# Set-theoretical solutions to the Hom-Yang-Baxter equation and Hom-cycle sets

Kaiqiang Zhang  and  Xiankun Du  K. Zhang: School of Mathematics, Jilin
University, Changchun 130012, China<EMAIL_ADDRESS>X. Du: School of
Mathematics, Jilin University, Changchun 130012, China<EMAIL_ADDRESS>
###### Abstract.

Set-theoretic solutions to the Yang-Baxter equation have been studied
extensively by means of related algebraic systems such as cycle sets and
braces, dynamical versions of which have also been developed. No work focuses
on set-theoretic solutions to the Hom-Yang-Baxter equation (HYBE for short).
This paper investigates set-theoretic solutions to HYBE and associated
algebraic system, called Hom-cycle sets. We characterize left non-degenerate
involutive set-theoretic solutions to HYBE and Hom-cycle sets, and establish
their relations. We discuss connections among Hom-cycle sets, cycle sets, left
non-degenerate involutive set-theoretic solutions to HYBE and the Yang-Baxter
equation.

###### Key words and phrases:

cycle set, Hom-cycle set, Hom-Yang-Baxter equation, left quasigroup, set-
theoretic solution

###### 1991 Mathematics Subject Classification:

16T25, 20N02, 20N05

Corresponding author: X Du

##  1\. Introduction

Let $V$ be a vector space. A solution to the Yang-Baxter equation (YBE
shortly) is a linear map $R:V\otimes V\rightarrow V\otimes V$ such that

$(R\otimes\operatorname{id}_{V})(\operatorname{id}_{V}\otimes
R)(R\otimes\operatorname{id}_{V})=(\operatorname{id}_{V}\otimes
R)(R\otimes\operatorname{id}_{V})(\operatorname{id}_{V}\otimes R).$

YBE first appeared in the work of Yang [39] and Baxter [1]. This equation is
fundamental to quantum groups. It is a central task to determine all solutions
to YBE. However, this is difficult to accomplish. In order to find new
solutions to YBE, Drinfeld [12] in 1992 suggested considering set-theoretic
solutions to YBE, that is, a map $r:X\times X\rightarrow X\times X$, where $X$
is a nonempty set, satisfying

$(r\times\operatorname{id}_{X})(\operatorname{id}_{X}\times
r)(r\times\operatorname{id}_{X})=(\operatorname{id}_{X}\times
r)(r\times\operatorname{id}_{X})(\operatorname{id}_{X}\times r).$ (1.1)

Etingof, Schedler and Soloviev [16], Gateva-Ivanova and Van den Bergh
[20], and Lu, Yan and Zhu [26] initially conducted a systematic study on this
subject. They studied set-theoretic solutions with invertibility,
nondegeneracy and involutivity by using group theory. Gateva-Ivanova in [17]
introduced a combinatorial approach to discuss set-theoretic solutions and she
conjectured that every square-free, non-degenerate involutive set-theoretic
solution $(X,r)$ is decomposable whenever $X$ is finite. This has been proved
by Rump in [30].

Left cycle sets were introduced by Rump [30] to study left non-degenerate
involutive set-theoretic solutions to YBE. Rump showed that there is a
bijective correspondence between left non-degenerate involutive set-theoretic
solutions to YBE and left cycle sets, and non-degenerate solutions correspond
to non-degenerate left cycle sets. He also prove that all finite left cycle
sets are non-degenerate. The theory of cycle sets has been proved to be very
useful in understanding the structure of solutions to YBE (see for example [2,
3, 4, 5, 7, 6, 10, 25]). This theory has been greatly developed, and inspires
theory of braces [8, 18, 21, 31, 36].

Another version of YBE, dynamical quantum Yang-Baxter equation, has been
studied [13, 14], which is closely related to dynamical quantum groups [15].
Their set-theoretic solution, called DYB maps, were proposed by Shibukawa in
[34] and received a lot of attention (see for example [24, 28, 35, 37]).
Dynamical braces and dynamical cycle sets were introduced and related to right
non-degenerate unitary DYB maps [27, 32].

The Hom-Yang-Baxter equation (HYBE shortly) was introduced by Yau [40]
motivated by Hom-Lie algebras, which is related to Hom-quantum groups [42].
Many researchers have devoted considerable attention to HYBE (see for example
[9, 23, 29, 38, 41, 43]). However, to our knowledge, no work concentrates on
set-theoretical solutions to GYBE.

The aim of this paper is to investigate left non-degenerate involutive set-
theoretic solutions to HYBE, corresponding algebraic systems, called Hom-cycle
sets, and their relationship.

The paper is organized as follows. In Section 2, we review some basic
definitions and results, and provide a general categorical framework for the
following discussion. In Section 3, we characterize left non-degenerate
involutive set-theoretic solutions to HYBE. In Section 4, we introduce the
notion of a Hom-cycle set, and prove that there exists a one to one
correspondence between left Hom-cycle sets and left non-degenerate involutive
set-theoretic solutions to HYBE. Section 5 is devoted to relationship among
Hom-cycle sets, cycle sets, left non-degenerate involutive solutions to HYBE
and YBE.

##  2\. Preliminaries

Let $X$ be a nonempty set and let $r:X\times X\rightarrow X\times X$ be a map.
We will write $r(x,y)=(\lambda_{x}(y),\rho_{y}(x))$, where $\lambda_{x}$ and
$\rho_{y}$ are maps from $X$ to itself for all $x,y\in X$. The pair $(X,r)$ is
referred to a quadratic set in [17].

A quadratic set $(X,r)$ (or a map $r$) is called

  1. (1)

left (respectively, right) non-degenerate if the map $\lambda_{x}$
(respectively, $\rho_{x}$) is bijective for all $x\in X$;

  2. (2)

non-degenerate if $r$ is both left and right non-degenerate;

  3. (3)

involutive if $r^{2}=\operatorname{id}$, the identify map;

  4. (4)

a set-theoretic solution to YBE if $r$ satisfies (1.1).

The following lemma comes from [16, Proposition 1.6] (see also [19, Lemma
2.4]).

######  Lemma 2.1.

  1. (1)

A quadratic set $(X,r)$ is involutive if and only if

$\displaystyle\lambda_{\lambda_{x}(y)}\rho_{y}(x)=x,$ (2.1)
$\displaystyle\rho_{\rho_{x}(y)}\lambda_{y}(x)=x,$ (2.2)

for all $x,y\in X$.

  2. (2)

A quadratic set $(X,r)$ is left non-degenerate and involutive if and only if
$\lambda_{x}$ is bijective for all $x\in X$ and

$\rho_{y}(x)=\lambda_{\lambda_{x}(y)}^{-1}(x),$ (2.3)

for all $x,y\in X$.

######  Theorem 2.2.

[16, Proposition 1.6] A quadratic set $(X,r)$ is a set-theoretic solution to
YBE if and only if

  1. (1)

$\lambda_{\lambda_{x}(y)}\lambda_{\rho_{y}(x)}(z)=\lambda_{x}\lambda_{y}(z)$,

  2. (2)

$\rho_{\lambda_{\rho_{y}(x)}(z)}\lambda_{x}(y)=\lambda_{\rho_{\lambda_{y}(z)}(x)}\rho_{z}(y)$,

  3. (3)

$\rho_{z}\rho_{y}(x)=\rho_{\rho_{z}(y)}\rho_{\lambda_{y}(z)}(x)$,

for all $x,y,z\in X$.

The following Theorem comes from [8, Proposition 2] (see also [22, Theorem
9.3.10]).

######  Theorem 2.3.

A quadratic set $(X,r)$ is a left non-degenerate involutive set-theoretic
solution to YBE if and only if the following hold.

  1. (1)

$\lambda_{x}$ is bijective for all $x\in X$;

  2. (2)

$r^{2}=\operatorname{id}$;

  3. (3)

$\lambda_{x}\lambda_{\lambda_{x}^{-1}(y)}=\lambda_{y}\lambda_{\lambda_{y}^{-1}(x)}$
for all $x,y\in X$.

Let $(X,r)$ and $(X^{\prime},r^{\prime})$ be quadratic sets. By a morphism
from $(X,r)$ to $(X^{\prime},r^{\prime})$ we mean a map $f:X\rightarrow
X^{\prime}$ satisfying $(f\times f)r=r^{\prime}(f\times f)$.

######  Lemma 2.4.

Given two quadratic sets $(X,r)$ and $(X^{\prime},r^{\prime})$, and a map
$f:X\rightarrow X^{\prime}$, the following are equivalent:

  1. (1)

$f$ is a morphism of quadratic sets;

  2. (2)

$f\lambda_{x}=\lambda_{f(x)}^{\prime}f~{}\text{and}~{}f\rho_{x}=\rho_{f(x)}^{\prime}f~{}\text{for
all}~{}x\in X$.

If $r$ and $r^{\prime}$ are both left non-degenerate and involutive, then both
conditions above are equivalent to one of the following conditions:

  1. (3)

$f\lambda_{x}=\lambda_{f(x)}^{\prime}f$ for all $x\in X$;

  2. (4)

$f\lambda_{x}^{-1}=\lambda_{f(x)}^{\prime-1}f$ for all $x\in X$.

