# Deformed solutions of the Yang-Baxter equation coming from dual weak braces
and unital near-trusses111This work is supported by the Dipartimento di
Matematica e Fisica “Ennio De Giorgi” - Università del Salento and EPSRC grant
EP/V008129/1. The first and the third authors are members of GNSAGA (INdAM).

Marzia MAZZOTTA<EMAIL_ADDRESS>Bernard RYBOŁOWICZ
<EMAIL_ADDRESS>Paola STEFANELLI<EMAIL_ADDRESS>Dipartimento di Matematica e Fisica “Ennio De Giorgi”
Università del Salento, Via Provinciale Lecce-Arnesano, 73100 Lecce (Italy)
Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS,
and Maxwell Institute for Mathematical Sciences, Edinburgh (UK)

###### Abstract

We show that a family of set-theoretic solutions comes from dual weak braces,
including the recent ones provided in the context of skew braces. In
particular, the solutions we give are obtained by “deforming” the classical
ones by certain parameters $z$. In any dual weak brace $(S,+,\circ)$, the
parameters giving rise to deformed solutions are only those belonging to the
_right distributor_ of $S$, i.e., $\mathcal{D}_{r}(S)=\\{z\in
S\,\mid\,\forall\,a,b\in S\quad(a+b)\circ z=a\circ z-z+b\circ z\\}$, that is a
full inverse subsemigroup of $\left(S,\circ\right)$. Particular attention is
posed to the class of skew braces and, in a natural way, of two-sided skew
braces. In addition, we construct new arbitrary solutions on the algebraic
structure of the unital near-truss determined by extending deformed solutions
coming from skew braces.

###### keywords:

Quantum Yang-Baxter equation , set-theoretic solution , inverse semigroup,
Clifford semigroup , skew brace , brace , weak brace , near-truss , heap

###### MSC:

[2020] 16T25, 81R50 , 20M18, 16Y99

## Introduction

The challenging problem of determining and classifying solutions arose in 1992
in the paper by Drinfel’d [12] and has attracted a lot of researchers for its
significant connections with diverse topics. A map $r:S\times S\to S\times S$
with $S$ a set, is said to be a _set-theoretic solution of the Yang-Baxter
equation_ , or shortly a _solution_ , if $r$ satisfies the identity

$\displaystyle\left(r\times\operatorname{id}_{S}\right)\left(\operatorname{id}_{S}\times
r\right)\left(r\times\operatorname{id}_{S}\right)=\left(\operatorname{id}_{S}\times
r\right)\left(r\times\operatorname{id}_{S}\right)\left(\operatorname{id}_{S}\times
r\right).$

Writing
$r\left(x,y\right)=\left(\lambda_{x}\left(y\right),\rho_{y}\left(x\right)\right)$,
with $\lambda_{x},\rho_{x}$ maps from $S$ into itself, then $r$ is _left
(resp. right) non-degenerate_ if $\lambda_{x}\in\operatorname{Sym}_{S}$ (resp.
$\rho_{x}\in\operatorname{Sym}_{S}$), for every $x\in S$, _non-degenerate_ if
it is both left and right non-degenerate.
Successful strategies for studying solutions consist of investigating
algebraic structures strictly related to them since their properties can be
described in terms of such structures. Among these, braces [21] and skew
braces [14] are the most studied. A recent generalization is the _weak (left)
brace_ [6], that is a triple $\left(S,+,\circ\right)$ such that
$\left(S,+\right)$ and $\left(S,\circ\right)$ are inverse semigroups and the
identities

$\displaystyle a\circ\left(b+c\right)=a\circ b-a+a\circ c\qquad\&\qquad a\circ
a^{-}=-a+a$

are satisfied, for all $a,b,c\in S$, where $-a$ and $a^{-}$ denote the
inverses of $a$ with respect to $+$ and $\circ$, respectively. Clearly, the
sets of the idempotents $\operatorname{E}(S,+)$ and
$\operatorname{E}(S,\circ)$ coincide. In particular, $(S,+)$ is a Clifford
semigroup, and if $(S,\circ)$ also is, $S$ is called _dual weak brace_.
Evidently, skew braces are dual weak braces since $(S,+)$ and $(S,\circ)$ are
groups having the same identity. Moreover, if $(S,+)$ is abelian, then $S$ is
a brace. Any weak brace $(S,+,\circ)$ gives rise to a solution $r:S\times S\to
S\times S$ defined by

$\displaystyle r\left(a,b\right)=\left(-a+a\circ b,\ \left(-a+a\circ
b\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$ (see [6, Theorem 11]). In the particular case of skew
braces, such a map $r$ is bijective and non-degenerate. Moreover, $r$ is
involutive, i.e., $r^{2}=\operatorname{id}_{S\times S}$, if and only if $S$ is
a brace. Up to here, it was not known that there were other solutions that can
be systematically constructed from skew braces. The paper [11] has shed new
light on the study of solutions in that sense. Indeed, the authors proved the
existence of several solutions coming from skew braces that can be obtained by
deforming the map $r$. In their work, one can also find the first hint that
the two-sided skew braces are crucial in such an investigation. Recall that a
skew brace $(S,+,\circ)$ is _two-sided_ if $\left(a+b\right)\circ c=a\circ
c-c+b\circ c$ holds, for all $a,b,c\in S$ (see [9, Definition 2.15]).

