# The first homology group with twisted coefficients for the mapping class
group of a non–orientable surface of genus three with two boundary components

Piotr Pawlak Michał Stukow  Institute of Mathematics, Faculty of Mathematics,
Physics and Informatics, University of Gdańsk, 80-308 Gdańsk, Poland
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We determine the first homology group with coefficients in
$H_{1}(N;\mathbb{Z})$ for the mapping class group of a non–orientable surface
$N$ of genus three with two boundary components.

###### Key words and phrases:

Mapping class group, Homology of groups, Non–orientable surface

###### 2020 Mathematics Subject Classification:

Primary 57K20; Secondary 20J06, 55N25, 22F50

##  1\. Introduction

Let $N_{g,s}^{n}$ be a smooth, non–orientable, compact surface of genus $g$
with $s$ boundary components and $n$ punctures. If $s$ and/or $n$ is zero,
then we omit it from the notation. If we do not want to emphasise the numbers
$g,s,n$, we simply write $N$ for a surface $N_{g,s}^{n}$. Recall that $N_{g}$
is a connected sum of $g$ projective planes and $N_{g,s}^{n}$ is obtained from
$N_{g}$ by removing $s$ open discs and specifying a set
$\Sigma=\\{z_{1},\ldots,z_{n}\\}$ of $n$ distinguished points in the interior
of $N$.

Let ${\textrm{Diff}}(N)$ be the group of all diffeomorphisms $h\colon N\to N$
such that $h$ is the identity on each boundary component. By
${\mathcal{M}}(N)$ we denote the quotient group of ${\textrm{Diff}}(N)$ by the
subgroup consisting of maps isotopic to the identity, where we assume that
isotopies are the identity on each boundary component. ${\mathcal{M}}(N)$ is
called the _mapping class group_ of $N$.

Let ${\mathcal{PM}}^{+}(N)$ be the subgroup of ${\mathcal{M}}(N)$ consisting
of elements which fix $\Sigma$ pointwise and preserve a local orientation
around the punctures $\\{z_{1},\ldots,z_{k}\\}$.

###  1.1. Main results

We showed in [2] that if $N_{g,s}$ is a non–orientable surface of genus $g\geq
3$ with $s\leq 1$ boundary components, then

$H_{1}({\mathcal{M}}(N_{g,s});H_{1}(N_{g,s};\mathbb{Z}))\cong\begin{cases}\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}&\text{if
$g\in\\{3,4,5,6\\},$}\\\ \mathbb{Z}_{2}\oplus\mathbb{Z}_{2}&\text{if $g\geq
7$.}\end{cases}$

It is quite natural to extend this result to surfaces with more than one
boundary component. However, this is quite challenging, because we do not have
a simple presentation for the mapping class group in this case. Nevertheless,
in the forthcoming paper [1] we managed (among other homological results) to
compute the first homology group
$H_{1}({\mathcal{M}}(N_{g,s});H_{1}(N_{g,s};\mathbb{Z}))$ for any $s\geq 0$ by
bounding it both from above and below. However, it turned out that the proof
of nontriviality of two important families of homology generators (denoted by
$a_{1,3+j}$ and $u_{1,3+j}$ in [1]) is beyond the methods used in [1]. By a
simple reduction, the required argument in [1] boils down to the exceptional
case of a non-orientable surface $N_{3}^{2}$ of genus 3 with two punctures.
The main purpose of this paper is to fill this gap, and to prove the following
two theorems

######  Theorem 1.1.

If $N_{3,2}$ is a non–orientable surface of genus $g=3$ with 2 boundary
components, then

$H_{1}({\mathcal{M}}(N_{3,2});H_{1}(N_{3,2};\mathbb{Z}))\cong\mathbb{Z}_{2}^{6}.$

######  Theorem 1.2.

If $N_{3}^{2}$ is a non–orientable surface of genus $g=3$ with 2 punctures,
then

$H_{1}({\mathcal{PM}^{+}}(N_{3}^{2});H_{1}(N_{3}^{2};\mathbb{Z}))\cong\mathbb{Z}_{2}^{5}.$

###  1.2. Homology of groups

Let us briefly review how to compute the first homology of a group with
twisted coefficients – for more details see Section 5 of [3] and references
there.

For a given group $G$ and $G$-module $M$ (that is $\mathbb{Z}G$-module) we
define $C_{2}(G)$ and $C_{1}(G)$ as the free $G$-modules generated
respectively by symbols $[h_{1}|h_{2}]$ and $[h_{1}]$, where $h_{i}\in G$. We
define also $C_{0}(G)$ as the free $G$-module generated by the empty bracket
$[\cdot]$. Then the first homology group $H_{1}(G;M)$ is the first homology
group of the complex

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where

(1.1) $\displaystyle\partial_{2}([h_{1}|h_{2}])$
$\displaystyle=h_{1}[h_{2}]-[h_{1}h_{2}]+[h_{1}],$
$\displaystyle\partial_{1}([h])$ $\displaystyle=h[\cdot]-[\cdot].$

For simplicity, we write $\otimes_{G}=\otimes$ and $\partial\otimes{\rm
id}=\overline{\partial}$ henceforth.

If the group $G$ has a presentation $G=\langle X\,|\,R\rangle$ and

$\langle\overline{X}\rangle=\langle[x]\otimes m\;|\;x\in X,m\in
M\rangle\subseteq C_{1}(G)\otimes M,$

then $H_{1}(G;M)$ is a quotient of
$\langle\overline{X}\rangle\cap\ker\overline{\partial}_{1}$.

The kernel of this quotient corresponds to relations in $G$ (that is elements
of $R$). To be more precise, if $r\in R$ has the form $x_{1}\cdots
x_{k}=y_{1}\cdots y_{n}$ and $m\in M$, then $r$ gives the relation (in
$H_{1}(G;M)$)

(1.2) $\overline{r}\otimes m\\!:\ \sum_{i=1}^{k}x_{1}\cdots
x_{i-1}[x_{i}]\otimes m=\sum_{i=1}^{n}y_{1}\cdots y_{i-1}[y_{i}]\otimes m.$

Then

$H_{1}(G;M)=\langle\overline{X}\rangle\cap\ker\overline{\partial}_{1}/\langle\overline{R}\rangle,$

where

$\overline{R}=\\{\overline{r}\otimes m\;|\;r\in R,m\in M\\}.$

##  2\. Presentations for the groups ${\mathcal{PM}}^{+}(N_{3}^{2})$ and
${\mathcal{M}}(N_{3,2})$

Represent the surface $N_{3,2}$ as a sphere with three crosscaps
$\mu_{1},\mu_{2},\mu_{3}$ and two boundary components (Figure 1). Let

$\alpha_{1},\alpha_{2},\varepsilon_{1},\varepsilon_{2},\delta_{1},\delta_{2},\beta_{1},\beta_{2},\beta_{3}$

be two–sided circles indicated in Figure 1.

Figure 1. Surface $N_{3,2}$ as a sphere with three crosscaps.

Small arrows in that figure indicate directions of Dehn twists

$a_{1},a_{2},e_{1},e_{2},d_{1},d_{2},b_{1},b_{2},b_{3}$

associated with these circles.

Let $u$ be a _crosscap transposition_ , that is the map which interchanges the
first two crosscaps (see Figure 2).

Figure 2. Crosscap transposition $u$.

######  Theorem 2.1.

The mapping class group ${\mathcal{M}}(N_{3,2})$ of a non–orientable surface
of genus 3 with two boundary components admits a presentation with generators
$\\{a_{1},a_{2},e_{1},e_{2},d_{1},d_{2},b_{1},b_{2},b_{3},u\\}$. The defining
relations are

  * (1a)

$a_{1}e_{1}=e_{1}a_{1}$,

  * (1b)

$a_{1}e_{2}=e_{2}a_{1}$,

  * (1c)

$e_{1}e_{2}=e_{2}e_{1}$,

  * (2a)

$a_{1}a_{2}a_{1}=a_{2}a_{1}a_{2}$,

  * (2b)

$e_{1}a_{2}e_{1}=a_{2}e_{1}a_{2}$,

  * (2c)

$e_{2}a_{2}e_{2}=a_{2}e_{2}a_{2}$,

  * (3)

$a_{1}ua_{1}=u$,

  * (4)

$ue_{2}=a_{2}ue_{2}a_{2}$,

  * (5)

$ub_{1}=b_{1}u$,

  * (6)

$ub_{3}=b_{3}u$,

  * (7)

$a_{2}b_{2}=b_{2}a_{2}$,

  * (8)

$(e_{1}u)^{2}=d_{1}b_{1}$,

  * (9a)

$b_{3}=(e_{2}u)^{2}$,

  * (9b)

$(e_{2}u)^{2}=(a_{2}e_{2}a_{1}^{2})^{3}$,

  * (10)

$ue_{1}u^{-1}b_{2}u=a_{2}b_{1}^{-1}a_{2}^{-1}ue_{1}$,

  * (11)

$b_{3}b_{1}(b_{2}u)^{2}=d_{2}^{2}d_{1}u^{2}$,

  * (12)

$(a_{2}e_{2}e_{1}a_{1})^{3}=d_{1}d_{2}$,

  * (13a)

$d_{1}a_{1}=a_{1}d_{1}$,

  * (13b)

$d_{1}a_{2}=a_{2}d_{1}$,

  * (13c)

$d_{1}e_{1}=e_{1}d_{1}$,

  * (13d)

$d_{1}u=ud_{1}$,

  * (13e)

$d_{2}u=ud_{2}$.

###### Proof.

By Theorem 7.18 of [4], the mapping class group ${\mathcal{M}}(N_{3,2})$
admits a presentation with generators
$\\{A_{1},A_{2},A_{3},B,D_{1},D_{2},D_{3},U,C_{1},C_{2}\\}$ and relations

  * (S1)

$A_{i}A_{j}=A_{j}A_{i}$, $i,j=1,2,3$,

  * (S2)

$A_{i}BA_{i}=BA_{i}B$, $i=1,2,3$,

  * (S3)

$UA_{1}U^{-1}=A_{1}^{-1}$,

  * (S4)

$UBU^{-1}=A_{3}^{-1}B^{-1}A_{3}$,

  * (S5)

$UD_{1}=D_{1}U$,

  * (S6)

$UD_{3}=D_{3}U$,

  * (S7)

$BD_{2}=D_{2}B$,

  * (S8)

$(UA_{2})^{2}=D_{1}C_{1}$,

  * (S9)

$(A_{1}^{2}A_{3}B)^{3}=(UA_{3})^{2}=D_{3}$,

  * (S10)

$A_{2}^{-1}UD_{2}U^{-1}A_{2}=UB^{-1}D_{1}^{-1}BU^{-1}$,

  * (S11)

$(UD_{2})^{2}D_{1}D_{3}=U^{2}C_{1}C_{2}^{2}$,

  * (S12)

$(A_{1}A_{2}A_{3}B)^{3}=C_{1}C_{2}=C_{2}C_{1}$,

  * (S13)

$C_{i}A_{j}=A_{j}C_{i}$, $C_{i}D_{j}=D_{j}C_{i}$, $C_{i}B=BC_{i}$,
$C_{i}U=UC_{i}$, $i=1,2$, $j=1,2,3$.

Moreover, the topological model for the surface $N_{3,2}$ can be chosen in
such way that

$\displaystyle A_{1}$ $\displaystyle=a_{1}^{-1},\ A_{2}=e_{1}^{-1},\
A_{3}=e_{2}^{-1},$ $\displaystyle B$ $\displaystyle=a_{2}^{-1},\ U=u^{-1},$
$\displaystyle D_{1}$ $\displaystyle=b_{1}^{-1},\ D_{2}=b_{2}^{-1},\
D_{3}=b_{3}^{-1},$ $\displaystyle C_{1}$ $\displaystyle=d_{1}^{-1},\
C_{2}=d_{2}^{-1}.$

To see this correspondence, simply reflect Figure 14 of [4] across the
vertical axis passing through the second crosscap.

Therefore, we can easily rewrite Relations (S1)–(S13) into our set of
generators. In most cases this is straightforward, but note the following
remarks.

  * •

Generators $A_{3},D_{1},D_{2},D_{3}$ and $C_{2}$ are superfluous in the above
presentation – they can be easily removed by use of Relations (S2)&(S4), (S8),
(S10), (S9) and (S12) respectively.

  * •

By the above remark, to ensure that $C_{1}$ and $C_{2}$ are central in
${\mathcal{M}}(N_{3,2})$ (Relation (S13)), it is enough to have commutativity
of these elements with $A_{1},A_{2},B$ and $U$.

  * •

In order to show that Relations (13a)–(13e) fully replace Relations (S13), we
still need to show that $d_{2}=d_{1}^{-1}(a_{2}e_{2}e_{1}a_{1})^{3}$ commutes
with $a_{1},a_{2}$ and $e_{1}$, or equivalently, that
$(a_{2}e_{2}e_{1}a_{1})^{3}$ commutes with $a_{1},a_{2}$ and $e_{1}$.

Let us first show this in the most difficult of these cases – the case of the
commutativity with $a_{2}$. The computations repeatedly make use of Relations
(1a)–(2c).

$\displaystyle a_{2}^{-1}(a_{2}e_{2}e_{1}a_{1})^{3}a_{2}$
$\displaystyle=\underset{\rightarrow}{[a_{2}^{-1}]}a_{2}e_{2}e_{1}a_{1}a_{2}\underset{\rightarrow}{[e_{2}]}e_{1}a_{1}a_{2}e_{2}e_{1}a_{1}a_{2}$
$\displaystyle=a_{2}a_{2}^{-1}e_{2}e_{1}a_{1}a_{2}\underset{\rightarrow}{[e_{1}]}a_{1}[e_{2}a_{2}e_{2}]e_{1}a_{1}a_{2}$
$\displaystyle=a_{2}a_{2}^{-1}e_{2}e_{1}[a_{1}a_{2}a_{1}]e_{1}a_{2}e_{2}a_{2}e_{1}a_{1}a_{2}$
$\displaystyle=a_{2}a_{2}^{-1}e_{2}e_{1}a_{2}a_{1}[a_{2}e_{1}a_{2}]e_{2}a_{2}e_{1}a_{1}a_{2}$
$\displaystyle=a_{2}a_{2}^{-1}e_{2}e_{1}a_{2}a_{1}\underset{\leftarrow}{[e_{1}]}a_{2}\underset{\rightarrow}{[e_{1}]}e_{2}a_{2}e_{1}a_{1}a_{2}$
$\displaystyle=a_{2}a_{2}^{-1}e_{2}[e_{1}a_{2}e_{1}]a_{1}a_{2}e_{2}[e_{1}a_{2}e_{1}]a_{1}a_{2}$
$\displaystyle=a_{2}[a_{2}^{-1}e_{2}a_{2}]e_{1}a_{2}a_{1}a_{2}e_{2}a_{2}e_{1}[a_{2}a_{1}a_{2}]$
$\displaystyle=a_{2}e_{2}a_{2}\underset{\rightarrow}{[e_{2}^{-1}]}e_{1}[a_{2}a_{1}a_{2}]e_{2}a_{2}\underset{\rightarrow}{[e_{1}]}a_{1}a_{2}a_{1}$
$\displaystyle=a_{2}e_{2}a_{2}e_{1}\underset{\rightarrow}{[e_{2}^{-1}]}a_{1}a_{2}a_{1}\underset{\leftarrow}{[e_{2}]}a_{2}a_{1}e_{1}a_{2}a_{1}$
$\displaystyle=a_{2}e_{2}a_{2}e_{1}a_{1}[e_{2}^{-1}a_{2}e_{2}][a_{1}a_{2}a_{1}]e_{1}a_{2}a_{1}$
$\displaystyle=a_{2}e_{2}a_{2}e_{1}a_{1}a_{2}e_{2}\underline{a_{2}}^{-1}\underline{a_{2}}\underset{\leftarrow}{[a_{1}]}[a_{2}e_{1}a_{2}]a_{1}$
$\displaystyle=a_{2}e_{2}a_{2}e_{1}[a_{1}a_{2}a_{1}]e_{2}e_{1}a_{2}e_{1}a_{1}$
$\displaystyle=a_{2}e_{2}[a_{2}e_{1}a_{2}]a_{1}a_{2}e_{2}\underset{\leftarrow}{[e_{1}]}a_{2}e_{1}a_{1}$
$\displaystyle=a_{2}e_{2}e_{1}a_{2}\underset{\rightarrow}{[e_{1}]}a_{1}a_{2}e_{1}e_{2}a_{2}e_{1}a_{1}$
$\displaystyle=a_{2}e_{2}e_{1}a_{2}a_{1}[e_{1}a_{2}e_{1}]e_{2}a_{2}e_{1}a_{1}$
$\displaystyle=a_{2}e_{2}e_{1}[a_{2}a_{1}a_{2}]e_{1}[a_{2}e_{2}a_{2}]e_{1}a_{1}$
$\displaystyle=(a_{2}e_{2}e_{1}a_{1})a_{2}\underset{\rightarrow}{[a_{1}]}e_{1}\underset{\leftarrow}{[e_{2}]}(a_{2}e_{2}e_{1}a_{1})=(a_{2}e_{2}e_{1}a_{1})^{3}.$

Now we show the commutativity of $(a_{2}e_{2}e_{1}a_{1})^{3}$ with $a_{1}$.