By a Hom-quadratic set we mean a triple $(X,r,\alpha)$ of a nonempty set $X$
with two maps $r:X\times X\to X\times X$ and $\alpha:X\to X$ such that
$r(\alpha\times\alpha)=(\alpha\times\alpha)r$. Thus a Hom-quadratic set is
exactly a quadratic set with an endomorphism.

We will identify the quadratic set $(X,r)$ with the Hom-quadratic set
$(X,r,\operatorname{id})$.

Given two Hom-quadratic sets $(X,r,\alpha)$ and
$(X^{\prime},r^{\prime},\alpha^{\prime})$, a map $f:X\rightarrow X^{\prime}$
is called a morphism of Hom-quadratic sets if

$(f\times f)r=r^{\prime}(f\times f)~{}\text{and}~{}f\alpha=\alpha^{\prime}f.$

Thus a morphism of Hom-quadratic sets is exactly a morphism of quadratic sets
satisfying $f\alpha=\alpha^{\prime}f$.

######  Corollary 2.5.

Given a quadratic set $(X,r)$ and a map $\alpha:X\to X$, the triple
$(X,r,\alpha)$ is a Hom-quadratic set if and only if

$\alpha\lambda_{x}=\lambda_{\alpha(x)}\alpha~{}\text{and}~{}\alpha\rho_{x}=\rho_{\alpha(x)}\alpha~{}\text{for
all}~{}x\in X.$

A Hom-quadratic set $(X,r,\alpha)$ is called left non-degenerate, non-
degenerate, and involutive, respectively, if $r$ has the same properties.

Denote by $\mathsf{QS}$ and $\mathsf{HQS}$ the categories of left non-
degenerate involutive quadratic sets and left non-degenerate involutive Hom-
quadratic sets, respectively. Then $\mathsf{QS}$ is a full subcategory of
$\mathsf{HQS}$ by identifying a quadratic set $(X,r)$ with a Hom-quadratic set
$(X,r,\operatorname{id})$.

By a groupoid we mean a set with a binary operation. For a groupoid $X$,
denote by $\sigma_{x}$ the left multiplication map by $x\in X$ defined by

$\sigma_{x}:X\to X,~{}~{}y\mapsto xy.$

By a left quasigroup we mean a groupoid $X$ such that the left multiplication
maps $\sigma_{x}$ are bijective for all $x\in X$ (see [33, Page 9]).

It should be pointed out that the image of an endomorphism of a left
quasigroup is a left quasigroup, though the image of a homomorphism from a
left quasigroup to a groupoid need not to be a left quasigroup (see [33, Page
15] and [33, Corollary 1.298]).

By a left Hom-quasigroup we mean a pair $(X,\alpha)$ of a left quasigroup $X$
with an endomorphism $\alpha$.

We also write a left Hom-quasigroup $(X,\alpha)$ as $(X,\cdot,\alpha)$ to
indicate the operation $\cdot$ of left quasigroup $X$.

We can identify a left quasigroup $X$ with the left Hom-quasigroup
$(X,\operatorname{id})$.

Let $(X,\alpha)$ and $(X^{\prime},\alpha^{\prime})$ be two left Hom-
quasigroups. A map $f:X\to X^{\prime}$ is called a morphism of left Hom-
quasigroups if $\alpha f=f\alpha^{\prime}$ and $f(xy)=f(x)f(y)$ for all
$x,y\in X$.

From Lemma 2.4, we have the following lemma.

######  Lemma 2.6.

Let $(X,r,\alpha)$ and $(X^{\prime},r^{\prime},\alpha^{\prime})$ be left non-
degenerate involutive Hom-quadratic sets, and let $(X,\cdot,\alpha)$ and
$(X^{\prime},*,\alpha^{\prime})$ be left Hom-quasigroups such that $x\cdot
y=\lambda_{x}^{-1}(y)$ for all $x,y\in X$ and
$x^{\prime}*y^{\prime}=\lambda^{\prime-1}_{x^{\prime}}(y^{\prime})$ for all
$x^{\prime},y^{\prime}\in X^{\prime}$. Then a map $f:X\rightarrow X^{\prime}$
is a morphism of Hom-quadratic sets if and only if it is a morphism of left
Hom-quasigroups.

Denote by $\mathsf{HQG}$ and $\mathsf{QG}$ the categories of left quasigroups
and left Hom-quasigroups, respectively. Then $\mathsf{HQG}$ is a full
subcategory of $\mathsf{QG}$ by identifying a left quasigroup $X$ with a left
Hom-quasigroup $(X,\operatorname{id})$.

Given a left non-degenerate involutive Hom-quadratic set $(X,r,\alpha)$, we
get a left Hom-quasigroups $(X,\cdot,\alpha)$, denoted by $G(X,r,\alpha)$,
with the operation defined by $x\cdot y=\lambda_{x}^{-1}(y)$ for all $x,y\in
X$. Then we have a functor $G:\mathsf{HQS}\to\mathsf{HQG}$ by associating
$(X,r,\alpha)$ with $G(X,r,\alpha)$, and a morphism $f$ in $\mathsf{HQS}$ with
$f$.

Conversely, given a left Hom-quasigroup $(X,\cdot,\alpha)$, we get a Hom-
quadratic set $(X,r,\alpha)$, denoted by $S(X,\cdot,\alpha)$, with
$\lambda_{x}(y)=\sigma_{x}^{-1}(y)$ and $\rho_{y}(x)=\sigma_{x}^{-1}(y)\cdot
x$ for all $x,y\in X$. It is routine to verify that $(X,r,\alpha)$ is left
non-degenerate and involutive. Then we have a functor
$S:\mathsf{HQG}\to\mathsf{HQS}$ by associating $(X,\cdot,\alpha)$ with
$S(X,\cdot,\alpha)$ and a morphism $f$ in $\mathsf{HQG}$ with $f$.

######  Theorem 2.7.

The functors $G$ and $S$ are mutually inverse, and so the categories
$\mathsf{HQS}$ and $\mathsf{HQG}$ are isomorphic.

###### Proof.

It is straightforward. ∎

The functor $G$ induces a functor from $\mathsf{QS}$ to $\mathsf{QG}$ and $S$
induces a functor from $\mathsf{QG}$ to $\mathsf{QS}$. We still denote the
induced functors by $G$ and $S$, respectively. By Theorem 2.7, we have the
following corollary.

######  Corollary 2.8.

The functors $G:\mathsf{QS}\to\mathsf{QG}$ and $S:\mathsf{QG}\to\mathsf{QS}$
are mutually inverse, and so the categories $\mathsf{QS}$ and $\mathsf{QG}$
are isomorphic.

Denote by $\mathsf{S_{ybe}}$ the category of left non-degenerate involutive
solutions to YBE. Then $\mathsf{S_{ybe}}$ is a full subcategory of
$\mathsf{QS}$.

A left quasigroup $X$ is called a left cycle set if $(xy)(xz)=(yx)(yz)$ for
all $x,y,z\in X$ [30].

Denote by $\mathsf{CS}$ the category of left cycle sets. Then $\mathsf{CS}$ is
a full subcategory of $\mathsf{QG}$.

By [30, Proposition 1], the functor $G$ induces a functor from
$\mathsf{S_{ybe}}$ to $\mathsf{CS}$ and $S$ induces a functor from
$\mathsf{CS}$ to $\mathsf{S_{ybe}}$. We still denote the induced functors by
$G$ and $S$, respectively. By Corollary 2.8, [30, Proposition 1] can be
restated as follows.

######  Theorem 2.9.

The functors $G:\mathsf{S_{ybe}}\to\mathsf{CS}$ and
$S:\mathsf{CS}\to\mathsf{S_{ybe}}$ are mutually inverse, and so the categories
$\mathsf{S_{ybe}}$ and $\mathsf{CS}$ are isomorphic.

A groupoid is called $\Delta$-bijective if the map $\Delta$ is bijective,
where $\Delta:X\times X\to X\times X,~{}~{}(x,y)\mapsto(xy,yx)$ [2].

The following lemma comes from [2, Lemma 2.10] (see also Lemma 1.28 in Chapter
XIII of [11]).

######  Lemma 2.10.

A groupoid $X$ is $\Delta$-bijective if and only if there exists an operation
$\circ$ on $X$ (called the dual operation) such that

$\displaystyle(x\cdot y)\circ(y\cdot x)=x,$ (2.4) $\displaystyle(x\circ
y)\cdot(y\circ x)=x,$ (2.5)

for all $x,y\in X$. Furthermore, if conditions hold, then the operation
$\circ$ is unique and the inverse of $\Delta$ is given by $(x,y)\mapsto(x\circ
y,y\circ x)$.

######  Lemma 2.11.

(see [2, Lemma 2.11]) If a groupoid $X$ is $\Delta$-bijective, then the square
map $q:X\to X,~{}~{}x\mapsto x^{2}$ is invertible.

We will say that a groupoid with extra structure is non-degenerate if the
underlying groupoid is $\Delta$-bijective.

Denote by $\mathsf{ndCS}$ the category of non-degenerate left cycle sets. Then
$\mathsf{ndCS}$ is a full subcategory of $\mathsf{QG}$.

Denote by $\mathsf{ndiS_{ybe}}$ the category of non-degenerate involutive
solution to YBE. Then $\mathsf{ndiS_{ybe}}$ is a full subcategory of
$\mathsf{QS}$.

By Corollary 2.8 and Theorem 2.9, Proposition 2 in [30] can be restated as
follows.