In this paper, we describe this class of deformed solutions directly in the
context of dual weak braces. To this end, in the third section, we introduce
and investigate the _right distributor_ of a dual weak brace $(S,+,\circ)$,
namely

$\displaystyle\mathcal{D}_{r}(S)=\\{z\in S\,\mid\,\forall\,a,b\in
S\quad(a+b)\circ z=a\circ z-z+b\circ z\\},$

that we prove to be an inverse subsemigroup of $\left(S,\circ\right)$ such
that $\operatorname{E}(S)\subseteq\mathcal{D}_{r}(S)$. Clearly, if $S$ is two-
sided, then $\mathcal{D}_{r}(S)=S$. The main result is contained in 2.9, where
we show that, fixed $z\in S$, the map $r_{z}:S\times S\to S\times S$ given by

$\displaystyle r_{z}(a,b)=\left(-a\circ z+a\circ b\circ z,\ \left(-a\circ
z+a\circ b\circ z\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$, is a solution if and only if $z\in\mathcal{D}_{r}(S)$. We
call the map $r_{z}$ _solution associated to $S$ deformed by $z$_. In such a
case, $r_{z}$ is not bijective in general, but it has a behavior close to
bijectivity and non-degeneracy, as we show in more detail in see 2.11. If $S$
is a skew brace and $z=0$ is the identity of the group $(S,\circ)$, then
$r_{z}$ coincides with the usual solution $r$. However, although any
idempotent determines a deformed solution in any dual weak brace, in general,
the map $r_{e}$, with $e\in\operatorname{E}(S)$, does not coincide with $r$.
More precisely, $r_{e}$ and $r$ are not equivalent in the sense of [13]. In
this regard, we raise the issue of studying under which conditions on the
parameters of two deformed solutions are equivalent and we give partial
answers in this sense. In the particular cases of two-sided skew braces, we
show that if two parameters $z$ and $w$ are in the same conjugacy classes,
then $r_{z}$ and $r_{w}$ are equivalent.
In the last section, we provide a method to obtain solutions by extending
solutions from skew braces to unital near-trusses. In 2018, Brzeziński [2]
introduced _trusses_ to describe brace and ring distributive laws in one
algebraic structure. A truss is a particular type of heap $(T,[-,-,-])$ (see
[1, 20]), together with an additional binary operation $\cdot:T\times T\to T$,
denoted by juxtaposition, such that $(T,\cdot)$ is a semigroup and the
following distributivity laws hold

$\displaystyle a[b,c,d]=[ab,ac,ad]\qquad\&\qquad[b,c,d]a=[ba,ca,da],$

for all $a,b,c,d\in T$. A few years later, authors of [4] generalized the
notion of truss to that of _near-truss_ , namely a heap in which only left
distributivity holds, with the aim of including structures of near-rings and
skew braces. In particular, if $(T,[-,-,-],\cdot)$ is a near-truss such that
$(T,\cdot)$ is a group, then $T$ is a skew brace. Here, we focus on _unital
near-trusses_ , that is, near-trusses for which $(T,\cdot)$ is a monoid with
the aim of introducing a method to construct a solution on a unital near-
truss. The biggest obstacle is the lack of invertibility of elements for the
binary operation. To overcome this, we consider a projection $\pi:T\to B$ of a
unital near-truss onto a skew brace. To assign an inverse of an invertible
element to every element of a near-truss, we assume the existence of
$\gamma:B\to T$ such that $\pi\gamma=\mathrm{id}_{B}$. Having that, in 4.6, we
construct a not-necessarily bijective solution on $T$. We obtain a sort of
“gluing” of the solution associated to a skew brace and the solution
$r(a,b)=(1,ab)$ on a monoid with identity $1$. The constructions introduced in
[21], generalised in [14] and further in [11], are a case for when $(T,\cdot)$
is a group and a projection $\pi=\mathrm{id}_{T}$ (see 4.8). In the end, in
4.12, we are showing conditions for the solution to be coming from the product
of a unital near-truss and a skew brace.

##  1 Preliminaries

###  1.1 Basics on dual weak braces

This section is devoted to introducing the structure of the weak brace and its
properties useful for our treatment.

Initially, for the ease of the reader, let us briefly recall some useful
notions on inverse semigroups theory (see [10, 15, 17, 19], for more details).
A semigroup $S$ is called _inverse semigroup_ if, for each $a\in S$, there
exists a unique element $a^{-1}$ of $S$ such that $a=aa^{-1}a$ and
$a^{-1}=a^{-1}aa^{-1}$, called the _inverse_ of $a$. The behaviour of inverse
elements in an inverse semigroup $S$ is similar to that in a group, since
$(ab)^{-1}=b^{-1}a^{-1}$ and $(a^{-1})^{-1}=a$, for all $a,b\in S$. Denoted by
$\operatorname{E}(S)$ the set of the idempotents of $S$, clearly, $e=e^{-1}$,
for every $e\in\operatorname{E}(S)$, and the idempotents of $S$ are exactly
the elements $aa^{-1}$ and $a^{-1}a$, for any $a\in S$. An inverse semigroup
such that $aa^{-1}=a^{-1}a$, for every $a\in S$, is named _Clifford
semigroup_. Equivalently, a Clifford semigroup $S$ is an inverse semigroup in
which the idempotents are central or, according to [15, Theorem 4.2.1], it is
a strong (lower) semilattice $Y$ of disjoint groups.

Below, we recall the definition of weak brace and dual weak brace contained in
[6, Definition 5] and in [8, Definition 2].

######  Definition 1.1.

Let $S$ be a set endowed with two binary operations $+$ and $\circ$ such that
$\left(S,+\right)$ and $\left(S,\circ\right)$ are inverse semigroups. Then,
$\left(S,+,\circ\right)$ is said to be a _weak (left) brace_ if the following
relations

$\displaystyle a\circ\left(b+c\right)=a\circ b-a+a\circ c\qquad\&\qquad a\circ
a^{-}=-a+a$

are satisfied, for all $a,b,c\in S$, where $-a$ and $a^{-}$ denote the
inverses of $a$ with respect to $+$ and $\circ$, respectively. Moreover, a
weak brace $\left(S,+,\circ\right)$ is said to be a _dual weak brace_ if
$\left(S,\circ\right)$ is a Clifford semigroup.

