# Fixed point behavior mapping of cumulants between the three-dimensional
Ising model and QCD

Xue Pan<EMAIL_ADDRESS>School of Electronic Information and Electrical
Engineering, Chengdu University, Chengdu 610106, China

###### Abstract

Fixed point behavior was found in the temperature dependence of normalized
cumulants of order parameter at different external magnetic fields in the
three-dimensional Ising model in my last work. In this paper, considering
possible existing QCD critical point belonging to the three-dimensional Ising
universality class and the non-universal mapping parameters between the Ising
model and QCD, effects of the mapping parameters on the fixed point behavior
in the net-baryon chemical potential dependence of normalized cumulants along
different experimental freeze-out curves is studied in this paper. We found
that when the directions of Ising variables, the reduced temperature and
external magnetic field, are orthogonal or not far from orthogonal after
mapping to the QCD temperature and net-baryon chemical potential plane, the
fixed point behavior exists in the net-baryon chemical potential dependence of
the normalized cumulants and is closer to the QCD critical point than the peak
in the cumulants.

###### pacs:

25.75.Nq, 24.60.Ky

##  I Introduction

One of the main goals of current relativistic heavy-ion collision experiments
is to reveal the phase diagram of quantum chromo-dynamics (QCD) maingoal ,
where the location of the critical point, which is a unique character of the
QCD phase diagram, is the most important. Recent years, high-order cumulants
of multiplicity distributions of conserved charges, such as net-baryon, net-
charge and net-strangeness, are suggested to search for the QCD critical point
stephanov-prl91 ; koch ; Stephanov-prl102 ; Karsch-EPJC71 .

In the vicinity of the critical point, non-monotonic behavior of the high-
order cumulants of conserved charges are found in a variety of QCD effective
models and is regarded as critical related signal Stephanov-prl102 ; Asakawa-
prl103 ; Fuweijie ; Vladi . Particularly, when the critical point is
approached on the crossover side, universal negative fourth-order cumulant of
the order parameter is predicted in Ref. Stephanov-prl107  and used to locate
the QCD critical point in experiments Phys.Rev.Lett.112.032302 . While in
Refs. Chin.Phys.C.43.033103 ; Chin.Phys.C.45.104103 , the authors argued that
the sign of high-order cumulants is not sufficient to prove the presence of
the critical point. Other work indicate that the peak structure of kurtosis
remains a solid feature and can be used as a clean signature of the critical
point Eur.Phys.J.C.79.245 ; Phys.Rev.C.103.034901 . Nevertheless the peak
structure of cumulants can also occurs in a first order phase transition or
crossover Nature ; JPG . Whether the non-monotonic or sign change behavior of
high-order cumulants of conserved charges is reliable, accurate or enough to
locate the critical point is uncertain.

In fact, the finite-size scaling, which implies a fixed point, can also be
employed to locate the critical point Phys.Rev.D.97.034015 ; JPG ;
Phys.Rev.E.100.052146 ; chenlz . Usually, the fixed point is obtained from the
scale transformation of the re-normalization group, resulting in the
independence of rescaled thermodynamics on the system sizes at the critical
point fixedpoint2 ; fixedpoint3 .

In my recent work, the fixed point behavior has been found in the temperature
dependence of normalized high-order cumulants of the order parameter at
different fixed external magnetic fields in the three-dimensional Ising model
cpc2022 . What is more, the fixed point is just at the critical temperature of
the Ising model.

The QCD critical point, if exists, is expected to belong to the same
universality class of the three-dimensional Ising model class1 ; class2 ;
class3 ; class4 . Critical behavior of the corresponding thermodynamics in
different systems that belongs to the same universality class is the same
which is supervised by the same critical exponents. Recently, many works have
been done to map the results of the three-dimensional Ising model to that of
the QCD PhysRevD102014505 ; Phys.Rev.C.103.034901 .

In order to map the results of the Ising model to that of the QCD, usually, a
linear ansatz between the QCD variables, temperature $T$ and net-baryon
chemical potential $\mu_{B}$, and the Ising variables, reduced temperature $t$
($t=(\tau-\tau_{c})/\tau_{c}$, where $\tau$ and $\tau_{c}$ are the temperature
and critical temperature of the three-dimensional Ising model, respectively)
and external magnetic field $h$ is suggested linearmap1 ; linearmap2 ;
linearmap3 ; NPA . Using three assumptions, firstly, the QCD system is in
equilibrium, secondly, the linear map keep Ising $t$-direction and
$h$-direction orthogonal to each other after mapped to the QCD $T-\mu_{B}$
phase diagram. At last, the freeze-out curves (FC) are below the
crossover/first-order phase transition line, and parallel to the $t$-direction
which has been mapped to the QCD phase diagram, the fixed point behavior has
been found in the energy (transformation from the net-baryon chemical
potential) dependence of normalized cumulants of order parameter in Ref.
cpc2022 . The fixed point is just at the critical energy (corresponding to the
net-baryon chemical potential at the QCD critical point).

However, the mapping parameters are non-universal and far unknown from the
Ising model to QCD. In this paper, on the assumption of the equilibrium of the
QCD system, using parametric representation of the three-dimensional Ising
model, we study and discuss the influences of the mapping parameters on the
fixed point behavior in the net-baryon chemical potential dependence of
normalized cumulants of order parameter along the experimental freeze-out
curves. The freeze-out curves are also supposed below the crossover/first-
order phase transition line, but are described by an empirical parametrization
of the heavy-ion-collision data Phys.Rev.C.73.034905 .

The paper is organized as follows. In section 2, the parametric representation
and corresponding parametric expressions of second- to fourth-order cumulants
of the three-dimensional Ising model is presented. The linear mapping from the
three-dimensional Ising model to QCD is discussed. The experimental freeze-out
curves are mapped to the Ising $t-h$ plane through different parameters. In
section 3, net-baryon chemical potential dependence of the second- to fourth-
order normalized cumulants is studied along the freeze-out curves. The effects
of different mapping parameters on the net-baryon chemical potential
dependence of the normalized cumulants are discussed. Finally, conclusions and
summary are given in section 4.

##  II The linear mapping from Ising to QCD

In the parametric representation of the three-dimensional Ising model,
magnetization $M$ and reduced temperature $t$ can be parameterized by two
variables $R$ and $\theta$ linearpara ; linearpara3 ,

$M=m_{0}R^{\beta}\theta,\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ t=R(1-\theta^{2}).$ (1)

The equation of state expressed by $R$ and $\theta$ is

$h=h_{0}R^{\beta\delta}\widetilde{h}(\theta).$ (2)

Where $m_{0}$ in Eq. (1) and $h_{0}$ in Eq. (2) are normalization constants.
They are fixed by imposing the normalization conditions $M(t=-1,h=+0)=1$ and
$M(t=0,h=1)=1$. $\beta$ and $\delta$ are critical exponents of the three-
dimensional Ising universality class with values 0.3267(10) and 4.786(14),
respectively Isingexponents .

One simple mean-field approximation of representation for
$\widetilde{h}(\theta)$ is as follows,

$\widetilde{h}(\theta)=\theta(3-2\theta^{2}).$ (3)

When taking the approximate values of the critical exponents $\beta=1/3$ and
$\delta=5$ (it is enough for our purpose), the first fourth order cumulants of
magnetization (order parameter of the three dimensional Ising model) in the
parametric representation are as follows,

$\begin{split}&\kappa_{1}(R,\theta)=m_{0}R^{1/3}\theta,\\\
&\kappa_{2}(R,\theta)=\frac{m_{0}}{h_{0}}\frac{1}{R^{4/3}(2\theta^{2}+3)},\\\
&\kappa_{3}(R,\theta)=\frac{m_{0}}{h_{0}^{2}}\frac{4\theta(\theta^{2}+9)}{R^{3}(\theta^{2}-3)(2\theta^{2}+3)^{3}},\\\
&\kappa_{4}(R,\theta)=12\frac{m_{0}}{h_{0}^{3}}\frac{(2\theta^{8}-5\theta^{6}+105\theta^{4}-783\theta^{2}+81)}{R^{14/3}(\theta^{2}-3)^{3}(2\theta^{2}+3)^{5}}.\\\
\end{split}$ (4)

The reduced temperature $t$ and external magnetic field $h$, are functions of
$R$ and $\theta$ provided by Eq. (1) and Eq. (2). Thus $\kappa_{n}(R,\theta)$
can be converted to $\kappa_{n}(t,h)$. At a negative $h$, the temperature
dependence of second- to fourth-order cumulants all have a positive peak
$\kappa_{n}^{max}(t,h)$ $(n=2,3,4)$. The cumulants can be normalized by their
maximum at the peak as follows,

$\kappa_{n}^{Norm}(t,h)=\kappa_{n}(t,h)/\kappa_{n}^{max}(t,h).$ (5)

The fixed point behavior has been found in the temperature dependence of
$\kappa_{n}^{Norm}(t,h)$ at different $h$ in the three-dimensional Ising
model. In order to map the results to that of the QCD, the following linear
mapping relations are assumed,

$\begin{split}&\frac{T-T_{c}}{\Delta T}=\sin\alpha_{1}\frac{t}{\Delta
t}+\sin\alpha_{2}\frac{h}{\Delta h},\\\
&\frac{\mu_{B}-\mu_{Bc}}{\Delta\mu_{B}}=-\cos\alpha_{1}\frac{t}{\Delta
t}-\cos\alpha_{2}\frac{h}{\Delta h}.\\\ \end{split}$ (6)

Where $T_{c}$, $\mu_{Bc}$ are the temperature and net-baryon chemical
potential at the QCD critical point. $\Delta T$ and $\Delta\mu_{B}$ denote the
width of the critical regime in the QCD $T-\mu_{B}$ plane. Because the
location of the critical point and the width of the critical regime for QCD
are not known, the suggestion that $\Delta\mu_{B}\approx 0.1$ GeV from model
calculations Phys.Rev.D.67.014028  and lattice QCD calculations
Phys.Rev.D.78.114503  is used. We set $\Delta\mu_{B}=0.1$ GeV and
$\mu_{Bc}=0.25$ GeV as was done in Ref. Phys.Rev.C.92.034912 . $T_{c}$ is set
as $0.165$ GeV and $\Delta T/T_{c}=1/8$.

$\Delta t$ and $\Delta h$ denote the width of the critical regime in the Ising
$t-h$ plane. For details of defining the critical region, see Ref.
Phys.Rev.C.92.034912 . After mapping $t$ and $h$ to the QCD $T-\mu_{B}$ phase
diagram, $\alpha_{1}$ is the angel between the $t$-direction and the opposite
direction of $\mu_{B}$ axis, $\alpha_{2}$ is the angel between the
$h$-direction and the opposite direction of $\mu_{B}$ axis. A simply sketch of
the mapping is shown in Fig. 1(a).

Figure 1: (Color online). A sketch of the mapping between QCD and Ising
variables described by Eq. (6) (a), the mapping of three different freeze-out
curves, FC I, FC II and FC III, given by Eqs. (7), (8) and (9) to the Ising
$t-h$ plane by Eq. (6) for $\Delta t=2$, $\Delta h=0.4$,
$\alpha_{1}=4.5^{\circ}$ and $\alpha_{2}=94.5^{\circ}$ (b), respectively.

The freeze-out curve is assumed below the crossover/first-order phase
transition line. An empirical parametrization of the heavy-ion-collision data
from Ref. Phys.Rev.C.73.034905  can be used to describe the freeze-out curve,

$T_{f}(\mu_{B})=a-b{\mu_{B}}^{2}-c{\mu_{B}}^{4}.$ (7)

Where $a=0.166$ GeV, $b=0.139$ GeV-1, $c=0.053$ GeV-3.

According to the results in Refs. Phys.Rev.C.71.054901 ;
AdvancesinHighEnergyPhysics.2021.6611394 , the chemical freeze-out temperature
and net-baryon chemical potential could change as the centrality in the
relativistic heavy-ion collisions. Although the temperature does not vary
much, the net-baryon chemical potential increases from peripheral to central
collisions AdvancesinHighEnergyPhysics.2021.6611394 . For simplicity, we move
the freeze-out curve described by Eq. (7) parallelly to the lower temperature
side to get the other two freeze-out curves, i.e.,

$T_{f1}(\mu_{B})=a-b{\mu_{B}}^{2}-c{\mu_{B}}^{4}-0.005,$ (8)

and

$T_{f2}(\mu_{B})=a-b{\mu_{B}}^{2}-c{\mu_{B}}^{4}-0.01.$ (9)

Based on the mapping relation and using Eq. (6), these three freeze-out curves
can be converted to three lines on the Ising $t-h$ plane. As an example,
supposing $\Delta t=2$, $\Delta h=0.4$, the angels $\alpha_{1}=4.5^{\circ}$
and $\alpha_{2}=94.5^{\circ}$, a sketch of the mapping is shown in Fig. 1(b).
The solid red, dashed blue and dotted black line from right to left represent
the mapping of the freeze-out curves FC I, FC II and FC III presented by Eqs.
(7), (8) and (9), respectively. It is clear that for each freeze-out curve
after mapping to the Ising $t-h$ plane, the external magnetic field is not
fixed as the cases in Ref. cpc2022 , but obvious fixed point behavior still
can be observed in the temperature dependence of $\kappa_{2}^{Norm}$,
$\kappa_{3}^{Norm}$ and $\kappa_{4}^{Norm}$ along these three freeze-out
curves, which are presented in Fig. 2.

