)>z_{i_{0}},\sigma\setminus\{f_{1}\}\in S_{z_{i_{0}}}$ by Lemma 4.13. So $\mu(y_{1})=\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{1}\}$. If $x\in[m]\setminus\{z_{1},\ldots,z_{n},f_{2}(k)\}$, then $\mu(y_{1})\setminus\{<x>\}\sim\ <x>$ and hence belongs to $S_{1}$. $\mu(y_{1})\setminus\{<x>\}=y_{1}$ if $x=z_{i_{0}}$. $\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{1},f_{2}\}\sim\ <z_{i}>$ for $i\neq i_{0},k$ and so $y_{2}$ has to be $\mu(y_{1})\setminus\{<x>\}$ where $x=f_{2}(k)$. By Theorem 4.8, $y_{2}\subset N(\tilde{z_{1}}\ldots\tilde{z_{n}})$, where $\tilde{z}_{k}=f_{2}(k),\tilde{z}_{i_{0}}=1$ and $\tilde{z}_{i}=z_{i},i\neq i_{0},k$, is a critical cell and the alternating path in this case is $\{\sigma,\sigma\setminus\{f_{1}\},\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{1}\},\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{1},<f_{2}(k)>\}\}.$

If $f_{2}(k)<z_{i_{0}},f_{1}(k)>z_{i_{0}}$ and $y_{1}=\sigma\setminus\{f_{2}\}$, then the alternating path is $\{\sigma,\sigma\setminus\{f_{2}\},\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{2}\},\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{2},<f_{1}(k)>\}\}.$

Case 2. At least one of $A_{1}$ or $A_{2}$ is non empty.

We prove some results necessary for the construction of the alternating paths $c$ from $\sigma$.

Lemma 4.16**.**

Let $\alpha=\{\sigma\cup\{<b>|\ b\in B\}\}\setminus\{<a>|\ a\in A\}$, where $B\neq\emptyset$. Then,

• (i)

$\alpha\setminus\{<x>\}\in S_{1}\ \forall\ x\in T_{1}.$**

• (ii)

If $B_{1}\cap B\neq\emptyset,B_{1}$ (or $B_{2}\cap B\neq\emptyset,B_{2}$), then $\alpha\setminus\{f_{1}\},\alpha\setminus\{f_{2}\}\in S_{1}$.

• (iii)

If $B=B_{1}\cup B_{2}$ and $B_{1},B_{2}\neq\emptyset$, then $\alpha\setminus\{f_{1}\},\alpha\setminus\{f_{2}\}\in S_{1}$.

• (iv)

If $B=B_{1}$ (or $B=B_{2}$), then $\alpha\setminus\{f_{1}\}\in S_{1}$ and $\alpha\setminus\{f_{2}\}\notin S_{1}$ (or $\alpha\setminus\{f_{1}\}\notin S_{1}$ and $\alpha\setminus\{f_{2}\}\in S_{1}$.)

Proof.

• (i)

Let $x\in T_{1}$. If $x\notin X_{\tau},$ then $x\notin X_{\alpha\setminus<x>}$ and therefore $\alpha\setminus\{<x>\}\sim\ <x>$, which implies $\alpha\setminus\{<x>\}\in S_{1}.$ If $x\in X_{\tau}$, then $x\in\{f_{1}(k),f_{2}(k)\}$ and $\alpha\setminus\{<x>\}\sim w_{1}\ldots w_{k-1}xw_{k+1}\ldots w_{n}$, where $\alpha\sim w_{1}w_{2}\dots w_{n}$, which implies that $\alpha\setminus\{<x>\}\in S_{1}$.

• (ii)

Since $B\cap B_{1}\neq B_{1}$, $\exists\ b\in B_{1}\setminus B$ such that $<b>\ \notin\alpha,b\in\text{Im}\,f_{2}$ and $b\notin\text{Im}\,f_{1}$. Thus $\alpha\setminus\{f_{2}\}\sim\ <b>$. $A\cap A_{1}\neq\emptyset$ because $B\cap B_{1}\neq\emptyset$. So if $x\in A\cap A_{1},<x>\ \notin\alpha$ and $<x>\ \sim f_{2}$. Hence, $\alpha\setminus\{f_{1}\}\sim\ <x>$.