$\displaystyle a_{1}^{-1}(a_{2}e_{2}e_{1}a_{1})^{3}a_{1}$
$\displaystyle=a_{1}^{-1}a_{2}e_{2}e_{1}\underset{\leftarrow}{[a_{1}]}a_{2}e_{2}e_{1}a_{1}a_{2}e_{2}e_{1}\underset{\leftarrow}{[a_{1}]}a_{1}$
$\displaystyle=[a_{1}^{-1}a_{2}a_{1}]\underset{\rightarrow}{[e_{2}]}e_{1}a_{2}e_{2}e_{1}[a_{1}a_{2}a_{1}]e_{2}e_{1}a_{1}$
$\displaystyle=a_{2}a_{1}a_{2}^{-1}e_{1}[e_{2}a_{2}e_{2}]e_{1}a_{2}a_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}a_{1}[a_{2}^{-1}e_{1}a_{2}]e_{2}[a_{2}e_{1}a_{2}]a_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}a_{1}e_{1}a_{2}\underline{e_{1}^{-1}}e_{2}\underline{e_{1}}a_{2}e_{1}a_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}a_{1}e_{1}[a_{2}e_{2}a_{2}]e_{1}a_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}\underset{\rightarrow}{[a_{1}]}e_{1}\underset{\leftarrow}{[e_{2}]}(a_{2}e_{2}e_{1}a_{1})(a_{2}e_{2}e_{1}a_{1})=(a_{2}e_{2}e_{1}a_{1})^{3}.$

And finally, we show the commutativity of $(a_{2}e_{2}e_{1}a_{1})^{3}$ with
$e_{1}$.

$\displaystyle e_{1}^{-1}(a_{2}e_{2}e_{1}a_{1})^{3}e_{1}$
$\displaystyle=e_{1}^{-1}a_{2}e_{2}\underset{\leftarrow}{[e_{1}]}a_{1}a_{2}e_{2}\underset{\rightarrow}{[e_{1}]}a_{1}a_{2}e_{2}e_{1}a_{1}\underset{\leftarrow}{[e_{1}]}$
$\displaystyle=[e_{1}^{-1}a_{2}e_{1}]\underset{\rightarrow}{[e_{2}]}a_{1}a_{2}e_{2}a_{1}[e_{1}a_{2}e_{1}]e_{2}e_{1}a_{1}$
$\displaystyle=a_{2}e_{1}a_{2}^{-1}a_{1}[e_{2}a_{2}e_{2}]a_{1}a_{2}e_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}e_{1}[a_{2}^{-1}a_{1}a_{2}]e_{2}[a_{2}a_{1}a_{2}]e_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}e_{1}a_{1}a_{2}\underline{a_{1}^{-1}}e_{2}\underline{a_{1}}a_{2}a_{1}\underset{\leftarrow}{[e_{1}]}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}e_{1}a_{1}[a_{2}e_{2}a_{2}]e_{1}a_{1}(a_{2}e_{2}e_{1}a_{1})$
$\displaystyle=a_{2}e_{1}\underset{\rightarrow}{[a_{1}]}\underset{\leftarrow}{[e_{2}]}(a_{2}e_{2}e_{1}a_{1})(a_{2}e_{2}e_{1}a_{1})=(a_{2}e_{2}e_{1}a_{1})^{3}.$

∎

######  Remark 2.2.

As we observed in the proof of Theorem 2.1, generators: $e_{2}$, $b_{1}$,
$b_{2}$, $b_{3}$, $d_{2}$ are superfluous in the above presentation. However,
we decided to leave these generators in order to have simpler relations. This
simplifies the computations in the proofs of Theorems 1.1 and 1.2.

######  Theorem 2.3.

The mapping class group ${\mathcal{PM}^{+}}(N_{3}^{2})$ of a non–orientable
surface of genus 3 with two punctures admits a presentation with generators
$\\{a_{1},a_{2},e_{1},e_{2},b_{1},b_{2},b_{3},u\\}$. The defining relations
are Relations (1a)–(7),(9),(10) from the statement of Theorem 2.1 and
additionally

  * (8’)

$(e_{1}u)^{2}=b_{1}$,

  * (11’)

$b_{3}b_{1}(b_{2}u)^{2}=u^{2}$,

  * (12’)

$(a_{2}e_{2}e_{1}a_{1})^{3}=1$.

###### Proof.

By Theorem 7.17 of [4], the mapping class group
${\mathcal{PM}^{+}}(N_{3}^{2})$ admits a presentations with Relations
(S1)–(S7), (S9), (S10) from the proof of Theorem 2.1 and additionally

  * (S8’)

$(UA_{2})^{2}=D_{1}$,

  * (S11’)

$(UD_{2})^{2}D_{1}D_{3}=U^{2}$,

  * (S12’)

$(A_{1}A_{2}A_{3}B)^{3}=1$.

If we rewrite Relations (S8’), (S11’) and (S12’) into our set of generators,
we get Relations (8’), (11’) and (12’) respectively. ∎

##  3\. Action of ${\mathcal{M}}(N_{3,s}^{n})$ on
$H_{1}(N_{3,s}^{n};\mathbb{Z})$, where $(s,n)=(0,2)$ or $(s,n)=(2,0)$

In the rest of the paper assume that $(s,n)=(0,2)$ or $(s,n)=(2,0)$. Let
$\gamma_{1},\gamma_{2},\gamma_{3},\delta_{1},\delta_{2}$ be circles indicated
in Figure 1. Note that $\gamma_{1},\gamma_{2},\gamma_{3}$ are one–sided,
$\delta_{1},\delta_{2}$ are two–sided and the $\mathbb{Z}$-module
$H_{1}(N_{3,s}^{n};\mathbb{Z})$ is freely generated by homology classes
$[\gamma_{1}],[\gamma_{2}],[\gamma_{3}],[\delta_{1}]$. In abuse of notation we
will not distinguish between the curves
$\gamma_{1},\gamma_{2},\gamma_{3},\delta_{1}$ and their cycle classes.

The mapping class group ${\mathcal{M}}(N_{3,s}^{n})$ acts on
$H_{1}(N_{3,s}^{n};\mathbb{Z})$, hence we have a representation

$\psi\colon{\mathcal{M}}(N_{3,s}^{n})\to\textrm{Aut}(H_{1}(N_{3,s}^{n};\mathbb{Z})).$

It is straightforward to check that

(3.1) $\displaystyle\psi(a_{1})$ $\displaystyle=\begin{bmatrix}0&1&0&0\\\
-1&2&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{bmatrix},\psi(a_{1}^{-1})=\begin{bmatrix}2&-1&0&0\\\ 1&0&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{bmatrix}$ $\displaystyle\psi(a_{2})$
$\displaystyle=\begin{bmatrix}1&0&0&0\\\ 0&0&1&0\\\ 0&-1&2&0\\\
0&0&0&1\end{bmatrix},\psi(a_{2}^{-1})=\begin{bmatrix}1&0&0&0\\\ 0&2&-1&0\\\
0&1&0&0\\\ 0&0&0&1\end{bmatrix}$ $\displaystyle\psi(u)$
$\displaystyle=\psi(u^{-1})=\begin{bmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{bmatrix}$ (3.2) $\displaystyle\psi(e_{1})$
$\displaystyle=\begin{bmatrix}0&1&0&0\\\ -1&2&0&0\\\ 0&0&1&0\\\
-1&1&0&1\end{bmatrix},\psi(e_{1}^{-1})=\begin{bmatrix}2&-1&0&0\\\ 1&0&0&0\\\
0&0&1&0\\\ 1&-1&0&1\end{bmatrix}$ $\displaystyle\psi(e_{2})$
$\displaystyle=\begin{bmatrix}2&-1&0&0\\\ 1&0&0&0\\\ 2&-2&1&0\\\
0&0&0&1\end{bmatrix},\psi(e_{2}^{-1})=\begin{bmatrix}0&1&0&0\\\ -1&2&0&0\\\
-2&2&1&0\\\ 0&0&0&1\end{bmatrix}$ $\displaystyle\psi(b_{j}^{\pm 1})$
$\displaystyle=\psi(d_{j}^{\pm 1})=I_{4}$

where $I_{4}$ is the identity matrix of rank 4.

##  4\. Computing $\langle\overline{X}\rangle\cap\ker\overline{\partial}_{1}$

Let $G={\mathcal{M}}(N_{3,s}^{n})$ and $M=H_{1}(N_{3,s}^{n};\mathbb{Z})$ as in
the previous section, and define

$\xi_{i}=\begin{cases}\gamma_{i}&\text{for $i=1,2,3$,}\\\ \delta_{1}&\text{for
$i=4$.}\end{cases}$

If $h\in G$, then

$\overline{\partial}_{1}([h]\otimes\xi_{i})=(h-1)[\cdot]\otimes\xi_{i}=(\psi(h)^{-1}-I_{4})\xi_{i},$

where we identified $C_{0}(G)\otimes M$ with $M$ by the map $[\cdot]\otimes
m\mapsto m$.

Let us denote

$[a_{j}]\otimes\xi_{i},\ [u]\otimes\xi_{i},\ [e_{j}]\otimes\xi_{i},\
[b_{j}]\otimes\xi_{i},\ [d_{j}]\otimes\xi_{i}$

respectively by

$a_{j,i},\ u_{i},\ e_{j,i},\ b_{j,i},\ d_{j,i}.$

Using the formulas (3.1) and (3.2), we obtain

$\displaystyle\overline{\partial}_{1}(a_{j,i})$
$\displaystyle=\begin{cases}\gamma_{j}+\gamma_{j+1}&\text{if $i=j$}\\\
-\gamma_{j}-\gamma_{j+1}&\text{if $i=j+1$}\\\ 0&\text{otherwise,}\end{cases}$
$\displaystyle\overline{\partial}_{1}(u_{i})$
$\displaystyle=\begin{cases}-\gamma_{1}+\gamma_{2}&\text{if $i=1$}\\\
\gamma_{1}-\gamma_{2}&\text{if $i=2$}\\\ 0&\text{otherwise,}\end{cases}$
$\displaystyle\overline{\partial}_{1}(e_{1,i})$
$\displaystyle=\begin{cases}\gamma_{1}+\gamma_{2}+\delta_{1}&\text{if
$i=1$}\\\ -\gamma_{1}-\gamma_{2}-\delta_{1}&\text{if $i=2$}\\\
0&\text{otherwise,}\end{cases}$
$\displaystyle\overline{\partial}_{1}(e_{2,i})$
$\displaystyle=\begin{cases}-\gamma_{1}-\gamma_{2}-2\gamma_{3}&\text{if
$i=1$}\\\ \gamma_{1}+\gamma_{2}+2\gamma_{3}&\text{if $i=2$}\\\
0&\text{otherwise,}\end{cases}$
$\displaystyle\overline{\partial}_{1}(b_{j,i})$
$\displaystyle=\overline{\partial}_{1}(d_{j,i})=0.$

The above formulas show that all of the following elements are contained in
$\ker\overline{\partial}_{1}$

  1. (K1)

$a_{j,i}$ for $j=1,2$ and $i\in\\{1,2,3,4\\}\setminus\\{j,j+1\\}$,

  2. (K2)

$a_{j,j}+a_{j,j+1}$ for $j=1,2$,

  3. (K3)

$u_{i}$ for $i=3,4$,

  4. (K4)

$u_{1}+u_{2}$,

  5. (K5)

$e_{j,i}$ for $j=1,2$ and $i=3,4$,

  6. (K6)

$e_{j,1}+e_{j,2}$ for $j=1,2$,

  7. (K7)

$e_{2,1}+2a_{2,2}-u_{1}$,

  8. (K8)

$b_{j,i}$ for $j=1,2,3$ and $i=1,2,3,4$,

  9. (K9)

$d_{j,i}$ for $j=1,2$ and $i=1,2,3,4$.

######  Proposition 4.1.

Let $G={\mathcal{M}}(N_{3,2})$. Then
$\langle\overline{X}\rangle\cap\ker\overline{\partial}_{1}$ is the abelian
group generated freely by Generators (K1)–(K9).

###### Proof.

By Theorem 2.1, $\langle\overline{X}\rangle$ is generated freely by
$\\{a_{j,i},u_{i},e_{j,i},b_{j,i},d_{j,i}\\}$. Suppose that
$h\in\langle\overline{X}\rangle\cap\ker\overline{\partial}_{1}$. We will show
that $h$ can be uniquely expressed as a linear combination of Generators
(K1)–(K9) specified in the statement of the theorem.

We decompose $h$ as follows:

  * •

$h=h_{0}=h_{1}+h_{2}$, where $h_{1}$ is a combination of Generators (K1)–(K2)
and $h_{2}$ does not contain $a_{j,i}$ with $i\neq j$;

  * •

$h_{2}=h_{3}+h_{4}$, where $h_{3}$ is a combination of Generators (K3)–(K4)
and $h_{4}$ does not contain $u_{i}$ with $i\neq 1$;

  * •

$h_{4}=h_{5}+h_{6}$, where $h_{5}$ is a combination of Generators (K5)–(K7)
and $h_{6}$ does not contain $e_{j,i}$ for $(i,j)\neq(1,1)$;

  * •

$h_{6}=h_{7}+h_{8}$, where $h_{7}$ is a combination of Generators (K8) and
$h_{8}$ does not contain $b_{j,i}$;

  * •

$h_{8}=h_{9}+h_{10}$, where $h_{9}$ is a combination of Generators (K9) and
$h_{10}$ does not contain $d_{j,i}$.