Figure 2: (Color online). Temperature dependence of $\kappa_{2}^{Norm}$,
$\kappa_{3}^{Norm}$, and $\kappa_{4}^{Norm}$ for $\Delta t=2$, $\Delta h=0.4$,
$\alpha_{1}=4.5^{\circ}$ and $\alpha_{2}=94.5^{\circ}$ along the three
different freeze-out curves, respectively. The vertical green dashed line
shows the critical temperature $t=0$.

It is clear that the fixed point behavior exists in each sub-figure of Fig. 2.
Although the temperature at the fixed point is not the critical temperature
and a little bigger than the critical one, i.e. the fixed point is at the
right side of the vertical green dashed line $t=0$, it is still in the
vicinity of the critical temperature. What is more, the fixed point is closer
to the vertical green dashed line than the peak in the cumulants.

##  III Fixed point behavior in $\mu_{B}$ dependence of the normalized
cumulants

Through the linear mapping relation between QCD and Ising variables
represented by Eq. (6), $t$ dependence of the cumulants in the Ising model can
be converted to $\mu_{B}$ dependence. There are four parameters $\alpha_{1}$,
$\alpha_{2}$, $\Delta t$ and $\Delta h$. Along the $t$-direction, the symmetry
is preserved. It is easy to realize that the $t$-direction should be
tangential to the QCD first-order phase transition line at the critical point.
As a result, the angel $\alpha_{1}$ is usually not set very large
Phys.Rev.C.103.034901 . Here it is fixed at $4.5^{\circ}$, which is close to
the value in Ref. Phys.Rev.C.103.034901 , where it is set as $4.6^{\circ}$.
For the $h$-direction after mapping to QCD, there is no general rule. To set
the $h$-direction orthogonal to the $t$-direction is the common and simply
choice in the literature. That is $\alpha_{2}=94.5^{\circ}$ here. To study the
influence of $h$-direction slightly deviated from the orthogonal direction of
$t$, the other two values are also chosen for $\alpha_{2}$. They are
$72^{\circ}$ and $117^{\circ}$, respectively. What is more, $\alpha_{2}$ and
$\alpha_{2}\pm\pi$ are equivalent for the mapping since the sign of $h$ is a
matter of convention.

After the values of $\Delta t$ and $\Delta h$ are set as $2$ and $0.4$,
respectively, the influence of $\alpha_{2}$ on the fixed point behavior in the
$\mu_{B}$ dependence of the normalized second- to fourth-order cumulants are
studied and shown in Fig. 3.

Figure 3: (Color online). Net-baryon chemical potential dependence of
$\kappa_{2}^{Norm}$, $\kappa_{3}^{Norm}$, and $\kappa_{4}^{Norm}$ at
$\alpha_{2}=72^{\circ}$ (up panels), $94.5^{\circ}$ (middle panels) and
$117^{\circ}$ (down panels) with $\alpha_{1}=4.5^{\circ}$, $\Delta t=2$ and
$\Delta h=0.4$ along three different freeze-out curves, respectively. The
green dashed line shows the net-baryon chemical potential at the QCD critical
point.

In each sub-figure of Fig. 3, the vertical dashed green line
$\mu_{B}=\mu_{Bc}$ shows the critical net-baryon chemical potential. The solid
red curve, dashed blue curve and dotted black curve represent the $\mu_{B}$
dependence of the normalized cumulants along the three freeze-out curves
represented by Eq. (7), Eq. (8) and Eq. (9), respectively. The qualitative
$\mu_{B}$ dependence of each order cumulant does not change along the
different freeze-out curves. They all have a peak when approaching $\mu_{Bc}$
from the smaller $\mu_{B}$ side. The peak gets sharper and closer to the
vertical dashed green line when the freeze-out curve gets closer to the phase
boundary. I.e. the peak is sharpest in the solid red curve and it is closest
to the vertical dashed green line. And the case is just opposite in the dotted
black curve.

It is clear that the $\mu_{B}$ dependence of normalized cumulants along three
different freeze-out curves intersect at a point. Except for
$\kappa_{4}^{Norm}$ at $\alpha_{2}=72^{\circ}$ in Fig. 3(c), the fixed points
are all at the right side of the peaks and closer to the vertical dashed green
line than the peaks.

With the three different values of $\alpha_{2}$ chosen here, the fixed point
behavior all exits, while the position of the fixed point moves a little. In
the top row of Fig. 3, $\alpha_{2}=72^{\circ}$, the fixed point is at the left
side of the vertical dashed green line. $\mu_{B}$ of the fixed point is
smaller than $\mu_{Bc}$. With $\alpha_{2}=94.5^{\circ}$, i.e. the middle row
of Fig. 3, the fixed point is nearly on the vertical dashed green line. Its
$\mu_{B}$ is very close to $\mu_{Bc}$. While with $\alpha_{2}=117^{\circ}$,
i.e. the bottom row of Fig. 3, the fixed point is at the right side of the
vertical dashed green line. Its $\mu_{B}$ is a little larger than $\mu_{Bc}$.
What is more, as the increase of $\alpha_{2}$, the value of
$\kappa_{2}^{Norm}$, $\kappa_{3}^{Norm}$ and $\kappa_{4}^{Norm}$ at the fixed
point all decreases. And the higher the order of the normalized cumulants, the
faster its value changes with $\alpha_{2}$.

Although it is not shown here, it should be clarified that if $\alpha_{2}$
continues to decrease, the fixed point will come to the left side of the peak,
just like the case in Fig. 3(c), then the peak will be closer to $\mu_{Bc}$.
In reverse, if $\alpha_{2}$ continues to increase, the value of the normalized
cumulant at the fixed point along the three freeze-out curves will approach to
zero, just like the case in Fig.3 (i). If $\alpha_{2}$ approaches to
$\alpha_{1}$ or $\pi\pm\alpha_{1}$, that is the $t$-direction and
$h$-direction are nearly parallel to each other after mapped to the QCD
$T-\mu_{B}$ plane, the three freeze-out curves will not pass through or even
reach the critical region of the Ising model. The fixed point behavior in the
$t$ dependence of the normalized cumulants along the freeze-out curves does
not exist any more.

On the whole, when the $h$-direction and $t$-direction after mapping to the
QCD phase diagram is orthogonal or not far from orthogonal, the fixed point
behavior exists in $\mu_{B}$ dependence of normalized cumulants along
different freeze-out curves and is closer to $\mu_{Bc}$ than the peak in the
cumulants. While the value of $\alpha_{2}$ will influence the position of the
fixed point. When $h$-direction and $t$-direction are nearly orthogonal to
each other, $\mu_{B}$ of the fixed point is closer to $\mu_{Bc}$.

Figure 4: (Color online). Net-baryon chemical potential dependence of
$\kappa_{2}^{Norm}$ at different $\Delta t$ and $\Delta h$ along three
different freeze-out curves. $\Delta t=1$, $2$ and $4$ for the up, middle and
down panels, respectively. $\Delta h=0.1$, $0.2$ and $0.4$ for the left,
middle and right panels, respectively. The green dashed line shows the net-
baryon chemical potential at the QCD critical point.

To discuss the effects of $\Delta t$ and $\Delta h$ on the fixed point
behavior in $\mu_{B}$ dependence of the normalized cumulants, $\Delta t$ is
taken three different values $1$, $2$ and $4$. $\Delta h$ is also taken three
different values $0.1$, $0.2$ and $0.4$. While the angles $\alpha_{1}$ and
$\alpha_{2}$ are fixed at $4.5^{\circ}$ and $94.5^{\circ}$, respectively. The
influence of the change of $\Delta t$ and $\Delta h$ on the fixed point
behavior in $\kappa_{2}^{Norm}$, $\kappa_{3}^{Norm}$ and $\kappa_{4}^{Norm}$
are the same. For simplicity, let us take $\kappa_{2}^{Norm}$ as an example.
The net-baryon chemical potential dependence of $\kappa_{2}^{Norm}$ is shown
in Fig. 4. Results along the three freeze-out curves are still represented by
the solid red curve, dashed blue curve and dotted black curve, respectively.
The vertical dashed green line shows the critical net-baryon chemical
potential.

In each column, the value of $\Delta h$ is fixed and the value of $\Delta t$
increases from top to bottom. It is clear that as $\Delta t$ gets bigger, the
peak in $\kappa_{2}^{Norm}$ gets sharper. The position of fixed point shift to
the bigger $\mu_{B}$ side a little. At the same time, the value of
$\kappa_{2}^{Norm}$ at fixed point gets a little smaller. In each row, the
value of $\Delta t$ is fixed while the value of $\Delta h$ increases from left
to right. It is clear that the peak in $\kappa_{2}^{Norm}$ gets flatter as
$\Delta h$ gets bigger. The position of fixed point shift to the smaller
$\mu_{B}$ side a little. Meanwhile the value of $\kappa_{2}^{Norm}$ at fixed
point gets a little bigger.

So the trend of the influence of $\Delta t$ and $\Delta h$ on the normalized
cumulants and position of fixed point is just opposite. Smaller $\Delta t$ and
bigger $\Delta h$ lead to flatter peak in $\mu_{B}$ dependence of
$\kappa_{2}^{Norm}$ and the move of the fixed point to the smaller $\mu_{B}$
side, i.e. the peak of $\kappa_{2}^{Norm}$ in Fig. 4(c) is the flattest and
the value of $\mu_{B}$ at the fixed point is the smallest. One more thing to
note is that the position of the fixed point is all close to the vertical
dashed green line in each sub-figure.

On the whole, the influence of $\Delta t$ and $\Delta h$ on the fixed point
behavior is not so great as that of $\alpha_{2}$. In all case, the fixed point
exists and is close to the critical net-baryon chemical potential when the
$t$-direction and $h$-direction are nearly orthogonal.

Turning to the heavy-ion collision experiments, the energy ($\sqrt{s}$)
dependence of the normalized cumulants can be got from the relation between
$\mu_{B}$ and $\sqrt{s}$ given in Ref. Phys.Rev.C.73.034905 , for details to
see Ref. cpc2022 . It is easy to realize that the fixed point behavior also
exists in the energy dependence of the normalized cumulants along the three
different freeze-out curves as long as the $t$-direction and $h$-direction is
not far from orthogonal to each other after mapped to the QCD $T-\mu_{B}$
plane. Based on the universality of the critical behavior, the fixed point
behavior may be used to locate the QCD critical point in the relativistic
heavy-ion collision experiments.

##  IV Summary

Through a linear mapping relation from the three-dimensional Ising model to
QCD, the experimental freeze-out curves are mapped to the Ising $t-h$ plane.
By using the parametric representation of the three-dimensional Ising model
and the linear mapping relation, the net-baryon chemical potential dependence
of normalized second- to fourth-order cumulants of the order parameter along
three different freeze-out curves are studied.

Fixed point behavior is observed in the net-baryon chemical potential
dependence of the normalized cumulants along three different freeze-out curves
when the $h$-direction and $t$-direction is orthogonal or not far from
orthogonal after mapping to the QCD $T-\mu_{B}$ phase plane. The fixed point
is closer to the net-baryon chemical potential of QCD critical point than the
peak of the cumulants.

The influence of three mapping parameters $\alpha_{2}$, $\Delta t$ and $\Delta
h$, on the net-baryon chemical potential dependence of the normalized
cumulants and the fixed point behavior is discussed. As the increase of
$\alpha_{2}$, that is from $72^{\circ}$ to $117^{\circ}$ here, the net-baryon
chemical potential at the fixed point increases, while the corresponding value
of normalized cumulants decreases. When the $t$-direction and $h$-direction
are nearly orthogonal to each other, the net-baryon chemical potential at the
fixed point is much closer to the one at the QCD critical point.

The two parameters describing the width of the critical region of the three-
dimensional Ising model, $\Delta t$ and $\Delta h$, have great influence on
the behavior of cumulants, but little influence on the fixed point behavior.
As the increase of $\Delta t$ and decrease of $\Delta h$, the peak in the net-
baryon chemical potential dependence of the cumulants gets sharper, while the
net-baryon chemical potential at the fixed point get a little bigger but not
so obvious. And the values of $\mu_{B}$ at the fixed point are all close to
that of $\mu_{Bc}$ when the $t$-direction and $h$-direction are orthogonal.

Fixed point behavior also exists in the energy dependence of the normalized
cumulants along the experimental freeze-out curves when the $t$-direction and
$h$-direction is orthogonal or not far from orthogonal after mapping to the
QCD $T-\mu_{B}$ plane. It may be used to locate the QCD critical point in the
relativistic heavy-ion collision experiments.