######  Theorem 2.12.

The functors $G:\mathsf{ndiS_{ybe}}\to\mathsf{ndCS}$ and
$S:\mathsf{ndCS}\to\mathsf{ndiS_{ybe}}$ are mutually inverse, and so the
categories $\mathsf{ndiS_{ybe}}$ and $\mathsf{ndCS}$ are isomorphic.

##  3\. Set-theoretic solutions to the Hom-Yang-Baxter equation

The Hom-Yang-Baxter equation was proposed by Yau [40] motivated by Hom-Lie
algebras.

######  Definition 3.1.

Given a vector space $V$ and two linear maps $R:V\otimes V\rightarrow V\otimes
V$ and $\alpha:V\rightarrow V$, the triple $(V,R,\alpha)$ is called a solution
to the Hom-Yang-Baxter equation, if

  1. (1)

$R(\alpha\otimes\alpha)=(\alpha\otimes\alpha)R$, and

  2. (2)

$(\alpha\otimes R)(R\otimes\alpha)(\alpha\otimes
R)=(R\otimes\alpha)(\alpha\otimes R)(R\otimes\alpha)$.

By analogy with set-theoretic solutions to YBE, we introduce set-theoretic
solutions to HYBE.

######  Definition 3.2.

Given a nonempty set $X$ and two maps $r:X\times X\rightarrow X\times X$ and
$\alpha:X\rightarrow X$, the triple $(X,r,\alpha)$ is called a set-theoretic
solution to HYBE, if

  1. (1)

$r\circ(\alpha\times\alpha)=(\alpha\times\alpha)\circ r$, and

  2. (2)

$(\alpha\times r)(r\times\alpha)(\alpha\times r)=(r\times\alpha)(\alpha\times
r)(r\times\alpha)$.

Clearly, $(X,r)$ is a set-theoretic solution to YBE if and only if
$(X,r,\operatorname{id})$ is a set-theoretic solution to HYBE.

Let $f:X\to X$ be a map and $Y$ a subset of $X$. Denote by $f|_{Y}$ the
restriction of $f$ to $Y$. When there is no ambiguity, we will write $f$ for
the restriction $f|_{Y}$.

By analogy with the relation between set-theoretic solutions to YBE and
solutions to YBE, we have the following theorem, and the proof is immediate.

######  Theorem 3.3.

Let $V$ be a vector space with a basis $X$.

  1. (1)

If $(V,R,\alpha)$ is a solution to HYBE such that $R(X\otimes X)\subseteq
X\otimes X$ and $\alpha(X)\subseteq X$, then $(X,R|_{X\otimes X},\alpha|_{X})$
is a set-theoretic solution to HYBE.

  2. (2)

Conversely, if $(X,r,\alpha)$ is a set-theoretic solution to HYBE, and
$R:V\rightarrow V$ and $\bar{\alpha}:V\to V$ are linear extensions of $r$ and
$\alpha$, respectively, then $(V,R,\bar{\alpha})$ is a solution to HYBE.

In what follows, a set-theoretic solution is simply called a solution.

######  Lemma 3.4.

A triple $(X,r,\alpha)$ with $r:X\times X\rightarrow X\times X$ and
$\alpha:X\rightarrow X$ is a solution to HYBE if and only if the following
conditions hold for all $x,y,z\in X$,

  1. (1)

$\alpha\lambda_{x}=\lambda_{\alpha(x)}\alpha,\text{and}~{}\alpha\rho_{x}=\rho_{\alpha(x)}\alpha$;

  2. (2)

$\alpha\lambda_{\alpha(x)}\lambda_{y}=\lambda_{\alpha\lambda_{x}(y)}\lambda_{\rho_{y}(x)}\alpha$;

  3. (3)

$\rho_{\lambda_{\rho_{y}(x)}\alpha(z)}\alpha\lambda_{x}(y)=\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\rho_{z}(y)$;

  4. (4)

$\alpha\rho_{\alpha(y)}\rho_{x}=\rho_{\alpha\rho_{y}(x)}\rho_{\lambda_{x}(y)}\alpha$.

###### Proof.

It is straightforward. ∎

Theorem 2.2 is a special case of Lemma 3.4.

######  Example 3.5.

A triple $(X,\operatorname{id}_{X\times X},\alpha)$ is a solution to HYBE if
and only if $\alpha^{2}=\alpha$. In this case, it is involutive, but neither
left nor right non-degenerate.

######  Example 3.6.

Let $(X,r,\alpha)$ be Hom-quadratic set. If $(X,r)$ is a solution to YBE, then
$(X,(\alpha\times\alpha)r,\alpha)$ is a solution to HYBE, and the converse
holds if additionally $\alpha$ is injective or surjective.

######  Example 3.7.

A triple $(X,\tau,\alpha)$ with $\tau(x,y)=(y,x)$ and arbitrary map
$\alpha:X\to X$ is a non-degenerate involutive solution to HYBE, called a
trivial solution.

######  Example 3.8.

A triple $(X,r,\alpha)$ with $r(x,y)=(f(y),g(x))$, where $f,g$ are maps from
$X$ to itself, is a solution to HYBE if and only if $\alpha,f,g$ commute.
Furthermore, the solution $(X,r,\alpha)$ is left non-degenerate and involutive
if and only if $f$ is bijective, and $g=f^{-1}$.

We are now in a position to characterize left non-degenerate involutive
solutions to HYBE.

######  Theorem 3.9.

A triple $(X,r,\alpha)$ with $r:X\times X\rightarrow X\times X$ and
$\alpha:X\rightarrow X$ is a left non-degenerate involutive solution to HYBE
if and only if the following conditions hold for all $x,y\in X$,

  1. (1)

$\lambda_{x}$ is bijective;

  2. (2)

$\rho_{y}(x)=\lambda_{\lambda_{x}(y)}^{-1}(x)$;

  3. (3)

$\alpha\lambda_{x}=\lambda_{\alpha(x)}\alpha$;

  4. (4)

$\alpha\lambda_{\alpha(x)}\lambda_{\lambda_{x}^{-1}(y)}=\lambda_{\alpha(y)}\lambda_{\lambda_{y}^{-1}(x)}\alpha$;

  5. (5)

$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}(y)}=\lambda_{\alpha(y)}\lambda_{\lambda_{y}^{-1}\alpha(x)}\alpha$;

  6. (6)

$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}=\lambda_{y}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha$.

###### Proof.

($\Rightarrow$) (1) and (2) follows from Lemma 2.1(2). (3) follows from Lemma
3.4(1).

To prove (4), replacing $y$ by $\lambda_{x}^{-1}(y)$ in (2) and Lemma 3.4(2),
we get

$\rho_{\lambda_{x}^{-1}(y)}(x)=\lambda_{y}^{-1}(x)~{}~{}\text{and}~{}~{}\alpha\lambda_{\alpha(x)}\lambda_{\lambda_{x}^{-1}(y)}=\lambda_{\alpha(y)}\lambda_{\rho_{\lambda_{x}^{-1}(y)}(x)}\alpha.$

Then (4) follows.

To prove (5), using (2) we can write Lemma 3.4(3) as

$\lambda_{\lambda_{\alpha\lambda_{x}(y)}\lambda_{\rho_{y}(x)}\alpha(z)}^{-1}\alpha\lambda_{x}(y)=\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\lambda_{\lambda_{y}(z)}^{-1}(y).$

It follows by Lemma 3.4(2) that

$\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}(y)=\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\lambda_{\lambda_{y}(z)}^{-1}(y),$

which implies that
$\alpha\lambda_{x}(y)=\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\lambda_{\lambda_{y}(z)}^{-1}(y)$.
Since $\lambda_{y}$ is bijective and $z$ is arbitrary, we can replace
$\lambda_{y}(z)$ by $z$ in the last equation to obtain
$\alpha\lambda_{x}(y)=\lambda_{\alpha\lambda_{\alpha(x)}(z)}\lambda_{\rho_{z}\alpha(x)}\alpha\lambda_{z}^{-1}(y)$.
Thus

$\alpha\lambda_{x}=\lambda_{\alpha\lambda_{\alpha(x)}(z)}\lambda_{\rho_{z}\alpha(x)}\alpha\lambda_{z}^{-1},$

and so
$\alpha\lambda_{x}\lambda_{z}=\lambda_{\alpha\lambda_{\alpha(x)}(z)}\lambda_{\rho_{z}\alpha(x)}\alpha$.
By (2) we have

$\alpha\lambda_{x}\lambda_{z}=\lambda_{\alpha\lambda_{\alpha(x)}(z)}\lambda_{\lambda_{\lambda_{\alpha(x)}(z)}^{-1}\alpha(x)}\alpha,$

in which replacing $z$ by $\lambda_{\alpha(x)}^{-1}(y)$ gives (5).

Now we prove (6). We first claim that the conditions (1) through (5) imply
that

$\alpha\lambda_{x}=\alpha\lambda_{\alpha^{2}(x)}.$ (3.1)

Indeed, replacing $x$ by $\alpha(x)$ in (4), we have

$\alpha\lambda_{\alpha^{2}(x)}\lambda_{\lambda_{\alpha(x)}^{-1}(y)}=\lambda_{\alpha(y)}\lambda_{\lambda_{y}^{-1}\alpha(x)}\alpha,$

which together with (5) gives
$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}(y)}=\alpha\lambda_{\alpha^{2}(x)}\lambda_{\lambda_{\alpha(x)}^{-1}(y)}$.
By (1) we get $\alpha\lambda_{x}=\alpha\lambda_{\alpha^{2}(x)}$.