• (iii)

If $B=B_{1}\cup B_{2}$, then $A=A_{1}\cup A_{2}$. Thus, $\alpha\setminus\{f_{1}\}\sim\ <x>$, where $x\in A_{1}$ and $\alpha\setminus\{f_{2}\}\sim\ <y>$, where $y\in A_{2}$.

• (iv)

Let $B=B_{1}$ and $B\cap B_{2}=\emptyset$.

As in the earlier cases, $\alpha\setminus\{f_{1}\}\sim\ <x>$ for any $x\in A_{1}$ and hence $\alpha\setminus\{f_{1}\}\in S_{1}$.

If $z\in B_{1}$, then $<z>\ \in\alpha$ and so $<z>$ is not a neighbor of $\alpha\setminus\{f_{2}\}$. If $z\in B_{2}$ or $W$, then $z\in\text{Im}f_{1}$ and so $<z>\ \nsim f_{1}$ and thus $<z>\ \nsim\alpha\setminus\{f_{2}\}.$ If $x\in A_{1}$, then $x\in\text{Im}f_{1}$ and $<x>\ \nsim f_{1}$.

If $x\in A_{2}$, then $<x>\ \in\alpha$. Thus, the only possible neighbors of $\alpha\setminus\{f_{2}\}$ are $<1>$ and $y_{1}\ldots y_{n},y_{i}\neq y_{j}$ for $i\neq j$.

Let $\alpha\setminus\{f_{2}\}\sim y_{1}\ldots y_{n}=y$. For $z\neq 1\in B_{2}\cup W\cup A_{1}$, since $y\sim f_{1}$, we see that $y_{k}\neq z$. If $x\in T_{1}\cup A_{2}\cup B_{1}$, then $<x>\ \in\alpha\setminus\{f_{2}\}$ and $y\sim\ <x>$ which implies $y_{k}\neq x$. The only possible choice for $y_{k}$ is $1$. Therefore, $\alpha\setminus\{f_{2}\}\notin S_{1}$.

∎

To construct the alternating path from the critical $p+1$-cell $\sigma$ to a critical $p$-cell in $\mathcal{N}(G)$, we first require the facet $y_{1}$ of $\sigma$ in $c$.

If $A_{1}=\emptyset$ and $A_{2}\neq\emptyset$, then $f_{1}=z_{1}\ldots z_{k-1}f_{1}(k)z_{k+1}\ldots z_{n}$. Hence, $\sigma\setminus\{f_{1}\}\sim\ <b>\ \forall\ b\in B_{2}$ and $\sigma\setminus\{<x>\}\sim\ <x>\ \forall\ x\in T_{1}\setminus\{f_{1}(k),f_{2}(k)\}$. For $x\in T\cap\{f_{1}(k),f_{2}(k)\}$, we observe that $x\neq 1$ and $\sigma\setminus\{<x>\}\sim z_{1}\ldots z_{k-1}xz_{k+1}\ldots z_{n}\sim\ <1>$ and thus $\sigma\setminus\{<x>\}\in S_{1}$. Therefore $y_{1}=\sigma\setminus\{f_{2}\}$ or $\sigma\setminus\{<x>\}$, where $x\in A_{2}$.

Similarly, in the case $A_{1}\neq\emptyset$ and $A_{2}=\emptyset$, we see that $y_{1}=\sigma\setminus\{f_{1}\}$ or $\sigma\setminus\{<x>\},x\in A_{1}$.

Let $i\neq j\in\{1,2\}$. Consider the case when $y_{1}=\sigma\setminus\{f_{i}\}$. Let $z_{i_{0}}=\text{min}\{z_{1},\ldots,\widehat{z_{k}},\ldots,$ $z_{n}\}$. If $A_{i}=\emptyset,A_{j}\neq\emptyset$, then $y_{1}=\sigma\setminus\{f_{j}\}$. Since $f_{i}(k)\neq z_{l}\ \forall\ l\in[n]\setminus\{k\},f_{i}(k)\neq z_{i_{0}}$ and thus $f_{i}(k)>z_{i_{0}}$ or $f_{i}(k)<z_{i_{0}}$.