## References

  * (1) J. Adams et al (STAR Collaboration), Nucl. Phys. A 757 (2005) 102.
  * (2) B. Friman, F. Karsch, K. Redlich et al, Eur. Phys. J. C 71 (2001) 1694.
  * (3) Y. Hatta, M. A. Stephanov, Phys. Rev. Lett. 91 (2003) 102003.
  * (4) V. Koch, arXiv:0810.2520.
  * (5) M. A. Stephanov, Phys. Rev. Lett. 102 (2009) 032301.
  * (6) M. Asakawa, S. Ejiri, M. Kitazawa, Phys. Rev. Lett. 103 (2009) 262301.
  * (7) V. Skokov, B. Friman and K. Redlich, Phys. Rev. C 83 (2011) 054904.
  * (8) Wei-jie Fu, Yu-Xin Liu, and Yue-Liang Wu, Phys. Rev. D 81 (2010) 014028.
  * (9) M. A. Stephanov, Phys. Rev. Lett. 107 (2011) 052301.
  * (10) S. Collaboration, L. Adamczyk, J. K. Adkins, et al., Phys. Rev. Lett. 112 (2014) 032302.
  * (11) W. Fan, X. Luo and H. Zong, Chin. Phys. C 43 (2019) 033103.
  * (12) Li-Zhu Chen, Ye-yin Zhao, Jin Wu, et al., Chin. Phys. C 45 (2021) 104103.
  * (13) Z. Li, K. Xu, X. Wang and M. Huang, Eur. Phys. J. C 79 (2019) 245.
  * (14) D. Mroczek, A. R. Nava Acuna, J. Noronha-Hostler et al., Phys. Rev. C 103 (2021) 034901.
  * (15) Y. Aoki, G. Endrödi, Z. Fodor et al, Nature 443 (2006) 675.
  * (16) Xue Pan, Mingmei Xu and Yuanfang Wu, J. Phys. G: Nucl. Part. phys. 42 (2015) 015104.
  * (17) N. G. Antoniou, F. K. Diakonos, X. N. Maintas and C. E. Tsagkarakis, Phys. Rev. D 97 (2018) 034015.
  * (18) Y. Zhang, Y. Zhao, L. Chen, X. Pan, M. Xu, Z. Li, Y. Zhou and Y. Wu, Phys. Rev. E 100 (2019) 052146.
  * (19) L. Chen, Y. Zhao, X. Li, Z. Li and Y. Wu, Int. Jour. Mod. Phys. E 30 (2021) 2150056.
  * (20) M. E. Fisher, Rev. Mod. Phys. 46 (1974) 597.
  * (21) K. Wilson, Phys. Rep. 12 (1974) 75.
  * (22) Xue Pan, Chin. Phys. C 46 (2022) 024104.
  * (23) P. de Forcrand, O. Philipsen, Phys. Rev. Lett. 105 (2010) 152001.
  * (24) M. Stephanov, K. Rajagopal, E. Shuryak, Phys. Rev. Lett. 81 (1998) 4816.
  * (25) M. Asakawa, J. Phys. G: Nucl. Part. Phys 36 (2009) 064042.
  * (26) R. D. Pisarski, F. Wilczek, Phys. Rev. D 29 (1984) 338.
  * (27) M. Caselle and M. Sorba, Phys. Rev. D 102 (2020) 014505.
  * (28) J.J. Rehr and N.D. Mermin, Phys. Rev. A 8 (1973) 472.
  * (29) N. B Wilding, J. Phys.: Condens. Matter 9 (1997) 585.
  * (30) C. Nonaka and M. Asakawa, Phys. Rev. C 71 (2005) 044904.
  * (31) Xue pan, Lizhu Chen, X.S. Chen and Yuanfang Wu, Nucl. Phys. A 913 (2013) 206.
  * (32) J. Cleymans, H. Oeschler, K. Redlich and S. Wheaton, Phys. Rev. C 73 (2006) 034905.
  * (33) P. Schofield, Phys. Rev. Lett. 22 (1969) 606.
  * (34) D. J. Wallace, R. K. P. Zia, J. Phys. C: Solid State Phys 7 (1974) 3480.
  * (35) H. W. J. Blöte, E. Luijten, J. R. Heringa, J. Phys. A: Math. Gen 28 (1995) 6289.
  * (36) Y. Hatta and T. Ikeda, Phys. Rev. D 67 (2003) 014028.
  * (37) R. V. Gavai and S. Gupta, Phys. Rev. D 78 (2008) 114503.
  * (38) S. Mukherjee, R. Venugopalan and Y. Yin, Phys. Rev. C 92 (2015) 034912.
  * (39) J. Cleymans, B. Kämpfer, M. Kaneta, S. Wheaton and N. Xu, Phys. Rev. C 71 (2005) 054901.
  * (40) D. Biswas, Advances in High Energy Physics 2021 (2021) 6611394.


* (70) Egron, E., Pellizzoni, A., Giroletti, M., Righini, S., Stagni, M., Orlati, A., Migoni, C., Melis, A., Concu, R., Barbas, L., Buttaccio, S., Cassaro, P., De Vicente, P., Gawroński, M.P., Lindqvist, M., Maccaferri, G., Stanghellini, C., Wolak, P., Yang, J., Navarrini, A., Loru, S., Pilia, M., Bachetti, M., Iacolina, M.N., Buttu, M., Corbel, S., Rodriguez, J., Markoff, S., Wilms, J., Pottschmidt, K., Cadolle Bel, M., Kalemci, E., Belloni, T., Grinberg, V., Marongiu, M., Vargiu, G.P., Trois, A.: Single-dish and VLBI observations of Cygnus X-3 during the 2016 giant flare episode. MNRAS 471(3), 2703–2714 (2017) arXiv:1707.03761 [astro-ph.HE]. https://doi.org/10.1093/mnras/stx1730
  * (71) Tsybulev, P.G., Nizhelskii, N.A., Dugin, M.V., Borisov, A.N., Kratov, D.V., Udovitskii, R.Y.: C-Band Radiometer for Continuum Observations at RATAN-600 Radio Telescope. Astrophysical Bulletin 73(4), 494–500 (2018). https://doi.org/10.1134/S1990341318040132
  * (72) Kale, R., Ishwara-Chandra, C.H.: CAPTURE: a continuum imaging pipeline for the uGMRT. Experimental Astronomy 51(1), 95–108 (2021) arXiv:2010.00196 [astro-ph.IM]. https://doi.org/10.1007/s10686-020-09677-6
  * (73) Marrone, D.P., Rao, R.: The submillimeter array polarimeter. In: Duncan, W.D., Holland, W.S., Withington, S., Zmuidzinas, J. (eds.) Millimeter and Submillimeter Detectors and Instrumentation for Astronomy IV. Society of Photo-Optical Instrumentation Engineers (SPIE) Conference Series, vol. 7020, p. 70202 (2008). https://doi.org/10.1117/12.788677
  * (74) Chandrasekhar, S.: Radiative Transfer. Dover, New York (1960)
  * (75) Goosmann, R.W., Gaskell, C.M.: Modeling optical and UV polarization of AGNs. I. Imprints of individual scattering regions. A&A 465(1), 129–145 (2007) arXiv:astro-ph/0507072 [astro-ph]. https://doi.org/10.1051/0004-6361:20053555
  * (76) Marin, F., Dovčiak, M., Muleri, F., Kislat, F.F., Krawczynski, H.S.: Predicting the X-ray polarization of type 2 Seyfert galaxies. MNRAS 473(1), 1286–1316 (2018) arXiv:1709.03304 [astro-ph.HE]. https://doi.org/10.1093/mnras/stx2382
  * (77) Asplund, M., Grevesse, N., Sauval, A.J.: The Solar Chemical Composition. In: Barnes, I. Thomas G., Bash, F.N. (eds.) Cosmic Abundances as Records of Stellar Evolution and Nucleosynthesis. Astronomical Society of the Pacific Conference Series, vol. 336, p. 25 (2005)
  * (78) Magdziarz, P., Zdziarski, A.A.: Angle-dependent Compton reflection of X-rays and gamma-rays. MNRAS 273(3), 837–848 (1995). https://doi.org/10.1093/mnras/273.3.837
  * (79) Miller-Jones, J.C.A., Bahramian, A., Orosz, J.A., Mandel, I., Gou, L., Maccarone, T.J., Neijssel, C.J., Zhao, X., Ziółkowski, J., Reid, M.J., Uttley, P., Zheng, X., Byun, D.-Y., Dodson, R., Grinberg, V., Jung, T., Kim, J.-S., Marcote, B., Markoff, S., Rioja, M.J., Rushton, A.P., Russell, D.M., Sivakoff, G.R., Tetarenko, A.J., Tudose, V., Wilms, J.: Cygnus X-1 contains a 21-solar mass black hole—Implications for massive star winds. Science 371(6533), 1046–1049 (2021) arXiv:2102.09091 [astro-ph.HE]. https://doi.org/10.1126/science.abb3363
  * (80) King, A.R.: Masses, beaming and Eddington ratios in ultraluminous X-ray sources. MNRAS 393(1), 41–44 (2009) arXiv:0811.1473 [astro-ph]. https://doi.org/10.1111/j.1745-3933.2008.00594.x
  * (81) Mushtukov, A.A., Portegies Zwart, S.: Bright X-ray pulsars: how outflows influence beaming, pulsations and pulse phase lags. MNRAS 518(4), 5457–5464 (2023) arXiv:2211.08952 [astro-ph.HE]. https://doi.org/10.1093/mnras/stac3431


# Unit-regular and semi-balanced elements in various semigroups of
transformations

Mosarof Sarkar Department of Mathematics, Central University of South Bihar,
Gaya, Bihar, India<EMAIL_ADDRESS>and  Shubh N. Singh Department of
Mathematics, Central University of South Bihar, Gaya, Bihar, India
<EMAIL_ADDRESS>
###### Abstract.

Let $T(X)$ be the full transformation semigroup on a set $X$, and let $L(V)$
be the semigroup under composition of all linear transformations on a vector
space $V$ over a field. For a subset $Y$ of $X$ and a subspace $W$ of $V$,
consider the semigroups $\overline{T}(X,Y)=\\{f\in T(X)\colon Yf\subseteq
Y\\}$ and $\overline{L}(V,W)=\\{f\in L(V)\colon Wf\subseteq W\\}$ under
composition. We describe unit-regular elements in $\overline{T}(X,Y)$ and
$\overline{L}(V,W)$. Using these, we determine when $\overline{T}(X,Y)$ and
$\overline{L}(V,W)$ are unit-regular. We prove that $f\in L(V)$ is unit-
regular if and only if
$\operatorname{\mbox{nullity}}(f)=\operatorname{\mbox{corank}}(f)$. We
alternatively prove that $L(V)$ is unit-regular if and only if $V$ is finite-
dimensional. A semi-balanced semigroup is a transformation semigroup whose all
elements are semi-balanced. We give necessary and sufficient conditions for
$\overline{T}(X,Y)$, $\overline{L}(V,W)$ and $L(V)$ to be semi-balanced.

###### Key words and phrases:

Transformations; Linear operators; Invariant subsets; Invariant subspaces;
Unit-regular elements; Semi-balanced transformations.

###### 2010 Mathematics Subject Classification:

20M17, 20M20, 15A03, 15A04.

##  1\. Introduction

An element $s$ of a semigroup $S$ with identity is _unit-regular_ in $S$ if
there exists a unit $u\in S$ such that $sus=s$. A _unit-regular semigroup_ is
a semigroup with identity in which every element is unit-regular. The concept
of unit-regularity, which was introduced by Ehrlich [8] within the context of
rings, has consecutively received much attention from many semigroup theorists
[1, 2, 3, 7, 9, 10, 18, 19, 28]. In fact, Alarcao [7, Proposition 1] proved
that any semigroup with identity is unit-regular if and only if it is
factorizable.

Let $X$ be a nonempty set. Denote by $T(X)$ the full transformation semigroup
on $X$. The semigroup $T(X)$ and its subsemigroups are extremely important,
since every semigroup can be embedded in some $T(Z)$ (cf. [14, Theorem 1.1.2].
In $1980$, Alarcao [7, Proposition 5] proved that $T(X)$ is unit-regular if
and only if $X$ is finite. For a fixed nonempty subset $Y$ of $X$, Magill [17]
first studied the semigroup $\overline{T}(X,Y)$ under composition consisting
of all transformations $f\in T(X)$ such that $Y$ is invariant under $f$. Using
symbols,

$\overline{T}(X,Y)=\\{f\in T(X)\colon Yf\subseteq Y\\}.$

If $Y=X$, then $\overline{T}(X,Y)=T(X)$. To this extent, the semigroup
$\overline{T}(X,Y)$ may be regarded as a generalization of $T(X)$. Many
interesting properties of the semigroup $\overline{T}(X,Y)$ and its certain
subsemigroups have been investigated [5, 6, 12, 20, 21, 25, 26, 27]. In
particular, Nenthein et al. [20, Theorem 2.3] described regular elements in
$\overline{T}(X,Y)$.

Let $V$ be a vector space over a field. Denote by $L(V)$ the semigroup under
composition consisting of all linear transformations on $V$. Ehrlich [8,
Theorem 4] proved that $L(V)$ is not unit-regular when $V$ is infinite-
dimensional. Jampachon et al. [15, Theorem 2] proved that $L(V)$ is
factorizable if and only if $V$ is finite-dimensional. If $V$ is finite-
dimensional, Kemprasit [16] directly proved that $L(V)$ is unit-regular. For a
fixed subspace $W$ of $V$, analogous to the semigroup $\overline{T}(X,Y)$,
Nenthein and Kemprasit [21] introduced the semigroup $\overline{L}(V,W)$ under
composition consisting of all linear transformations $f\in L(V)$ such that $W$
is invariant under $f$. Using symbols,

$\overline{L}(V,W)=\\{f\in L(V)\colon Wf\subseteq W\\}.$

If $W$ is trivial, then $\overline{L}(V,W)=L(V)$. To this extent, the
semigroup $\overline{L}(V,W)$ may be regarded as a generalization of $L(V)$.
The authors [21, Proposition 3.1] described regular elements in
$\overline{L}(V,W)$ and proved that $\overline{L}(V,W)$ is regular if and only
if $W$ is trivial. Chaiya [2, Theorem 11] determined when $\overline{L}(V,W)$
is a unit-regular semigroup. Several other properties of the semigroup
$\overline{L}(V,W)$ have also been studied [2, 4, 13, 22].