Observe also that for each $k=0,\ldots,8$, $h_{k+1}$ and $h_{k+2}$ are
uniquely determined by $h_{k}$. Element $h_{10}$ has the form

$h_{10}=k_{1}a_{1,1}+k_{2}a_{2,2}+le_{1,1}+mu_{1}$

for some integers $k_{1},k_{2},l,m$. Hence

$\displaystyle 0=$
$\displaystyle\overline{\partial}_{1}(h_{10})=k_{1}(\gamma_{1}+\gamma_{2})+k_{2}(\gamma_{2}+\gamma_{3})+l(\gamma_{1}+\gamma_{2}+\delta_{1})+m(-\gamma_{1}+\gamma_{2}).$

This implies that $l=0,k_{2}=0$, and then $k_{1}=m=0$ and thus $h_{10}=0$. ∎

By an analogous argument, Theorem 2.3 implies that

######  Proposition 4.2.

Let $G={\mathcal{PM}}^{+}(N_{3}^{2})$. Then
$\langle\overline{X}\rangle\cap\ker\overline{\partial}_{1}$ is the abelian
group generated freely by Generators (K1)–(K8).

##  5\. Computing $H_{1}({\mathcal{M}}(N_{3,2});H_{1}(N_{3,2};\mathbb{Z}))$

The goal of this section is to prove that

######  Proposition 5.1.

The abelian group
$H_{1}({\mathcal{PM}^{+}}(N_{3}^{2});H_{1}(N_{3}^{2};\mathbb{Z}))$ has a
presentation with generators

$a_{1,3},\ a_{1,4},\ a_{1,1}+a_{1,2},\ u_{3},\ u_{4},\ d_{1,1},$

and relations

$2a_{1,3}=2a_{1,4}=2\left(a_{1,1}+a_{1,2}\right)=2u_{3}=2u_{4}=2d_{1,1}=0.$

###### Proof.

We start with the generators provided by Proposition 4.1, and using the
formula (1.2), we rewrite relations from Theorem 2.1 as relations in
$H_{1}({\mathcal{M}}(N_{3,2});H_{1}(N_{3,2};\mathbb{Z}))$.

### (3)

Relation (3) gives

$\displaystyle 0$
$\displaystyle=([a_{1}]+a_{1}[u]+a_{1}u[a_{1}]-[u])\otimes\xi_{i}$
$\displaystyle=[a_{1}]\otimes(I_{4}+\psi(u^{-1}a_{1}^{-1}))\xi_{i}+[u]\otimes(\psi(a_{1}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[a_{1}]\otimes\begin{bmatrix}2&0&0&0\\\ 2&0&0&0\\\ 0&0&2&0\\\
0&0&0&2\end{bmatrix}\xi_{i}+[u]\otimes\begin{bmatrix}1&-1&0&0\\\ 1&-1&0&0\\\
0&0&0&0\\\ 0&0&0&0\end{bmatrix}\xi_{i}.$

We conclude from this relation that Generator (K4) is trivial:

(5.1) $u_{1}+u_{2}=0,$

and each of the generators $a_{1,3},\ a_{1,4},\ a_{1,1}+a_{1,2}$ is of order
at most 2:

(5.2) $2a_{1,3}=2a_{1,4}=2(a_{1,1}+a_{1,2})=0.$

### (2a)–(2c)

Relation (2a) gives

$\displaystyle 0=$
$\displaystyle([a_{1}]+a_{1}[a_{2}]+a_{1}a_{2}[a_{1}]-[a_{2}]-a_{2}[a_{1}]-a_{2}a_{1}[a_{2}])\otimes\xi_{i}$
$\displaystyle=$
$\displaystyle[a_{1}]\otimes(I_{4}+\psi(a_{2}^{-1}a_{1}^{-1})-\psi(a_{2}^{-1}))\xi_{i}$
$\displaystyle+[a_{2}]\otimes(\psi(a_{1}^{-1})-I_{4}-\psi(a_{1}^{-1}a_{2}^{-1}))\xi_{i}$
$\displaystyle=$ $\displaystyle[a_{1}]\otimes\begin{bmatrix}2&-1&0&0\\\
2&-1&0&0\\\ 1&-1&1&0\\\
0&0&0&1\end{bmatrix}\xi_{i}+[a_{2}]\otimes\begin{bmatrix}-1&1&-1&0\\\
0&-1&0&0\\\ 0&-1&0&0\\\ 0&0&0&-1\end{bmatrix}\xi_{i}.$

We conclude from this relation, that

(5.3) $\displaystyle a_{2,4}$ $\displaystyle=a_{1,4},$ $\displaystyle a_{2,1}$
$\displaystyle=a_{1,3}.$

Hence generators $a_{2,1}$ and $a_{2,4}$ are superfluous. Moreover,

(5.4) $a_{2,2}+a_{2,3}=a_{1,1}+a_{1,2},$

which together with the formula (5.2) imply that Generators (K2) generate a
cyclic group of order at most 2.

Relation (2b) gives

$\displaystyle 0=$
$\displaystyle([e_{1}]+e_{1}[a_{2}]+e_{1}a_{2}[e_{1}]-[a_{2}]-a_{2}[e_{1}]-a_{2}e_{1}[a_{2}])\otimes\xi_{i}$
$\displaystyle=$
$\displaystyle[e_{1}]\otimes(I_{4}+\psi(a_{2}^{-1}e_{1}^{-1})-\psi(a_{2}^{-1}))\xi_{i}$
$\displaystyle+[a_{2}]\otimes(\psi(e_{1}^{-1})-I_{4}-\psi(e_{1}^{-1}a_{2}^{-1}))\xi_{i}$
$\displaystyle=$ $\displaystyle[e_{1}]\otimes\begin{bmatrix}2&-1&0&0\\\
2&-1&0&0\\\ 1&-1&1&0\\\
1&-1&0&1\end{bmatrix}\xi_{i}+[a_{2}]\otimes\begin{bmatrix}-1&1&-1&0\\\
0&-1&0&0\\\ 0&-1&0&0\\\ 0&1&-1&-1\end{bmatrix}\xi_{i}.$

This, the formulas (5.2), (5.3) and (5.4) imply that

(5.5) $\displaystyle e_{1,4}$ $\displaystyle=a_{2,4}=a_{1,4},$ $\displaystyle
e_{1,3}$ $\displaystyle=a_{2,1}+a_{2,4}=a_{1,3}+a_{1,4},$ $\displaystyle
e_{1,1}+e_{1,2}$
$\displaystyle=(a_{2,2}+a_{2,3})+a_{2,4}=(a_{1,1}+a_{1,2})+a_{1,4}$

and Generators (K5) and (K6) are superfluous for $j=1$.

Relation (2c) gives

$\displaystyle 0=$
$\displaystyle([e_{2}]+e_{2}[a_{2}]+e_{2}a_{2}[e_{2}]-[a_{2}]-a_{2}[e_{2}]-a_{2}e_{2}[a_{2}])\otimes\xi_{i}$
$\displaystyle=$
$\displaystyle[e_{2}]\otimes(I_{4}+\psi(a_{2}^{-1}e_{2}^{-1})-\psi(a_{2}^{-1}))\xi_{i}$
$\displaystyle+[a_{2}]\otimes(\psi(e_{2}^{-1})-I_{4}-\psi(e_{2}^{-1}a_{2}^{-1}))\xi_{i}$
$\displaystyle=$ $\displaystyle[e_{2}]\otimes\begin{bmatrix}0&1&0&0\\\
0&1&0&0\\\ -1&1&1&0\\\
0&0&0&1\end{bmatrix}\xi_{i}+[a_{2}]\otimes\begin{bmatrix}-1&-1&1&0\\\
0&-3&2&0\\\ 0&-3&2&0\\\ 0&0&0&-1\end{bmatrix}\xi_{i}.$

This relation, the formulas (5.2), (5.3) and (5.4) imply that

(5.6) $\displaystyle e_{2,4}$ $\displaystyle=a_{2,4}=a_{1,4},$ $\displaystyle
e_{2,3}$ $\displaystyle=-a_{2,1}=a_{1,3},$ $\displaystyle e_{2,1}+e_{2,2}$
$\displaystyle=a_{2,2}+a_{2,3}=a_{1,1}+a_{1,2}.$

Thus Generators (K5) and (K6) are superfluous for $j=2$.

### (1a)–(1c)

Relation (1a) gives

$\displaystyle 0$
$\displaystyle=([a_{1}]+a_{1}[e_{1}]-[e_{1}]-e_{1}[a_{1}])\otimes\xi_{i}$
$\displaystyle=[a_{1}]\otimes(I_{4}-\psi(e_{1}^{-1}))\xi_{i}+[e_{1}]\otimes(\psi(a_{1}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[a_{1}]\otimes\begin{bmatrix}-1&1&0&0\\\ -1&1&0&0\\\ 0&0&0&0\\\
-1&1&0&0\end{bmatrix}\xi_{i}+[e_{1}]\otimes\begin{bmatrix}1&-1&0&0\\\
1&-1&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{bmatrix}\xi_{i}.$

This relation gives no new information.

Relation (1b) gives

$\displaystyle 0$
$\displaystyle=([a_{1}]+a_{1}[e_{2}]-[e_{2}]-e_{2}[a_{1}])\otimes\xi_{i}$
$\displaystyle=[a_{1}]\otimes(I_{4}-\psi(e_{2}^{-1}))\xi_{i}+[e_{2}]\otimes(\psi(a_{1}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[a_{1}]\otimes\begin{bmatrix}1&-1&0&0\\\ 1&-1&0&0\\\
2&-2&0&0\\\
0&0&0&0\end{bmatrix}\xi_{i}+[e_{2}]\otimes\begin{bmatrix}1&-1&0&0\\\
1&-1&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{bmatrix}\xi_{i}.$

This relation gives no new information.

Relation (1c) gives

$\displaystyle 0$
$\displaystyle=([e_{1}]+e_{1}[e_{2}]-[e_{2}]-e_{2}[e_{1}])\otimes\xi_{i}$
$\displaystyle=[e_{1}]\otimes(I_{4}-\psi(e_{2}^{-1}))\xi_{i}+[e_{2}]\otimes(\psi(e_{1}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[e_{1}]\otimes\begin{bmatrix}1&-1&0&0\\\ 1&-1&0&0\\\
2&-2&0&0\\\
0&0&0&0\end{bmatrix}\xi_{i}+[e_{2}]\otimes\begin{bmatrix}1&-1&0&0\\\
1&-1&0&0\\\ 0&0&0&0\\\ 1&-1&0&0\end{bmatrix}\xi_{i}.$

This relation gives no new information.

### (4)

Relation (4) gives

$\displaystyle 0=$
$\displaystyle([u]+u[e_{2}]-[a_{2}]-a_{2}[u]-a_{2}u[e_{2}]-a_{2}ue_{2}[a_{2}])\otimes\xi_{i}$
$\displaystyle=$
$\displaystyle[u]\otimes(I_{4}-\psi(a_{2}^{-1}))\xi_{i}+[e_{2}]\otimes(\psi(u^{-1})-\psi(u^{-1}a_{2}^{-1}))\xi_{i}$
$\displaystyle+[a_{2}]\otimes(-I_{4}-\psi(e_{2}^{-1}u^{-1}a_{2}^{-1}))\xi_{i}$
$\displaystyle=$ $\displaystyle[u]\otimes\begin{bmatrix}0&0&0&0\\\ 0&-1&1&0\\\
0&-1&1&0\\\
0&0&0&0\end{bmatrix}\xi_{i}+[e_{2}]\otimes\begin{bmatrix}0&-1&1&0\\\
0&0&0&0\\\ 0&-1&1&0\\\ 0&0&0&0\end{bmatrix}\xi_{i}$
$\displaystyle+[a_{2}]\otimes\begin{bmatrix}-2&0&0&0\\\ -2&1&-1&0\\\
-2&3&-3&0\\\ 0&0&0&-2\end{bmatrix}\xi_{i}.$

This relation, the formulas (5.1), (5.2), (5.4) and (5.6) give

(5.7) $\displaystyle e_{2,1}+2a_{2,2}-u_{1}$
$\displaystyle=3(a_{2,2}+a_{2,3})-(u_{1}+u_{2})-u_{3}-e_{2,3}$
$\displaystyle=(a_{1,1}+a_{1,2})-u_{3}+a_{1,3}$

which implies that Generator (K7) is superfluous.

### (9b)

If we denote

$\displaystyle M$
$\displaystyle=I+\psi(u^{-1}e_{2}^{-1})=\begin{bmatrix}0&2&0&0\\\ 0&2&0&0\\\
-2&2&2&0\\\ 0&0&0&2\end{bmatrix}$ $\displaystyle N$
$\displaystyle=I+\psi(a_{1}^{-2}e_{2}^{-1}a_{2}^{-1})+\psi(a_{1}^{-2}e_{2}^{-1}a_{2}^{-1})^{2}=\begin{bmatrix}3&-1&1&0\\\
3&-1&1&0\\\ 3&-1&1&0\\\ 0&0&0&3\end{bmatrix}$

then Relation (9b) gives

$\displaystyle 0=$
$\displaystyle[e_{2}]\otimes\left(M-\psi(a_{2}^{-1})N\right)\xi_{i}+[u]\otimes\psi(e_{2}^{-1})M\xi_{i}-[a_{2}]\otimes
N\xi_{i}$
$\displaystyle-[a_{1}]\otimes\left(\psi(e_{2}^{-1}a_{2}^{-1})N+\psi(a_{1}^{-1}e_{2}^{-1}a_{2}^{-1})N\right)\xi_{i}$
$\displaystyle=$ $\displaystyle[e_{2}]\otimes\begin{bmatrix}-3&3&-1&0\\\
-3&3&-1&0\\\ -5&3&1&0\\\
0&0&0&-1\end{bmatrix}\xi_{i}+[u]\otimes\begin{bmatrix}0&2&0&0\\\ 0&2&0&0\\\
-2&2&2&0\\\ 0&0&0&2\end{bmatrix}\xi_{i}$
$\displaystyle+[a_{2}]\otimes\begin{bmatrix}-3&1&-1&0\\\ -3&1&-1&0\\\
-3&1&-1&0\\\
0&0&0&-3\end{bmatrix}\xi_{i}+[a_{1}]\otimes\begin{bmatrix}-6&2&-2&0\\\
-6&2&-2&0\\\ -6&2&-2&0\\\ 0&0&0&-6\end{bmatrix}\xi_{i}.$

The last two columns of this relation, the formulas (5.2), (5.3), (5.4) and
(5.6) imply that

(5.8) $\displaystyle 2u_{3}=$
$\displaystyle(e_{2,1}+e_{2,2}+a_{2,2}+a_{2,3})+(a_{2,1}-e_{2,3})+$
$\displaystyle+2(a_{1,1}+a_{1,2})+2a_{1,3}=0,$ $\displaystyle 2u_{4}=$
$\displaystyle e_{2,4}+3a_{2,4}+6a_{1,4}=0.$

As for the first two columns of this relation

$\displaystyle-2u_{3}=$ $\displaystyle
3(e_{2,1}+e_{2,2}+a_{2,2}+a_{2,3})+5(a_{2,1}+e_{2,3})+$
$\displaystyle+6(a_{1,1}+a_{1,2})+6a_{1,3}-2a_{2,1}=0,$
$\displaystyle-2u_{3}=$
$\displaystyle(3e_{2,1}+3e_{2,2}+2a_{1,1}+2a_{1,2}+a_{2,2}+a_{2,3})+3(a_{2,1}+e_{2,3})+$
$\displaystyle+2(u_{1,1}+u_{1,2})+2a_{1,3}-2a_{2,1}=0,$

they do not give any additional information.