Subcase 1. $f_{i}(k)<z_{i_{0}}$.

Since $f_{i}=z_{1}\ldots z_{k-1}f_{i}(k)z_{k+1}\ldots z_{n},\sigma\setminus\{f_{j}\}$ satisfies all the criteria of Theorem 4.8 and thus a critical $p$-cell, i.e. $c=\{\sigma,\sigma\setminus\{f_{j}\}\}$.

Subcase 2. $f_{i}(k)>z_{i_{0}}$.

By Lemma 4.13, $\mu(y_{1})=\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{j}\}$. If $<x>\ \in\mu(y_{1})$, $x\in T_{1}\cup A_{j}\cup\{z_{i_{0}}\}$. Since $y_{2}\neq y_{1},y_{2}\neq\mu(y_{1})\setminus\{<z_{i_{0}}>\}$. If $x\in T_{1}\cup A_{j}\setminus\{f_{i}(k)\}$, then $\mu(y_{1})\setminus\{<x>\}\sim\ <x>$ and $\mu(y_{1})\setminus\{f_{i}\}\sim\ <z_{l}>,l\neq i_{0},k$. Therefore, $y_{2}=\mu(y_{1})\setminus\{<f_{i}(k)>\}$ and $y_{2}\sim w_{1}\ldots w_{n}$, where $w_{k}=f_{i}(k),w_{i_{0}}=1$ and $\forall\ l\neq i_{0},k,w_{l}=z_{l}$. By Theorem 4.8, $y_{2}$ is a critical cell and $c=\{\sigma,\sigma\setminus\{f_{j}\},\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{j}\},\{\sigma\,\cup<z_{i_{0}}>\}\setminus\{f_{j},<f_{i}(k)>\}\}$.

If $A_{1},A_{2}$ are both non empty, then $\sigma\setminus\{f_{i}\}\sim\ <b>\forall\ b\in B_{j}$. If $x\in T_{1}\setminus\{f_{1}(k),f_{2}(k)\}$, then $\sigma\setminus\{<x>\}\sim\ <x>$ and $\sigma\setminus\{<x>\}\sim z_{1}\ldots z_{k-1}xz_{k+1}\ldots z_{n}$ for $x\in\{f_{1}(k),f_{2}(k)\}$. Therefore, $y_{1}=\sigma\setminus\{<x>\}$, where $x\in A_{1}\cup A_{2}$.

Let $b\in B\subset B_{1}\cup B_{2}$. If $b=z_{l}\in B_{i}$, then $f_{i}(l)\neq z_{l},f_{j}(l)=z_{l}$ and $f_{i}(l)\in A_{i}$. The element $f_{i}(l)$ in $A_{i}$ is called the element corresponding to $b$ and is denoted by $x_{b}$. Similarly, if $x\in A_{1}\cup A_{2}$, then $b_{x}$ is called the element of $B$ corresponding to $x$.

Let $B_{1}=\{b_{11}<b_{12}<\ldots<b_{1r}\},B_{2}=\{b_{21}<b_{22}<\ldots<b_{2l}\}$ and $B=\{b_{1}<b_{2}<\ldots<b_{q}\}$ be ordered sets and the corresponding sets $A_{1},A_{2}$ and $A$ be $\{x_{11},x_{12},\ldots,x_{1r}\}$, $\{x_{21},\ldots,x_{2l}\}$ and $\{x_{1},\ldots,x_{q}\}$ respectively i.e. $x_{1i}=x_{b_{1i}},x_{2j}=x_{b_{2j}}$ and $x_{t}=x_{b_{t}}\ \forall\ 1\leq i\leq r,1\leq j\leq l,1\leq t\leq q$. Let $\alpha=\{\sigma\,\cup\{<b>|\ b\in B\}\}\setminus\{<x>|\ x\in A\}$. By Lemma 4.11, $\alpha\in\mu_{b_{1}}(S_{b_{1}})$. Here, $B$ and at least one of $B_{1}$ or $B_{2}$ are always non empty sets. Let $b_{0}=\text{min}\{b\ |\ b\in B_{1}\cup B_{2}\}$.