Let $f\in T(X)$. The _collapse_ (resp. _defect_) of $f$ is the cardinality of
$X\setminus T_{f}$ (resp. $X\setminus Xf$), where $T_{f}$ is any transversal
of the equivalence relation $\ker(f)$ on $X$ defined by $(x,y)\in\ker(f)$ if
$xf=yf$. In $1998$, Higgins et al. [11, p. 1356] called an element of $T(X)$
to be _semi-balanced_ if its collapse and defect are equal, and denoted the
set of all semi-balanced elements of $T(X)$ by $B$. We say that a subsemigroup
of $T(X)$ is _semi-balanced_ if all its elements are semi-balanced. It is
clear that every subsemigroup of $T(X)$ is semi-balanced when $X$ is finite.
The authors [24, Theorem 3.3] proved that an element of $T(X)$ is unit-regular
if and only if it is semi-balanced.

The aim of this paper is to describe unit-regular and semi-balanced elements
in $\overline{T}(X,Y)$, $\overline{L}(V,W)$, and $L(V)$. The rest of this
paper is organized as follows. In the next section, we define concepts,
introduce notation, and list some necessary results. In Section $3$, we
describe unit-regular elements in $\overline{T}(X,Y)$ and determine when
$\overline{T}(X,Y)$ is unit-regular. In Section $4$, we describe unit-regular
elements in $\overline{L}(V,W)$. Using this, we prove that $f\in L(V)$ is
unit-regular if and only if
$\operatorname{\mbox{nullity}}(f)=\operatorname{\mbox{corank}}(f)$. We also
alternatively deduce that $L(V)$ is unit-regular if and only if $V$ is finite-
dimensional. We further determine when $\overline{L}(V,W)$ is unit-regular. In
Section $5$, we give necessary and sufficient conditions for the semigroups
$\overline{T}(X,Y)$, $\overline{L}(V,W)$, and $L(V)$ to be semi-balanced.

##  2\. Preliminaries and Notation

The cardinality of a set $A$ is denoted by $|A|$. Given any sets $A$ and $B$,
let $A\setminus B\colonequals\\{x\in A\colon x\notin B\\}$. The set of all
positive integers (resp. rational numbers) is denoted by $\mathbb{N}$ (resp.
$\mathbb{Q}$), and let $\aleph_{0}\colonequals|\mathbb{N}|$. A _transversal_
of an equivalence relation $\rho$ on a nonempty set $X$ is a subset of $X$
that contains exactly one element from each $\rho$-class. Given a semigroup
$S$ with identity, the set of all units of $S$ is denoted by $U(S)$; the set
of all unit-regular elements in $S$ is denoted by
$\operatorname{\mbox{ureg}}(S)$.

We compose maps from left to right and denote their composition by
juxtaposition. Let $f\colon X\to Y$ be a map. The domain, codomain, and range
of $f$ are denoted by $\operatorname{\mbox{dom}}(f)$,
$\operatorname{\mbox{codom}}(f)$, and $R(\alpha)$, respectively. We use $xf$
to denote the image of an element $x$ of $X$ under $f$. Given any
$A\subseteq\operatorname{\mbox{dom}}(f)$ (resp.
$B\subseteq\operatorname{\mbox{codom}}(f)$), let $Af$ (resp. $Bf^{-1}$) denote
the set $\\{af\colon a\in A\\}$ (resp.
$\\{x\in\operatorname{\mbox{dom}}(f)\colon xf\in B\\}$). Moreover, we write
$bf^{-1}$ instead of $\\{b\\}f^{-1}$ if
$B=\\{b\\}\subseteq\operatorname{\mbox{codom}}(f)$.

Let $f\colon X\to X$ be a map and $A$ be a nonempty subset of
$\operatorname{\mbox{dom}}(f)$. We say that $A$ is _invariant_ under $f$ if
$Af\subseteq A$. The _restriction_ of $f$ to $A$ is the map $f_{|_{A}}\colon
A\to X$ defined by $a(f_{|_{A}})=af$ for all $a\in A$. If
$B\subseteq\operatorname{\mbox{codom}}(f)$ such that $Xf\subseteq B$, then the
_corestriction_ of $f$ to $B$ is the map from $X$ to $B$ that agrees with $f$.
Moreover, we use $f_{\upharpoonright_{B}}$ to denote the corestriction of the
map $f_{|_{B}}$ to $B$ if $B$ is invariant under $f$. Note that if $f\in
U(\overline{T}(X,Y))$, then $f_{\upharpoonright_{Y}}\in U(T(Y))$.

Let $f\colon X\to X$ be a map. Set $D(f)\colonequals X\setminus R(f)$. The
_defect_ $\textnormal{d}(f)$ of $f$ is the cardinality of $D(f)$. The _kernel_
of $f$, denoted by $\ker(f)$, is an equivalence relation on $X$ defined by
$\ker(f)\colonequals\\{(x,y)\in X\times X\colon xf=yf\\}$. The symbol $\pi(f)$
denotes the partition of $X$ induced by $\ker(f)$, and $T_{f}$ denotes any
transversal of the equivalence relation $\ker(f)$. Note that $|X\setminus
T_{f}|$ is independent of the choice of transversal of $\ker(f)$ (cf. [11, p.
1356]), and $|T_{f}|=|R(f)|$. The _collapse_ $\textnormal{c}(f)$ of $f$ is the
cardinality of $X\setminus T_{f}$ (cf. [11, p. 1356]). It is clear that
$\textnormal{c}(f)=0$ if and only if $f$ is injective. If
$\textnormal{c}(f)=\textnormal{d}(f)$, then we say that $f$ is _semi-balanced_
(cf. [11, p. 1356]). Let $B(S)$ denote the set of all semi-balanced elements
in a subsemigroup $S$ of $T(X)$. Notice that $B(S)=S\cap B$.

Throughout this paper, unless otherwise noted, let $V$ denote a vector space
over an arbitrary field. We denote by $0$ the zero vector of $V$. The
subspaces $\\{0\\}$ and $V$ of $V$ are called _trivial_ subspaces. The
subspace spanned by a subset $T$ of $V$ is denoted by $\langle T\rangle$. Let
$U$ be a subspace of $V$. We say that $U$ is a _proper_ subspace of $V$ if
$U\neq V$. The dimensions of $U$ and the quotient space $V/U$ are denoted by
$\dim(U)$ and $\operatorname{\mbox{codim}}_{V}(U)$, respectively. We denote by
$V=U\oplus W$ the (internal) direct sum of subspaces $U$ and $W$ of $V$. If
$V=U\oplus W$, then we say that $W$ is a _complement_ of $U$ in $V$.

Let $f\in L(V)$. Set $R(f)\colonequals\\{vf\colon v\in V\\}$ and
$N(f)\colonequals\\{v\in V\colon vf=0\\}$. Note that $N(f)$ is a subspace of
$\operatorname{\mbox{dom}}(f)$, and $R(f)$ is a subspace of
$\operatorname{\mbox{codom}}(f)$. We write $\operatorname{\mbox{nullity}}(f)$,
$\operatorname{\mbox{rank}}(f)$, and $\operatorname{\mbox{corank}}(f)$ to
denote $\dim(N(f))$, $\dim(R(f))$, and $\dim(V/R(f))$, respectively. If $V$
and $U$ are isomorphic vector spaces, then we write $V\approx U$. Let $B$ and
$B^{\prime}$ be bases for $V$. We write $\bar{g}$ to denote the unique linear
transformation on $V$ obtained by linear extension of a map $g\colon B\to V$
or a map $g\colon B\to B^{\prime}$ to the entire space $V$ (cf. [23, Theorem
2.2]). Note that if $g\colon B\to B^{\prime}$ is a bijective map, then
$\bar{g}\in U(L(V))$. If $f\in\overline{L}(V,W)$, then
$f_{\upharpoonright_{W}}\in L(W)$ (cf. [23, p. 73]).

We refer the reader to [14] and [23] for undefined terminology in semigroup
theory and linear algebra, respectively. In the rest of the paper, $Y$ is a
nonempty subset of a set $X$, and $W$ is a subspace of a vector space $V$ over
a field. We end this section by stating a list of results needed in the
sequel.

######  Proposition 2.1.

[7, Proposition 1] A monoid is unit-regular if and only if it is factorizable.

######  Proposition 2.2.

[7, Proposition 5] The semigroup $T(X)$ is unit-regular if and only if $X$ is
finite.

######  Theorem 2.3.

[15, Theorem 2] The semigroup $L(V)$ is factorizable if and only if $V$ is
finite-dimensional.

######  Lemma 2.4.

[24, Lemma 3.1] Let $f\colon A\to B$ and $g\colon B\to A$ be maps. If $fgf=f$,
then $R(fg)$ is a transversal of the equivalence relation $\ker(f)$ on $A$.

######  Theorem 2.5.

[24, Theorem 3.3] Let $f\in T(X)$. Then $f$ is unit-regular in $T(X)$ if and
only if $f$ is semi-balanced.

######  Lemma 2.6.

[24, Lemma 3.5] Every element of $T(X)$ is semi-balanced if and only if $X$ is
finite.

##  3\. Unit-regular elements in $\overline{T}(X,Y)$

In this section, we describe unit-regular elements in $\overline{T}(X,Y)$.
Using this, we determine when $\overline{T}(X,Y)$ is unit-regular. We begin
with the following lemma.

######  Lemma 3.1.

Let $f\in\overline{T}(X,Y)$. If $f\in\textnormal{ureg}(\overline{T}(X,Y))$,
then $f_{\upharpoonright_{Y}}\in\textnormal{ureg}(T(Y))$.

###### Proof.

If $f\in\operatorname{\mbox{ureg}}(\overline{T}(X,Y))$, then there exists
$g\in U(\overline{T}(X,Y))$ such that $fgf=f$. Therefore we get
$f_{\upharpoonright_{Y}}g_{\upharpoonright_{Y}}f_{\upharpoonright_{Y}}=f_{\upharpoonright_{Y}}$
and $g_{\upharpoonright_{Y}}\in U(T(Y))$. Hence
$f_{\upharpoonright_{Y}}\in\operatorname{\mbox{ureg}}(T(Y))$. ∎

Note that $\overline{T}(X,Y)$ contains the identity map on $X$. The next
theorem describes unit-regular elements in $\overline{T}(X,Y)$.

######  Theorem 3.2.

Let $f\in\overline{T}(X,Y)$. Then $f\in\textnormal{ureg}(\overline{T}(X,Y))$
if and only if

  1. (i)

$f_{\upharpoonright_{Y}}\in\textnormal{ureg}(T(Y))$;

  2. (ii)

$R(f_{\upharpoonright_{Y}})=Y\cap R(f)$;

  3. (iii)

$|C(f)\setminus C(f_{\upharpoonright_{Y}})|=|D(f)\setminus
D(f_{\upharpoonright_{Y}})|$, where $C(f)\colonequals X\setminus T_{f}$ and
$C(f_{\upharpoonright_{Y}})\colonequals Y\setminus
T_{(f_{\upharpoonright_{Y}})}$ for some transversals $T_{f}$ and
$T_{(f_{\upharpoonright_{Y}})}$ of $\ker(f)$ and
$\ker(f_{\upharpoonright_{Y}})$, respectively, such that
$T_{(f_{\upharpoonright_{Y}})}=Y\cap T_{f}$.

###### Proof.

Suppose that $f\in\operatorname{\mbox{ureg}}(\overline{T}(X,Y))$. Then $fgf=f$
for some $g\in U(\overline{T}(X,Y))$.

  1. (i)

By Lemma 3.1, we get
$f_{\upharpoonright_{Y}}\in\operatorname{\mbox{ureg}}(T(Y))$.

  2. (ii)

It is clear that $R(f_{\upharpoonright_{Y}})\subseteq Y\cap R(f)$, since
$Yf\subseteq Y$. For the reverse inclusion, let $y\in Y\cap R(f)$. Then $xf=y$
for some $x\in X$. Therefore, since $f=fgf$, $y\in Y$, and $Yg\subseteq Y$, we
obtain

$y=xf=(xf)gf=(yg)f\in R(f_{\upharpoonright_{Y}}).$

This implies $Y\cap R(f)\subseteq R(f_{\upharpoonright_{Y}})$ as required.