In any weak brace the sets of idempotents $E\left(S,+\right)$ and
$E\left(S,\circ\right)$ coincide, thus we simply denote them by
$E\left(S\right)$. More in general, any idempotent
$e\in\operatorname{E}\left(S\right)$ satisfies the following

$\displaystyle e+a=e\circ a=\lambda_{e}\left(a\right),$ (1)

for every $a\in S$.

As proved in [6, Theorem 8], the additive structure of any weak brace is
necessarily a Clifford semigroup. An example of a weak brace that is not dual
is contained in [6, Example 2-3.].
Clearly, _skew braces_ [14] are instances of weak braces, since the additive
and the multiplicative structures are groups having the same identities.
Besides, _braces_ [21] are skew braces in which the additive group is abelian.
Moreover, weak braces are _inverse semi-braces_ [7] in which $(S,+)$ is an
inverse semigroup and $a\circ\left(a^{-}+b\right)=-a+a\circ b$, for all
$a,b\in S$ (see [6, Proposition 16]). Easy examples of dual weak braces can be
obtained starting from any Clifford semigroup $\left(S,\circ\right)$, by
setting $a+b:=a\circ b$ or $a+b:=b\circ a$, for all $a,b\in S$. These are the
_trivial weak brace_ and the _almost trivial weak brace_ , respectively.

######  Theorem 1.2.

[6, Theorem 11] Let $\left(S,+,\circ\right)$ be a weak brace. Then, the map
$r:S\times S\to S\times S$ defined by

$\displaystyle r\left(a,b\right)=\left(-a+a\circ b,\ \left(-a+a\circ
b\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$, is a solution.

Such a map $r$ has a behavior close to bijectivity since there exists the
solution $r^{op}$ associated to the _opposite weak brace_
$\left(S,+^{op},\circ\right)$ of $S$, where $a+^{op}b:=b+a$, for all $a,b\in
S$, such that

$\displaystyle r\,r^{op}\,r=r,\qquad r^{op}\,r\,r^{op}=r^{op},\qquad\&\qquad
rr^{op}=r^{op}r,$

namely, $r$ is a completely regular element in $\operatorname{Map}(S\times
S)$. In particular, if $S$ is a skew brace, then $r^{op}=r^{-1}$, see [16]. We
are used to denote the components of $r$ by introducing the maps
$\lambda_{a},\rho_{b}:S\to S$ defined by

$\displaystyle\lambda_{a}\left(b\right)=-a+a\circ
b\qquad\&\qquad\rho_{b}\left(a\right)=\lambda_{a}\left(b\right)^{-}\circ
a\circ b,$

for all $a,b\in S$. The components of the map $r^{op}$ are given by

$\lambda^{op}_{a}\left(b\right)=a\circ
b-a=\left(\rho_{a^{-}}\left(b^{-}\right)\right)^{-}\quad\&\quad\rho^{op}_{b}\left(a\right)=\left(a\circ
b-a\right)^{-}\circ a\circ
b=\left(\lambda_{b^{-}}\left(a^{-}\right)\right)^{-}.$

If $S$ is a dual weak brace, the map $r$ also is close to being non-
degenerate, since

$\displaystyle\lambda_{a}\lambda_{a^{-}}\lambda_{a}=\lambda_{a},\qquad\lambda_{a^{-}}\lambda_{a}\lambda_{a^{-}}=\lambda_{a^{-}},\qquad$
$\displaystyle\&\qquad\lambda_{a}\lambda_{a^{-}}=\lambda_{a^{-}}\lambda_{a},$
$\displaystyle\rho_{a}\rho_{a^{-}}\rho_{a}=\rho_{a},\qquad\rho_{a^{-}}\rho_{a}\rho_{a^{-}}=\rho_{a^{-}},\qquad$
$\displaystyle\&\qquad\rho_{a}\rho_{a^{-}}=\rho_{a^{-}}\rho_{a},$

for every $a\in S$. Clearly, if $S$ is a skew brace, such maps are bijective.
By [6, Lemma 3], it holds that
$\lambda_{a}\left(b\right)\circ\rho_{b}\left(a\right)=a\circ b$, for all
$a,b\in S$. In addition, one has that the map
$\lambda:S\to\operatorname{End}\left(S,+\right),a\mapsto\lambda_{a}$ is a
homomorphism of the inverse semigroup $\left(S,\circ\right)$ into the
endomorphism semigroup of $\left(S,+\right)$ and the map
$\rho:S\to\operatorname{Map}(S),b\mapsto\rho_{b}$ is a semigroup anti-
homomorphism of the inverse semigroup $\left(S,\circ\right)$ to the monoid
$\operatorname{Map}(S)$ of the maps from $S$ into itself.

The following are instances of dual weak braces obtained as strong
semilattices of braces according to the construction in [8, Proposition 5].

######  Examples 1.

  1. 1.

Let $(B,+,\circ)$ be a brace and $\\{I_{\alpha}\\}_{\alpha\in\mathbb{N}}$ a
family of ideals (see [14, Definition 2.1]) such that $I_{\alpha}\subseteq
I_{\alpha+1}$, for every $\alpha$, and $I_{0}=\\{0\\}$. Then, considering the
following sequence of canonical projections of braces

$B\xrightarrow{\pi_{1}}B/I_{1}\xrightarrow{\pi_{2}}\cdots\xrightarrow{\pi_{\alpha}}B/I_{\alpha}\xrightarrow{\pi_{\alpha+1}}B/I_{\alpha+1}\xrightarrow{\pi_{\alpha+2}}\cdots,$

we obtain that the strong semilattice $S$ of braces
$[\mathbb{N},B/I_{\alpha},\pi_{\alpha}]$ is a dual weak brace.