### (9a)

Let $M=I+\psi(u^{-1}e_{2}^{-1})$ as above. Then Relation (9a) gives

$\displaystyle 0$ $\displaystyle=[e_{2}]\otimes
M\xi_{i}+[u]\otimes(\psi(e_{2}^{-1})M)\xi_{i}-[b_{3}]\otimes\xi_{i}=$
$\displaystyle=[e_{2}]\otimes\begin{bmatrix}0&2&0&0\\\ 0&2&0&0\\\ -2&2&2&0\\\
0&0&0&2\end{bmatrix}\xi_{i}+[u]\otimes\begin{bmatrix}0&2&0&0\\\ 0&2&0&0\\\
-2&2&2&0\\\ 0&0&0&2\end{bmatrix}\xi_{i}-[b_{3}]\otimes\xi_{i}.$

This, the formulas (5.1), (5.2), (5.6) and (5.8) imply that generators

(5.9) $\displaystyle b_{3,4}$ $\displaystyle=2e_{2,4}+2u_{4}=0,$
$\displaystyle b_{3,3}$ $\displaystyle=2e_{2,3}+2u_{3}=0,$ $\displaystyle
b_{3,1}$ $\displaystyle=-2e_{2,3}-2u_{3}=0,$ $\displaystyle b_{3,2}$
$\displaystyle=2(e_{2,1}+e_{2,2})+2e_{2,3}+2(u_{1}+u_{2})+2u_{3}=0$

are trivial.

### (12)

Let

$M=I+\psi(a_{1}^{-1}e_{1}^{-1}e_{2}^{-1}a_{2}^{-1})+\psi(a_{1}^{-1}e_{1}^{-1}e_{2}^{-1}a_{2}^{-1})^{2}=\begin{bmatrix}3&-1&1&0\\\
3&-1&1&0\\\ 3&-1&1&0\\\ 0&-1&1&3\end{bmatrix}.$

Relation (12), the formulas (5.2), (5.3), (5.4), (5.5) and (5.6) give

$\displaystyle 0=$ $\displaystyle[a_{2}]\otimes
M\xi_{i}+[e_{2}]\otimes(\psi(a_{2}^{-1})M)\xi_{i}+[e_{1}]\otimes(\psi(e_{2}^{-1}a_{2}^{-1})M)\xi_{i}$
$\displaystyle+[a_{1}]\otimes(\psi(e_{1}^{-1}e_{2}^{-1}a_{2}^{-1})M)\xi_{i}-[d_{1}]\otimes\xi_{i}-[d_{2}]\otimes\psi(d_{1}^{-1})\xi_{i}$
$\displaystyle=$
$\displaystyle([a_{2}]+[e_{2}]+[e_{1}]+[a_{1}])\otimes\begin{bmatrix}3&-1&1&0\\\
3&-1&1&0\\\ 3&-1&1&0\\\
0&-1&1&3\end{bmatrix}\xi_{i}-[d_{1}]\otimes\xi_{i}-[d_{2}]\otimes\xi_{i}=$
$\displaystyle=$ $\displaystyle-[d_{1}]\otimes\xi_{i}-[d_{2}]\otimes\xi_{i}.$

This implies that

(5.10) $d_{2,i}=-d_{1,i},\quad\text{for $i=1,2,3,4$.}$

### (13a)–(13e)

Relation (13a) gives

$\displaystyle 0$
$\displaystyle=([d_{1}]+d_{1}[a_{1}]-[a_{1}]-a_{1}[d_{1}])\otimes\xi_{i}$
$\displaystyle=[d_{1}]\otimes(I_{4}-\psi(a_{1}^{-1}))\xi_{i}+[a_{1}]\otimes(\psi(d_{1}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[d_{1}]\otimes\begin{bmatrix}-1&1&0&0\\\ -1&1&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{bmatrix}\xi_{i}.$

Thus

$d_{1,2}=-d_{1,1}.$

By a similar argument, Relation (13d) gives

$\displaystyle 0$ $\displaystyle=[d_{1}]\otimes\begin{bmatrix}1&-1&0&0\\\
-1&1&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{bmatrix}\xi_{i}.$

Hence

(5.11) $d_{1,2}=d_{1,1},$

and

(5.12) $2d_{1,1}=0.$

Analogously, Relation (13b) gives

$\displaystyle 0$ $\displaystyle=[d_{1}]\otimes\begin{bmatrix}0&0&0&0\\\
0&-1&1&0\\\ 0&-1&1&0\\\ 0&0&0&0\end{bmatrix}\xi_{i},$

which implies that

$d_{1,3}=-d_{1,2}.$

This together with the formulas (5.11) and (5.12) imply that

(5.13) $d_{1,3}=d_{1,1}.$

Relation (13c) gives

$\displaystyle 0$
$\displaystyle=([d_{1}]+d_{1}[e_{1}]-[e_{1}]-e_{1}[d_{1}])\otimes\xi_{i}$
$\displaystyle=[d_{1}]\otimes(I_{4}-\psi(e_{1}^{-1}))\xi_{i}+[e_{1}]\otimes(\psi(d_{1}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[d_{1}]\otimes\begin{bmatrix}-1&1&0&0\\\ -1&1&0&0\\\ 0&0&0&0\\\
-1&1&0&0\end{bmatrix}\xi_{i}.$

Hence, by the formulas (5.11) and (5.12)

(5.14) $d_{1,4}=-d_{1,1}-d_{1,2}=0$

and we conclude that Generators (K9) generate a cyclic group of order at most
2.

Relation (13e)

$\displaystyle 0$ $\displaystyle=[d_{2}]\otimes\begin{bmatrix}1&-1&0&0\\\
-1&1&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{bmatrix}\xi_{i}$

gives no new information.

### (8)

Let

$M=I+\psi(u^{-1}e_{1}^{-1})=\begin{bmatrix}2&0&0&0\\\ 2&0&0&0\\\ 0&0&2&0\\\
1&-1&0&2\end{bmatrix}.$

Relation (8) gives

$\displaystyle 0$ $\displaystyle=[e_{1}]\otimes
M\xi_{i}+[u]\otimes(\psi^{-1}(e_{1}^{-1})M)\xi_{i}-[d_{1}]\otimes\xi_{i}-[b_{1}]\otimes\xi_{i}$
$\displaystyle=[e_{1}]\otimes\begin{bmatrix}2&0&0&0\\\ 2&0&0&0\\\ 0&0&2&0\\\
1&-1&0&2\end{bmatrix}\xi_{i}+[u]\otimes\begin{bmatrix}2&0&0&0\\\ 2&0&0&0\\\
0&0&2&0\\\
1&-1&0&2\end{bmatrix}\xi_{i}-[d_{1}]\otimes\xi_{i}-[b_{1}]\otimes\xi_{i}.$

This, the formulas (5.1), (5.2), (5.4), (5.5), (5.8), (5.13) and (5.14) imply
that generators

(5.15) $\displaystyle b_{1,1}$
$\displaystyle=e_{1,4}+u_{4}-d_{1,1}=a_{1,4}+u_{4}+d_{1,1},$ $\displaystyle
b_{1,2}$ $\displaystyle=-e_{1,4}-u_{4}-d_{1,2}=a_{1,4}+u_{4}+d_{1,1},$
$\displaystyle b_{1,3}$ $\displaystyle=2e_{1,3}+2u_{3}-d_{1,3}=d_{1,1},$
$\displaystyle b_{1,4}$ $\displaystyle=2e_{1,4}+2u_{4}-d_{1,4}=0$

are superfluous.

### (5), (6)

Relations (5) and (6) give

$\displaystyle 0$
$\displaystyle=([u]+u[b_{k}]-[b_{k}]-b_{k}[u])\otimes\xi_{i}$
$\displaystyle=[u]\otimes(I_{4}-\psi(b_{k}^{-1}))\xi_{i}+[b_{k}]\otimes(\psi(u^{-1})-I_{4})\xi_{i}$
$\displaystyle=[b_{k}]\otimes\begin{bmatrix}-1&1&0&0\\\ 1&-1&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{bmatrix}\xi_{i}$

for $k=1,3$. This relation gives no new information.

### (10)

Observe first that if $g\in G$, then the formulas (1.1), (1.2) and the
relation $g^{-1}g=1$ imply that

$[g^{-1}]\otimes\xi_{i}=-g^{-1}[g]\otimes\xi_{i}.$

Hence Relation (10) gives

$\displaystyle 0=$
$\displaystyle[u]\otimes\left(I_{4}+\psi(b_{2}^{-1}ue_{1}^{-1}u^{-1})-\psi(a_{2}b_{1}a_{2}^{-1})\right)\xi_{i}+[u^{-1}]\otimes\psi(e_{1}^{-1}u^{-1})\xi_{i}$
$\displaystyle+[e_{1}]\otimes\left(\psi(u^{-1})-\psi(u^{-1}a_{2}b_{1}a_{2}^{-1})\right)\xi_{i}$
$\displaystyle+[b_{2}]\otimes\psi(ue_{1}^{-1}u^{-1})\xi_{i}-[b_{1}^{-1}]\otimes\psi(a_{2}^{-1})\xi_{i}$
$\displaystyle-[a_{2}]\otimes\xi_{i}-[a_{2}^{-1}]\otimes\psi(b_{1}a_{2}^{-1})\xi_{i}$
$\displaystyle=$
$\displaystyle[u]\otimes\left(I_{4}+\psi(b_{2}^{-1}ue_{1}^{-1}u^{-1})-\psi(a_{2}b_{1}a_{2}^{-1})-\psi(ue_{1}^{-1}u^{-1})\right)\xi_{i}$
$\displaystyle+[e_{1}]\otimes\left(\psi(u^{-1})-\psi(u^{-1}a_{2}b_{1}a_{2}^{-1})\right)\xi_{i}$
$\displaystyle+[b_{2}]\otimes\psi(ue_{1}^{-1}u^{-1})\xi_{i}+[b_{1}]\otimes\psi(b_{1}a_{2}^{-1})\xi_{i}$
$\displaystyle+[a_{2}]\otimes\left(\psi(a_{2}b_{1}a_{2}^{-1})-I_{4}\right)\xi_{i}.$

Recall that $\psi(b_{j}^{\pm 1})=I_{4}$, hence the above formula takes form

$\displaystyle 0=$
$\displaystyle[b_{2}]\otimes\psi(ue_{1}^{-1}u^{-1})\xi_{i}+[b_{1}]\otimes\psi(a_{2}^{-1})\xi_{i}$
$\displaystyle=$ $\displaystyle[b_{2}]\otimes\begin{bmatrix}0&1&0&0\\\
-1&2&0&0\\\ 0&0&1&0\\\
-1&1&0&1\end{bmatrix}\xi_{i}+[b_{1}]\otimes\begin{bmatrix}1&0&0&0\\\
0&2&-1&0\\\ 0&1&0&0\\\ 0&0&0&1\end{bmatrix}\xi_{i}.$

This, the formulas (5.2), (5.8) and (5.15) imply that generators

(5.16) $\displaystyle b_{2,4}$ $\displaystyle=-b_{1,4}=0,$ $\displaystyle
b_{2,3}$ $\displaystyle=b_{1,2}=a_{1,4}+u_{4}+d_{1,1},$ $\displaystyle
b_{2,2}$ $\displaystyle=b_{1,1}-b_{2,4}=a_{1,4}+u_{4}+d_{1,1},$ $\displaystyle
b_{2,1}$ $\displaystyle=-2b_{1,2}-b_{1,3}-2b_{2,2}-b_{2,4}=-d_{1,1}$

are superfluous.

### (7)

Relation (7) gives

$\displaystyle 0$
$\displaystyle=([a_{2}]+a_{2}[b_{2}]-[b_{2}]-b_{2}[a_{2}])\otimes\xi_{i}$
$\displaystyle=[a_{2}]\otimes(I_{4}-\psi(b_{2}^{-1}))\xi_{i}+[b_{2}]\otimes(\psi(a_{2}^{-1})-I_{4})\xi_{i}$
$\displaystyle=[b_{2}]\otimes\begin{bmatrix}0&0&0&0\\\ 0&1&-1&0\\\ 0&1&-1&0\\\
0&0&0&0\end{bmatrix}\xi_{i}.$

This relation gives no new information.

### (11)

Relation (11) gives

$\displaystyle 0=$
$\displaystyle[b_{3}]\otimes\xi_{i}+[b_{1}]\otimes(\psi(b_{3}^{-1}))\xi_{i}+[b_{2}]\otimes(\psi(b_{1}^{-1}b_{3}^{-1})+\psi(u^{-1}b_{2}^{-1}b_{1}^{-1}b_{3}^{-1}))\xi_{i}$
$\displaystyle+[u]\otimes(\psi(b_{2}^{-1}b_{1}^{-1}b_{3}^{-1})+\psi(b_{2}^{-1}u^{-1}b_{2}^{-1}b_{1}^{-1}b_{3}^{-1}))-\psi(d_{1}^{-1}d_{2}^{-2})-\psi(u^{-1}d_{1}^{-1}d_{2}^{-2}))\xi_{i}$
$\displaystyle-[d_{2}]\otimes(I_{4}+\psi(d_{2}^{-1}))\xi_{i}-[d_{1}]\otimes(\psi(d_{2}^{-2}))\xi_{i}$
$\displaystyle=$
$\displaystyle[b_{3}]\otimes\xi_{i}+[b_{1}]\otimes\xi_{i}+[b_{2}]\otimes(I_{4}+\psi(u^{-1}))\xi_{i}-2[d_{2}]\otimes\xi_{i}-[d_{1}]\otimes\xi_{i}$
$\displaystyle=$
$\displaystyle[b_{3}]\otimes\xi_{i}+[b_{1}]\otimes\xi_{i}+[b_{2}]\otimes\begin{bmatrix}1&1&0&0\\\
1&1&0&0\\\ 0&0&2&0\\\
0&0&0&2\end{bmatrix}\xi_{i}-2[d_{2}]\otimes\xi_{i}-[d_{1}]\otimes\xi_{i}.$

This relation gives no new information.

In order to sum up the above computation, recall that
$H_{1}({\mathcal{M}}(N_{3,2});H_{1}(N_{3,2};\mathbb{Z}))$ is the quotient of
the free abelian group with basis specified in the statement of Proposition
4.1 by the relations obtained above. If we remove the superfluous generators

  * •

$u_{1}+u_{2}$ by the formula (5.1),

  * •

$a_{2,1}$ and $a_{2,4}$ by the formula (5.3),

  * •

$a_{2,2}+a_{2,3}$ by the formula (5.4),

  * •

$e_{1,1}+e_{1,2}$, $e_{1,3}$ and $e_{1,4}$ by the formula (5.5),

  * •

$e_{2,1}+e_{2,2}$, $e_{2,3}$ and $e_{2,4}$ by the formula (5.6),

  * •

$e_{2,1}+2a_{2,2}-u_{1}$ by the formula (5.7)

  * •

$d_{1,2}$, $d_{1,3}$, $d_{1,4}$, $d_{2,1}$, $d_{2,2}$, $d_{2,3}$ and $d_{2,4}$
by the formulas (5.11), (5.13), (5.14) and (5.10),

  * •

$b_{1,1}$, $b_{1,2}$, $b_{1,3}$ and $b_{1,4}$ by the formula (5.15),

  * •

$b_{2,1}$, $b_{2,2}$, $b_{2,3}$ and $b_{2,4}$ by the formula (5.16),

  * •

$b_{3,1}$, $b_{3,2}$, $b_{3,3}$ and $b_{3,4}$ by the formula (5.9)

then $H_{1}({\mathcal{M}}(N_{3,2});H_{1}(N_{3,2};\mathbb{Z}))$ is generated by
homology classes

$a_{1,3},\ a_{1,4},\ a_{1,1}+a_{1,2},\ u_{3},\ u_{4},\ d_{1,1},$

with respect to the relations obtained in the formulas (5.2), (5.8) and (5.12)

$2a_{1,3}=2a_{1,4}=2\left(a_{1,1}+a_{1,2}\right)=2u_{3}=2u_{4}=2d_{1,1}=0.$

∎ This concludes the proof of Theorem 1.1 which is an immediate consequence of
Proposition 5.1.