  3. (iii)

Recall first that $fgf=f$. Then
$f_{\upharpoonright_{Y}}g_{\upharpoonright_{Y}}f_{\upharpoonright_{Y}}=f_{\upharpoonright_{Y}}$.
Therefore by Lemma 2.4, the sets $T_{f}\colonequals R(fg)$ and
$T_{(f_{\upharpoonright_{Y}})}\colonequals
R(f_{\upharpoonright_{Y}}g_{\upharpoonright_{Y}})$ are transversals of
$\ker(f)$ and $\ker(f_{\upharpoonright_{Y}})$, respectively. Notice that
$T_{(f_{\upharpoonright_{Y}})}=Y\cap T_{f}$. Set $C(f)\colonequals X\setminus
T_{f}$ and $C(f_{\upharpoonright_{Y}})\colonequals Y\setminus
T_{(f_{\upharpoonright_{Y}})}$. Since $g$ and $g_{\upharpoonright_{Y}}$ are
bijections, we obtain

$D(f)g=(X\setminus R(f))g=Xg\setminus R(fg)=X\setminus
T_{f}=C(f)\quad\text{and}$
$D(f_{\upharpoonright_{Y}})g_{\upharpoonright_{Y}}=(Y\setminus
R(f_{\upharpoonright_{Y}}))g_{\upharpoonright_{Y}}=Yg_{\upharpoonright_{Y}}\setminus
R(f_{\upharpoonright_{Y}}g_{\upharpoonright_{Y}})=Y\setminus
T_{(f_{\upharpoonright_{Y}})}=C(f_{\upharpoonright_{Y}}).$

Therefore

$(D(f)\setminus D(f_{\upharpoonright_{Y}}))g=D(f)g\setminus
D(f_{\upharpoonright_{Y}})g=D(f)g\setminus
D(f_{\upharpoonright_{Y}})g_{\upharpoonright_{Y}}=C(f)\setminus
C(f_{\upharpoonright_{Y}}).$

This implies $|C(f)\setminus C(f_{\upharpoonright_{Y}})|=|D(f)\setminus
D(f_{\upharpoonright_{Y}})|$ as required.

Conversely, suppose that $f$ satisfies the given conditions. To prove
$f\in\textnormal{ureg}(\overline{T}(X,Y))$, we will construct $g\in
U(\overline{T}(X,Y))$ such that $fgf=f$. For this, we first note from (iii)
that $T_{f}\subseteq X$ and $T_{(f_{\upharpoonright_{Y}})}\subseteq Y$ are
transversals of $\ker(f)$ and $\ker(f_{\upharpoonright_{Y}})$, respectively.
Therefore the corestrictions of the maps $f_{|_{T_{f}}}$ and
${(f_{\upharpoonright_{Y}})}_{|_{T_{(f_{\upharpoonright_{Y}})}}}$ to $R(f)$
and $R(f_{\upharpoonright_{Y}})$, respectively, are bijections. Let $g_{0}$ be
the inverse of the corestriction map of $f_{|_{T_{f}}}$ to $R(f)$, and let
$h_{0}$ be the inverse of the corestriction map of
${(f_{\upharpoonright_{Y}})}_{|_{T_{(f_{\upharpoonright_{Y}})}}}$ to
$R(f_{\upharpoonright_{Y}})$. It is routine to verify that $xg_{0}=xh_{0}$ for
all $x\in R(f_{\upharpoonright_{Y}})$, since
$T_{(f_{\upharpoonright_{Y}})}=Y\cap T_{f}$ by (iii). Therefore
$R(f_{\upharpoonright_{Y}})g_{0}=T_{(f_{\upharpoonright_{Y}})}$. Set
$C(f)\colonequals X\setminus T_{f}$ and
$C(f_{\upharpoonright_{Y}})\colonequals Y\setminus
T_{(f_{\upharpoonright_{Y}})}$. Combining (i) and Theorem 2.5, we see that
$f_{\upharpoonright_{Y}}$ is semi-balanced. Therefore there exists a bijection
$g_{1}\colon D(f_{\upharpoonright_{Y}})\to C(f_{\upharpoonright_{Y}})$. Also
by (iii), there exists a bijection $g_{2}\colon D(f)\setminus
D(f_{\upharpoonright_{Y}})\to C(f)\setminus C(f_{\upharpoonright_{Y}})$. Using
these three bijections $g_{0},g_{1}$, and $g_{2}$, define a map $g\colon X\to
X$ by

$\displaystyle xg=\begin{cases}xg_{0}&\text{if $x\in R(f)$}\\\ xg_{1}&\text{if
$x\in D(f_{\upharpoonright_{Y}})$}\\\ xg_{2}&\text{if $x\in D(f)\setminus
D(f_{\upharpoonright_{Y}})$}.\end{cases}$

It is easy to check that $g$ is bijective. Moreover, we obtain

$Yg=(R(f_{\upharpoonright_{Y}})\cup
D(f_{\upharpoonright_{Y}}))g=R(f_{\upharpoonright_{Y}})g_{0}\cup
D(f_{\upharpoonright_{Y}})g_{1}=T_{(f_{\upharpoonright_{Y}})}\cup
C(f_{\upharpoonright_{Y}})\subseteq Y.$

Therefore $g\in U(\overline{T}(X,Y))$. We can also verify in a routine manner
that $fgf=f$. Hence $f\in\operatorname{\mbox{ureg}}(\overline{T}(X,Y))$. ∎

To describe the unit-regularity of $\overline{T}(X,Y)$, we need the following
lemma.

######  Lemma 3.3.

We have $R(f_{\upharpoonright_{Y}})=Y\cap R(f)$ for all
$f\in\overline{T}(X,Y)$ if and only if $|Y|=1$ or $Y=X$.

###### Proof.

Suppose that $R(f_{\upharpoonright_{Y}})=Y\cap R(f)$ for all
$f\in\overline{T}(X,Y)$. For $|X|\leq 2$, the statement is trivial. Let us now
assume that $|X|\geq 3$. Suppose to the contrary that $Y$ is a proper subset
of $X$ such that $|Y|\geq 2$. Let $a$ and $b$ be distinct elements of $Y$.
Define a map $f\colon X\to X$ by

$\displaystyle xf=\begin{cases}a&\text{if $x\in Y$}\\\
b&\text{otherwise}.\end{cases}$

It is clear that $Y$ is invariant under $f$, since $a\in Y$. Therefore
$f\in\overline{T}(X,Y)$. However, we see that
$R(f_{\upharpoonright_{Y}})=Yf=\\{a\\}$ while $Y\cap R(f)=\\{a,b\\}$, a
contradiction. Hence $|Y|=1$ or $Y=X$.

The proof of the converse part is immediate. ∎

The following theorem describes the unit-regularity of $\overline{T}(X,Y)$.

######  Theorem 3.4.

The semigroup $\overline{T}(X,Y)$ is unit-regular if and only if

  1. (i)

$|Y|=1$ or $Y=X$;

  2. (ii)

$X$ is finite.

###### Proof.

Suppose that $\overline{T}(X,Y)$ is unit-regular.

  1. (i)

Then every element of $\overline{T}(X,Y)$ is unit-regular. Therefore by
Theorem 3.2, we have $R(f_{\upharpoonright_{Y}})=Y\cap R(f)$ for all
$f\in\overline{T}(X,Y)$. Hence $|Y|=1$ or $Y=X$ by Lemma 3.3.

  2. (ii)

Recall from (i) that $Y=X$ or $|Y|=1$. If $Y=X$, then
$\overline{T}(X,Y)=T(X)$. Therefore $X$ is finite by Proposition 2.2. Let us
now assume that $|Y|=1$. In this case, it suffices to prove that $X\setminus
Y$ is finite. Suppose to the contrary that $X\setminus Y$ is infinite. Then
there exists a map $\alpha\colon X\setminus Y\to X\setminus Y$ that is
injective but not surjective. Define a map $f\colon X\to X$ by

$\displaystyle xf=\begin{cases}x&\text{if $x\in Y$}\\\
x\alpha&\text{otherwise.}\end{cases}$

Clearly $f\in\overline{T}(X,Y)$. Moreover, since $\alpha$ is injective but not
surjective, the map $f$ is injective but not surjective. This gives
$\textnormal{c}(f)=0$ and $\textnormal{d}(f)\geq 1$. Therefore $f$ is not
semi-balanced, so $f\notin\operatorname{\mbox{ureg}}(T(X))$ by Theorem 2.5.
This implies $f\notin\operatorname{\mbox{ureg}}(\overline{T}(X,Y))$, a
contradiction. Hence $X\setminus Y$ is finite, as required.

Conversely, suppose that the given conditions hold, and let
$f\in\overline{T}(X,Y)$. In view of Theorem 3.2, in order to show that
$f\in\operatorname{\mbox{ureg}}(\overline{T}(X,Y))$, it suffices to prove that
$f$ satisfies all the conditions in Theorem 3.2. First, we note from (ii) that
$Y$ is finite. Therefore
$f_{\upharpoonright_{Y}}\in\operatorname{\mbox{ureg}}(T(Y))$ by Proposition
2.2, so Theorem 3.2(i) holds. Next, we recall (i). Then
$R(f_{\upharpoonright_{Y}})=Y\cap R(f)$ by Lemma 3.3, so Theorem 3.2(ii)
holds. Finally, we note from (ii) that both $X$ and $Y$ are finite. Therefore
$\mbox{c}(f)=\mbox{d}(f)$ and
$\mbox{c}(f_{\upharpoonright_{Y}})=\mbox{d}(f_{\upharpoonright_{Y}})$ by Lemma
2.6. We further notice that there exist transversals $T_{f}$ and
$T_{(f_{\upharpoonright_{Y}})}$ of $\ker(f)$ and
$\ker(f_{\upharpoonright_{Y}})$, respectively, such that
$T_{(f_{\upharpoonright_{Y}})}=Y\cap T_{f}$. Set $C(f)\colonequals X\setminus
T_{f}$ and $C(f_{\upharpoonright_{Y}})\colonequals Y\setminus
T_{(f_{\upharpoonright_{Y}})}$. Then
$|C(f)|=\textnormal{c}(f)=\textnormal{d}(f)=|D(f)|$ and
$|C(f_{\upharpoonright_{Y}})|=\mbox{c}(f_{\upharpoonright_{Y}})=\mbox{d}(f_{\upharpoonright_{Y}})=|D(f_{\upharpoonright_{Y}})|$.
Therefore, since $X$ is finite by (i), we obtain

$|C(f)\setminus
C(f_{\upharpoonright_{Y}})|=|C(f)|-|C(f_{\upharpoonright_{Y}})|=|D(f)|-|D(f_{\upharpoonright_{Y}})|=|D(f)\setminus
D(f_{\upharpoonright_{Y}})|,$

so Theorem 3.2(iii) holds. Thus
$f\in\operatorname{\mbox{ureg}}(\overline{T}(X,Y))$ by Theorem 3.2. Hence,
since $f\in\overline{T}(X,Y)$ is arbitrary, we conclude that
$\overline{T}(X,Y)$ is unit-regular. ∎

##  4\. Unit-regular elements in $\overline{L}(V,W)$

In this section, we describe unit-regular elements in $\overline{L}(V,W)$.
Using this, we prove that an element $f$ of $L(V)$ is unit regular if and only
if $\operatorname{\mbox{nullity}}(f)=\operatorname{\mbox{corank}}(f)$.
Moreover, we alternatively prove that $L(V)$ is unit-regular if and only if
$V$ is finite-dimensional. We also give a new proof of [2, Theorem 11], which
determine the unit regularity of $\overline{L}(V,W)$. We first prove a list of
lemmas, which assist to describe unit-regular elements in $\overline{L}(V,W)$.

######  Lemma 4.1.

Let $f,g\in L(V)$. If $g$ is injective and $N(f)\subseteq R(g)$, then
$N(gf)\approx N(f)$.

###### Proof.

First, we prove that $vg\in N(f)$ whenever $v\in N(gf)$. For this, let $v\in
N(gf)$. Then $v(gf)=0$. Therefore $(vg)f=v(gf)=0$, so $vg\in N(f)$.

Now, define a map $\varphi\colon N(gf)\to N(f)$ by $v\varphi=vg$ for all $v\in
N(gf)$. It is clear that $\varphi$ is also a monomorphism, since $g$ is a
monomorphism. To prove $\varphi$ is surjective, let $w\in N(f)$. Then, since
$N(f)\subseteq R(g)$, we have $w\in R(g)$. Since $g$ is injective, we see that
the corestriction of $g$ to $R(g)$ is bijective. Therefore there exists a
unique $u\in V$ such that $wg^{-1}=u$. It remains to show that $u\in N(gf)$
and $u\varphi=w$. These two conditions can easily verify in a routine manner.
Hence $\varphi$ is surjective. Thus $N(gf)\approx N(f)$. ∎

Let $U$ and $V$ be isomorphic vector spaces. If $U^{\prime}$ is a subspace of
$U$ and $V^{\prime}$ is a subspace of $V$ such that $U^{\prime}\approx
V^{\prime}$, then it is not necessarily true that $U/U^{\prime}\approx
V/V^{\prime}$ (cf. [23, p. 93, line 7]). However, we have the following lemma.

######  Lemma 4.2.

Let $U^{\prime}$ and $V^{\prime}$ be subspaces of vector spaces $U$ and $V$,
respectively. If there exists an isomorphism $f\colon U\to V$ such that
$U^{\prime}f=V^{\prime}$, then $U/U^{\prime}\approx V/V^{\prime}$.