  2. 2.

Let us consider the following sequence of rings for some fixed
$n\in\mathbb{N}$:

$\mathbb{Z}/2^{n}\mathbb{Z}\xrightarrow{\pi_{n-1}}\mathbb{Z}/2^{n-1}\mathbb{Z}\xrightarrow{\pi_{n-2}}\cdots\xrightarrow{\pi_{2}}\mathbb{Z}/2\mathbb{Z}.$

This induces the following sequence of braces

$U\left(\mathbb{Z}/2^{n}\mathbb{Z}\right)\xrightarrow{\pi^{\prime}_{n-1}}U\left(\mathbb{Z}/2^{n-1}\mathbb{Z}\right)\xrightarrow{\pi^{\prime}_{n-2}}\cdots\xrightarrow{\pi^{\prime}_{2}}U\left(\mathbb{Z}/2\mathbb{Z}\right),$

where $U(R)$ denotes the set of units of a ring $R$,
$\pi^{\prime}_{i}=\pi_{i|_{U(\mathbb{Z}/2^{i+1}\mathbb{Z})}}$, with addition
defined by $a+_{1}b:=a-1+b\pmod{2}^{i}$, for any $i\in\\{1,\cdots,n\\}$ and
for all $a,b\in U\left(\mathbb{Z}/2^{i}\mathbb{Z}\right)$, and multiplication
given by modulo $2^{i}$. Moreover, $\pi^{\prime}_{i}$ is a homomorphism of
braces, so the strong semilattice $S$ of braces $[\\{1,\ldots,n\\},\
\mathbb{Z}/2^{i}\mathbb{Z},\ \pi^{\prime}_{i}]$ is a dual weak brace.

In the following lemma, we collect some properties contained in [6, 8] that we
will use throughout the paper.

######  Lemma 1.3.

Let $(S,+,\circ)$ be a weak brace. Then, the following hold:

  1. 1.

$a\circ b=a+\lambda_{a}(b)$,

  2. 2.

$a+b=a\circ\lambda_{a^{-}}\left(b\right)$,

  3. 3.

$\lambda_{a}(b)=a\circ b\circ\rho_{b}(a)^{-}$,

  4. 4.

$a\circ(-b)=a-a\circ b+a$,

for all $a,b\in S$.

###  1.2 Basics on near-trusses

This section is devoted to giving some preliminaries on the structures of the
near-trusses useful for the last section.

Near-trusses are heaps with an additional binary operation, thus let us start
with introducing heaps by referring to [1, 20].

######  Definition 1.4.

A pair $(H,[-,-,-])$ with $H$ a set and $[-,-,-]:H\times H\times H\to H$ a
ternary operation on $H$ is a heap if, for all $a,b,c,d,e\in H$,

  1. 1.

$[a,b,[c,d,e]]=[[a,b,c],d,e],$

  2. 2.

$[a,a,b]=b=[b,a,a]$

are satisfied.

######  Remark 1.5.

In a heap $(H,[-,-,-])$ the following equality

$[a,b,[c,d,e]]=[a,[d,c,b],e]=[[a,b,c],d,e]$ (2)

holds, for all $a,b,c,d,e\in H$.

There is a deep link between heaps and groups, as we explain below.

######  Remark 1.6.

  1. 1.

If $H$ is a heap, $e\in H$, and $+_{e}:=[-,e,-]:H\times H\to H$ , then
$\left(H,+_{e}\right)$ is a group called an $e$-retract of $H$. For all
$e,f\in H$, $\left(H,+_{e}\right)$ and $\left(H,+_{f}\right)$ are isomorphic
as groups.

  2. 2.

If $(G,\cdot)$ is a group, then $\mathcal{H}(G):=\left(G,[-,-,-]\right)$ is a
heap, where $[a,b,c]:=a\cdot b^{-1}\cdot c$, for all $a,b,c\in G$. Moreover,
for any heap $H$, $\mathcal{H}(H,+_{e})=H$ and, for any $e\in\mathcal{H}(G)$,
$\left(\mathcal{H}(G),+_{e}\right)$ and $(G,\cdot)$ are isomorphic groups.

######  Examples 2.

  1. 1.

Considered the set $\mathrm{Odd}(\mathbb{Z}):=\\{2n+1\ |\ n\in\mathbb{Z}\\}$,
one can define the ternary operation on $\mathrm{Odd}(\mathbb{Z})$, given by,
for all $a,b,c\in\mathrm{Odd}(\mathbb{Z})$, $[a,b,c]:=a-b+c$, where $+$ is the
addition of integers. Then, the pair $(\mathrm{Odd}(\mathbb{Z}),[-,-,-])$ is a
heap.

  2. 2.

Similarly, consider the set
$\mathrm{Odd}(\mathbb{Q}):=\biggl{\\{}\dfrac{2n+1}{2m+1}\ |\
m,n\in\mathbb{Z}\biggr{\\}}$ and define the ternary operation as in 1., where
here $+$ is the addition of rationals. Then, the pair
$(\mathrm{Odd}(\mathbb{Q}),[-,-,-])$ is a heap.

######  Definition 1.7.

[4, Definition 3.1] A near-truss is a heap $(T,[-,-,-])$ together with an
associative binary operation $\cdot:T\times T\to T$ such that

$a\cdot[b,c,d]=[a\cdot b,a\cdot c,a\cdot d],$

for all $a,b,c,d\in T$. If there exists $1_{T}\in T$ such that $1_{T}\cdot
a=a\cdot 1_{T}=a$, for any $a\in T$, we say that $T$ is a _unital near-truss_.
We will write $ab:=a\cdot b$, for all $a,b\in T$.