Note that Lemma 3.4(2) implies that

$\lambda_{\rho_{x}(z)}\alpha=\lambda_{\alpha\lambda_{z}(x)}^{-1}\alpha\lambda_{\alpha(z)}\lambda_{x}.$
(3.2)

By (2), we can write 3.4(4) as

$\displaystyle\alpha\lambda_{\lambda_{\rho_{x}(z)}\alpha(y)}^{-1}\rho_{x}(z)=\lambda_{\lambda_{\rho_{\lambda_{x}(y)}\alpha(z)}\alpha\rho_{y}(x)}^{-1}\rho_{\lambda_{x}(y)}\alpha(z),$

for any $z\in X$. It follows by (2) that

$\displaystyle\alpha\lambda_{\lambda_{\rho_{x}(z)}\alpha(y)}^{-1}\lambda_{\lambda_{z}(x)}^{-1}(z)=\lambda_{\lambda_{\rho_{\lambda_{x}(y)}\alpha(z)}\alpha\lambda_{\lambda_{x}(y)}^{-1}(x)}^{-1}\lambda_{\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\alpha(z).$

By using (3.2) and substituting $\lambda_{\rho_{x}(z)}\alpha$ and
$\lambda_{\rho_{\lambda_{x}(y)}\alpha(z)}\alpha$ into the last equation, we
obtain

$\alpha\lambda_{\lambda_{\alpha\lambda_{z}(x)}^{-1}\alpha\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\lambda_{\lambda_{z}(x)}^{-1}(z)=\lambda_{\lambda_{\alpha\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\alpha\lambda_{\alpha^{2}(z)}(x)}^{-1}\lambda_{\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\alpha(z).$

Thus by (3.1), we have

$\alpha\lambda_{\lambda_{\alpha\lambda_{z}(x)}^{-1}\alpha\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\lambda_{\lambda_{z}(x)}^{-1}(z)=\lambda_{\lambda_{\alpha\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\alpha\lambda_{z}(x)}^{-1}\lambda_{\lambda_{\alpha(z)}\lambda_{x}(y)}^{-1}\alpha(z).$

Replacing $x$ by $\lambda_{z}^{-1}(x)$ in the previous equation, we have

$\alpha\lambda_{\lambda_{\alpha(x)}^{-1}\alpha\lambda_{\alpha(z)}\lambda_{\lambda_{z}^{-1}(x)}(y)}^{-1}\lambda_{x}^{-1}(z)=\lambda_{\lambda_{\alpha\lambda_{\alpha(z)}\lambda_{\lambda_{z}^{-1}(x)}(y)}^{-1}\alpha(x)}^{-1}\lambda_{\lambda_{\alpha(z)}\lambda_{\lambda_{z}^{-1}(x)}(y)}^{-1}\alpha(z).$

Noting that $\lambda_{\alpha(z)}\lambda_{\lambda_{z}^{-1}(x)}^{-1}(y)$ is an
arbitrary element of $X$, we may simply denote it by $y$. Then the last
equation can be written as

$\alpha\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}^{-1}\lambda_{x}^{-1}(z)=\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}^{-1}\lambda_{y}^{-1}\alpha(z),$

that is,
$\alpha\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}^{-1}\lambda_{x}^{-1}=\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}^{-1}\lambda_{y}^{-1}\alpha$,
which implies

$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}=\lambda_{y}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha.$

This proves (6).

($\Leftarrow$) The nondegeneracy and involutivity of $(X,r,\alpha)$ follows
from (1) and (2) by Lemma 2.1(2). We now prove that $(X,r,\alpha)$ satisfies
the four conditions in Lemma 3.4.

Lemma 3.4(1) follows from Lemma 2.4.

By (2) and (4), we have

$\lambda_{\alpha\lambda_{x}(y)}\lambda_{\rho_{y}(x)}\alpha=\lambda_{\alpha\lambda_{x}(y)}\lambda_{\lambda_{\lambda_{x}(y)}^{-1}(x)}\alpha=\alpha\lambda_{\alpha(x)}\lambda_{\lambda_{x}^{-1}\lambda_{x}(y)}=\alpha\lambda_{\alpha(x)}\lambda_{y},$

which proves Lemma 3.4(2).

To prove Lemma 3.4(3), replacing $x$ by $\alpha(x)$ and $y$ by
$\lambda_{y}(z)$ in Lemma 3.4(2) and using (3.1) we get

$\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha=\alpha\lambda_{\alpha^{2}(x)}\lambda_{\lambda_{y}(z)}=\alpha\lambda_{x}\lambda_{\lambda_{y}(z)}.$

It follows that

$\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}=\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\lambda_{\lambda_{y}(z)}^{-1}.$
(3.3)

Thus by Lemma 3.4(2), (3.3) and (2), we have

$\displaystyle\rho_{\lambda_{\rho_{y}(x)}\alpha(z)}\alpha\lambda_{x}(y)$
$\displaystyle=\lambda_{\lambda_{\alpha\lambda_{x}(y)}\lambda_{\rho_{y}(x)}\alpha(z)}^{-1}\alpha\lambda_{x}(y)$
$\displaystyle=\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}(y)$
$\displaystyle=\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\lambda_{\lambda_{y}(z)}^{-1}(y)$
$\displaystyle=\lambda_{\rho_{\lambda_{y}(z)}(\alpha(x))}\alpha\rho_{z}(y),$

which proves Lemma 3.4(3).

We now prove Lemma 3.4(4). By Lemma 3.4(2), we have

$\displaystyle\lambda_{\rho_{y}(x)}\alpha$
$\displaystyle=\lambda_{\alpha\lambda_{x}(y)}^{-1}\alpha\lambda_{\alpha(x)}\lambda_{y}.$
(3.4)

And by (3.3), we have

$\displaystyle\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha$
$\displaystyle=\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}\lambda_{\lambda_{y}(z)}.$
(3.5)

Then using (2) and (3.4) we obtain that

$\displaystyle\alpha\rho_{\alpha(z)}\rho_{y}(x)$
$\displaystyle=\alpha\lambda_{\lambda_{\rho_{y}(x)}\alpha(z)}^{-1}\rho_{y}(x)$
$\displaystyle=\alpha\lambda_{\lambda_{\alpha\lambda_{x}(y)}^{-1}\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\lambda_{\lambda_{x}(y)}^{-1}(x).$
(3.6)

By (2) and (3.5) we get

$\displaystyle\rho_{\alpha\rho_{z}(y)}\rho_{\lambda_{y}(z)}\alpha(x)$
$\displaystyle=\lambda_{\lambda_{\rho_{\lambda_{y}(z)}\alpha(x)}\alpha\lambda_{\lambda_{y}(z)}^{-1}(y)}^{-1}\rho_{\lambda_{y}(z)}\alpha(x)$
$\displaystyle=\lambda_{\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}\lambda_{\lambda_{y}(z)}\lambda_{\lambda_{y}(z)}^{-1}(y)}^{-1}\rho_{\lambda_{y}(z)}\alpha(x)$
$\displaystyle=\lambda_{\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}(y)}^{-1}\lambda_{\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha(x).$
(3.7)

Substituting $y$ for $\alpha(y)$ in (5), we have
$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}=\lambda_{\alpha^{2}(y)}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha$,
which together with (6) yields

$\lambda_{\alpha^{2}(y)}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha=\lambda_{y}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha.$
(3.8)

Replacing $x$ by $\alpha\lambda_{x}(y)$ and $y$ by
$\alpha\lambda_{\alpha(x)}\lambda_{y}(z)$ in (4), respectively, we have

$\alpha\lambda_{\alpha^{2}\lambda_{x}(y)}\lambda_{\lambda_{\alpha\lambda_{x}(y)}^{-1}\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}=\lambda_{\alpha^{2}\lambda_{\alpha(x)}\lambda_{y}(z)}\lambda_{\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}(y)}\alpha.$

By (3.1) and (3.8), we get that

$\alpha\lambda_{\lambda_{x}(y)}\lambda_{\lambda_{\alpha\lambda_{x}(y)}^{-1}\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}=\lambda_{\lambda_{\alpha(x)}\lambda_{y}(z)}\lambda_{\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}(y)}\alpha,$

whence

$\alpha\lambda_{\lambda_{\alpha\lambda_{x}(y)}^{-1}\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\lambda_{\lambda_{x}(y)}^{-1}=\lambda_{\lambda_{\alpha\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha\lambda_{x}(y)}^{-1}\lambda_{\lambda_{\alpha(x)}\lambda_{y}(z)}^{-1}\alpha.$
(3.9)

Lemma 3.4(4) follows from (3.6), (3.7) and (3.9). ∎

######  Corollary 3.10.

Let $(X,r,\alpha)$ be a left non-degenerate involutive solution to HYBE. Then

  1. (1)

$\lambda_{x}\alpha=\lambda_{\alpha^{2}(x)}\alpha$ for all $x\in X$;

  2. (2)

$\lambda_{x}\alpha^{2}=\alpha^{2}\lambda_{x}$ for all $x\in X$.