Lemma 4.17**.**

Let $B\neq B_{1},B_{2},\eta=\{\sigma\,\cup\{<b_{i}>|\ 2\leq i\leq q\}\}\setminus\{<x_{i}>|\ 1\leq i\leq q\}\in S_{b_{1}}$ and $\alpha=\mu_{b_{1}}(\eta)=\eta\ \cup<b_{1}>\ \in\mu_{b_{1}}(S_{b_{1}})$. If

• (i)

$\alpha\in c$* and $b_{0}\notin B$, then the only facets $y$ of $\alpha$ which can belong to $c$ are of the form $\alpha\setminus\{<x>\}$, where $b_{x}<b_{1}$ and $x\in\{A_{1}\cup A_{2}\}\setminus A$.*

• (ii)

$b_{0}=b_{1}$, then no facet of $\alpha$ belongs to $c$ and thus $\alpha\notin c$.

Proof.

By Lemma 4.16, $\forall\ x\in T_{1},\alpha\setminus\{<x>\},\alpha\setminus\{f_{1}\}$ and $\alpha\setminus\{f_{2}\}$ belong to $S_{1}$. For $b\in B\setminus\{b_{1}\}$, by Lemma 4.11, $\alpha\setminus\{<b>\}\in\mu_{b_{1}}(S_{b_{1}})$ and $\alpha\setminus\{<b_{1}>\}=\eta$. Since $\alpha\in c$ implies that $\eta\in c$, the facet $y\neq\alpha\setminus\{<b_{1}>\}$. For $x\in A_{1}\cup A_{2}$ such that $b_{x}>b_{1}$, by Lemma 4.11, $\alpha\setminus\{<x>\}\in\mu_{b_{1}}(S_{b_{1}})$. Thus, the only possible facets of $\alpha$ which can belong to $c$ are of the type $\alpha\setminus\{<x>\}$, where $b_{x}<b_{1}$.

If $b_{0}=b_{1}$, then no facet of $\alpha$ can belong to $c$. Since $\alpha\in\mu_{b_{1}}(S_{b_{1}})$, it is not a critical cell and thus $\alpha\notin c$.

∎

From the above Lemma, we conclude that no cell $\alpha=\{\sigma\ \cup\{<b>|\ b\in B\}\}\setminus\{<x>|\ x\in A\}$ belongs to $c$, if $B\cap B_{1}$ and $B\cap B_{2}$ are both non empty. This follows by an inductive argument. If $b_{1}\neq b_{0}$, then by Lemma 4.17 $(i)$, $\alpha_{1}=\{\alpha\ \cup<b_{x}>\}\setminus\{<x>\}\in c$, where $b_{x}<b_{1}$, if $\alpha\in c$. Inductively $\alpha_{t}=\{\alpha_{t-1}\cup<b_{0}>\}\setminus\{<x_{b_{0}}>\}\in c$ which contradicts Lemma 4.17 $(ii)$. Similarly if $b_{0}\in B\cap B_{i}$ and $B\subsetneq B_{i}$ for $i\in\{1,2\}$, then $\alpha\notin c$. We now show that if $b_{11}$ or $b_{21}\in B$, then $B=B_{1}$ or $B_{2}$.

Lemma 4.18**.**

Let $\alpha=\{\sigma\,\cup\{<b>|\ b\in B\}\}\setminus\{<x>|\ x\in A\}$ and $b_{11}\in B$. If $\alpha\in c$, then $B=B_{1}$.

Proof.

By Lemma 4.11, $\alpha\in\mu_{b_{11}}(S_{b_{11}})$ and is thus not a critical cell. If $B\subsetneq B_{1}$, then by Lemma 4.17 $(i)$, the facet of $\alpha\in c$ is of the form $\alpha\setminus\{<x>\}$, where $b_{x}<b_{11}$. Since $b_{11}$ is the least element of $B_{1},b_{x}\in B_{2}$, a contradiction. Thus $B=B_{1}$.