######  Example 3.

The heaps $\mathrm{Odd}(\mathbb{Z})$ and $\mathrm{Odd}(\mathbb{Q})$ in 2 are
trusses with binary operation given by multiplication of integers and
rationals, respectively.

######  Remark 1.8.

[11, Lemma 2.1] Let $T$ be a unital near-truss, then $[b,c,d]a=[ba,ca,da]$ if
and only if $[b,1,c]a=[ba,a,ca]$, for all $a,b,c,d\in T$.

The following links between near-trusses and various algebraic structures can
be found in [4, Section 3].

######  Examples 4.

  1. 1.

Let $(B,+,\circ)$ be a skew brace. Then,
$\mathrm{T}(B)=\left(B,[-,-,-],\circ\right)$ is a near-truss, where
$[a,b,c]:=a-b+c$, for all $a,b,c\in B$.

  2. 2.

Let $(R,+,\cdot)$ be a (left) near-ring. Then,
$\mathrm{T}(R)=\left(R,[-,-,-],\cdot\right)$ is a near-truss, where
$[a,b,c]:=a-b+c$, for all $a,b,c\in R$.

######  Lemma 1.9.

Let $T$ be a near-truss such that $(T,\cdot)$ is a group with neutral element
$1$. Then, the triple $(T,+_{1},\cdot)$ is a skew brace, with
$a+_{1}b:=[a,1,b]$, for all $a,b\in T$.

Following [4, p. 689], we recall that a map $f:T\to S$ between two near-
trusses is called a near-truss homomorphism if

$f\left([a,b,c]\right)=[f(a),f(b),f(c)]\qquad\&\qquad
f\left(ab\right)=f(a)f(b),$

for all $a,b,c\in T$. If $T$ and $S$ are unital near-trusses and
$f(1_{T})=1_{S}$, then we say that $f$ is a homomorphism of unital near-
trusses.

######  Remark 1.10.

[3, Lemma 3.4] If $f:T\to S$ is a homomorphism of near-trusses, then for any
$a\in S$, the set

$\mathrm{ker}_{a}(f):=\\{b\in T\ |\ f(b)=a\\}$

is a subheap. Moreover, if $a\in\operatorname{E}(S,\cdot)$, then
$\mathrm{ker}_{a}(f)$ is a subnear-truss of $T$.

##  2 Deformed solutions on dual weak braces

This section aims to describe deformed solutions associated to any dual weak
brace, namely solutions obtained by deforming the classical ones. These novel
solutions have been introduced for the first time in the context of skew
braces in [11].

######  Theorem 2.1.

[11, Theorem 2.4] Let $(B,+,\circ)$ be a skew brace and $z\in S$ such that

$\displaystyle\left(a-b+c\right)\circ z=a\circ z-b\circ z+c\circ z,$ (3)

for all $a,b,c\in B$. Then, the map $\check{r}_{z}:B\times B\to B\times B$
given by

$\displaystyle\check{r}_{z}(a,b)=\left(a\circ b-a\circ z+z,\ \left(a\circ
b-a\circ z+z\right)^{-}\circ a\circ b\right),$

is a non-degenerate and bijective solution, called _deformed solution by $z$
on $B$_.

We denote the components of $\check{r}_{z}$ by introducing the maps

$\check{\sigma}^{z}_{a}(b)=a\circ b-a\circ
z+z\qquad\&\qquad\check{\tau}^{z}_{b}(a)=\left(a\circ b-a\circ
z+z\right)^{-}\circ a\circ b,$

for all $a,b\in B$.

######  Remark 2.2.

Let $(B,+,\circ)$ be a skew brace. Note that $\check{r}_{0}$ coincides with
the inverse solution of the solution $r$ associated to $B$, namely
$\check{r}_{0}=r^{op}$. In general, if $z\in B$ satisfies (3), one can check
that $\check{r}_{z}^{-1}$ is the map $r_{z^{-}}:B\times B\to B\times B$ given
by

$\displaystyle r_{z^{-}}(a,b)=\left(-a\circ z^{-}+a\circ b\circ
z^{-},\left(-a\circ z^{-}+a\circ b\circ z^{-}\right)^{-}\circ a\circ
b\right),$

for all $a,b\in B$. Such a map $r_{z^{-}}$ clearly is non-degenerate as
$\check{r}_{z}$ is. Indeed, for all $a,b\in B$, if we consider the maps
$\sigma_{a}^{z},\tau_{b}^{z}:B\to B$ defined by

$\displaystyle\sigma_{a}^{z}(b)=-a\circ z+a\circ b\circ
z\qquad\&\qquad\tau_{b}^{z}(a)=\left(-a\circ z+a\circ b\circ z\right)^{-}\circ
a\circ b,$

the components of $r_{z^{-}}$ are $\sigma_{a}^{z^{-}}(b)$ and
$\tau_{b}^{z^{-}}(a)$, respectively, and are such that

$\displaystyle\left(\sigma_{a}^{z^{-}}\right)^{-1}(b)$
$\displaystyle=\sigma_{a^{-}}^{z}(b)=\left(\check{\tau}_{a}^{z}\left(b^{-}\right)\right)^{-}$
$\displaystyle\left(\tau^{z^{-}}_{b}\right)^{-1}(a)$
$\displaystyle=\tau_{b^{-}}^{z^{-}}(a)=\left(\check{\sigma}_{b}^{z^{-}}\left(a^{-}\right)\right)^{-},$

for all $a,b\in B$.