###### Proof.

Define a map $\varphi\colon U\to V/V^{\prime}$ by $u\varphi=uf+V^{\prime}$. By
using linearity of $f$, we see that $\varphi$ is linear. To prove $\varphi$ is
surjective, let $v+V^{\prime}\in V/V^{\prime}$. Then, since $v\in V$ and
$f\colon U\to V$ is bijective, there exists a unique $u\in U$ such that
$uf=v$. Therefore $u\varphi=uf+V^{\prime}=v+V^{\prime}$, so $\varphi$ is
surjective. Thus $U/N(\varphi)\approx V/V^{\prime}$ by the first isomorphism
theorem.

Finally, we prove that $N(\varphi)=U^{\prime}$. Let $u\in N(\varphi)$. Then
$uf+V^{\prime}=u\varphi=V^{\prime}$, so $uf\in V^{\prime}$. Therefore, since
$f$ is bijective and $U^{\prime}f=V^{\prime}$, we get $u\in U^{\prime}$. Thus
$N(\varphi)\subseteq U^{\prime}$. For the reverse inclusion, let $u\in
U^{\prime}$. Then, since $U^{\prime}f=V^{\prime}$, we have $uf\in V^{\prime}$.
Therefore $u\varphi=uf+V^{\prime}=V^{\prime}$, so $u\in N(\varphi)$. Thus
$U^{\prime}\subseteq N(\varphi)$. Hence $N(\varphi)=U^{\prime}$ and
consequently $U/U^{\prime}\approx V/V^{\prime}$. ∎

######  Lemma 4.3.

Let $f\in\overline{L}(V,W)$. Then there exists a subspace $U$ of $V$ such that
$U$ and $U\cap W$ are transversals of $\ker(f)$ and
$\ker(f_{\upharpoonright_{W}})$, respectively.

###### Proof.

Let $B_{1}$ be a basis for $R(f_{\upharpoonright_{W}})$. If
$B_{1}\neq\varnothing$, then we have $uf^{-1}\cap W\neq\varnothing$ for all
$u\in B_{1}$. Therefore fix $u^{\prime}\in uf^{-1}\cap W$ for each $u\in
B_{1}$, and let $C_{1}\colonequals\\{u^{\prime}\colon u\in B_{1}\\}$. For
$B_{1}=\varnothing$, we let $C_{1}\colonequals\varnothing$.

In either case, we prove that $U_{1}\colonequals\langle C_{1}\rangle$ is a
transversal of $\ker(f_{\upharpoonright_{W}})$. If $C_{1}=\varnothing$, then
$U_{1}=\\{0\\}$, and we are done. Assume that $C_{1}\neq\varnothing$, and let
$w\in R(f_{\upharpoonright_{W}})$. Then, since $B_{1}$ is a basis for
$R(f_{\upharpoonright_{W}})$, we have $w=c_{1}u_{1}+\cdots+c_{m}u_{m}$ for
some $u_{1},\ldots,u_{m}\in B_{1}$, where $m\geq 0$. Let
$w^{\prime}=c_{1}u_{1}^{\prime}+\cdots+c_{m}u_{m}^{\prime}\in U_{1}$, where
$u_{1}^{\prime},\ldots,u_{m}^{\prime}\in C_{1}$. Notice that $w^{\prime}f=w$,
so $w^{\prime}\in wf^{-1}\cap U_{1}$. Therefore, by construction of basis
$C_{1}$ for $U_{1}$, we see that $wf^{-1}\cap U_{1}=\\{w^{\prime}\\}$ and
consequently $|wf^{-1}\cap U_{1}|=1$. Hence, since $w$ is arbitrary, we
conclude that the subspace $U_{1}$ of $V$ is a transversal of
$\ker{(f_{\upharpoonright_{W}})}$.

Now, consider $R(f)\setminus R(f_{\upharpoonright_{W}})$. Then there are two
possibilities to consider.

Case 1: Suppose $R(f)\setminus R(f_{\upharpoonright_{W}})=\varnothing$. Set
$U\colonequals U_{1}$. Then $U\cap W=U_{1}$. Therefore $U$ and $U\cap W$ are
transversals of $\ker(f)$ and $\ker(f_{\upharpoonright_{W}})$, respectively.

Case 2: Suppose $R(f)\setminus R(f_{\upharpoonright_{W}})\neq\varnothing$.
Recall that $B_{1}$ is a basis for $R(f_{\upharpoonright_{W}})$. Extend
$B_{1}$ to a basis $B_{1}\cup B_{2}$ for $R(f)$, where $B_{2}\subseteq
R(f)\setminus R(f_{\upharpoonright_{W}})$. Clearly $B_{2}\neq\varnothing$.
Note that $vf^{-1}\neq\varnothing$ for all $v\in B_{2}$. Therefore fix
$\bar{v}\in vf^{-1}$ for each $v\in B_{2}$. Set $C\colonequals
C_{1}\cup\\{\bar{v}\in vf^{-1}\colon v\in B_{2}\\}$ and $U\colonequals\langle
C\rangle$. It is easy to see that $U\cap W=U_{1}$. Therefore $U\cap W$ is a
transversal of $\ker(f_{\upharpoonright_{W}})$. It remains to prove that $U$
is a transversal of $\ker(f)$. For this, let $w\in R(f)$. Then, since
$B_{1}\cup B_{2}$ is a basis for $R(f)$, we have
$w=c_{1}u_{1}+\cdots+c_{m}u_{m}+d_{1}v_{1}+\cdots+d_{n}v_{n}$ for some
$u_{1},\ldots,u_{m}\in B_{1}$ and $v_{1},\ldots,v_{n}\in B_{2}$, where
$m,n\geq 0$. Let
$w^{\prime}=c_{1}u_{1}^{\prime}+\cdots+c_{m}u_{m}^{\prime}+d_{1}\bar{v_{1}}+\cdots+d_{n}\bar{v_{n}}\in
U$, where
$u_{1}^{\prime},\ldots,u_{m}^{\prime},\bar{v_{1}},\ldots,\bar{v_{n}}\in C$.
Notice that $w^{\prime}f=w$, so $w^{\prime}\in wf^{-1}\cap U$. Therefore, by
construction of basis $C$ for $U$, we get $wf^{-1}\cap U=\\{w^{\prime}\\}$ and
consequently $|wf^{-1}\cap U|=1$. Hence, since $w$ is arbitrary, the subspace
$U$ of $V$ is a transversal of $\ker{(f)}$. ∎

######  Lemma 4.4.

Let $f\in L(V)$ and $T_{f}$ be a transversal of $\ker(f)$. If $B_{0}$ is a
basis for $N(f)$ and $B$ is a basis for $R(f)$, then $B_{0}\cup(T_{f}\cap
Bf^{-1})$ is a basis for $V$.

###### Proof.

We first note that $B_{0}\cap(T_{f}\cap Bf^{-1})=\varnothing$. It is easy to
see that $T_{f}\cap Bf^{-1}$ is linearly independent. Combining the previous
two statements shows that $B_{0}\cup(T_{f}\cap Bf^{-1})$ is linearly
independent. It remains to prove that $\langle B_{0}\cup(T_{f}\cap
Bf^{-1})\rangle=V$. For this, consider a complement $U$ of $N(f)$ in $V$. Then
$B_{0}\cup B(f_{|_{U}})^{-1}$ is a basis for $V$, where the isomorphism
$f_{|_{U}}$ also denotes the corestriction of $f_{|_{U}}$ to $R(f)$. Now, let
$u\in B(f_{|_{U}})^{-1}$. Clearly $u\in U$. Then $v\colonequals
u+u^{\prime}\in T_{f}\cap Bf^{-1}$ for some $u^{\prime}\in N(f)$. Therefore
$u=v-u^{\prime}\in\langle B_{0}\cup(T_{f}\cap Bf^{-1})\rangle$. Hence, since
$u$ is arbitrary, we have $B(f_{|_{U}})^{-1}\subseteq\langle
B_{0}\cup(T_{f}\cap Bf^{-1})\rangle$ and consequently $B_{0}\cup
B(f_{|_{U}})^{-1}\subseteq\langle B_{0}\cup(T_{f}\cap Bf^{-1})\rangle$. Thus
$\langle B_{0}\cup(T_{f}\cap Bf^{-1})\rangle=V$ as required. ∎

######  Lemma 4.5.

Let $f\in\overline{L}(V,W)$. If $f\in\textnormal{ureg}(\overline{L}(V,W))$,
then $f_{\upharpoonright_{W}}\in\textnormal{ureg}(L(W))$.

###### Proof.

If $f\in\operatorname{\mbox{ureg}}(\overline{L}(V,W))$, then $fgf=f$ for some
$g\in U(\overline{L}(V,W))$. Therefore we get
$f_{\upharpoonright_{W}}g_{\upharpoonright_{W}}f_{\upharpoonright_{W}}=f_{\upharpoonright_{W}}$
and $g_{\upharpoonright_{W}}\in U(L(W))$. Hence
$f_{\upharpoonright_{W}}\in\operatorname{\mbox{ureg}}(L(W))$. ∎

Note that $\overline{L}(V,U)$ contains the identity map on $V$. The following
theorem describes unit-regular elements in $\overline{L}(V,W)$.

######  Theorem 4.6.

Let $f\in\overline{L}(V,W)$. Then $f\in\textnormal{ureg}(\overline{L}(V,W))$
if and only if

  1. (i)

$R(f_{\upharpoonright_{W}})=W\cap R(f)$;

  2. (ii)

$\textnormal{nullity}(f_{\upharpoonright_{W}})=\textnormal{corank}(f_{\upharpoonright_{W}})$;

  3. (iii)

$\textnormal{codim}_{V}(W+T_{f})=\textnormal{codim}_{V}(W+R(f))$ for some
subspace $T_{f}$ of $V$ such that $T_{f}$ and $W\cap T_{f}$ are transversals
of $\ker(f)$ and $\ker(f_{\upharpoonright_{W}})$, respectively.

###### Proof.

Suppose that $f\in\operatorname{\mbox{ureg}}(\overline{L}(V,W))$. Then $fgf=f$
for some $g\in U(\overline{L}(V,W))$.

  1. (i)

It is clear that $R(f_{\upharpoonright_{W}})\subseteq W\cap R(f)$, since
$R(f_{\upharpoonright_{W}})\subseteq W$. For the reverse inclusion, let $w\in
W\cap R(f)$. Then there exists $v\in V$ such that $vf=w$. Therefore, since
$w\in W$, $f=fgf$, and $Wg\subseteq W$, we obtain

$w=vf=(vf)gf=(wg)f\in R(f_{\upharpoonright_{W}}).$

This implies $W\cap R(f)\subseteq R(f_{\upharpoonright_{W}})$ as required.

  2. (ii)

By Lemma 4.5, we have $f_{\upharpoonright_{W}}\in\textnormal{ureg}(L(W))$.
Therefore there exists $g_{\upharpoonright_{W}}\in U(L(W))$ such that
$f_{\upharpoonright_{W}}g_{\upharpoonright_{W}}f_{\upharpoonright_{W}}=f_{\upharpoonright_{W}}$.
Note that
$R(g_{\upharpoonright_{W}}f_{\upharpoonright_{W}})=R(f_{\upharpoonright_{W}})$
and $g_{\upharpoonright_{W}}f_{\upharpoonright_{W}}$ is an idempotent of
$L(W)$. Combining this with [23, Theorem 2.22], we obtain
$N(g_{\upharpoonright_{W}}f_{\upharpoonright_{W}})\oplus
R(f_{\upharpoonright_{W}})=W$. Therefore
$N(g_{\upharpoonright_{W}}f_{\upharpoonright_{W}})\approx
W/R(f_{\upharpoonright_{W}})$. Since $g_{\upharpoonright_{W}}$ is bijective,
it follows from Lemma 4.1 that $N(f_{\upharpoonright_{W}})\approx
N(g_{\upharpoonright_{W}}f_{\upharpoonright_{W}})$ and subsequently
$N(f_{\upharpoonright_{W}})\approx W/R(f_{\upharpoonright_{W}})$. Hence
$\operatorname{\mbox{nullity}}(f_{\upharpoonright_{W}})=\operatorname{\mbox{corank}}(f_{\upharpoonright_{W}})$.

  3. (iii)

Note that $R(fg)$ and $R(f_{\upharpoonright_{W}}g_{\upharpoonright_{W}})$ are
subspaces of $V$ and $W$, respectively. Let $T_{f}\colonequals R(fg)$ and
$T_{(f_{\upharpoonright_{W}})}\colonequals
R(f_{\upharpoonright_{W}}g_{\upharpoonright_{W}})$. Recall that $fgf=f$ and
$f_{\upharpoonright_{W}}g_{\upharpoonright_{W}}f_{\upharpoonright_{W}}=f_{\upharpoonright_{W}}$.
Therefore by Lemma 2.4, it follows that $T_{f}$ and
$T_{(f_{\upharpoonright_{W}})}$ are transversals of $\ker(f)$ and
$\ker(f_{\upharpoonright_{W}})$, respectively. Moreover, we establish that
$T_{(f_{\upharpoonright_{W}})}=W\cap T_{f}$. Clearly
$T_{(f_{\upharpoonright_{W}})}\subseteq W\cap T_{f}$. For the reverse
inclusion, let $v\in W\cap T_{f}$. Then $u(fg)=v$ for some $u\in V$.
Therefore, since $g$ is bijective, we get $uf\in W$. Hence by (i), there
exists $u^{\prime}\in W$ such that $u^{\prime}f=uf$. This implies
$u^{\prime}(fg)=u(fg)=v$, so $v\in T_{(f_{\upharpoonright_{W}})}$ as required.
Also, since $g$ is bijective, we see that $(W+R(f))g=W+T_{f}$. Thus
$V/(W+R(f))\approx V/(W+T_{f})$ by Lemma 4.2. Hence
$\operatorname{\mbox{codim}}_{V}(W+T_{f})=\operatorname{\mbox{codim}}_{V}(W+R(f))$.