∎

Hence, we can conclude that $y_{1}=\sigma\setminus\{<x_{1r}>\}$ or $\sigma\setminus\{<x_{2l}>\}$.

Subcase 3. $y_{1}=\sigma\setminus\{<x_{1r}>\}$.

Using Lemmas 4.17 and 4.18, $y_{i}=\{\sigma\,\cup\{<b_{1t}>|\ r-i+1<t\leq r\}\}\setminus\{<x_{1r}>,\ldots,<x_{1r-i+1}>\},1\leq i\leq r$ and $\mu(y_{r})=\sigma\cup\{<b>|\ b\in B_{1}\}\setminus\{<a>|\ a\in A_{1}\}$. From Lemma 4.16 and Lemma 4.17, $y_{r+1}=\mu(y_{r})\setminus\{f_{2}\}$ or $\mu(y_{r})\setminus\{<x>\},x\in A_{2}$ and $b_{x}<b_{11}$. If $y_{r+1}=\mu(y_{r})\setminus\{<x>\}$, then $\mu(y_{r+1})=y_{r+1}\cup<b_{x}>,B\cap B_{1},B\cap B_{2}\neq\emptyset$ and therefore $\mu(y_{r+1})\notin c$. Thus, $y_{r+1}=\mu(y_{r})\setminus\{f_{2}\}$. The vertex $w_{1}\ldots w_{n}$ defined by $w_{i}=f_{1}(i)\ \forall\ i\neq k$ and $w_{k}=1$ is a neighbor of $y_{r+1}$. Let $w_{i_{0}}=\text{min}\{w_{1},\ldots,\widehat{w_{k}},\ldots,w_{n}\}$.

1. (a)

$f_{1}(k)<w_{i_{0}}$.

$y_{r+1}=\mu(y_{r})\setminus\{f_{2}\}$ satisfies all the properties of Theorem 4.8 and is hence a critical $p$-cell. The alternating path $c=\{\sigma,\sigma\setminus\{<x_{1r}>\},\{\sigma\,\cup<b_{1r}>\}\setminus\{<x_{1r}>\},\ldots,y_{r},\mu(y_{r}),\mu(y_{r})\setminus\{f_{2}\}\}$. 2. (b)

$f_{1}(k)>w_{i_{0}}$.

Since $f_{1}(i)=w_{i}\ \forall\ i\neq k,f_{1}(k)>w_{i_{0}}$, by Lemma 4.13, $y_{r+1}=\mu(y_{r})\setminus\{f_{2}\}\in S_{w_{i_{0}}}$, i.e. $\mu(y_{r+1})=\{\mu(y_{r})\,\cup<w_{i_{0}}>\}\setminus\{f_{2}\}$ ($<w_{i_{0}}>\ \notin\mu(y_{r})$, since $w_{i_{0}}=z_{i}\Rightarrow w_{i_{0}}\in B_{2}$ and $<w_{i_{0}}>\ \notin\mu(y_{r})$ and $w_{i_{0}}\in A_{1}\Rightarrow<w_{i_{0}}>\ \notin\mu(y_{r})$).

Claim 3**.**

$y_{r+2}=\mu(y_{r+1})\setminus\{<f_{1}(k)>\}$.

If $<x>\ \in\mu(y_{r+1})$, then $x\in T_{1}\cup B_{1}\cup A_{2}\cup\{w_{i_{0}}\}$. If $x\in T_{1}\cup B_{1}\cup A_{2}$ and $x\neq f_{1}(k)$, then $x\notin\text{Im}\,f_{1}$ and hence $\mu(y_{r+1})\setminus\{<x>\}\in S_{1}$. Further, $\mu(y_{r+1})\setminus\{f_{1}\}\sim\ <x>$ for any $x\in A_{1}$. So, $y_{r+2}=\mu(y_{r+1})\setminus\{<f_{1}(k)>\}$. The claim is thus proved.