Throughout our work, we lay the groundwork to prove that the set of elements
$z$ which gives rise to a deformed solution is a subgroup of $(B,\circ)$. This
fact will allow studying the map

$\displaystyle r_{z}(a,b)=\left(-a\circ z+a\circ b\circ z,\ \left(-a\circ
z+a\circ b\circ z\right)^{-}\circ a\circ b\right),$

avoiding the use of $z^{-}$, that in the case $z=0$ exactly coincides with the
usual solution $r$ associated to $B$. For this reason, hereinafter, we will
study such a map $r_{z}$ directly in the context of a dual weak brace
$(S,+,\circ)$ and prove that it is a solution.
Note that using (3), it can be also written as

$\displaystyle r_{z}\left(a,b\right)=\left(z^{-}\circ\lambda_{z\circ
a}\left(b\right)\circ z,\ z^{-}\circ\rho_{b}\left(z\circ a\right)\right),$
($\star$)

for all $a,b\in S$. Below, we provide an identity that is equivalent to the
relation (3).

######  Lemma 2.3.

Let $(S,+,\circ)$ be a dual weak brace and $z\in S$. Then, (3) is equivalent
to

$\displaystyle\left(a+b\right)\circ z=a\circ z-z+b\circ z,$ ($D$)

for all $a,b\in S$.

###### Proof.

The identity ($D$) can be trivially obtained by (3) and (1) taking $b=z\circ
z^{-}$. Conversely, let $a,b,c\in S$. Firstly, if $x\in S$, we get

$\displaystyle\left(-x\right)\circ z=z-x\circ z+z.$ (4)

Indeed, $x\circ x^{-}\circ z=\left(x-x\right)\circ z=x\circ
z-z+\left(-x\right)\circ z$, thus

$\left(-x\right)\circ z\underset{\eqref{pro_dual_idemp}}{=}z-x\circ z+x\circ
z-z+\left(-x\right)\circ z=z-x\circ z+x\circ x^{-}\circ z=z-x\circ z+z.$

Now, applying (4) and (1), we obtain

$\displaystyle\left(a-b+c\right)\circ z$ $\displaystyle=a\circ z-z+(-b+c)\circ
z=a\circ z-z+(-b)\circ z-z+c\circ z$ $\displaystyle=a\circ z-b\circ z+c\circ
z,$

i.e., (3) is satisfied. ∎

######  Remark 2.4.

Observe that if $(S,+,\circ)$ is a dual weak brace and $z\in S$, then ($D$) is
also equivalent to the equality

$\displaystyle(a+b)\circ z=a\circ z+\left(z^{-}+b\right)\circ z,$
($D^{\prime}$)

for all $a,b\in S$. Indeed, if $b\in S$ and ($D$) is satisfied, then

$\left(z^{-}+b\right)\circ z=z^{-}\circ z-z+b\circ z=-z+b\circ z,$

and so ($D^{\prime}$) holds. Conversely, if ($D^{\prime}$) is satisfied we
have that

$\displaystyle\left(z^{-}+b\right)\circ z=-z+\left(z^{-}\circ z\right)\circ
z+\left(z^{-}+b\right)\circ z=-z+\left(z^{-}\circ z+b\right)\circ
z\underset{\eqref{pro_dual_idemp}}{=}-z+b\circ z,$

i.e., ($D$) holds.

In light of Lemma 2.3, we introduce the following set.

######  Definition 2.5.

Let $(S,+,\circ)$ be a dual weak brace. Then, we call the set

$\displaystyle\mathcal{D}_{r}(S)=\\{z\in S\,\mid\,\forall\,a,b\in
S\quad(a+b)\circ z=a\circ z-z+b\circ z\\},$

_right distributor_ of $S$.

It immediately follows by (1) that
$\operatorname{E}(S)\subseteq\mathcal{D}_{r}(S)$. Moreover, if $(B,+,\circ)$
is a _two-sided skew brace_ , that is, $\left(a+b\right)\circ c=a\circ
c-c+b\circ c$ holds, for all $a,b,c\in B$ (see [9, Definition 2.15]), then
$\mathcal{D}_{r}(B)$ coincides with all $B$. We can give an analogous
definition in the case of dual weak braces, by calling also such structures
_two-sided_.

######  Remark 2.6.

A dual weak brace $(S,+,\circ)$ is two-sided if and only if
$\mathcal{D}_{r}(S)=S$.

We aim to show that the map $r_{z}$ is a solution in any dual brace if and
only if $z\in\mathcal{D}_{r}(S)$. In the next section, we will deepen the
algebraic structure of $\mathcal{D}_{r}(S)$. To prove the main result of this
section, we need the following preliminary lemmas.

######  Lemma 2.7.

Let $(S,+,\circ)$ be a dual weak brace and $z\in\mathcal{D}_{r}(S)$. Then,
they hold:

  1. 1.

$\sigma^{z}_{a}\left(b\right)\circ\tau^{z}_{b}(a)=a\circ b\circ z\circ z^{-}$,

  2. 2.

$\tau^{z}:(S,\circ)\to\operatorname{Map}(S)$ is an anti-homomorphism,

  3. 3.

$\sigma_{a}^{z}(b)=a\circ b\circ b^{-}\circ\left(a^{-}\circ
z^{-}+b\right)\circ z$,

  4. 4.

$\sigma_{a}^{z}(b)\circ\sigma_{a}^{z}(b)^{-}=a\circ a^{-}+b\circ b^{-}+z\circ
z^{-}$,

for all $a,b\in S$.