###### Proof.

(1) Replacing $y$ by $\alpha(y)$ in Theorem 3.9(5) we have

$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}=\lambda_{\alpha^{2}(y)}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha,$

which together with Theorem 3.9(6) yields

$\lambda_{\alpha^{2}(y)}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha=\lambda_{y}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha.$

By Theorem 3.9(3),
$\lambda_{\alpha^{2}(y)}\alpha\lambda_{\lambda_{y}^{-1}(x)}=\lambda_{y}\alpha\lambda_{\lambda_{y}^{-1}(x)}$.
Thus $\lambda_{\alpha^{2}(y)}\alpha\ =\lambda_{y}\alpha$.

(2) By Theorem 3.9(3) and Corollary 3.10(1) we have

$\alpha\lambda_{\alpha(x)}=\lambda_{\alpha^{2}(x)}\alpha=\lambda_{x}\alpha.$

By Theorem 3.9(3), we have
$\alpha^{2}\lambda_{x}=\alpha\lambda_{\alpha(x)}\alpha=\lambda_{x}\alpha^{2}$.
∎

######  Theorem 3.11.

A triple $(X,r,\alpha)$ with $r:X\times X\rightarrow X\times X$ and
$\alpha:X\rightarrow X$ is a left non-degenerate involutive solution to HYBE
if and only if the following statements are true for all $x,y\in X$,

  1. (1)

$\lambda_{x}\ $is bijective;

  2. (2)

$\rho_{y}(x)=\lambda_{\lambda_{x}(y)}^{-1}(x)$;

  3. (3)

$\alpha\lambda_{x}=\lambda_{\alpha(x)}\alpha$;

  4. (4)

$\lambda_{x}\alpha=\alpha\lambda_{\alpha(x)}$;

  5. (5)

$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}(y)}=\lambda_{\alpha(y)}\lambda_{\lambda_{y}^{-1}\alpha(x)}\alpha$.

###### Proof.

($\Rightarrow$) By Theorem 3.9 we need only prove (4). By (3) and Corollary
3.10,
$\alpha\lambda_{\alpha(x)}=\lambda_{\alpha^{2}(x)}\alpha=\lambda_{x}\alpha$,
as desired.

($\Leftarrow$) It suffices to prove (4) and (6) of Theorem 3.9. Replacing $y$
by $\alpha(y)$ in (5) and using (3) and (4) we get

$\alpha\lambda_{x}\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}=\lambda_{\alpha^{2}(y)}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha=\lambda_{y}\lambda_{\lambda_{\alpha(y)}^{-1}\alpha(x)}\alpha,$

which is Theorem 3.9(6).

Replacing $x$ by $\alpha(x)$ in (5) we get
$\alpha\lambda_{\alpha(x)}\lambda_{\lambda_{\alpha^{2}(x)}^{-1}(y)}=\lambda_{\alpha(y)}\lambda_{\lambda_{y}^{-1}\alpha^{2}(x)}\alpha$.
It follows from (3) and (4) that

$\lambda_{\alpha(y)}\lambda_{\lambda_{y}^{-1}\alpha^{2}(x)}\alpha=\alpha\lambda_{\alpha(x)}\lambda_{\lambda_{\alpha^{2}(x)}^{-1}(y)}=\lambda_{\alpha^{2}(x)}\lambda_{\lambda_{\alpha^{3}(x)}^{-1}\alpha(y)}\alpha\\\
=\lambda_{\alpha^{2}(x)}\lambda_{\lambda_{\alpha(x)}^{-1}\alpha(y)}\alpha=\alpha\lambda_{\alpha(x)}\lambda_{\lambda_{x}^{-1}(y)}.$

This proves Theorem 3.9(4). ∎

When $\alpha=\operatorname{id}$, Theorem 3.11 gives Theorem 2.3.

######  Example 3.12.

A triple $(X,r,\alpha)$ such that $\alpha(X)=\\{\theta\\}$ is a left non-
degenerate involutive solution to HYBE if and only if $\lambda_{x}$ is
bijective, $\lambda_{x}(\theta)=\theta$ and
$\rho_{y}(x)=\lambda_{\lambda_{x}(y)}^{-1}(x)$ for all $x,y\in X$.

##  4\. Hom-cycle sets

In this section, we introduce a Hom-version of cycle sets and relate them to
left non-degenerate involutive solutions to HYBE.

######  Definition 4.1.

A left Hom-quasigroup $(X,\alpha)$ is called a left Hom-cycle set if the
following conditions are satisfied for all $x,y,z\in X$,

$\displaystyle\alpha((xy)(\alpha(x)z))=(yx)(\alpha(y)\alpha(z)),$ (4.1)
$\displaystyle\alpha((\alpha(x)y)(xz))=(y\alpha(x))(\alpha(y)\alpha(z)),$
(4.2)
$\displaystyle\alpha((\alpha(x)\alpha(y))(xz))=(\alpha(y)\alpha(x))(y\alpha(z)).$
(4.3)

In what follows, left cycle sets and left Hom-cycle sets are referred to cycle
sets and Hom-cycle sets.

Clearly, $X$ is a cycle set if and only if $(X,\operatorname{id}_{X})$ is a
Hom-cycle set. Hence we will identify the cycle set $X$ with the Hom-cycle set
$(X,\operatorname{id}_{X})$.

Denote by $\mathsf{HCS}$ the category of Hom-cycle sets, which is a full
subcategory of $\mathsf{HQG}$.

Denote by $\mathsf{S_{hybe}}$ the category of left non-degenerate involutive
solution to HYBE, which is a full subcategory of $\mathsf{HQS}$.

######  Theorem 4.2.

The functors $G:\mathsf{S_{hybe}}\to\mathsf{HCS}$ and
$S:\mathsf{HCS}\to\mathsf{S_{hybe}}$ are mutually inverse, and so the
categories $\mathsf{S_{hybe}}$ and $\mathsf{HCS}$ are isomorphic.

###### Proof.

Let $(X,r,\alpha)$ be a left non-degenerate involutive solution to HYBE.
$G(X,r,\alpha)$ is a left Hom-quasigroup. Furthermore, since $x\cdot
y=\lambda_{x}^{-1}(y)$, Theorem 3.9(4)–(6) imply the conditions (4.1)–(4.3).
Thus $G(X,r,\alpha)$ is a Hom-cycle set.

Conversely, let $(X,\alpha)$ be a Hom-cycle set and
$S(X,\alpha)=(X,r,\alpha)$. Then $(X,r,\alpha)$ is a left non-degenerate
involutive Hom-quadratic set. By Lemma 2.1(2) and Corollary 2.5,
$(X,r,\alpha)$ satisfies Theorem 3.9(1)–(3). Furthermore, the conditions
(4.1)–(4.3) imply Theorem 3.9(4)–(6). Thus $S(X,\alpha)$ is a left non-
degenerate involutive solution to HYBE. ∎

Theorem 4.2 generalizes [30, Proposition 1].

######  Theorem 4.3.

A left Hom-quasigroup $(X,\alpha)$ is a Hom-cycle set if and only if the
following conditions hold for all $x,y\in X$,

$\displaystyle\alpha^{2}(x)\alpha(y)=x\alpha(y),$ (4.4)
$\displaystyle(x\alpha(y))(\alpha(x)\alpha(z))=(y\alpha(x))(\alpha(y)\alpha(z)).$
(4.5)

###### Proof.

Suppose $S(X,\alpha)=(X,r,\alpha)$.

($\Rightarrow$) By Theorem 4.2, $(X,r,\alpha)$ is a left non-degenerate
involutive solution to HYBE. Thus (4.4) follows from Theorem 3.11(4), and
(4.5) follows from (4.2).

($\Leftarrow$) By Theorem 4.2 it suffices to prove that $(X,r,\alpha)$ is a
left non-degenerate involutive solution to HYBE. We need to verify that
$(X,r,\alpha)$ satisfies (1)–(5) of Theorem 3.11. In fact, Lemma 2.1(2) and
Corollary 2.5 imply that $(X,r,\alpha)$ satisfies (1)–(3) in Theorem 3.11, and
the conditions (4.4) and (4.5) imply that $(X,r,\alpha)$ satisfies (4)–(5) in
Theorem 3.11, as desired. ∎

######  Example 4.4.

Let $X$ be a nonempty set with a map $\alpha:X\to X$, and let $f:X\to X$ be a
bijection such that $f\alpha=\alpha f$. Define an operation on $X$ by
$xy=f(y)$ for all $x,y\in X$. Then $(X,\alpha)$ is a Hom-cycle set, which
corresponds to the left non-degenerate involutive solution to HYBE defined in
Example 3.8.

######  Example 4.5.

Let $(X,\alpha)$ be a left Hom-quasigroup with $\alpha(X)=\\{\theta\\}$. Then
$(X,\alpha)$ is a Hom-cycle set if and only if $\theta$ is a right zero, i.e.,
$x\theta=\theta$ for all $x\in X$. In this case, $(X,\alpha)$ corresponds to
the solution defined in Example 3.12.

######  Example 4.6.

A trivial solution to HYBE corresponds to a right zero groupoid with an
endomorphism.