$y_{r+2}\subset N(v_{1}\ldots v_{n})$, where $v_{i_{0}}=1,v_{k}=f_{1}(k)$ and $v_{i}=w_{i}\ \forall\ i\neq i_{0},k$, satisfies all the conditions of Theorem 4.8 and is thus a critical $p$-cell. The alternating path $c=\{\sigma,\sigma\setminus\{<x_{1r}>\},\{\sigma\,\cup<b_{1r}\}\setminus\{<x_{1r}>\},\ldots,\{\sigma\,\cup\{<b_{11}>,\ldots,<b_{1r}>\}\}\setminus\{<x_{11}>,\ldots,<x_{1r}>\},\mu(y_{r})\setminus\{f_{2}\},\{\mu(y_{r})\,\cup<w_{i_{0}}>\}\setminus\{f_{2}\},\{\mu(y_{r})\,\cup<w_{i_{0}}>\}\setminus\{f_{2},<f_{1}(k)>\}\}$.

Subcase 4. $y_{1}=\sigma\setminus\{<x_{2l}>\}$.

Here, $y_{i}=\{\sigma\,\cup\{<b_{2t}>|\ l-i+1<t\leq l\}\}\setminus\{<x_{2l}>,\ldots,<x_{2l-i+1}>\},1\leq i\leq l$ and $y_{l+1}=\mu(y_{l})\setminus\{f_{1}\}$. The vertex $u_{1}\ldots u_{n}$, where $u_{i}=f_{2}(i),i\neq k$ and $u_{k}=1$ is a neighbor of $y_{l+1}$. Let $u=\text{min}\{u_{1},\ldots,u_{n}\}\setminus\{1\}$.

1. (a)

$f_{2}(k)<u$.

The alternating path $c$ is $\{\sigma,\sigma\setminus\{<x_{2l}>\},\ldots,\mu(y_{l})=\sigma\cup\{<b>|\ b\in B_{2}\}\setminus\{<a>|\ a\in A_{2}\},\mu(y_{l})\setminus\{f_{1}\}\}.$ 2. (b)

$f_{2}(k)>u$.

In this case, $c=\{\sigma,\sigma\setminus\{<x_{2l}>\},\ldots,\mu(y_{l}),\mu(y_{l})\setminus\{f_{1}\},\{\mu(y_{l})\,\cup<u>\}\setminus\{f_{1}\},\{\mu(y_{l})$ $\,\cup<u>\}\setminus\{f_{1},<f_{2}(k)>\}\}.$

∎

Consider the critical cell $\tau=\{<2>,<3>,\ldots,<m>\}\subset N(<1>)$. Since every facet of $\tau$ is in $S_{1}$, there exists no alternating path from $\tau$.

Our objective now is to first study the $\mathbb{Z}_{2}$ homology groups of the Morse complex corresponding to the acyclic matching $\mu$ on $P$.

Let the Discrete Morse Complex corresponding to the acyclic matching $\mu$ be $\mathcal{M}=(\mathcal{M}_{n},\partial_{n})$, $n\geq 0$ where $\mathcal{M}_{i}$ denotes the free abelian groups over $\mathbb{Z}_{2}$ generated by the critical $i$-cells. For any two critical cells $\tau$ and $\sigma$ such that dim($\tau$) = dim($\sigma)$ +1, the incidence number $[\tau:\sigma]$ is either 0 or 1.

Let $C_{i}$ denote the set of critical cells of dimension $i$. Since $<1>$ is the only [math]-dimensional critical cell, $C_{0}=\{<1>\}$. If $n\geq 3$, using Theorem 4.8 and Lemma 4.9, we can conclude that $C_{i}=\emptyset$ for $0<i\leq p-1$. Let $C_{p}=\{\alpha_{1},\ldots,\alpha_{r_{1}}\}$ and $C_{p+1}=\{\tau_{1},\ldots,\tau_{r_{2}}\}$. Let $A=[a_{ij}]$ be a matrix of order $|C_{p}|\times|C_{p+1}|$, where $a_{ij}=1$, if there exists an alternating path from $\tau_{j}$ to $\alpha_{i}$ and [math] if no such path exists. Using Theorem 4.15, each column of $A$ contains exactly two non zero elements which are $1$ (except, when $\{<2>,\ldots,<m>\}\in C_{p+1}$ and in this case the column of $A$ corresponding to this cell is zero, as there is no alternating path from $\{<2>,\ldots,<m>\}$ to any critical cell).