##  6\. Computing
$H_{1}({\mathcal{PM}^{+}}(N_{3}^{2});H_{1}(N_{3}^{2};\mathbb{Z}))$

By Theorems 2.1 and 2.3, the presentation of the group
${\mathcal{PM}^{+}}(N_{3}^{2})$ can be obtained from the presentation of
${\mathcal{M}}(N_{3,2})$ by adding relations $d_{1}=d_{2}=1$. Hence, it is
straightforward to conclude from Proposition 4.2 and the proof of Proposition
5.1 that

######  Proposition 6.1.

The abelian group
$H_{1}({\mathcal{PM}^{+}}(N_{3}^{2});H_{1}(N_{3}^{2};\mathbb{Z}))$ has a
presentation with generators

$a_{1,3},\ a_{1,4},\ a_{1,1}+a_{1,2},\ u_{3},\ u_{4},$

and relations

$2a_{1,3}=2a_{1,4}=2\left(a_{1,1}+a_{1,2}\right)=2u_{3}=2u_{4}=0.\qed$

This concludes the proof of Theorem 1.2.

## Acknowledgements

The author wishes to thank the referee for his/her helpful suggestions.

## References

  * [1] P. Pawlak and M. Stukow, The first homology group with twisted coefficients for the mapping class group of a non–orientable surface with boundary.  arXiv:2206.13642, 2022.
  * [2] M. Stukow, The first homology group of the mapping class group of a nonorientable surface with twisted coefficients, Topology Appl. 178 (2014), 417–437.
  * [3] M. Stukow, A finite presentation for the hyperelliptic mapping class group of a nonrientable surface, Osaka J. Math. 52 (2015), 495–515.
  * [4] B. Szepietowski, A presentation for the mapping class group of a non-orientable surface from the action on the complex of curves, Osaka J. Math. 45 (2008), 283–326.


# The maximal subsemigroups of the ideals on a monoid of partial injections

Apatsara Sareeto  Corresponding author. Institue of Mathematics, University of
Potsdam, Potsdam, 14476, Germany (E-mail<EMAIL_ADDRESS> Jörg Koppitz
Institute of Mathematics and Informatics, Bulgarian Academy of Sciences,
Sofia, 1113, Bulgaria (E-mail<EMAIL_ADDRESS>
###### Abstract

In the present paper, a submonoid of the well studied monoid $POI_{n}$ of all
order-preserving partial injections on an $n$-element chain is studied. The
set $IOF_{n}^{par}$ of all partial transformations in $POI_{n}$ which are
fence-preserving as well as parity-preserving form a submonoid of $POI_{n}$.
We describe the Green’s relations and ideals of $IOF_{n}^{par}$. For each
ideal of $IOF_{n}^{par}$, we characterize the maximal subsemigroups. We will
observe that there are three different types of maximal subsemigroups.

2020 Mathematics Subject Classification: 20M12, 20M18, 20M20

Keywords: Symmetric inverse monoid, Order-preserving, Maximal subsemigroups,
Ideals, Green’s relations

##  1 Introduction and Preliminaries

Let $\overline{n}$ be a finite chain with $n$ elements (where $n$ is a
positive integer), denoted as $\overline{n}=\\{1<2<\cdots<n\\}$. We denote by
$PT_{n}$ the monoid (under composition) of all partial transformations on
$\overline{n}$. A partial transformation $\alpha$ on the set $\overline{n}$ is
a mapping from a subset $A$ of $\overline{n}$ into $\overline{n}$. The
domain(respectively, image) of $\alpha$ is denoted by
$dom(\alpha)(\mbox{respectively},\ im(\alpha))$. The empty transformation is
symbolized as $\varepsilon$, and it is the transformation with
$dom(\varepsilon)=im(\varepsilon)=\emptyset$. Let $Id_{\overline{n}}$ be the
set of all partial identities on $\overline{n}$, where $id$ is the identity
mapping on $\overline{n}$. A transformation $\alpha\in PT_{n}$ is called
order-preserving if $x<y$ implies $x\alpha<y\alpha$ for all $x,y\in
dom(\alpha)$. An injective $\alpha\in PT_{n}$ is called partial injection. The
set $I_{n}$ (under composition) of all partial injections on $\overline{n}$
forms a monoid, referred to as a symmetric inverse semigroup, which was
introduced by Wagner [23]. We denoted by $POI_{n}$ the monoid of all partial
order-preserving injections on $\overline{n}$.
Recall, a subsemigroup $T$ of a semigroup $S$ is called maximal subsemigroup
of $S$ if $T$ is contained in no other proper subsemigroup of $S$. A left
ideal of $S$ is a subset $I$ of $S$ such that $SI=\\{sx:s\in S,x\in
I\\}\subseteq I$. A right ideal is defined analogously, and an ideal of $S$ is
a subset of $S$ that is both a left ideal and a right ideal. For more general
background on semigroups and standard notations, we refer the reader to [1,
14].
There has been a growing interest in the study of maximal subsemigroups within
transformation semigroups. Notably, several researchers have made significant
contributions. In [24], Yang characterized the maximal subsemigroups of the
semigroup $O_{n}$ of all full order-preserving transformations. Dimitrova and
Koppitz classified the maximal subsemigroups of the ideals of $O_{n}$ in [6].
Ganyushkin and Mazorchuk provided a description of the maximal subsemigroups
of the semigroup $POI_{n}$ in [11]. Dimitrova and Koppitz offered a
characterization of the maximal subsemigroups of the ideals of the semigroup
$POI_{n}$ [5]. In [8], Dimitrova and Mladenova explored the maximal
subsemigroups of the semigroup of all partial order-preserving
transformations. Recently, Zhao and Hu have determined both the maximal
subsemigroups and the maximal subsemibands of the ideals of the monoid of all
orientation-preserving and extensive full transformations on $\overline{n}$
[25]. Additionally, in [12], Graham, Graham, and Rhodes have demonstrated that
every maximal subsemigroup of a finite semigroup has certain features, and
that every maximal subsemigroup must be one of a small number of types. As is
often the case for semigroups, this classification depends on the description
of maximal subgroups of certain finite groups. It is worth noting that maximal
subsemigroups in many other families of transformation monoids have already
been described or quantified, primarily through the work by Dimitrova, East,
Fernandes, and other co-authors, as detailed in references such as [3, 4, 9,
13] and the associated literature.
A non-linear order that is closed to a linear order in some sense is the so-
called zig-zag order. The pair $(\overline{n},\preceq)$ is called a zig-zag
poset or fence if

$1\prec 2\succ\cdot\cdot\cdot\prec n-1\succ n$ if n is odd and $1\prec
2\succ\cdot\cdot\cdot\succ n-1\prec n$ if n is even, respectively.

The definition of the partial order $\preceq$ is self-explanatory. The number
of order-preserving maps of fences and crowns, as well as transformations on
fences, was first considered by Currie and Visentin [2] and Rutkowski [18].
The formula for the number of order-preserving self-mappings of a fence was
also introduced by Rutkowski [18]. We observe that every element in a fence is
either minimal or maximal, and for all $x,y\in\overline{n}$ with $x\prec y$,
it follows $y\in\\{x-1,x+1\\}$. We say that a transformation $\alpha\in I_{n}$
is fence-preserving if $x\prec y$ implies $x\alpha\prec y\alpha$, for all
$x,y\in dom(\alpha)$. We denote by $PFI_{n}$ the submonoid of $I_{n}$ of all
fence-preserving partial injections on $\overline{n}$. Fernandes et al.
determined the rank and a minimal generating set of the monoid of all order-
preserving transformations on an $n$-element zig-zag ordered set [10]. It is
worth mentioning that several other properties of monoids of fence-preserving
full transformations were also studied in [7, 15, 16, 17, 20, 21, 22]. We
denote by $IF_{n}$ the inverse subsemigroup of $PFI_{n}$ of all regular
elements in $PFI_{n}$. It is easy to see that $IF_{n}$ is the set of all
$\alpha\in PFI_{n}$ with $\alpha^{-1}\in PFI_{n}$. In the present paper, we
consider a submonoid of both monoids $IF_{n}$ and $POI_{n}$, i.e. a submonoid
of $IOF_{n}=IF_{n}\cap POI_{n}$. Let $a\in dom(\alpha)$ for some $\alpha\in
IOF_{n}$. If $a+1\in dom(\alpha)$ or $a-1\in dom(\alpha)$ then it is easy to
verify that $a$ and $a\alpha$ have the same parity, that is, $a$ is odd if and
only if $a\alpha$ is odd. However, if $a-1$ and $a+1$ are not in
$dom(\alpha)$, then $a$ and $a\alpha$ can have different parity. In order to
exclude this case, we require that the image of any $a\in dom(\alpha)$ has the
same parity as $a$. In this scenario, we refer to $\alpha$ as parity-
preserving. Our focus lies on the set $IOF_{n}^{par}$ of all parity-preserving
transformations in $IOF_{n}$. Notably, for any $\alpha\in IOF_{n}^{par}$, the
inverse partial injection $\alpha^{-1}$ exists and possesses order-preserving,
fence-preserving, and parity-preserving. This observation implies that
$IOF_{n}^{par}$ can be considered as an inverse submonoid of $I_{n}$, as
explained in [19].
In Section 2, we will repeat a charactrization of the elements within
$IOF_{n}^{par}$. We introduce a relation, denoted as $\sim$, on the power set
$\mathcal{P}(\overline{n})$ of $\overline{n}$. This relation offers an
alternative characterization the monoid $IOF_{n}^{par}$, and furthermore, this
characterization leads us to an immediate descriptions of the Green’s relation
$\mathcal{J}$. Note that the Green’s relations $\mathcal{R},\mathcal{L}$, and
$\mathcal{H}$ are already known, given that $IOF_{n}^{par}$ is an inverse
submonoid of $I_{n}$. However, this paper deals mainly with the ideals of
$IOF_{n}^{par}$. Of course, the sets $IOF_{n}^{par}$ and $\\{\varepsilon\\}$
are ideals of $IOF_{n}^{par}$, which are often referred to as the trivial
ideals. We will demonstrate that the ideals of $IOF_{n}^{par}$ are of the
form:

$I_{P^{*}}=\\{\alpha\in IOF_{n}^{par}:dom(\alpha)\in P^{*}\\}$

for particular subsets $P^{*}\subseteq\mathcal{P}(\overline{n})$. For each
ideal $I\neq\\{\varepsilon\\}$ of $IOF_{n}^{par}$, we characterize the maximal
subsemigroups of $I$. Our investigation will reveal that there are three
distinct types of maximal subsemigroups within an ideal
$I\neq\\{\varepsilon\\}$ of $IOF_{n}^{par}$. Therefore, the characterization
of the ideals of $IOF_{n}^{par}$ (Section 2) and the description of the
maximal subsemigroups of these ideals (Section 3) constitute the main results
of this paper.

##  2 The ideals

In this section, we will describe the ideals of $IOF_{n}^{par}$. First, we
will provide a characterization of the elements within $IOF_{n}^{par}$. For
the sake of completeness, we will recall the proof of the following
Proposition, which describes the partial injections in $IOF_{n}^{par}$.

######  Proposition 1.

[19] Let $p\leq n$ and let
$\alpha=\bigl{(}\begin{smallmatrix}d_{1}&<&d_{2}&<&\cdots&<&d_{p}\\\
m_{1}&&m_{2}&&\cdots&&m_{p}\end{smallmatrix}\bigr{)}\in I_{n}$. Then
$\alpha\in IOF_{n}^{par}$ if and only if the following four conditions hold:
(i) $m_{1}<m_{2}<\cdots<m_{p}$;
(ii) $d_{1}$ and $m_{1}$ have the same parity;
(iii) $d_{i+1}-d_{i}=1$ if and only if $m_{i+1}-m_{i}=1$ for all
$i\in\\{1,...,p-1\\}$;
(iv) $d_{i+1}-d_{i}$ is even if and only if $m_{i+1}-m_{i}$ is even for all
$i\in\\{1,...,p-1\\}$.

###### Proof.

$(\Rightarrow)$: (i) and (ii) hold since $\alpha$ is order- and parity-
preserving, respectively. (iii): Since $\alpha\in IF_{n}$, we have
$d_{i+1}\alpha-d_{i}\alpha=1$, i.e. $m_{i+1}-m_{i}=1$, if and only if
$d_{i+1}-d_{i}=1$, for all $i\in\\{1,...,p-1\\}$. (iv): Suppose
$d_{i+1}-d_{i}$ is even. Then $d_{i+1}$ and $d_{i}$ have the same parity.
Moreover, $\alpha$ is parity-preserving. This implies $d_{i+1}\alpha$ and
$d_{i}\alpha$ have the same parity, i.e. $m_{i+1}-m_{i}$ is even. The converse
direction can be proved dually.

$(\Leftarrow)$: By (i), we can conclude that $\alpha$ is order-preserving. Let
$i\in\\{1,...,p-1\\}$ and suppose $d_{i}$ and $m_{i}$ have the same parity.
Then $d_{i}-m_{i}=2k$ for some integer $k$. According to (iv), we have
$(d_{i+1}-d_{i})-(m_{i+1}-m_{i})=2l$ for some integer $l$. We obtain
$2l=d_{i+1}-m_{i+1}-(d_{i}-m_{i})=d_{i+1}-m_{i+1}-2k$, i.e.
$d_{i+1}-m_{i+1}=2(l+k)$. This implies that $d_{i+1}$ and $m_{i+1}$ have the
same parity. Together with (ii), we can conclude that $\alpha$ is parity-
preserving. Now, let $x\prec y$. This provides $\lvert x-y\rvert=1$. We have
$\lvert x\alpha-y\alpha\rvert=1$ by (iii). Since $\alpha$ is parity-
preserving, $\lvert x\alpha-y\alpha\rvert=1$ and $x\prec y$ give $x\alpha\prec
y\alpha$. Therefore, $\alpha\in PFI_{n}$. Similarly, we can demonstrate that
$\alpha^{-1}\in PFI_{n}$, i.e., $\alpha\in IF_{n}$. Hence, we can conclude
that $\alpha\in IOF_{n}^{par}$. ∎

A set $X\subseteq\mathcal{P}(\overline{n})$ is called convex if, for all
$A,B\in X$ with $A\subseteq B$ and for all $C\in\mathcal{P}(\overline{n})$,
the following condition holds: if $A\subseteq C\subseteq B$, then $C\in X$.
Here, $\mathcal{P}(A)$ denotes the power set of $A$, for any
$A\subseteq\overline{n}$. The following is easy to verify:

######  Remark 1.

If the empty set is contained in $X\subseteq\mathcal{P}(\overline{n})$, i.e.
$\emptyset\in X$, then $X$ is convex if and only if $\mathcal{P}(A)\subseteq
X$ for all $A\subseteq X$.