######  Example 4.7.

Let $(X,+)$ be an Abelian group with endomorphisms $\varphi,\psi,\alpha$ and
define $x\cdot y=\varphi(x)+\psi(y)$ for all $x,y\in X$. Then
$(X,\cdot,\alpha)$ is a Hom-cycle set if and only if the following hold:

  1. (1)

$\psi$ is bijective;

  2. (2)

$\varphi\alpha=\alpha\varphi$ and $\psi\alpha=\alpha\psi$;

  3. (3)

$\varphi\alpha^{2}=\varphi$;

  4. (4)

$\varphi^{2}=\varphi\psi\alpha-\psi\varphi\alpha$.

In fact, (1) is equivalent to saying that $(X,\cdot)$ is a left quasigroup by
[2, Section 4.1]; (2) is equivalent to the assertion that $\alpha$ is an
endomorphism of $(X,\cdot)$; (3) and (4) are equivalent to (4.4) and (4.5),
respectively.

######  Lemma 4.8.

If $(X,\alpha)$ is a left Hom-quasigroup satisfying (4.4), then

$(x\alpha^{2}(y))\alpha(z)=(xy)\alpha(z),$ (4.6)

for all $x,y,z\in X$.

###### Proof.

By using (4.4) we have

$(xy)\alpha(z)=\alpha^{2}(xy)\alpha(z)=(\alpha^{2}(x)\alpha^{2}(y))\alpha(z)=(x\alpha^{2}(y))\alpha(z),$

for all $x,y,z\in X$. ∎

######  Lemma 4.9.

If $(X,\alpha)$ is a left Hom-quasigroup satisfying (4.4), then (4.1), (4.2)
and (4.3) are equivalent.

###### Proof.

Replacing $x$ by $\alpha(x)$ in (4.1) gives (4.2).

Replacing $y$ by $\alpha(y)$ in (4.2) gives (4.3).

Suppose (4.3) holds. Replacing $x$ by $\alpha(x)$ and $y$ by $\alpha(y)$ in
(4.3) gives
$\alpha((\alpha^{2}(x)\alpha^{2}(y))(\alpha(x)z))=(\alpha^{2}(y)\alpha^{2}(x))(\alpha(y)\alpha(z))$.
It follows that

$\alpha((xy)(\alpha(x)z))=(y\alpha^{2}(x))(\alpha(y)\alpha(z)),$

which together with (4.6) gives (4.1). ∎

######  Theorem 4.10.

Let $(X,\alpha)$ be a left Hom-quasigroup. Then $(X,\alpha)$ is a Hom-cycle
set if and only if it satisfies (4.4) and one of (4.1), (4.2) and (4.3).

###### Proof.

It follows from Lemma 4.9 and Theorem 4.3. ∎

######  Theorem 4.11.

A Hom-cycle set is non-degenerate if and only if the corresponding solution to
HYBE is non-degenerate.

###### Proof.

Let $(X,\alpha)$ be a Hom-cycle set and $(X,r,\alpha)$ the corresponding
solution to HYBE.

($\Rightarrow$) Suppose $(X,\alpha)$ is non-degenerate and $\circ$ is the dual
operation defined in Lemma 2.10. Then (2.5) can be rewritten as
$\lambda_{x\circ y}^{-1}(y\circ x)=x$ for all $x,y\in X$. Thus $y\circ
x=\lambda_{x\circ y}(x)$ for all $x,y\in X$. It follows by interchanging $x$
and $y$ that $x\circ y=\lambda_{y\circ x}(y)$ for all $x,y\in X$. Substituting
the last equation into (2.5), we get

$\lambda_{y\circ x}(y)\cdot\newline (y\circ x)=x.$

Denote by $\tau_{y}$ the left multiplication by $y$ with respect to the
operation $\circ$. Then $\rho_{y}\tau_{y}(x)=\rho_{y}(y\circ
x)=\lambda_{y\circ x}(y)\cdot(y\circ x)=x$. Noting that $x\lambda_{x}(y)=y$,
by (2.4) we have

$\tau_{y}\rho_{y}(x)=(x\lambda_{x}(y))\circ(\lambda_{x}(y)x)=x.$

Thus $\rho_{y}$ is bijective, and so $(X,r,\alpha)$ is non-degenerate.

($\Leftarrow$) Suppose $(X,r,\alpha)$ is non-degenerate. Define $x\circ
y=\rho_{x}^{-1}(y)$ for all $x,y\in X$. Replacing $y$ by $\lambda_{x}^{-1}(y)$
in (2.1) we obtain $\rho_{\lambda_{x}^{-1}(y)}(x)=\lambda_{y}^{-1}(x)$, whence
$\rho_{\lambda_{x}^{-1}(y)}^{-1}\lambda_{y}^{-1}(x)=x$, which gives (2.4).
Similarly, replacing $y$ by $\rho_{x}^{-1}(y)$ in (2.2) we can get (2.5). Thus
$(X,\alpha)$ is non-degenerate by Lemma 2.10. ∎

Denote by $\mathsf{ndHCS}$ the category of non-degenerate left Hom-cycle sets.
Then $\mathsf{ndHCS}$ is a full subcategory of $\mathsf{HCS}$.

Denote by $\mathsf{ndiS_{hybe}}$ the category of non-degenerate involutive
solutions to HYBE. Then $\mathsf{ndiS_{hybe}}$ is a full subcategory of
$\mathsf{S_{hybe}}$.

Theorem 4.2 and Theorem 4.11 have the following corollary, which generalize
[30, Proposition 2].

######  Corollary 4.12.

The functors $G:\mathsf{ndiS_{hybe}}\to\mathsf{ndHCS}$ and
$S:\mathsf{ndHCS}\to\mathsf{ndiS_{hybe}}$ are mutually inverse, and so the
categories $\mathsf{ndiS_{hybe}}$ and $\mathsf{ndHCS}$ are isomorphic.

######  Theorem 4.13.

Let $(X,\alpha)$ be a non-degenerate Hom-cycle set with the dual operation
$\circ$. Then $(X,\circ,\alpha)$ is a non-degenerate Hom-cycle set.

###### Proof.

Suppose $S(X,\alpha)=(X,r,\alpha)$. By Theorem 4.11, $(X,r,\alpha)$ is a non-
degenerate involutive solution to HYBE. Replacing $y$ by $\lambda_{x}(y)$ in
(2.4), we have $(x\lambda_{x}(y))\circ(\lambda_{x}(y)x)=x$, whence
$y\circ\rho_{y}(x)=x$. Replacing $x$ by $\rho_{y}^{-1}(x)$ in the last
equation gives $y\circ x=\rho_{y}^{-1}(x)$, and so $\rho_{y}(y\circ x)=x$.
Thus $(X,\circ)$ is a left quasigroup, and we have

$\alpha(x)=\alpha\rho_{y}(y\circ x)=\rho_{\alpha(y)}\alpha(y\circ x),$

whence $\alpha(y\circ
x)=\rho_{\alpha(y)}^{-1}\alpha(x)=\alpha(y)\circ\alpha(x)$. Hence
$(X,\circ,\alpha)$ is a left Hom-quasigroup. Suppose
$S(X,\circ,\alpha)=(X,r^{\circ},\alpha)$. By Corollary 4.12, it suffices to
prove that $(X,r^{\circ},\alpha)$ is a non-degenerate involutive solution to
HYBE. Since $\lambda_{y}^{\circ-1}(x)=y\circ x=\rho_{y}^{-1}(x)$, we have
$\lambda_{y}^{\circ}=\rho_{y}$. Replacing $y$ by $\rho_{x}(y)$ in (2.5), we
have $(x\circ\rho_{x}(y))(\rho_{x}(y)\circ x)=x$. Thus

$y(\rho_{x}(y)\circ x)=x=y\lambda_{y}(x),$

and so we have $\lambda_{y}(x)=\rho_{x}(y)\circ x=\lambda_{x}^{\circ}(y)\circ
x=\rho_{y}^{\circ}(x)$, whence $\rho_{y}^{\circ}=\lambda_{y}$. Thus
$r^{\circ}(x,y)=(\rho_{x}(y),\lambda_{y}(x))$. Clearly $r^{\circ}=\tau r\tau$,
where $\tau(x,y)=(y,x)$. Since $(X,r,\alpha)$ is a non-degenerate involutive
solution to HYBE, so is $(X,r^{\circ},\alpha)$, as desired. ∎

Finite cycle sets and cycle sets with bijective square maps, especially
square-free cycle sets, are non-degenerate [30], but Hom-cycle sets are not
the case.

######  Example 4.14.

Let $X=\\{1,2,3,4\\}$ with a map $\alpha:X\rightarrow X$ such that
$\alpha(X)=\\{1\\}$. Define an operation on $X$ by the following
multiplication table:

$\begin{array}[]{c|cccc}\cdot&1&2&3&4\\\ \hline\cr 1&1&2&3&4\\\ 2&1&2&3&4\\\
3&1&4&3&2\\\ 4&1&3&2&4\end{array}$

Then $(X,\cdot,\alpha)$ is a square-free Hom-cycle set, but not non-degenerate
since $\Delta(3,4)=(2,2)=\Delta(2,2)$.