Theorem 4.19**.**

Let $m-n=p\geq 1$. Then $H_{p}(\mathcal{N}(G);\mathbb{Z}_{2})\neq 0.$

Proof.

$\mathcal{M}_{p}\cong\mathbb{Z}_{2}^{|C_{p}|}$ and $\mathcal{M}_{p+1}\cong\mathbb{Z}_{2}^{|C_{p+1}|}$. Since each column of $A$ is either zero (when $\{<2>,\ldots,<m>\}\in C_{p+1}$) or contains exactly two non zero elements, both being $1$, the column sum is zero (mod $2$). Therefore, rank($A$) $<|C_{p}|$. In particular, rank of the boundary map $\partial_{p+1}:\mathbb{Z}_{2}^{|C_{p+1}|}\rightarrow\mathbb{Z}_{2}^{|C_{p}|}$ is strictly less than $|C_{p}|$.

Since $\text{Hom}(K_{2}\times K_{n},K_{m})\simeq\text{Hom}(K_{2},K_{m}^{K_{n}})\simeq\mathcal{N}(K_{m}^{K_{n}})\simeq\mathcal{N}(G)$ and the maximum degree of $K_{2}\times K_{n}$ is $n-1$, $\text{conn}(\text{Hom}(K_{2}\times K_{n},K_{m}))=\text{conn}(\mathcal{N}(G))\geq m-n-1$. Hence $\mathcal{N}(G)$ is path connected and therefore $H_{0}(\mathcal{N}(G);\mathbb{Z}_{2})\cong\mathbb{Z}_{2}$.

If $p=1$, then since $\mathcal{M}_{0}\cong\mathbb{Z}_{2}$, Ker($\partial_{1}$) $\cong\mathbb{Z}_{2}^{|C_{1}|}$, where $\partial_{1}:\mathcal{M}_{1}\longrightarrow\mathcal{M}_{0}$. Hence $H_{1}(\mathcal{N}(G);\mathbb{Z}_{2})\neq 0$. If $p>1$, then $C_{p-1}=\emptyset$ implies that $\mathcal{M}_{p-1}=0$. Thus, Ker($\partial_{p}$) $\cong\mathbb{Z}_{2}^{|C_{p}|}$, where $\partial_{p}:\mathcal{M}_{p}\longrightarrow\mathcal{M}_{p-1}$ is the boundary map. Since rank($\partial_{p+1})<|C_{p}|$, we see that $H_{p}(\mathcal{N}(G);\mathbb{Z}_{2})\neq 0$. ∎

We have developed all the necessary tools to prove the main result. We recall the following results to prove Theorem 1.1.

Proposition 4.20**.**

(Theorem 3A.3, [7])

If $C$ is a chain complex of free abelian groups, then there exist short exact sequences

$0\longrightarrow H_{n}(C;\mathbb{Z})\otimes\mathbb{Z}_{2}\longrightarrow H_{n}(C;\mathbb{Z}_{2})\longrightarrow$* Tor*$(H_{n-1}(C;\mathbb{Z}),\mathbb{Z}_{2})\longrightarrow 0$**

for all n and these sequences split.

Proposition 4.21**.**

*(The Hurewicz Theorem)

If a space $X$ is $(n-1)$ connected, $n\geq 2$, then $\tilde{H_{i}}(X;\mathbb{Z})=0$ for $i<n$ and $\pi_{n}(X)\cong H_{n}(X;\mathbb{Z})$.*

Proof of Theorem 1.1.