###### Proof.

Let $a,b,c\in S$. Initially, by ($\star$ ‣ 2) and Lemma 2.3, we have that

$\sigma^{z}_{a}\left(b\right)\circ\tau^{z}_{b}(a)=z^{-}\circ\lambda_{z\circ
a}\left(b\right)\circ\rho_{b}\left(z\circ a\right)=z^{-}\circ z\circ a\circ
b.$

Moreover,

$\displaystyle\tau^{z}_{b\circ c}\left(a\right)$
$\displaystyle=z^{-}\circ\rho_{b\circ c}\left(z\circ
a\right)=z^{-}\circ\rho_{c}\rho_{b}\left(z\circ
a\right)=z^{-}\circ\rho_{c}\left(z\circ\tau^{z}_{b}\left(a\right)\right)=\tau^{z}_{c}\tau^{z}_{b}\left(a\right).$

Furthermore, since, by (1), $\lambda_{a}(b)=a\circ b\circ
b^{-}\circ\left(a^{-}+b\right)$, we obtain

$\sigma_{a}^{z}(b)=z^{-}\circ z\circ a\circ b\circ b^{-}\circ\left(a^{-}\circ
z^{-}+b\right)\circ z=a\circ b\circ b^{-}\circ\left(a^{-}\circ
z^{-}+b\right)\circ z.$

Finally,

$\displaystyle\sigma_{a}^{z}(b)\circ\sigma_{a}^{z}(b)^{-}$
$\displaystyle=z^{-}\circ\lambda_{z\circ a}\left(b\right)\circ\lambda_{z\circ
a}\left(b\right)^{-}\circ z$ $\displaystyle=z^{-}\circ z\circ
a\circ\left(a^{-}\circ z^{-}+b\right)\circ\left(a^{-}\circ
z^{-}+b\right)^{-}\circ a^{-}\circ z^{-}\circ z$ $\displaystyle=a\circ
a^{-}\circ\left(a^{-}\circ z^{-}+b-b-a^{-}\circ z^{-}\right)\circ z\circ
z^{-}$ by (1) $\displaystyle=a\circ a^{-}+b\circ b^{-}+z\circ z^{-}$ by (1)

which completes the proof. ∎

######  Lemma 2.8.

Given a set $X$, a function $r:X\times X\to X\times X,\
(a,b)\mapsto(\sigma_{a}(b),\tau_{b}(a))$ is a solution if and only if the
following three fundamental equalities hold

$\displaystyle\sigma_{a}\sigma_{b}(c)=\sigma_{\sigma_{a}\left(b\right)}\sigma_{\tau_{b}\left(a\right)}\left(c\right),$
(Y1)
$\displaystyle\sigma_{\tau_{\sigma_{b}\left(c\right)}\left(a\right)}\tau_{c}\left(b\right)=\tau_{\sigma_{\tau_{b}\left(a\right)}\left(c\right)}\sigma_{a}\left(b\right),$
(Y2)
$\displaystyle\tau_{c}\tau_{b}(a)=\tau_{\tau_{c}\left(b\right)}\tau_{\sigma_{b}\left(c\right)}\left(a\right),$
(Y3)

for all $a,b,c\in X$.

######  Theorem 2.9.

Let $S$ be a dual weak brace and $z\in S$. Then, the map $r_{z}:S\times S\to
S\times S$ given by

$\displaystyle r_{z}(a,b)=\left(-a\circ z+a\circ b\circ z,\left(-a\circ
z+a\circ b\circ z\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$, is a solution if and only if $z\in\mathcal{D}_{r}(S)$. We
call such a map $r_{z}$ _solution associated to $S$ deformed by $z$_.

###### Proof.

Let $a,b,c\in S$. To prove (Y1), we observe that

$\displaystyle\sigma^{z}_{a}\sigma^{z}_{b}(c)$ $\displaystyle=-a\circ
z+\left(a-a\circ b\circ z+a\circ b\circ c\circ z\right)\circ z$ by Lemma 1.3-4
$\displaystyle=-a\circ z+a\circ z-z+\left(-a\circ b\circ z+a\circ b\circ
c\circ z\right)\circ z$ $z\in\mathcal{D}_{r}(S)$ $\displaystyle=a\circ
a^{-}-z+\sigma^{z}_{a\circ b}(c)\circ z$ by (1) $\displaystyle=a\circ
a^{-}+b\circ b^{-}-z+\sigma^{z}_{a\circ b}(c)\circ z$ by (1)

and

$\displaystyle\sigma^{z}_{\sigma^{z}_{a}\left(b\right)}\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)$
$\displaystyle=-\sigma^{z}_{a}\left(b\right)\circ
z+\left(\sigma^{z}_{a}\left(b\right)-a\circ b\circ z+a\circ b\circ c\circ
z\right)\circ z$ by Lemma 1.3-4 & Lemma 2.7-1
$\displaystyle=-\sigma^{z}_{a}\left(b\right)\circ
z+\sigma^{z}_{a}\left(b\right)\circ z-z+\sigma^{z}_{a\circ b}(c)\circ z$
$z\in\mathcal{D}_{r}(S)$
$\displaystyle=\sigma^{z}_{a}\left(b\right)\circ\sigma^{z}_{a}\left(b\right)^{-}-z+\sigma^{z}_{a\circ
b}(c)\circ z$ by (1) $\displaystyle=a\circ a^{-}+b\circ
b^{-}-z+\sigma^{z}_{a\circ b}(c)\circ z$ by Lemma 2.7-4