##  5\. Twists

Let $(X,\alpha)$ be a left Hom-quasigroup. Define an operation on $X$ by
$x\cdot^{\prime}y=\alpha(x)y$ for all $x,y\in X$. Then $(X,\cdot^{\prime})$ is
a left quasigroup, and

$\alpha(x\cdot^{\prime}y)=\alpha\left(\alpha(x)y\right)=\alpha^{2}(x)\alpha(y)=\alpha(x)\cdot^{\prime}\alpha(y).$

Thus $\alpha$ is an endomorphism of $(X,\cdot^{\prime})$. Hence
$(X,\cdot^{\prime},\alpha)$ is a left Hom-quasigroup. We call
$(X,\cdot^{\prime},\alpha)$ the twist of $(X,\alpha)$, denoted by
$T(X,\alpha)$.

Using twist we can define a functor $T:\mathsf{HQG}\to\mathsf{HQG}$ in a
natural way.

A left Hom-quasigroup $(X,\alpha)$ is called an im-cycle set if the following
are satisfied:

$\displaystyle\alpha^{3}(x)\alpha(y)=\alpha(x)\alpha(y),$ (5.1)
$\displaystyle(\alpha(x)\alpha(y))(\alpha(x)\alpha(z))=(\alpha(y)\alpha(x))(\alpha(y)\alpha(z)),$
(5.2)

for all $x,y,z\in X$.

It is clear that $(X,\alpha)$ is an im-cycle set if and only if $\alpha(X)$ is
a cycle set and (5.1) holds.

######  Theorem 5.1.

A left Hom-quasigroup is an im-cycle set if and only if its twist is a Hom-
cycle set.

###### Proof.

Let $(X,\cdot,\alpha)$ be a left Hom-quasigroup. Then its twist
$(X,\cdot^{\prime},\alpha)$ is also a left Hom-quasigroup.

($\Rightarrow$) By (5.1) we have
$\alpha^{2}(x)\cdot^{\prime}\alpha(y)=x\cdot^{\prime}\alpha(y)$ for all
$x,y\in X$. Replacing $x$ by $\alpha(x)$ and $y$ by $\alpha(y)$ in (5.2), we
obtain

$(\alpha^{2}(x)\alpha^{2}(y))(\alpha^{2}(x)\alpha(z))=(\alpha^{2}(y)\alpha^{2}(x))(\alpha^{2}(y)\alpha(z)),$

which implies that
$(x\cdot^{\prime}\alpha(y))\cdot^{\prime}(\alpha(x)\cdot^{\prime}\alpha(z))=(y\cdot^{\prime}\alpha(x))\cdot^{\prime}(\alpha(y)\cdot^{\prime}\alpha(z))$.
By Theorem 4.3, $(X,\cdot^{\prime},\alpha)$ is a Hom-cycle set.

($\Leftarrow$) Since $(X,\cdot^{\prime},\alpha)$ is a Hom-cycle set, applying
Theorem 4.3 to $(X,\cdot^{\prime},\alpha)$ yields (5.1) and

$\alpha(\alpha(x)\alpha(y))(\alpha^{2}(x)\alpha(z))=\alpha(\alpha(y)\alpha(x))(\alpha^{2}(y)\alpha(z)).$

Replacing $x$ by $\alpha(x)$ and $y$ by $\alpha(y)$ in the last equation, we
have

$(\alpha^{3}(x)\alpha^{3}(y))(\alpha^{3}(x)\alpha(z))=(\alpha^{3}(y)\alpha^{3}(x))(\alpha^{3}(y)\alpha(z)).$

By (5.1), we see that $(\alpha(X),\cdot)$ is a cycle set. Thus $(X,\alpha)$ is
an im-cycle set. ∎

######  Corollary 5.2.

The twist of an im-cycle set is a Hom-cycle set.

######  Corollary 5.3.

Let $X$ be a cycle set with an endomorphism $\alpha$ such that (5.1) holds.
Then the twist of $(X,\alpha)$ is a Hom-cycle set.

######  Theorem 5.4.

The twist of a Hom-cycle set is an im-cycle set.

###### Proof.

Let $(X,\cdot,\alpha)$ be a Hom-cycle set with the twist
$(X,\cdot^{\prime},\alpha)$. By (4.4), we can rewrite (4.5) as

$\alpha^{2}(x\alpha(y))(\alpha(x)\alpha(z))=\alpha^{2}(y\alpha(x))(\alpha(y)\alpha(z)),$

that is,

$(\alpha^{2}(x)\alpha^{3}(y))(\alpha(x)\alpha(z))=(\alpha^{2}(y)\alpha^{3}(x))(\alpha(y)\alpha(z)),$

for all $x,y,z\in X$. Thus by (4.6), we have

$(\alpha^{2}(x)\alpha(y))(\alpha(x)\alpha(z))=(\alpha^{2}(y)\alpha(x))(\alpha(y)\alpha(z)),$

which implies
$(x\cdot^{\prime}{y})\cdot^{\prime}(x\cdot^{\prime}\alpha(z))=(y\cdot^{\prime}x)\cdot^{\prime}(y\cdot^{\prime}\alpha(z))$
for all $x,y,z\in X$. Hence $(\alpha(X),\cdot^{\prime})$ is a cycle set. By
(4.4), we have

$\alpha^{4}(x)\cdot\alpha(y)=\alpha^{2}(x)\cdot\alpha(y)$

for all $x,y\in X$, whence
$\alpha^{3}(x)\cdot^{\prime}\alpha(y)=\alpha(x)\cdot^{\prime}\alpha(y)$ for
all $x,y\in X$. Thus $(X,\cdot^{\prime},\alpha)$ is an im-cycle set. ∎

######  Theorem 5.5.

If $(X,\alpha)$ be a non-degenerate Hom-cycle set, then its twist is non-
degenerate.

###### Proof.

Let $T(X,\alpha)=(X,\cdot^{\prime},\alpha)$. By Theorem 4.13,
$(X,\circ,\alpha)$ is a Hom-cycle set. Applying Theorem 4.3 to
$(X,\circ,\alpha)$ we obtain

$\alpha^{2}(x)\circ\alpha(y)=x\circ\alpha(y),$ (5.3)

for all $x,y\in X$. Define $x\circ^{\prime}y=\alpha(x)\circ y$ for all $x,y\in
X$. Then by (4.4), (5.3) and Lemma 2.10 we have

$\displaystyle(x\cdot^{\prime}y)\circ^{\prime}(y\cdot^{\prime}x)=\alpha(\alpha(x)y)\circ(\alpha(y)x)=(x\alpha(y))\circ(\alpha(y)x)=x,$
$\displaystyle(x\circ^{\prime}y)\cdot^{\prime}(y\circ^{\prime}x)=\alpha(\alpha(x)\circ
y)(\alpha(y)\circ x)=(x\circ\alpha(y))(\alpha(y)\circ x)=x.$

Thus $(X,\cdot^{\prime},\alpha)$ is non-degenerate by Lemma 2.10. ∎

######  Lemma 5.6.

Let $(X,\alpha)$ be a left Hom-quasigroup with $\alpha(X)$ singleton. Then its
twist is a non-degenerate left Hom-quasigroup.

###### Proof.

Let $\alpha(X)=\\{\theta\\}$. Then $\theta$ is a right zero of the left
quasigroup $X$. Let $T(X,\alpha)=(X,\cdot^{\prime},\alpha)$. Thus

$\Delta(x,y)=(x\cdot^{\prime}y,y\cdot^{\prime}x)=(\alpha(x)y,\alpha(y)x)=(\theta
y,\theta x)$

for all $x,y\in X$. It follows $\Delta=\sigma_{\theta}\times\sigma_{\theta}$.
Since $\sigma_{\theta}$ is bijective, so is $\Delta$. Hence
$(X,\cdot^{\prime},\alpha)$ is non-degenerate. ∎

######  Remark 5.7.

A Hom-cycle set is not necessarily the twist of an im-cycle set, and vice
versa. For example, let $(X,\alpha)$ be as in Example 4.14. It is degenerate
and both a Hom-cycle set and an im-cycle set, but it is isomorphic to neither
the twist of a Hom-cycle set nor the twist of an im-cycle set by Lemma 5.6.

The following example shows that twist of a Hom-cycle set is not necessarily a
Hom-cycle set.

######  Example 5.8.

Let $F$ be a field of characteristic $\neq 2$, $X$ the vector space ${F}^{3}$
and $\varphi,\psi,\alpha$ the linear endomorphisms of $X$ defined under the
natural basis by the following matrices, respectively,

$A=\begin{pmatrix}0&1&0\\\ 0&0&1\\\
0&0&0\end{pmatrix},~{}B=\begin{pmatrix}1&0&0\\\ 0&1&1\\\
0&0&1\end{pmatrix},~{}C=\begin{pmatrix}-1&0&1\\\ 0&-1&0\\\
0&0&-1\end{pmatrix}.$

Then it is easy to check that $\psi$ is bijective and

$\
\varphi\alpha=\alpha\varphi,~{}\psi\alpha=\alpha\psi,~{}\varphi\alpha^{2}=\varphi,~{}\varphi^{2}=\varphi\psi-\psi\varphi,~{}\varphi^{2}\alpha\neq\varphi^{2}.$

Define $x\cdot y=\varphi(x)+\psi(y)$ for all $x,y\in X$. Then $(X,\cdot)$ is a
cycle set by [2, Section 4.1] and $\alpha$ is an endomorphism of $(X,\cdot)$
satisfying $\alpha^{2}(x)y=xy$ for all $x,y\in X$.