We will observe that the domains of all partial transformations within an
ideal form a convex set with an additional requirement. In order to describe
this requirement, we define a partial order $\sim$ on
$\mathcal{P}(\overline{n})$. Let $k_{1},k_{2}\in\mathcal{P}(\overline{n})$
with $k_{1}=\\{i_{1}<i_{2}<\cdots<i_{k}\\}$ and
$k_{2}=\\{j_{1}<j_{2}<\cdots<j_{l}\\}$ for some positive integers $k,l$. We
put $k_{1}\sim k_{2}$, if the following three properties are satisfied:
(i) $k=l$;
(ii) $i_{r}$ and $j_{r}$ have the same parity for all $r\in\\{1,...,k\\}$;
(iii) $i_{r}-i_{r-1}=1$ if and only if $j_{r}-j_{r-1}=1$ for all
$r\in\\{2,...,k\\}$.

It is worth mentioning that here (iii) (above) corresponds with (iii) in
Proposition 1. In fact, for $A,B\subseteq\overline{n}$, there is $\alpha\in
IOF_{n}^{par}$ with $A=dom(\alpha)$ and $B=im(\alpha)$ if and only if $A\sim
B$. So any $\alpha\in IOF_{n}^{par}$ is uniquely determine by domain and
image, i.e. by $dom(\alpha)$ and $im(\alpha)$. In particular, we have
$dom(\alpha)\sim im(\alpha)$ for any $\alpha\in IOF_{n}^{par}$. Using this
description of the elements in $IOF_{n}^{par}$ , we obtain immediately the
Green’s relations $\mathcal{L},\mathcal{R},\mathcal{H}$, and $\mathcal{J}$
(see [14]) for the inverse submonoid $IOF_{n}^{par}$ of $PT_{n}$:

######  Proposition 2.

Let $a,b\in IOF_{n}^{par}$ then
1) $a\mathcal{L}b$ if and only if $im(a)=im(b)$;
2) $a\mathcal{R}b$ if and only if $dom(a)=dom(b)$;
3) $a\mathcal{H}b$ if and only if $a=b$;
4) $a\mathcal{J}b$ if and only if $dom(a)\sim dom(b)$ (or $im(a)\sim im(b)).$

For a set $X\subseteq\mathcal{P}(\overline{n})$ let

$I_{X}=\\{\alpha\in IOF_{n}^{par}:dom(\alpha)\in X\\}$.

Now, we are able to characterize the ideals of $IOF_{n}^{par}$.

######  Proposition 3.

Any ideal $I$ of $IOF_{n}^{par}$ is of the form $I=I_{P^{*}}$, where $P^{*}$
is a convex subset of $\mathcal{P}(\overline{n})$ with $\emptyset\in P^{*}$
such that $y\in P^{*}$ implies $z\in P^{*}$ for all
$z\in\mathcal{P}(\overline{n})$ with $z\sim y$.

###### Proof.

Let $P^{*}$ be a convex subset of $\mathcal{P}(\overline{n})$ as defined in
the statement. Additionally, let $a\in I_{P^{*}}$ and $b\in IOF_{n}^{par}$. We
observe that $im(a)\sim dom(a)$ by Proposition 1. Consequently, $im(a)\in
P^{*}$. Further, we have $dom(ab)\subseteq dom(a)$, i.e. $dom(ab)\in P^{*}$ by
Remark 1. This implies $ab\in I_{P^{*}}$. Moreover, we observe that $im(ba)\in
P^{*}$ since $im(ba)\subseteq im(a)$ and Remark 1. This gives $dom(ba)\in
P^{*}$ since $dom(ba)\sim im(ba)$. So $ba\in I_{P^{*}}$, i.e. $I_{P^{*}}$ is
an ideal.

Conversely, let $I$ be an ideal and let $P^{*}=\\{dom(a):a\in I\\}$. Note that
any ideal contains the empty transformation $\varepsilon$. This gives
$\emptyset\in P^{*}$. Now, let $A\in P^{*}$. This means that there exists
$a\in I_{P^{*}}=I$ such that $A=dom(\alpha)$. Let $B\subseteq A$ and $c\in
Id_{\overline{n}}$ with $dom(c)=B$. Then $c\in IOF_{n}^{par}$, where
$B=dom(ca)$ and we observe that $ca\in I$. This provides $B=dom(ca)\in P^{*}$.
So, we have shown that $\mathcal{P}(A)\subseteq P^{*}$. Thus, $P^{*}$ is a
convex set as established in Remark 1.
Let $y\in P^{*}$. Then there is $a\in I$ with $dom(a)=y$. Further, let
$z\in\mathcal{P}(\overline{n})$ with $z\sim y$. Then, there is $b\in
IOF_{n}^{par}$ with $dom(b)=z$ amd $im(b)=y$. We get that $dom(ba)=dom(b)$ and
$ba\in I$ because of $a\in I$. So $z=dom(b)=dom(ba)\in P^{*}$. It clear that
$I\subseteq I_{P^{*}}$ by the definition of the sets $P^{*}$ and $I_{P^{*}}$.
Let $b\in I_{P^{*}}$. Then there is $\gamma\in I$ with $dom(b)=dom(\gamma)$
and then $im(\gamma)\sim im(b)$. Furthermore, there are
$\alpha_{1},\alpha_{2}\in IOF_{n}^{par}$ with
$dom(\alpha_{1})=dom(b),im(\alpha_{1})=dom(\gamma),dom(\alpha_{2})=im(\gamma)$,
and $im(\alpha_{2})=im(b)$. So, we see that $\alpha_{1}\gamma\alpha_{2}=b$,
i.e. $b\in I$. Consequently, $I=I_{P^{*}}$. ∎

Clearly, $\\{\emptyset\\}$ is a convex set and
$I_{\\{\emptyset\\}}=\\{\varepsilon\\}$ is the least ideal of $IOF_{n}^{par}$.
Now, we determine the minimal ideals of $IOF_{n}^{par}$. An ideal $I$ is
called minimal if $I\neq\\{\varepsilon\\}$ and $M\subseteq I$ implies $M=I$,
for all non-trivial ideals $M$ of $IOF_{n}^{par}$.

######  Proposition 4.

Let $I$ be a non-trivial ideal of $IOF_{n}^{par}$. Then $I$ is a minimal ideal
of $IOF_{n}^{par}$ if and only if $I=I_{P^{*}}$, where either
$P^{*}=\\{\emptyset,\\{1\\},\\{3\\},...,\\{n-1\\}\\}$ or
$P^{*}=\\{\emptyset,\\{2\\},\\{4\\},...,\\{n\\}\\}$ if $n$ is even and either
$P^{*}=\\{\emptyset,\\{1\\},\\{3\\},...,\\{n\\}\\}$ or
$P^{*}=\\{\emptyset,\\{2\\},\\{4\\},...,\\{n-1\\}\\}$ if $n$ is odd,
respectively.

###### Proof.

Without loss of generality, we can assume that $n$ is even. The proof for $n$
is odd is similar.
Note that $\\{1\\}\sim\\{3\\}\sim\cdots\sim\\{n-1\\}$ and
$\\{2\\}\sim\\{4\\}\sim\cdots\sim\\{n\\}$. By Proposition 3, we get that
$I_{P^{*}}$ is a minimal ideal, whenever
$P^{*}=P^{O}=\\{\emptyset,\\{1\\},\\{3\\},...,\\{n-1\\}\\}$ or
$P^{*}=P^{E}=\\{\emptyset,\\{2\\},\\{4\\},...,\\{n\\}\\}$.

Conversely, let $I$ be a minimal ideal of $IOF_{n}^{par}$. Then, there is
$P^{*}\subseteq\mathcal{P}(\overline{n})$, satisfying the conditions in
Proposition 3, such that $I=I_{P^{*}}$. Assume there is $a\in I$ such that
$rank(a)\geq 2$. This provides that there exists $b\in I$ with $rank(b)=1$ and
$dom(b)\in\mathcal{P}(dom(a))$. Let $I^{\prime}=\\{b^{\prime}\in
I:rank(b^{\prime})\leq 1\\}$. It is obvious that $I^{\prime}$ is an ideal.
This gives $\\{\varepsilon\\}\neq I^{\prime}\subset I$, a contradiction to $I$
is a minimal ideal. So, $rank(a)\leq 1$ for all $a\in I$. We have now
$P^{*}=\\{dom(a):a\in
I\\}\subseteq\\{\emptyset,\\{1\\},\\{2\\},...,\\{n\\}\\}$. Assume $P^{*}\neq
P^{O}$ and $P^{*}\neq P^{E}$. Moreover, assumex $P^{E}\cap
P^{*}\neq\\{\emptyset\\}$. Then $P^{E}\subseteq P^{*}$ since $P^{*}$ satisfies
the conditions in Proposition 3. Since $I_{P^{*}}$ as well as $I_{P^{E}}$ are
minimal ideals, we obtain $P^{*}=P^{E}$, a contradiction to $P^{*}\neq P^{E}$.
Hence $P^{E}\cap P^{*}=\\{\emptyset\\}$. Similarly, we have $P^{O}\cap
P^{*}=\\{\emptyset\\}$. But $P^{E}\cap P^{*}=P^{O}\cap P^{*}=\\{\emptyset\\}$
gives $P^{*}=\\{\emptyset\\}$, a contradiction to
$I_{P^{*}}\neq\\{\varepsilon\\}$. Thus, $I=I_{P^{*}}$ with $P^{*}=P^{O}$ or
$P^{*}=P^{E}$. ∎

##  3 The maximal subsemigroups of the ideals on $IOF_{n}^{par}$

In this section, we determine the maximal subsemigroups of the ideals on
$IOF_{n}^{par}$. First, we need a few technical tools. Let
$y=\\{i_{1}<i_{2}<\cdots<i_{k}\\}\in P^{*}$ for some positive integer $k\geq
2$ and let $t\in\\{1,2,...,k\\}$. Then we put
$y^{[t]}=y\backslash\\{i_{t}\\}$. For $x,y\in P^{*}$, we write $x\sqsubset y$
if there is $t\in\\{1,...,|y|\\}$, such that $x=y^{[t]}$. Otherwise, we write
$x\not\sqsubset y$.

######  Lemma 1.

Let $y_{1},y_{2},z_{1},z_{2}\in P^{*}$ with $|y_{1}|=|z_{1}|\geq 2$ and let
$r,s\in\\{1,2,...,|y_{1}|\\}$ such that $y_{1}^{[r]}=z_{1}^{[s]}=y_{1}\cap
z_{1}$, $y_{2}\sim y_{1}$, $z_{2}\sim z_{1}$, and $y_{2}^{[r]}\sim
z_{2}^{[s]}\sim y_{1}\cap z_{1}$. Then there are $\theta,\delta\in I_{P^{*}}$
with $rank(\theta)=rank(\delta)=|y_{1}|$ such that
$dom(\theta\delta)=y_{2}^{[r]}$ and $im(\theta\delta)=z_{2}^{[s]}$.

###### Proof.

Let $y_{1},y_{2},z_{1},z_{2}\in P^{*}$ with $|y_{1}|=|z_{1}|\geq 2$ and let
$r,s\in\\{1,2,...,|y_{1}|\\}$ such that $y_{1}^{[r]}=z_{1}^{[s]}=y_{1}\cap
z_{1}$, $y_{2}\sim y_{1}$, $z_{2}\sim z_{1}$, and $y_{2}^{[r]}\sim
z_{2}^{[s]}\sim y_{1}\cap z_{1}$. Then there are $\theta,\delta\in I_{P^{*}}$
with $dom(\theta)=y_{2},im(\theta)=y_{1},dom(\delta)=z_{1}$, and
$im(\delta)=z_{2}$. Because of $y_{1}^{[r]}=z_{1}^{[s]}=y_{1}\cap z_{1}$, then
$rank(\theta\delta)=|y_{1}|-1$. Because of $y_{2}\sim y_{1}$ and $z_{2}\sim
z_{1}$ with $y_{2}^{[r]}\sim z_{2}^{[s]}\sim y_{1}\cap z_{1}$, then
$dom(\theta\delta)=y_{2}^{[r]}$ and $im(\theta\delta)=z_{2}^{[s]}$. ∎

Let $I_{P^{*}}^{u}$ be the set of all $\alpha\in I_{P^{*}}$ with
$dom(\alpha)\neq y_{2}^{[r]}$ or $im(\alpha)\neq z_{2}^{[s]}$, whenever
$y_{2},z_{2}\in P^{*}$ and $r,s\in\\{1,2,...,|y_{1}|\\}$ with $y_{2}\sim
y_{1},z_{2}\sim z_{1},y_{1}^{[r]}=z_{1}^{[s]}=y_{1}\cap z_{1}$, and
$y_{2}^{[r]}\sim z_{2}^{[s]}\sim y_{1}\cap z_{1}$ for some $y_{1},z_{1}\in
P^{*}$ with $|y_{1}|=|z_{1}|\geq 2$. Directly from Lemma 1, we obtain:
$\alpha\notin I_{P^{*}}^{u}$ if and only if there are $\theta,\delta\in
I_{P^{*}}$ with $rank(\theta)=rank(\delta)=rank(\theta\delta)+1$ such that
$\alpha=\theta\delta$. Since $P^{*}$ is a convex subset of
$\mathcal{P}(\overline{n})$, we can conclude:

######  Corollary 1.

Let $\alpha\in I_{P^{*}}$. Then $\alpha\notin I_{P^{*}}^{u}$ if and only if
there are $\theta,\delta\in I_{P^{*}}$ with
$rank(\theta),rank(\delta)>rank(\theta\delta)$ such that
$\alpha=\theta\delta$.

Our initial observation is that all maximal subsemigroups of an ideal $I$ have
the form $I\backslash T$, where all transformations in $T$ have the same rank.

######  Lemma 2.

Let $J$ be a maximal subsemigroup of $I_{P^{*}}$ and let $\alpha\notin J$.
Then $\beta\in J$ for all $\beta\in I_{P^{*}}$ with $rank(\beta)\neq
rank(\alpha)$.

###### Proof.

Assume there is $\beta\in I_{P^{*}}$ such that $\beta\notin J$ with
$rank(\beta)\neq rank(\alpha)$. Suppose $rank(\beta)>rank(\alpha)$. We observe
that $\langle J,\alpha\rangle$ is semigroup, where $rank(a\alpha),rank(\alpha
a)\leq rank(\alpha)$ for all $a\in J$ and we see that $\langle
J,\alpha\rangle\neq I_{P^{*}}$ since $\beta\in I_{P^{*}}$ but
$\beta\notin\langle J,\alpha\rangle$. Moreover, $J\subset\langle
J,\alpha\rangle\neq I_{P^{*}}$, a contradiction to $J$ is a maximal
subsemigroup of $I_{P^{*}}$. Suppose $rank(\beta)<rank(\alpha)$. Then we can
show by contradiction that $J\subset\langle J,\beta\rangle\neq I_{P^{*}}$ in
the same way. ∎

For the remainder of this section, let $P^{*}$ be a convex subset of $X$,
satisfying the conditions in Proposition 3, with
$I_{P^{*}}\neq\\{\varepsilon\\}$. Now, we determine several subsemigroups of
$I_{P^{*}}$ and will show that they are exactly the maximal ones.
Let $C=\\{\\{g\\}:g\in P^{*},g\sqsubset y\ \text{for some}\ y\in
P^{*}\text{with}\ y\nsim z\ \text{for all}\ z\in P^{*}\backslash\\{y\\}\
\text{or}\ g\not\sqsubset z\ \text{for all}\ z\in P^{*}\\}$.
For $\Gamma_{1}\in P^{*}$ with $|\Gamma_{1}|\geq 2$, we define
$A_{\Gamma_{1}}=\\{\Gamma\in P^{*}:\Gamma\sim\Gamma_{1}\\}$ and
$B_{\Gamma_{1}}=\\{\Gamma^{t}:\Gamma\in
A_{\Gamma_{1}},t\in\\{1,2,...,|\Gamma|\\}$.
We put $A=\\{A_{\Gamma}:\Gamma\in P^{*},|\Gamma|\geq 2,|A_{\Gamma}|\geq 2\\}$
and $B=\\{B_{\Gamma}:\Gamma\in P^{*},|\Gamma|\geq 2,|A_{\Gamma}|\geq 2\\}$.
Let $\Delta_{1}\in B\cup C$. We define

$BC_{\Delta_{1}}=\\{\Delta\in B\cup C:x\sim y\ \text{for all}\
x\in\Delta,y\in\Delta_{1}\\}$.