Conversely, suppose that the given conditions hold for
$f\in\overline{L}(V,W)$. To prove
$f\in\operatorname{\mbox{ureg}}(\overline{L}(V,W))$, we will construct $g\in
U(\overline{L}(V,W))$ such that $fgf=f$. For this, let $B_{1}$ be a basis for
$R(f)\cap W$. Extend $B_{1}$ to bases $B_{1}\cup B_{2}$ and $B_{1}\cup B_{3}$
for $R(f)$ and $W$, respectively, where $B_{2}\subseteq R(f)\setminus\langle
B_{1}\rangle$ and $B_{3}\subseteq W\setminus\langle B_{1}\rangle$. It is easy
to see that $B_{1}\cup B_{2}\cup B_{3}$ is a basis for $W+R(f)$. Now, extend
$B_{1}\cup B_{2}\cup B_{3}$ to a basis $B\colonequals B_{1}\cup B_{2}\cup
B_{3}\cup B_{4}$ for $V$, where $B_{4}\subseteq V\setminus(W+R(f))$.

Note from (iii) that the subspace $T_{f}$ of $V$ is a transversal of
$\ker(f)$. Therefore the corestriction of $f_{|_{T_{f}}}$ to $R(f)$ is an
isomorphism. Denote by $g_{0}$ the inverse of this corestriction map. Clearly
$g_{0}\colon R(f)\to T_{f}$ is an isomorphism. Set $B_{1}g_{0}\colonequals
C_{1}$ and $B_{2}g_{0}\colonequals C_{2}$. It is evident that $C_{1}\cup
C_{2}$ is a basis for $T_{f}$, since $B_{1}\cup B_{2}$ is a basis for $R(f)$.
Next, note from (iii) that the subspace
$T_{(f_{\upharpoonright_{W}})}\colonequals W\cap T_{f}$ of $W$ is a
transversal of $\ker(f_{\upharpoonright_{W}})$. Therefore the corestriction of
${(f_{\upharpoonright_{W}})}_{|_{T_{(f_{\upharpoonright_{W}})}}}$ to
$R(f_{\upharpoonright_{W}})$ is an isomorphism. Notice that the inverse of
this isomorphism agrees with $g_{0}$ on $R(f_{\upharpoonright_{W}})$. Recall
that $B_{1}$ is a basis for $W\cap R(f)$. Therefore $C_{1}$ is a basis for
$T_{(f_{\upharpoonright_{W}})}$, since $R(f_{\upharpoonright_{W}})=W\cap R(f)$
by (i).

Let $C_{3}$ be a basis for $N(f_{\upharpoonright_{W}})$. Then $C_{1}\cup
C_{3}$ is a basis for $W$ by Lemma 4.4, and subsequently $C_{1}\cup C_{2}\cup
C_{3}$ is a basis for $W+T_{f}$. Extend $C_{1}\cup C_{2}\cup C_{3}$ to a basis
$C\colonequals C_{1}\cup C_{2}\cup C_{3}\cup C_{4}$ for $V$, where
$C_{4}\subseteq V\setminus(W+T_{f})$. Recall that $B_{1}$ and $B_{1}\cup
B_{3}$ are bases for $W\cap R(f)$ and $W$, respectively. Since
$R(f_{\upharpoonright_{W}})=W\cap R(f)$ by (i), it follows that
$\operatorname{\mbox{corank}}(f_{\upharpoonright_{W}})=|B_{3}|$. Therefore
$|C_{3}|=\operatorname{\mbox{nullity}}(f_{\upharpoonright_{W}})=|B_{3}|$ by
(ii). Thus there exists a bijection $\alpha\colon B_{3}\to C_{3}$. Next,
recall that $B$ and $C$ are bases for $V$ such that $B_{1}\cup B_{2}\cup
B_{3}\subseteq B$ and $C_{1}\cup C_{2}\cup C_{3}\subseteq C$. Since $B_{1}\cup
B_{2}\cup B_{3}$ is a basis for $W+R(f)$ and $C_{1}\cup C_{2}\cup C_{3}$ is
basis for $W+T_{f}$, we get $\operatorname{\mbox{codim}}_{V}(W+R(f))=|B_{4}|$
and $\operatorname{\mbox{codim}}_{V}(W+T_{f})=|C_{4}|$. Therefore
$|B_{4}|=|C_{4}|$ by (iii). Thus there exists a bijection $\beta\colon
B_{4}\to C_{4}$.

Now, define a map $g\colon B\to C$ by setting

$\displaystyle vg=\begin{cases}vg_{0}&\text{if $v\in B_{1}\cup B_{2}$}\\\
v\alpha&\text{if $v\in B_{3}$}\\\ v\beta&\text{if $v\in B_{4}$}.\end{cases}$

It is easy to check that $g$ is bijective. Therefore the unique linear map
$\bar{g}\colon V\to V$ is bijective. Moreover, we obtain $W\bar{g}=\langle
B_{1}\cup B_{3}\rangle\bar{g}=\langle B_{1}g_{0}\cup
B_{3}\alpha\rangle=\langle C_{1}\cup C_{3}\rangle\subseteq W$. Thus
$\bar{g}\in U(\overline{L}(V,W))$. We can also verify in a routine manner that
$f\bar{g}f=f$. Hence $f\in\operatorname{\mbox{ureg}}(\overline{L}(V,W))$. ∎

If $W=V$, then conditions (i) and (iii) of Theorem 4.6 trivially hold. The
following is an immediate corollary of Theorem 4.6 for the case when $W=V$,
which describes unit-regular elements in $L(V)$.

######  Corollary 4.7.

Let $f\in L(V)$. Then $f\in\textnormal{ureg}(L(V))$ if and only if
$\textnormal{nullity}(f)=\textnormal{corank}(f)$.

We need the following lemma to give an alternative proof of Theorem 4.9, which
may also be obtained by combining Proposition 2.1 and Theorem 2.3.

######  Lemma 4.8.

We have $\textnormal{nullity}(f)=\textnormal{corank}(f)$ for all $f\in L(V)$
if and only if $V$ is finite-dimensional.

###### Proof.

We prove the necessity part of the lemma by contraposition. Suppose that $V$
is infinite-dimensional. Let $v$ be a nonzero vector of $V$. Clearly
$\dim(\langle v\rangle)=1$. Note that $V=\langle v\rangle\oplus U$ for some
complement $U$ of $\langle v\rangle$ in $V$. Then by [23, Theorem 1.14], we
get $\dim(V)=\dim(U)$. Therefore there is an isomorphism $\varphi\colon V\to
U$. Since $U$ is a proper subspace of $V$, it follows that there is an
injective linear map $\tilde{\varphi}\in L(V)$ that agrees with $\varphi$ on
$V$. Thus $\operatorname{\mbox{nullity}}(\tilde{\varphi})=0$. It is also
immediate that $\operatorname{\mbox{corank}}(\tilde{\varphi})=1$. Hence
$\textnormal{nullity}(\tilde{\varphi})\neq\textnormal{corank}(\tilde{\varphi})$.

Conversely, suppose that $V$ is finite-dimensional, and let $f\in L(V)$. Then
$\operatorname{\mbox{corank}}(f)=\dim(V)-\operatorname{\mbox{rank}}(f)$.
Therefore by rank-nullity theorem, we obtain
$\operatorname{\mbox{nullity}}(f)=\dim(V)-\operatorname{\mbox{rank}}(f)=\operatorname{\mbox{corank}}(f)$
as required. ∎

The following known theorem is straightforward from Corollary 4.7 and Lemma
4.8.

######  Theorem 4.9.

The semigroup $L(V)$ is unit-regular if and only if $V$ is finite-dimensional.

Next, we need the following lemma to prove Theorem 4.11

######  Lemma 4.10.

We have $R(f_{\upharpoonright_{W}})=W\cap R(f)$ for all
$f\in\overline{L}(V,W)$ if and only if $W$ is trivial.

###### Proof.

We prove the necessity part of the lemma by contraposition. Suppose that $W$
is nontrivial. Then there exists a nontrivial subspace $U$ of $V$ such that
$V=W\oplus U$. Let $B$ and $C$ be bases for $W$ and $U$, respectively. Then
$B\cup C$ is a basis for $V$. Fix $w\in B$ and $u\in C$. Define a map $f\colon
B\cup C\to V$ by

$vf=\begin{cases}w&\text{if $v=u$}\\\ 0&\text{if $v\in
B\cup(C\setminus\\{u\\})$}.\end{cases}$

It is easy to see that $B\bar{f}=Bf=\\{0\\}$. Therefore
$W\bar{f}=\\{0\\}\subseteq W$, so $\bar{f}\in\overline{L}(V,W)$. We further
see that $W\cap R(\bar{f})=\langle w\rangle$. Therefore
$R(\bar{f}_{\upharpoonright_{W}})\neq W\cap R(\bar{f})$.

The proof of the converse part is immediate. ∎

We can now obtain a new proof of the following known theorem (cf. [2, Theorem
11]).

######  Theorem 4.11.

The semigroup $\overline{L}(V,W)$ is unit-regular if and only if

  1. (i)

$W$ is trivial;

  2. (ii)

$V$ is finite-dimensional.

###### Proof.

Suppose that $\overline{L}(V,W)$ is unit-regular.

  1. (i)

By Theorem 4.6, we have $R(f_{\upharpoonright_{W}})=W\cap R(f)$ for all
$f\in\overline{L}(V,W)$. Therefore $W$ is trivial by Lemma 4.10.

  2. (ii)

From (i), it is clear that $\overline{L}(V,W)=L(V)$. Then by assumption, the
semigroup $L(V)$ is unit-regular. Therefore $V$ is finite-dimensional by
Theorem 4.9.

Conversely, suppose that the given conditions hold. Then by (i), we have
$\overline{L}(V,W)=L(V)$. Therefore by combining (ii) and Theorem 2.3, the
semigroup $\overline{L}(V,W)$ is factorizable. Hence $\overline{L}(V,W)$ is
unit-regular by Proposition 2.1. ∎

##  5\. Semi-Balanced Semigroups

Recall that an element $f$ of $T(X)$ is semi-balanced if
$\textnormal{c}(f)=\textnormal{d}(f)$; a subsemigroup of $T(X)$ is semi-
balanced if all its elements are semi-balanced. In this section, we give
necessary and sufficient conditions for the semigroups $\overline{T}(X,Y)$,
$L(V)$, and $\overline{L}(V,W)$ to be semi-balanced. We begin with the
following trivial remark.

######  Remark 5.1.

Every subsemigroup of a semi-balanced semigroup is semi-balanced.

The following theorem is immediate from Proposition 2.2 and Lemma 2.6.

######  Theorem 5.2.

The following statements are equivalent.

  1. (i)

The semigroup $T(X)$ is unit-regular.

  2. (ii)

The semigroup $T(X)$ is semi-balanced.

  3. (iii)

The set $X$ is finite.

Using Theorem 2.5, we prove the next proposition.

######  Proposition 5.3.

If $M$ is a submonoid of $T(X)$, then $\textnormal{ureg}(M)\subseteq B(M)$.

###### Proof.

Let $f\in\textnormal{ureg}(M)$. Then $f\in\textnormal{ureg}(T(X))$, since $M$
is a submonoid of $T(X)$. Therefore $f\in B$ by Theorem 2.5. Hence $f\in M\cap
B=B(M)$ as required. ∎

The following is an immediate corollary of Proposition 5.3.

######  Corollary 5.4.

Every unit-regular submonoid of $T(X)$ is semi-balanced.

The converse of Corollary 5.4 is not necessarily true as seen from the
following example.

######  Example 5.5.

Let $X$ be a finite set such that $|X|\geq 3$, and let $Y$ be a proper subset
of $X$ such that $|Y|\geq 2$. It is clear that $\overline{T}(X,Y)$ is semi-
balanced, since $X$ is finite. However, the semigroup $\overline{T}(X,Y)$ is
not regular by [21, Proposition 2.1(ii)]. Therefore $\overline{T}(X,Y)$ is not
unit-regular.

In the next theorem, we give a necessary and sufficient condition for the
semigroup $\overline{T}(X,Y)$ to be semi-balanced.

######  Theorem 5.6.

The semigroup $\overline{T}(X,Y)$ is semi-balanced if and only if $X$ is
finite.

###### Proof.

We prove the necessity part of the theorem by contraposition. Suppose that $X$
is infinite. Then there are two possibilities to consider.

Case 1: Suppose $X\setminus Y$ is finite. Then $Y$ is infinite. Therefore
there exists a map $\alpha\colon Y\to Y$ that is injective but not surjective.
Define a map $f\colon X\to X$ by

$\displaystyle xf=\begin{cases}x\alpha&\text{if $x\in Y$}\\\ x&\text{if $x\in
X\setminus Y$.}\end{cases}$

It is routine to verify that $f\in\overline{T}(X,Y)$. However, since $\alpha$
is injective but not surjective, we see that $f$ is injective but not
surjective.