The twist $T(X,\cdot,\alpha)$ of $(X,\cdot,\alpha)$ is a Hom-cycle set By
Corollary 5.3. Clearly, the twist of $T(X,\cdot,\alpha)$ equals
$(X,\cdot,\alpha)$, which is not a Hom-cycle set since (4) in Example 4.7 does
not hold.

######  Corollary 5.9.

If $(X,\alpha)$ is a non-degenerate Hom-cycle set, then the map
$q^{\prime}:X\to X,~{}~{}x\mapsto\alpha(x)x$ is bijective.

###### Proof.

By Theorem 5.5 the twist $(X,\cdot^{\prime},\alpha)$ is non-degenerate. Since
$q^{\prime}$ is the square map of $(X,\cdot^{\prime})$, $q^{\prime}$ is
bijective by Lemma 2.11. ∎

######  Remark 5.10.

The converses of Theorem 5.5 and Corollary 5.9 do not hold. Example 4.14
provides a counterexample to the both cases.

We now apply Theorem 5.1, Corollary 5.3 and Theorem 5.4 to solutions to HYBE.

Let $(X,r,\alpha)$ be a left non-degenerate involutive Hom-quadratic set. We
call $STG(X,r,\alpha)$ the twist of $(X,r,\alpha)$. The twist of
$(X,r,\alpha)$ is also a left non-degenerate involutive Hom-quadratic set.

######  Theorem 5.11.

Let $(X,r,\alpha)$ be a left non-degenerate involutive Hom-quadratic set with
the twist $(X,r^{\prime},\alpha)$. Then

$r^{\prime}(x,y)=(\lambda_{\alpha(x)}(y),\lambda_{\lambda_{\alpha^{2}(x)}\alpha(y)}^{-1}(x))$
(5.4)

for all $x,y\in X$. If $(X,r,\alpha)$ is additionally a solution to HYBE, then

$r^{\prime}(x,y)=(\lambda_{\alpha(x)}(y),\rho_{\alpha(y)}(x))$ (5.5)

for all $x,y\in X$.

###### Proof.

Since $STG(X,r,\alpha)=(X,r^{\prime},\alpha)$, by Theorem 2.7 we have

$TG(X,r,\alpha)=G(X,r^{\prime},\alpha).$

Thus $G(X,r^{\prime},\alpha)$ is the twist of $G(X,r,\alpha)$. Let
$G(X,r,\alpha)=(X,\cdot,\alpha)$. Then
$G(X,r^{\prime},\alpha)=(X,\cdot^{\prime},\alpha)$, where
$x\cdot^{\prime}y=\alpha(x)y$ for all $x,y\in X$. Thus
$\lambda^{\prime}_{x}=\lambda_{\alpha(x)}$, and so

$\rho^{\prime}_{y}(x)=\lambda^{\prime-1}_{\lambda^{\prime}_{x}(y)}(x)=\lambda^{-1}_{\alpha\lambda_{\alpha(x)}(y)}(x)=\lambda^{-1}_{\lambda_{\alpha^{2}(x)}\alpha(y)}(x).$

Therefore (5.4) holds.

Furthermore, if $(X,r,\alpha)$ is a solution to HYBE, then (5.5) follows from
(5.4), Corollary 3.10(1) and (2.3). ∎

######  Remark 5.12.

By Example 5.8 and Theorem 4.2, the twist of a left non-degenerate involutive
solution to HYBE is not necessarily a solution to HYBE.

######  Lemma 5.13.

Let $(X,r,\alpha)$ be a left non-degenerate involutive Hom-quadratic set and
$i,j$ be nonnegative integers such that $0\leq j-i\leq 2$. Then the following
are equivalent.

  1. (1)

$r(\alpha^{i+2}\times\alpha^{j})=(1\times\alpha^{2})r(\alpha^{i}\times\alpha^{j})$;

  2. (2)

$\lambda_{\alpha^{i+2}(x)}\alpha^{j}=\lambda_{\alpha^{i}(x)}\alpha^{j}$ for
all $x\in X$.

###### Proof.

We first note that for all $x,y\in X$,

$\displaystyle
r(\alpha^{i+2}\times\alpha^{j})(x,y)=(\lambda_{\alpha^{i+2}(x)}\alpha^{j}(y),\rho_{\alpha^{j}(y)}\alpha^{i+2}(x)),$
$\displaystyle(1\times\alpha^{2})r(\alpha^{i}\times\alpha^{j})(x,y)=(\lambda_{\alpha^{i}(x)}\alpha^{j}(y),\alpha^{2}\rho_{\alpha^{j}(y)}\alpha^{i}(x)).$

It suffices to prove that (2) implies
$\rho_{\alpha^{j}(y)}\alpha^{i+2}(x)=\alpha^{2}\rho_{\alpha^{j}(y)}\alpha^{i}(x)$
for all $x,y\in X$. Since $0\leq j-i\leq 2$, we have
$\lambda_{\alpha^{i}(x)}\alpha^{j}=\alpha^{j}\lambda_{\alpha^{i+2-j}(x)}$, and
so
$\alpha^{j}\lambda_{\alpha^{i+2-j}(x)}^{-1}=\lambda_{\alpha^{i}(x)}^{-1}\alpha^{j}$.
Thus
$\lambda_{\alpha^{i+2}(x)}^{-1}\alpha^{j}=\lambda_{\alpha^{i}(x)}^{-1}\alpha^{j}$.
Consequently,

$\alpha^{2}\rho_{\alpha^{j}(y)}\alpha^{i}(x)=\lambda_{\lambda_{\alpha^{i+2}(x)}\alpha^{j+2}(y)}^{-1}\alpha^{i+2}(x)=\lambda_{\alpha^{i+2}(\lambda_{x}\alpha^{j-i}(y))}^{-1}\alpha^{i+2}(x)\\\
=\lambda_{\alpha^{i}(\lambda_{x}\alpha^{j-i}(y))}^{-1}\alpha^{i+2}(x)=\lambda_{\lambda_{\alpha^{i}(x)}\alpha^{j}(y)}^{-1}\alpha^{i+2}(x)\\\
=\lambda_{\lambda_{\alpha^{i+2}(x)}\alpha^{j}(y)}^{-1}\alpha^{i+2}(x)=\rho_{\alpha^{j}(y)}\alpha^{i+2}(x),$

as desired. ∎

######  Corollary 5.14.

Let $(X,r,\alpha)$ be a left non-degenerate involutive solution to HYBE. Then
$r(\alpha^{2}\times\alpha)=(\operatorname{id}\times\alpha^{2})r(\operatorname{id}\times\alpha)$.

###### Proof.

It follows from Corollary 3.10 and Lemma 5.13. ∎

######  Theorem 5.15.

Let $(X,r,\alpha)$ be left non-degenerate involutive Hom-quadratic set with
the twist $(X,r^{\prime},\alpha)$.

  1. (1)

If $(X,r,\alpha)$ is a solution to HYBE, then $(\alpha(X),r^{\prime})$ is a
solution to YBE and
$r^{\prime}(\alpha^{2}\times\operatorname{id})=(\operatorname{id}\times\alpha^{2})r^{\prime}$
on $\alpha(X)$.

  2. (2)

The twist $(X,r^{\prime},\alpha)$ is a solution to HYBE if and only if
$(\alpha(X),r)$ is a solution to YBE and
$r(\alpha^{2}\times\operatorname{id})=(\operatorname{id}\times\alpha^{2})r$ on
$\alpha(X)$.

  3. (3)

If $(X,r)$ is a solution to YBE and
$r(\alpha^{3}\times\alpha)=(\alpha\times\alpha^{3})r$, then
$(X,r^{\prime},\alpha)$ is a solution to HYBE.

###### Proof.

Let $S(X,r,\alpha)=(X,\cdot,\alpha)$. Then
$S(X,r^{\prime},\alpha)=(X,\cdot^{\prime},\alpha)$, the twist of
$(X,\cdot,\alpha)$.

(1) Theorem 4.2 implies that $(X,\cdot,\alpha)$ is a Hom-cycle set. By Theorem
5.4, $(\alpha(X),\cdot^{\prime})$ is a cycle set and (5.1) holds with respect
to the operation $\cdot^{\prime}$. It follows that
$\alpha^{2}(x)\cdot^{\prime}y=x\cdot^{\prime}y$ for all $x,y\in\alpha(X)$,
which implies
$r^{\prime}(\alpha^{2}\times\operatorname{id})=(\operatorname{id}\times\alpha^{2})r^{\prime}$
on $\alpha(X)$ by Lemma 5.13. By Theorem 2.9, $(\alpha(X),r^{\prime})$ is a
solution to YBE.

(2) By Theorem 4.2, $(X,r^{\prime},\alpha)$ is a left non-degenerate
involutive solution to HYBE if and only if $(X,\cdot^{\prime},\alpha)$ is a
Hom-cycle set. Equivalently, by Theorem 5.1 $(\alpha(X),\cdot)$ is a left
cycle set satisfying (5.1). This is equivalent to that $(\alpha(X),r)$ is a
solution to YBE satisfying
$r(\alpha^{2}\times\operatorname{id})=(\operatorname{id}\times\alpha^{2})r$ on
$\alpha(X)$ by Theorem 2.9 and Lemma 5.13.

(3) follows from the sufficiency in (2). ∎

## Acknowledgements

This work is supported by NSF of China (No.12171194, No.11971289).

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