For $\Delta\in BC_{\Delta_{1}}$, we also define

$T_{\Delta}=\\{c\in I_{P^{*}}:dom(c),im(c)\in\Delta\\}$.

And for a partition $Q=(Q_{1},Q_{2})$ of $BC_{\Delta_{1}}$, let

$T_{Q}=\\{c\in I_{P^{*}}:dom(c)\in\tilde{\Delta},im(c)\in\hat{\Delta}\
\text{with}\ \tilde{\Delta}\in Q_{1},\hat{\Delta}\in Q_{2}\\}$.

######  Lemma 3.

Let $g\in P^{*}$ with $\\{g\\}\in C$ and $|BC_{\\{g\\}}|=1$. Then
$I_{P^{*}}\backslash T_{\\{g\\}}$ with $T_{\\{g\\}}\subseteq I_{P^{*}}^{u}$ is
a semigroup.

###### Proof.

Let $\alpha\in T_{\\{g\\}}$. Because of $\\{g\\}\in C$ and $|BC_{\\{g\\}}|=1$,
we get that $\alpha\in Id_{\overline{n}}$ and $T_{\\{g\\}}=\\{\alpha\\}$. Let
$a,b\in I_{P^{*}}\backslash\\{\alpha\\}$. If $rank(a)\leq rank(\alpha)$ or
$rank(b)\leq rank(\alpha)$ then $ab\neq\alpha$ because $dom(a)\nsim
dom(\alpha)$ and $dom(b)\nsim im(\alpha)$. If $rank(a),rank(b)>rank(\alpha)$
and $rank(ab)=rank(\alpha)$ then by Corollary 1, we have $ab\neq\alpha$ since
$\alpha\in I_{P^{*}}^{u}$. This shows $ab\in I_{P^{*}}\backslash\\{\alpha\\}$.
Consequently, $I_{P^{*}}\backslash\\{\alpha\\}$ is a semigroup. ∎

######  Lemma 4.

Let $y_{1}\in P^{*}$ with $|y_{1}|\geq 2$, $B_{y_{1}}\in B$, and
$|BC_{B_{y_{1}}}|=1$. Then $I_{P^{*}}\backslash T_{B_{y_{1}}}$ with
$T_{B_{y_{1}}}\subseteq I_{P^{*}}^{u}$ is a semigroup.

###### Proof.

Let $\alpha\in T_{B_{y_{1}}}$ and let $a,b\in I_{P^{*}}\backslash
T_{B_{y_{1}}}$. Note that if $rank(a)\leq rank(\alpha)$ or $rank(b)\leq
rank(\alpha)$ then we can immediately deduce that $dom(ab)\nsim dom(\alpha)$.
Thus, $ab\notin B_{y_{1}}$ and $ab\in I_{P^{*}}\backslash T_{B_{y_{1}}}$.
Suppose $rank(a),rank(b)>rank(\alpha)$ and $rank(ab)=rank(\alpha)$. Then by
Corollary 1, we have $ab\notin I_{P^{*}}^{u}$. This shows $ab\in
I_{P^{*}}\backslash T_{B_{y_{1}}}$. Consequently, $I_{P^{*}}\backslash
T_{B_{y_{1}}}$ is a semigroup. ∎

######  Lemma 5.

Let $Q=(Q_{1},Q_{2})$ be a partition of $BC_{\Delta_{1}}$, where
$\Delta_{1}\in B\cup C$. Then $I_{P^{*}}\backslash T_{Q}$ with $T_{Q}\subseteq
I_{P^{*}}^{u}$ is a semigroup.

###### Proof.

Let $a,b\in I_{P^{*}}\backslash T_{Q}$ and let $c\in T_{Q}$. If
$rank(a)<rank(c)$ or $rank(b)<rank(c)$ then $rank(ab)<rank(c)$. So, $ab\in
I_{P^{*}}\backslash T_{Q}$. If $rank(a)=rank(b)=rank(c)$ then we need to
consider only the case that $dom(a)\in\Delta^{\prime}$ for some
$\Delta^{\prime}\in Q_{1}$ and $im(b)\in\Delta^{\prime\prime}$ for some
$\Delta^{\prime\prime}\in Q_{2}$. We get that $im(a)\in Y$ for some $Y\in
Q_{1}$ and $dom(b)\in Y^{\prime}$ for some $Y^{\prime}\in Q_{2}$ because
$a,b\in I_{P^{*}}\backslash T_{Q}$. We see that $rank(ab)<rank(c)$ because
$Q_{1}\cap Q_{2}=\emptyset$. Suppose that $rank(a)>rank(b)=rank(c)$ and
$rank(ab)=rank(c)$. Note if $im(b)\in\Delta\in Q_{1}$ then $im(ab)\in\Delta\in
Q_{1}$. So, $ab\in I_{P^{*}}\backslash T_{Q}$. Suppose $im(b)\in\Delta$ for
some $\Delta\in Q_{2}$. Then $dom(b)\in\Delta^{\prime}$ for some
$\Delta^{\prime}\in Q_{2}$ because of $b\in I_{P^{*}}\backslash T_{Q}$ and we
have $dom(b)\subset im(a)$. We put $dom(a)=y_{1},im(a)=y_{2}$, and
$dom(b)=y_{2}^{[t]}$ for some $t\in\\{1,2,...,|y_{2}|\\}$. This gives
$dom(ab)=y_{1}^{[t]}$. So, we see that $y_{1}^{[t]}\in\Delta^{\prime}\in
Q_{2}$ because of $y_{1}\sim y_{2}$. Then $ab\in I_{P^{*}}\backslash T_{Q}$.
If $rank(b)>rank(a)=rank(\alpha)$ then we obtain $ab\in I_{P^{*}}\backslash
T_{Q}$ dually. If $rank(a),rank(b)>rank(\alpha)$ with $rank(ab)=rank(\alpha)$
then by Corollary 1, we have $ab\notin I_{P^{*}}^{u}$. This shows $ab\in
I_{P^{*}}\backslash T_{Q}$. Consequently, $I_{P^{*}}\backslash T_{Q}$ is a
semigroup. ∎

Now, we are able to characterize the maximal subsemigroups of $IOF_{n}^{par}$,
which is the main result of this section.

######  Theorem 1.

Let $J$ be a subsemigroup of $I_{P^{*}}$. Then $J$ is a maximal subsemigroup
of $I_{P^{*}}$ if and only if $J$ has one of the following forms.
1) $J=I_{P^{*}}\backslash T_{\\{g\\}}$ with $|BC_{\\{g\\}}|=1$ for some $g\in
P^{*}$ such that $\\{g\\}\in C$ and $T_{\\{g\\}}\subseteq I_{P^{*}}^{u}$;
2) $J=I_{P^{*}}\backslash T_{B_{y_{1}}}$ with $|BC_{B_{y_{1}}}|=1$ for some
$y_{1}\in P^{*}$ such that $|y_{1}|\geq 2$, $B_{y_{1}}\in B$, and
$T_{B_{y_{1}}}\subseteq I_{P^{*}}^{u}$;
3) $J=I_{P^{*}}\backslash T_{Q}$ for some partition $Q=(Q_{1},Q_{2})$ of
$BC_{\Delta_{1}}$, where $\Delta_{1}\in B\cup C$ and $T_{Q}\subseteq
I_{P^{*}}^{u}$.

###### Proof.

Let $J$ be a maximal subsemigroup of $I_{P^{*}}$ and let $\alpha\in
I_{P^{*}}\backslash J$.
Assume $\alpha\notin I_{P^{*}}^{u}$. Then by the definition of
$I_{P^{*}}^{u}$, we get that $dom(\alpha)=y_{2}^{[r]}$ and
$im(\alpha)=z_{2}^{[s]}$, where $y_{2},z_{2}\in P^{*}$ and
$r,s\in\\{1,2,...,|y_{1}|\\}$ with $y_{2}\sim y_{1},z_{2}\sim
z_{1},y_{1}^{[r]}=z_{1}^{[s]}=y_{1}\cap z_{1}$, and $y_{2}^{[r]}\sim
z_{2}^{[s]}\sim y_{1}\cap z_{1}$ for some $y_{1},z_{1}\in P^{*}$ with
$|y_{1}|=|z_{1}|\geq 2$. By Lemma 1, we get that there are $\theta,\delta\in
I_{P^{*}}$ with $rank(\theta)=rank(\delta)=|y_{2}|$ such that
$\theta\delta=\alpha$. Then $\theta,\delta\in J$ by Lemma 2, a contradiction
to $\alpha\notin J$. So, $\alpha\in I_{P^{*}}^{u}$.
Suppose that $m=dom(\alpha)$ and $m=im(\alpha)$ for all $m\in P^{*}$ with
$m\sim dom(\alpha)$. So, we can put $m=dom(\alpha)=im(\alpha)$. This provides
$\\{m\\}\in C$, i.e. $BC_{\\{m\\}}=\\{\\{m\\}\\}$. So, $\alpha\in
Id_{\overline{n}}$ and by the definition of $T_{\\{m\\}}$, we have that
$T_{\\{m\\}}=\\{\alpha\\}$. It is easy to see that
$J\cap\\{\alpha\\}=\emptyset$, this means $J\subseteq
I_{P^{*}}\backslash\\{\alpha\\}$ and we have $I_{P^{*}}\backslash\\{\alpha\\}$
is a semigroup by Lemma 3. Together with $J$ is maximal subsemigroup of
$I_{P^{*}}$, we have $J=I_{P^{*}}\backslash\\{\alpha\\}$.
Suppose there is $m\in P^{*}$ with $m\sim dom(\alpha)$ such that $m\neq
dom(\alpha)$ or $m\neq im(\alpha)$ and there are $y_{1}\in P^{*}$ with
$|y_{1}|\geq 2$ and $t\in\\{1,2,...,|y_{1}|\\}$ such that for all $k\in P^{*}$
with $k\sim dom(\alpha)$ there is $y_{3}\in P^{*}$ with $y_{3}\sim y_{1}$ and
$k=y_{3}^{[t]}$. This implies that there exists $y_{2}\in P^{*}$ with
$y_{2}\sim y_{1}\neq y_{1}$. Then $A_{y_{1}}\in A$ and thus, $B_{y_{1}}\in B$.
We can conclude that $\\{B_{y_{1}}\\}=BC_{B_{y_{1}}}$ and $|B_{y_{1}}|\geq 2$.
So, we get $dom(\alpha),im(\alpha)\in B_{y_{1}}$. Then there are
$y_{4},y_{5}\in A_{y_{1}}$ such that $dom(\alpha)=y_{4}^{[t]}$ and
$im(\alpha)=y_{5}^{[t]}$.
Assume there is $\theta\in J$ with $dom(\theta),im(\theta)\in B_{y_{1}}$.
There are $y_{6},y_{7}\in A_{y_{1}}$ such that
$dom(\theta)=y_{6}^{[t]},im(\theta)=y_{7}^{[t]}$. We have
$\gamma_{1},\gamma_{2}\in I_{P^{*}}$ with
$dom(\gamma_{1})=y_{4},im(\gamma_{1})=y_{6},dom(\gamma_{2})=y_{7}$, and
$im(\gamma_{2})=y_{5}$. This gives $\gamma_{1},\gamma_{2}\in J$ because of
$rank(\gamma_{1}),rank(\gamma_{2})>rank(\alpha)$ together with Lemma 2. We get
that $\gamma_{1}\theta\gamma_{2}=\alpha$, a contradiction to $\alpha\notin J$
and $J$ is semigroup. Thus, $\theta\notin J$ for all $\theta\in I_{P^{*}}$
with $dom(\theta),im(\theta)\in B_{y_{1}}$, i.e. we have $\theta\notin J$ for
all $\theta\in T_{B_{y_{1}}}$. This means $J\cap T_{B_{y_{1}}}=\emptyset$. So
$J\subseteq I_{P^{*}}\backslash T_{B_{y_{1}}}$ and by Lemma 4, we have that
$I_{P^{*}}\backslash T_{B_{y_{1}}}$ is a semigroup. Together with $J$ is
maximal subsemigroup of $I_{P^{*}}$, we have $J=I_{P^{*}}\backslash
T_{B_{y_{1}}}$.
Assume there is $\theta\in T_{B_{y_{1}}}$ with $\theta\notin I_{P^{*}}^{u}$.
By Corollary 1, there are $a_{1},a_{2}\in I_{P^{*}}$ with
$rank(a_{1}),rank(a_{2})>rank(a_{1}a_{2})$ such that $\theta=a_{1}a_{2}$. This
provides, $a_{1},a_{2}\in J$ by Lemma 2, i.e. $\theta\in J$, a contradiction
to $T_{B_{y_{1}}}\cap J=\emptyset$. Thus, $T_{B_{y_{1}}}\subseteq
I_{P^{*}}^{u}$.
Suppose for all $y_{1}\in P^{*}$ with $|y_{1}|\geq 2$ and for all
$t\in\\{1,2,...,|y_{1}|\\}$, there is $k\in P^{*}$ with $k\sim dom(\alpha)$
such that $k\neq y_{3}^{[t]}$ for all $y_{3}\in P^{*}$ with $y_{3}\sim y_{1}$,
and there is $m\in P^{*}$ with $m\sim dom(\alpha)$ such that $m\neq
dom(\alpha)$ or $m\neq im(\alpha)$. Let $\lambda\in I_{P^{*}}\backslash J$
with $dom(\lambda)\sim dom(\alpha)$. Then we have the following four cases:

1\. $dom(\lambda)=y_{3}^{[t]}$ for some $t\in\\{1,2,...,|y_{3}|\\}$ and
$im(\lambda)=z_{3}^{[s]}$ for some $s\in\\{1,2,...,|z_{3}|\\}$ and
$y_{3},z_{3}\in P^{*}$ and there are $y_{1},y_{2},z_{1},z_{2}\in P^{*}$ with
$y_{3}\sim y_{1}\sim y_{2}\neq y_{1}$ and $z_{3}\sim z_{1}\sim z_{2}\neq
z_{1}$.
Then there is $k\in P^{*}$ with $k\sim dom(\alpha)$ such that $k\neq
y_{3}^{[t]}$. Assume $y_{1}\sim z_{1}$ and $t=s$. For $\theta\in I_{P^{*}}$,
with $dom(\theta)=y_{4}^{[t]},im(\theta)=z_{4}^{[s]}$, and $y_{4}\sim
z_{4}\sim y_{1}$, we have $\theta\notin J$. Otherwise, let
$\gamma_{1},\gamma_{2}\in I_{P^{*}}$ with
$dom(\gamma_{1})=y_{3},im(\gamma_{1})=y_{4},dom(\gamma_{2})=z_{4}$, and
$im(\gamma_{2})=z_{3}$. This gives $\gamma_{1},\gamma_{2}\in J$ because of
Lemma 2. We get that $\gamma_{1}\theta\gamma_{2}=\lambda$, a contradiction to
$\lambda\notin J$ because $J$ is semigroup. Moreover, there are
$\beta_{1},\beta_{2}\in I_{P^{*}}$ with
$dom(\beta_{1})=y_{3}^{[t]},im(\beta_{1})=dom(\beta_{2})=k$, and
$im(\beta_{2})=z_{3}^{[s]}$. So, we observe that $\beta_{1}\beta_{2}=\lambda$.
This show $\beta_{1}\notin J$ or $\beta_{2}\notin J$. Suppose $\beta_{1}\notin
J$. Let $\theta\in I_{P^{*}}$ with
$dom(\theta)=z_{3}^{[s]},im(\theta)=y_{3}^{[t]}$, where $y_{3},z_{3}\in P^{*}$
with $y_{3}\sim z_{3}\sim y_{1}$. As we have shown above, we have
$\theta\notin J$. Note that $id\in J$ by Lemma 2 and $rank(\theta)<n$. So
$\beta_{1}\in\langle J\cup\\{\theta\\}\rangle$ and we observe that
$\beta_{1}=\alpha^{\prime}\theta\rho$ with $\alpha^{\prime}\in\langle
J\cup\\{\theta\\}\rangle$ and $\rho\in J$. Since $y_{3}^{[t]}\neq k$, we can
conclude that $\rho=\beta_{1}$, a contradiction to $\beta_{1}\notin J$.
Dually, we can prove that $\beta_{2}\notin J$ is not possible. Therefore, if
$y_{1}\sim z_{1}$ then $t\neq s$.
Furthermore, for all $\beta\in I_{P^{*}}$ with $dom(\beta)=y_{4}^{[t]}$ and
$im(\beta)=z_{4}^{[s]}$ for some $y_{4},z_{4}\in P^{*}$ such that $y_{4}\sim
y_{1}$ and $z_{4}\sim z_{1}$, we have $\beta\notin J$. Otherwise,
$\theta_{1}\beta\theta^{\prime}=\lambda$, where
$dom(\theta_{1})=y_{3},im(\theta_{1})=y_{4},dom(\theta^{\prime})=z_{4}$, and
$im(\theta^{\prime})=z_{3}$, i.e. $\theta_{1},\theta^{\prime}\in J$ by Lemma
2, a contradiction to $\lambda\notin J$.