Case 2: Suppose $X\setminus Y$ is infinite. Then there exists a map
$\beta\colon X\setminus Y\to X\setminus Y$ that is injective but not
surjective. Define a map $f\colon X\to X$ by

$\displaystyle xf=\begin{cases}x&\text{if $x\in Y$}\\\ x\beta&\text{if $x\in
X\setminus Y$.}\end{cases}$

It is routine to verify that $f\in\overline{T}(X,Y)$. However, since $\beta$
is injective but not surjective, we see that $f$ is injective but not
surjective.

Thus, in either case, there exists a map $f\in\overline{T}(X,Y)$ that is
injective but not surjective. Therefore $\mbox{c}(f)=0$ but $\mbox{d}(f)\geq
1$. This shows that $f$ is not semi-balanced. Hence $\overline{T}(X,Y)$ is not
semi-balanced.

Conversely, suppose that $X$ is finite. Then $T(X)$ is semi-balanced by
Theorem 5.2. Therefore $\overline{T}(X,Y)$ is semi-balanced by Remark 5.1. ∎

######  Remark 5.7.

Let $X$ be a finite set. Then $\overline{T}(X,Y)$ is semi-balanced by Theorem
5.6. However, the semigroup $\overline{T}(X,Y)$ is not necessarily unit-
regular (cf. Theorem 3.4).

From Proposition 5.3, we know that every unit-regular element in a submonoid
$M$ of $T(X)$ is also a semi-balanced element of $M$. Moreover, we note from
Theorem 2.5 that the concepts of unit-regular and semi-balanced elements are
equivalent in $T(X)$. However, these concepts are not necessarily equivalent
in $L(V)$ as seen from the following example.

######  Example 5.8.

Let $V\colonequals\mathbb{Q}[x]$ be the vector space of all polynomials in $x$
over the field $\mathbb{Q}$. Clearly $\\{1,x,x^{2},\ldots\\}$ is a basis for
$V$. Consider $f\in L(V)$ such that

$\displaystyle p(x)f=\begin{cases}0&\text{if $p(x)=1$}\\\ xp(x)&\text{if
$p(x)\in\\{x,x^{2},x^{3}\ldots\\}$}.\end{cases}$

It is routine to verify that $N(f)=\langle 1\rangle=\mathbb{Q}$, $R(f)=\langle
x^{2},x^{3},x^{4},\ldots\rangle$, and $V/R(f)=\langle 1+R(f),x+R(f)\rangle$.
Then $\operatorname{\mbox{nullity}}(f)=1$ and
$\operatorname{\mbox{corank}}(f)=2$. Therefore $f$ is not unit-regular in
$L(V)$ by Corollary 4.7.

Now, we prove that $f$ is semi-balanced. Since $|\mathbb{Q}|=\aleph_{0}$, we
see that $|V|=\aleph_{0}$ and $|V/R(f)|=\aleph_{0}$. To see $f$ is semi-
balanced, we first prove that $\textnormal{d}(f)=\aleph_{0}$. For this, fix
$u\in V\setminus R(f)$ and define a map $\varphi\colon V\setminus R(f)\to
V/R(f)$ by

$\displaystyle v\varphi=\begin{cases}R(f)&\text{if $v=u$}\\\
v+R(f)&\text{otherwise}.\end{cases}$

It is clear that $\varphi$ is surjective, since $(u+v)\varphi=u+R(f)$ for all
nonzero vector $v$ of $R(f)$. This yields $|V\setminus R(f)|\geq|V/R(f)|$.
Therefore, since $|V/R(f)|=\aleph_{0}$, we get $\textnormal{d}(f)=\aleph_{0}$.
Next, we prove that $\textnormal{c}(f)=\aleph_{0}$. For this, let $T_{f}$ be a
transversal of the equivalence relation $\ker(f)$. Then we may write
$\pi(f)=\\{p(x)+N(f)\colon p(x)\in T_{f}\\}$. Note for every $p(x)\in T_{f}$
that $|p(x)+N(f)|=|N(f)|=\aleph_{0}$, so
$|p(x)+N(f)\setminus\\{p(x)\\}|=\aleph_{0}$. Since $|T_{f}|\leq|V|$ and
$V\setminus T_{f}=\displaystyle\bigcup_{p(x)\in
T_{f}}\big{(}p(x)+N(f)\setminus\\{p(x)\\}\big{)}$, we get

$\textnormal{c}(f)=\sum_{p(x)\in
T_{f}}|p(x)+N(f)\setminus\\{p(x)\\}|=|T_{f}|\times\aleph_{0}=\aleph_{0}.$

Thus $\textnormal{c}(f)=\textnormal{d}(f)$ and consequently $f$ is semi-
balanced.

In the next theorem, we give a necessary and sufficient condition for the
semigroup $L(V)$ to be semi-balanced.

######  Theorem 5.9.

The semigroup $L(V)$ is semi-balanced if and only if $V$ is finite-
dimensional.

###### Proof.

We prove the necessity part of the theorem by contraposition. Suppose that $V$
is infinite-dimensional. Let $B$ be a basis for $V$. Then, since $B$ is
infinite, there exists a map $\alpha\colon B\to B$ that is injective but not
surjective. Define a linear map $f\colon V\to V$ by $vf=v\alpha$ for all $v\in
B$. It is clear that $f$ is injective but not surjective, since $\alpha$ is
injective but not surjective. This gives $\textnormal{c}(f)=0$ and
$\textnormal{d}(f)\geq 1$. Therefore $f$ is not semi-balanced. Hence $L(V)$ is
not semi-balanced.

Conversely, suppose that $V$ is finite-dimensional. Then $L(V)$ is unit-
regular by combining Proposition 2.1 and Theorem 2.3. Therefore $L(V)$ is
semi-balanced by Corollary 5.4. ∎

Thus, we have the following from Theorems 4.9 and 5.9

######  Theorem 5.10.

The following statements are equivalent.

  1. (i)

The semigroup $L(V)$ is unit-regular.

  2. (ii)

The semigroup $L(V)$ is semi-balanced.

  3. (iii)

The space $V$ is finite-dimensional.

In the next theorem, we give a necessary and sufficient condition for the
semigroup $\overline{L}(V,W)$ to be semi-balanced.

######  Theorem 5.11.

The semigroup $\overline{L}(V,W)$ is semi-balanced if and only if $V$ is
finite-dimensional.

###### Proof.

We prove the necessity part of the theorem by contraposition. Suppose that $V$
is infinite-dimensional. Then there are two possibilities to consider.

Case 1: Suppose $V/W$ is finite-dimensional. Then $W$ is infinite-dimensional.
Therefore there exists a linear map $\alpha\colon W\to W$ that is injective
but not surjective. Define a map $f\colon V\to V$ by

$\displaystyle vf=\begin{cases}v\alpha&\text{if $v\in W$}\\\ v&\text{if $v\in
V/W$.}\end{cases}$

It is routine to verify that $f\in\overline{L}(V,W)$. However, since $\alpha$
is injective but not surjective, we see that $f$ is injective but not
surjective.

Case 2: Suppose $V/W$ is infinite-dimensional. Then there exists a linear map
$\beta\colon V/W\to V/W$ that is injective but not surjective. Define a map
$f\colon V\to V$ by

$\displaystyle vf=\begin{cases}v&\text{if $v\in W$}\\\ v^{\prime}&\text{if
$v\in V\setminus W$, where $(v+W)\beta=v^{\prime}+W$.}\end{cases}$

It is routine to verify that $f\in\overline{L}(V,W)$. However, since $\beta$
is injective but not surjective, we see that $f$ is injective but not
surjective.

Thus, in either case, there exists a map $f\in\overline{L}(V,W)$ that is
injective but not surjective. This gives $\textnormal{c}(f)=0$ but
$\textnormal{d}(f)\geq 1$. Therefore $f$ is not semi-balanced. Hence
$\overline{L}(V,W)$ is not semi-balanced.

Conversely, suppose that $V$ is finite-dimensional. Then $L(V)$ is semi-
balanced by Theorem 5.9. Therefore $\overline{L}(V,W)$ is semi-balanced by
Remark 5.1. ∎

######  Remark 5.12.

Let $V$ be a finite-dimensional vector space. Then $\overline{L}(V,W)$ is
semi-balanced by Theorem 5.11. However, the semigroup $\overline{L}(V,W)$ is
not necessarily unit-regular (cf. Theorem 4.11).

## References

  * [1] Blyth, T. S. and McFadden, R.  Unit orthodox semigroups.  Glasgow Mathematical Journal, 24(1): 39–42, 1983.
  * [2] Chaiya, Y.  Natural partial order and finiteness conditions on semigroups of linear transformations with invariant subspaces.  Semigroup Forum, 99: 579–590, 2019.
  * [3] Chen, S. Y. and Hsieh, S. C.  Factorizable inverse semigroups.  Semigroup Forum, 8: 283–297, 1974.
  * [4] Chinram, R. and Baupradist, S.  Magnifying elements in semigroups of linear transformations with invariant subspaces.  Journal of Interdisciplinary Mathematics, 21(6): 1457–1462, 2018.
  * [5] Chinram, R. and Baupradist, S.  Magnifying elements of semigroups of transformations with invariant set.  Asian-European Journal of Mathematics, 12(4): 1950056, 2019.
  * [6] Choomanee, W. and Honyam, P. and Sanwong, J.  Regularity in semigroup of transformations with invariant sets.  International Journal of Pure and Applied Mathematics, 87(1): 289–300, 2013.
  * [7] D’Alarcao, H.  Factorizable as a finiteness condition.  Semigroup Forum, 20: 281–282, 1980.
  * [8] Ehrlich, G.  Unit-regular rings.  Portugaliae Mathematica, 27(4): 209–212, 1968.
  * [9] Fountain, J.  An introduction to covers for semigroups.  In Semigroups, Algorithms, Automata and Languages, pages 155–194.  World Scientific, Singapore, 2002.
  * [10] Hickey, J. B. and Lawson, M. V.  Unit regular monoids.  Proceedings of the Royal Society of Edinburgh: Section A Mathematics, 127(1): 127–143, 1997.
  * [11] Higgins, P. M., Howie, J. M. and Ruškuc, N.  Generators and factorisations of transformation semigroups.  Proceedings of the Royal Society of Edinburgh: Section A Mathematics, 128(6): 1355–1369, 1998.
  * [12] Honyam, P. and Sanwong, J.  Semigroups of transformations with invariant set.  Journal of the Korean Mathematical Society, 48(2): 289–300, 2011.
  * [13] Honyam, P. and Sanwong, J.  Semigroups of linear transformations with invariant subspaces.  International Journal of Algebra, 6(8): 375–386, 2012.
  * [14] Howie, J. M.  Fundamentals of Semigroup Theory.  Volume 12 of London Mathematical Society Monographs. New Series.  The Clarendon Press, Oxford University Press, New York, 1995.
  * [15] Jampachon, P. and Saichalee, M. and Sullivan, R. P.  Locally factorisable transformation semigroups.  Southeast Asian Bulletin of Mathematics, 25: 233–-244, 2001.
  * [16] Kemprasit, Y.  Regularity and unit-regularity of generalized semigroups of linear transformations.  Southeast Asian Bulletin of Mathematics, 25: 617–-622, 2002.
  * [17] Magill Jr., K. D.  Subsemigroups of S(X).  Mathematica Japonica, 11: 109–115, 1966.
  * [18] McAllister, D. B.  Some covering and embedding theorems for inverse semigroups.  Journal of the Australian Mathematical Society, 22(2): 188–211, 1976.
  * [19] McFadden, R. B.  Unit-regular orthodox semigroups.  Glasgow Mathematical Journal, 25(2): 229–240, 1984.
  * [20] Nenthein, S. and Youngkhong, P. and Kemprasit, Y.  Regular elements of some transformation semigroups  Pure Mathematics and Applications, 16(3): 307–314, 2005.
  * [21] Nenthein, S. and Kemprasit, Y.  On transformation semigroups which are $\mathcal{B}\mathcal{Q}$-semigroup.  International Journal of Mathematics and Mathematical Sciences, 2006: 1–10, 2006.
  * [22] Pei, H.  A note on semigroups of linear transformations with invariant subspace.  International Journal of Algebra, 6(27): 1319–-1324, 2012.
  * [23] Roman, S.  Advanced Linear Algebra.  Volume 135 of Graduate Texts in Mathematics.  Springer-Verlag, New York, 3rd edition, 2007.
  * [24] Sarkar, M. and Singh, S. N.  Regular and unit-regular elements in various monoids of transformations.  Journal of Algebra and Its Applications. DOI: 10.1142/S0219498823502444
  * [25] Sarkar, M. and Singh, S. N.  On certain semigroups of transformations with an invariant set.  Asian-European Journal of Mathematics, 15(11): 2250198, 2022.
  * [26] Sun, L. and Wang, L.  Natural partial order in semigroups of transformations with invariant set.  Bulletin of the Australian Mathematical Society, 87(1): 94–107, 2013.
  * [27] Symons, J. S. V.  Some results concerning a transformation semigroup.  Journal of the Australian Mathematical Society, 19(4): 413–425, 1975.
  * [28] Tirasupa, Y.  Factorizable transformation semigroups.  Semigroup Forum, 18(1): 15–19: 1979.