If $n=2$, then $\text{Hom}(K_{2}\times K_{2},K_{m})\simeq\text{Hom}(K_{2}\sqcup K_{2},K_{m})\simeq$ Hom $(K_{2},K_{m})$ $\times$ Hom $(K_{2},K_{m})\simeq S^{m-2}\times S^{m-2}$. Hence $\text{conn}(\text{Hom}(K_{2}\times K_{2},$ $K_{m}))=\text{conn}(S^{m-2}\times S^{m-2})=m-3$.

Let $n\geq 3$. If $m<n$, 4.1 shows that $K_{m}^{K_{n}}$ can be folded to the graph $G^{\prime}$, where $V(G^{\prime})=\{<x>|\ x\in[m]\}$. Then $N(<x>)=\{<y>|\ y\in[m]\setminus\{x\}\}$, for all $<x>\ \in V(G^{\prime})$ and therefore $\mathcal{N}(G^{\prime})$ is homotopic to the simplicial boundary of $(m-1)$-simplex. Hence $\text{Hom}(K_{2}\times K_{n},K_{m})\simeq\mathcal{N}(K_{m}^{K_{n}})\simeq\mathcal{N}(G^{\prime})\simeq S^{m-2}$. Therefore $\text{conn}(\text{Hom}(K_{2}\times K_{n},K_{m}))=m-3$.

If $m=n$, then for any $f\in V(K_{n}^{K_{n}})$ with $\text{Im}\,f=[n]$, $N(f)=\{f\}$. Since $n\geq 2$, $\mathcal{N}(K_{n}^{K_{n}})$ is disconnected.

Assume $m-n=p\geq 1.$ Since $\text{conn}(\text{Hom}(K_{2}\times K_{n},K_{m}))\geq m-n-1$, $\tilde{H_{i}}(\text{Hom}(K_{2}\times K_{n},K_{m});\mathbb{Z})=0\ \forall\ 0\leq i\leq p-1.$ Since $\text{Hom}(K_{2}\times K_{n},K_{m})\simeq\mathcal{N}(G),\tilde{H_{i}}(\mathcal{N}(G);\mathbb{Z})=0\ \forall\ 0\leq i\leq p-1$ and $H_{p}(\text{Hom}(K_{2}\times K_{n},K_{m});\mathbb{Z}_{2})\neq 0$, by Theorem 4.19. By Proposition 4.20, $H_{p}(\mathcal{N}(G);\mathbb{Z}_{2})\cong H_{p}(\mathcal{N}(G);\mathbb{Z})\otimes\mathbb{Z}_{2}\oplus\text{Tor}(H_{p-1}(\mathcal{N}(G);\mathbb{Z}),\mathbb{Z}_{2})$. Since $p\geq 1,\mathcal{N}(G)$ is path connected and hence $\text{Tor}(H_{0}(\mathcal{N}(G);\mathbb{Z}),\mathbb{Z}_{2})\cong\text{Tor}(\mathbb{Z},\mathbb{Z}_{2})=0$. Further, since for any $0<q<p$, $\tilde{H_{i}}(\mathcal{N}(G);\mathbb{Z})=0$, $H_{p}(\mathcal{N}(G);\mathbb{Z})=0$ implies $H_{p}(\mathcal{N}(G);\mathbb{Z}_{2})=0$, which is a contradiction. Hence $H_{p}(\text{Hom}(K_{2}\times K_{n},K_{m});\mathbb{Z})\cong H_{p}(\mathcal{N}(G);\mathbb{Z})\neq 0$. If $p=1$, then since the abelinazitation of $\pi_{1}(\text{Hom}(K_{2}\times K_{n},K_{m}))$ is $H_{1}(\text{Hom}$ $(K_{2}\times K_{n},K_{m});\mathbb{Z})\neq 0$, $\pi_{1}(\text{Hom}(K_{2}\times K_{n},K_{m}))\neq 0$. For $p>1$, since $\text{Hom}(K_{2}\times K_{n},K_{m})$ is simply connected and $H_{p}(\text{Hom}(K_{2}$ $\times K_{n},K_{m});\mathbb{Z})\neq 0$, the result follows from Proposition 4.21.