2\. $dom(\lambda)=y_{3}^{[t]}$ for some $t\in\\{1,2,...,|y_{3}|\\}$ with
$y_{3}\in P^{*}$ and there are $y_{1},y_{2}\in P^{*}$ with $y_{3}\sim
y_{1}\sim y_{2}\neq y_{1}$ and $im(\lambda)\not\sqsubset z$ for all $z\in
P^{*}$ (or $im(\lambda)\sqsubset y\in P^{*}$ with $y\nsim z$ for all $z\in
P^{*}\backslash\\{y\\}$).
Then for all $\beta\in I_{P^{*}}$ with $dom(\beta)=y_{4}^{[t]}$ for some
$y_{4}\in P^{*}$ with $y_{4}\sim y_{1}$ and $im(\beta)=im(\lambda)$, we have
$\beta\notin J$. Otherwise, $\theta\beta=\lambda$, where $dom(\theta)=y_{3}$
and $im(\theta)=y_{4}$ for some $\theta\in J$ by Lemma 2, a contradiction to
$\lambda\notin J$.

3\. $dom(\lambda)\not\sqsubset y$ for all $y\in P^{*}$ (or
$dom(\lambda)\sqsubset z\in P^{*}$ with $z\nsim y$ for all $y\in
P^{*}\backslash\\{z\\}$) and $im(\lambda)=z_{3}^{[s]}$ for some
$s\in\\{1,2,...|z_{3}|\\}$, $z_{3}\in P^{*}$, and there are $z_{1},z_{2}\in
P^{*}$ with $z_{3}\sim z_{1}\sim z_{2}\neq z_{1}$.
Then for all $\beta\in I_{P^{*}}$with $dom(\beta)=dom(\lambda)$ and
$im(\beta)=z_{4}^{[s]}$ for some $z_{4}\in P^{*},z_{4}\sim z_{1}$, we have
$\beta\notin J$. Otherwise, $\beta\theta^{\prime}=\lambda$ by Lemma 2, where
$dom(\theta^{\prime})=z_{4}$ and $im(\theta^{\prime})=z_{3}$ for some
$\theta^{\prime}\in J$, a contradiction to $\lambda\notin J$.

4\. $dom(\lambda)\not\sqsubset y\in P^{*}$ for all $y\in P^{*}$ (or
$dom(\lambda)\sqsubset z\in P^{*}$ with $z\nsim y$ for all $y\in
P^{*}\backslash\\{z\\}$) and $im(\lambda)\not\sqsubset y$ for all $y\in P^{*}$
(or $im(\lambda)\sqsubset z\in P^{*}$ with $z\nsim y$ for all $y\in
P^{*}\backslash\\{z\\}$).

Note that: $B_{y_{3}},B_{z_{3}}\in B$ with $dom(\alpha)\in B_{y_{3}}$ and
$im(\alpha)\in B_{z_{3}}$, if $\alpha=\lambda$ is of form 1;
$B_{y_{3}}\in B$ with $dom(\alpha)\in B_{y_{3}}$ and $\\{im(\alpha)\\}\in C$,
if $\alpha=\lambda$ is of form 2;
$B_{z_{3}}\in B$ with $im(\alpha)\in B_{z_{3}}$ and $\\{dom(\alpha)\\}\in C$,
if $\alpha=\lambda$ is of form 3;
$\\{dom(\alpha)\\},\\{im(\alpha)\\}\in C$, if $\alpha=\lambda$ is of form 4.
Hence, there is $\Delta_{1}\in B\cup C$ with $dom(\alpha)\in\Delta_{1}$. We
define,

$\tilde{B}\tilde{C}=\\{\Delta\in BC_{\Delta_{1}}:\text{there is}\ \beta\in
I_{P^{*}}\backslash J\ \text{with}\ dom(\beta)\in\Delta\\}$;

$\hat{B}\hat{C}=\\{\Delta\in BC_{\Delta_{1}}:\text{there is}\ \beta\in
I_{P^{*}}\backslash J\ \text{with}\ im(\beta)\in\Delta\\}$;

$T_{BC}=\\{c\in I_{P^{*}}:dom(c)\in\tilde{\Delta},im(c)\in\hat{\Delta}\
\text{with}\
\tilde{\Delta}\in\tilde{B}\tilde{C},\hat{\Delta}\in\hat{B}\hat{C}\\}$.

Assume there are $\theta_{1},\theta_{2}\notin J$ with
$\theta_{1}\neq\theta_{2}$ and $im(\theta_{1}),dom(\theta_{2})\in\Delta$ for
some $\Delta\in BC_{\Delta_{1}}$. So, we observe that
$\gamma\theta_{1}\rho=\theta_{2}$ with $\gamma\in\langle
J\cup\\{\theta_{1}\\}\rangle$ and $\rho\in J$. Then we have
$im(\theta_{2})\subset im(\rho)$ or $im(\theta_{2})=im(\rho)$. Because of
$im(\theta_{1}),dom(\theta_{2})\in\Delta$ and 1-3, then
$im(\theta_{2})=im(\rho)$. This gives $dom(\rho)=im(\theta_{1})$. Then
$\rho\notin J$ because of 2, a contradiction. Thus,
$\tilde{B}\tilde{C}\cap\hat{B}\hat{C}=\emptyset$.
Clearly, by the definition of $\tilde{B}\tilde{C}$ and $\hat{B}\hat{C}$, we
have $\tilde{B}\tilde{C}\ \cup\hat{B}\hat{C}\subseteq BC_{\Delta_{1}}$. Let
$\Delta\in BC_{\Delta_{1}}$. Further, let $h\in\Delta$, i.e. $h\sim
dom(\alpha)$, and let $\gamma_{1},\gamma_{2}\in I_{P^{*}}$ with
$h=im(\gamma_{1})=dom(\gamma_{2}),dom(\gamma_{1})=dom(\alpha)$, and
$im(\gamma_{2})=im(\alpha)$. We have that $\gamma_{1}\gamma_{2}=\alpha$. This
gives $\gamma_{1}\notin J$ or $\gamma_{2}\notin J$. So by the definition of
$\hat{B}\hat{C}$($\tilde{B}\tilde{C}$), we see that if $\gamma_{1}\notin J$(
$\gamma_{2}\notin J$), then
$\Delta\in\hat{B}\hat{C}$($\Delta\in\tilde{B}\tilde{C})$. This means
$\tilde{B}\tilde{C}\ \cup\hat{B}\hat{C}\supseteq BC_{\Delta_{1}}$ and together
with $\tilde{B}\tilde{C}\ \cup\hat{B}\hat{C}\subseteq BC_{\Delta_{1}}$, we
obtain $\tilde{B}\tilde{C}\ \cup\hat{B}\hat{C}=BC_{\Delta_{1}}$. Consequently,
$BC=(\tilde{B}\tilde{C},\hat{B}\hat{C})$ is a partition of $BC_{\Delta_{1}}$.
Let $\gamma\in T_{BC}$. Then there are $\Delta\in\tilde{B}\tilde{C}$ and
$\Delta^{\prime}\in\hat{B}\hat{C}$ such that $dom(\gamma)\in\Delta$ and
$im(\gamma)\in\Delta^{\prime}$. By the definition of $\tilde{B}\tilde{C}$ and
$\hat{B}\hat{C}$, there are $\delta_{1},\delta_{2}\in I_{P^{*}}\backslash J$
with $dom(\delta_{1})\in\Delta$ and $im(\delta_{2})\in\Delta^{\prime}$. If
$\gamma=\delta_{1}$ or $\gamma=\delta_{2}$ then we have $\gamma\notin J$.
Suppose $\gamma\neq\delta_{1}$ and $\gamma\neq\delta_{2}$. Recall, we have
$dom(\gamma),dom(\delta_{1})\in\Delta$. By 1-4, we get that there is
$\theta_{1}\notin J$ with $dom(\theta_{1})=dom(\gamma)$. There is
$\gamma^{\prime}\in I_{P^{*}}$ with $dom(\gamma^{\prime})=im(\gamma)$ and
$im(\gamma^{\prime})=im(\theta_{1})$. We observe that
$\gamma\gamma^{\prime}=\theta_{1}$. This means, $\gamma\notin J$ or
$\gamma^{\prime}\notin J$. Assume that $\gamma^{\prime}\notin J$. We have
$dom(\gamma^{\prime})=im(\gamma)\in\Delta^{\prime}$. Then
$\Delta^{\prime}\in\tilde{B}\tilde{C}$ and we have
$\tilde{B}\tilde{C}\cap\hat{B}\hat{C}\neq\emptyset$, a contradiction. Thus,
$\gamma\notin J$. We can conclude that $T_{BC}\cap J=\emptyset$, this means
$J\subseteq I_{P^{*}}\backslash T_{BC}$ and by Lemma 5, we have
$I_{P^{*}}\backslash T_{BC}$ is a semigroup. Together with $J$ is maximal
subsemigroup of $I_{P^{*}}$, we have $J=I_{P^{*}}\backslash T_{BC}$.
Assume there is $\theta\in T_{BC}$ with $\theta\notin I_{P^{*}}^{u}$. By
Corollary 1, there are $a_{1},a_{2}\in I_{P^{*}}$ with
$rank(a_{1}),rank(a_{2})>rank(a_{1}a_{2})$ such that $\theta=a_{1}a_{2}$. This
provides, $a_{1},a_{2}\in J$ by Lemma 2, i.e. $\theta\in J$, a contradiction
to $T_{BC}\cap J=\emptyset$. Thus, $T_{BC}\subseteq I_{P^{*}}^{u}$.

Conversely, let $J=I_{P^{*}}\backslash T_{\\{g\\}}$ with $|BC_{\\{g\\}}|=1$
for some $g\in P^{*}$ such that $\\{g\\}\in C$ and $T_{\\{g\\}}\subseteq
I_{P^{*}}^{u}$. Then $I_{P^{*}}\backslash T_{\\{g\\}}$ is a semigroup by Lemma
3. Since $|T_{\\{g\\}}|=1$, we can conclude that $J$ is a maximal subsemigroup
of $I_{P^{*}}$.
Let $J=I_{P^{*}}\backslash T_{B_{y_{1}}}$ with $|BC_{B_{y_{1}}}|=1$ for some
$y_{1}\in P^{*}$ such that $|y_{1}|\geq 2$, $B_{y_{1}}\in B$, and
$T_{B_{y_{1}}}\subseteq I_{P^{*}}^{u}$. Then $I_{P^{*}}\backslash
T_{B_{y_{1}}}$ is a semigroup by Lemma 4. Moreover, we can conclude that
$\\{B_{y_{1}}\\}=BC_{B_{y_{1}}}$. Let $\alpha,\beta\in T_{B_{y_{1}}}$. There
are $y_{3},y_{4},y_{5},y_{6}\in A_{y_{1}}$ and some
$t\in\\{1,2,...,|y_{1}|\\}$ such that
$dom(\alpha)=y_{3}^{[t]},im(\alpha)=y_{4}^{[t]},dom(\beta)=y_{5}^{[t]}$, and
$im(\beta)=y_{6}^{[t]}$. Further, there are $\theta_{1},\theta_{2}\in
I_{P^{*}}$ with
$dom(\theta_{1})=y_{5},im(\theta_{1})=y_{3},dom(\theta_{2})=y_{4}$, and
$im(\theta_{2})=y_{6}$. This shows $\theta_{1},\theta_{2}\in J$ because of
$rank(\theta_{1})=rank(\theta_{2})>rank(\alpha)$. So, we have
$\theta_{1}\alpha\theta_{2}=\beta$. Thus, we get $\beta\in\langle
J\cup\\{\alpha\\}\rangle$. Consequently, $J$ is a maximal subsemigroup of
$I_{P^{*}}$.
Let $J=I_{P^{*}}\backslash T_{Q}$ for some partition $Q=(Q_{1},Q_{2})$ of
$BC_{\Delta_{1}}$, where $\Delta_{1}\in B\cup C$ and $T_{Q}\subseteq
I_{P^{*}}^{u}$. We have that $I_{P^{*}}\backslash T_{Q}$ is a semigroup by
Lemma 5. Let $\alpha,\beta\in T_{Q}$. Then
$dom(\alpha)\in\Delta_{2},im(\alpha)\in\Delta_{3},dom(\beta)\in\Delta_{4}$,
and $im(\beta)\in\Delta_{5}$, where $\Delta_{2},\Delta_{4}\in Q_{1}$ and
$\Delta_{3},\Delta_{5}\in Q_{2}$. There are $\theta_{1},\theta_{2}\in
I_{P^{*}}$ with $dom(\theta_{1})=dom(\beta),im(\theta_{1})=dom(\alpha)$,
$dom(\theta_{2})=im(\alpha)$, and $im(\theta_{2})=im(\beta)$. We get that
$\theta_{1},\theta_{2}\in J$ because of
$im(\theta_{1})=dom(\alpha)\in\Delta_{2}\in Q_{1}$ and
$dom(\theta_{2})=im(\alpha)\in\Delta_{3}\in Q_{2}$. So, we have
$\theta_{1}\alpha\theta_{2}=\beta$. Thus, we get $\beta\in\langle
J\cup\\{\alpha\\}\rangle$. Consequently, we have that $J$ is a maximal
subsemigroup of $I_{P^{*}}$. ∎

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