Besides, (Y3) follows by

$\displaystyle\tau^{z}_{\tau^{z}_{c}\left(b\right)}\tau^{z}_{\sigma^{z}_{b}\left(c\right)}\left(a\right)$
$\displaystyle=\tau^{z}_{\sigma^{z}_{b}\left(c\right)\circ\tau^{z}_{c}\left(b\right)}\left(a\right)$
by Lemma 2.7-2 $\displaystyle=\tau^{z}_{z^{-}\circ z\circ b\circ
c}\left(a\right)$ by Lemma 2.7-1 $\displaystyle=z^{-}\circ\rho_{b\circ
c}\rho_{z^{-}\circ z}\left(z\circ a\right)$
$\displaystyle=z^{-}\circ\rho_{b\circ c}\left(z\circ a\right)$ by (1)
$\displaystyle=z^{-}\circ\rho_{c}\rho_{b}\left(z\circ a\right)$
$\displaystyle=z^{-}\circ\rho_{c}\left(z\circ\tau^{z}_{b}\left(a\right)\right)$
$\displaystyle=\tau^{z}_{c}\tau^{z}_{b}\left(a\right).$

Furthermore,

$\displaystyle\sigma^{z}_{\tau^{z}_{\sigma^{z}_{b}\left(c\right)}\left(a\right)}\tau^{z}_{c}\left(b\right)$
$\displaystyle=\tau^{z}_{\sigma^{z}_{b}\left(c\right)}\left(a\right)\circ\tau^{z}_{c}(b)\circ\tau^{z}_{c}(b)^{-}\circ\left(\tau^{z}_{\sigma^{z}_{b}\left(c\right)}\left(a\right)^{-}\circ
z^{-}+\tau^{z}_{c}(b)\right)\circ z$ by Lemma 2.7-3
$\displaystyle=\left(\sigma^{z}_{a}\sigma^{z}_{b}(c)\right)^{-}\circ
a\circ\sigma^{z}_{b}(c)\circ\tau^{z}_{c}(b)\circ\left(\tau^{z}_{\tau^{z}_{c}\left(b\right)}\tau^{z}_{\sigma^{z}_{b}\left(c\right)}\left(a\right)\right)^{-}$
by Lemma 2.7-3
$\displaystyle=\left(\sigma^{z}_{\sigma^{z}_{a}\left(b\right)}\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)\right)^{-}\circ
a\circ\sigma^{z}_{b}(c)\circ\tau^{z}_{c}(b)\circ\left(\tau^{z}_{c}\tau^{z}_{b}\left(a\right)\right)^{-}$
by (Y1)-(Y3)
$\displaystyle=\left(\sigma^{z}_{\sigma^{z}_{a}\left(b\right)}\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)\right)^{-}\circ
a\circ b\circ c\circ z^{-}\circ
z\circ\left(\tau^{z}_{c}\tau^{z}_{b}\left(a\right)\right)^{-}$ by Lemma 2.7-1
$\displaystyle=\left(\sigma^{z}_{\sigma^{z}_{a}\left(b\right)}\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)\right)^{-}\circ\sigma^{z}_{a}(b)\circ\tau^{z}_{b}(a)\circ
c\circ\left(\tau^{z}_{c}\tau^{z}_{b}\left(a\right)\right)^{-}$ by Lemma 2.7-1
$\displaystyle=\left(\sigma^{z}_{\sigma^{z}_{a}\left(b\right)}\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)\right)^{-}\circ\sigma^{z}_{a}(b)\circ\tau^{z}_{b}(a)\circ
c\circ c^{-}\circ\left(\tau^{z}_{b}(a)^{-}\circ z^{-}+c\right)\circ z$
$\displaystyle=\left(\sigma^{z}_{\sigma^{z}_{a}\left(b\right)}\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)\right)^{-}\circ\sigma^{z}_{a}(b)\circ\sigma^{z}_{\tau^{z}_{b}(a)}(c)$
by Lemma 2.7-3
$\displaystyle=\tau^{z}_{\sigma^{z}_{\tau^{z}_{b}\left(a\right)}\left(c\right)}\sigma^{z}_{a}\left(b\right),$

i.e., (Y2) holds. Therefore, $r_{z}$ is a solution. Finally, we show that (Y1)
implies ($D$), namely $z\in\mathcal{D}_{r}(S)$. Indeed, if $a,b\in S$, by
choosing $x=a$, $y=a^{-}\circ a$, and $w=a^{-}\circ(a\circ z+b)\circ z^{-}$,
we obtain

$\displaystyle\sigma^{z}_{x}\sigma^{z}_{y}(w)$ $\displaystyle=-x\circ
z+x\circ(-y\circ z+y\circ w\circ z)\circ z$ $\displaystyle=-a\circ
z+a\circ\left(-z+a\circ a^{-}+a^{-}\circ(a\circ z+b)\circ z^{-}\circ
z\right)\circ z$ by (1) $\displaystyle=-a\circ z+a\circ\left(-z+a^{-}\circ
a\circ z-a^{-}+a^{-}\circ b\right)\circ z$ by (1) $\displaystyle=-a\circ
z+a\circ\lambda_{a^{-}}(b)\circ z$ by (1) $\displaystyle=-a\circ z+(a+b)\circ
z$ by Lemma 1.3-2

and, using Lemma 2.7-1, Lemma 1.3-4, and (1), we get

$\displaystyle\sigma^{z}_{\sigma^{z}_{x}\left(y\right)}\sigma^{z}_{\tau^{z}_{w}\left(a\right)}\left(w\right)$
$\displaystyle=-\sigma^{z}_{x}\left(y\right)\circ
z+\left(\sigma^{z}_{x}\left(y\right)-x\circ y\circ z+x\circ y\circ w\circ
z\right)\circ z$ $\displaystyle=-\left(-a\circ z+a\circ z\right)\circ
z+\left(\left(-a\circ z+a\circ z\right)-a\circ z+a\circ a^{-}\circ(a\circ
z+b)\circ z^{-}\circ z\right)\circ z$ $\displaystyle=-\left(a\circ a^{-}\circ
z\circ z^{-}\right)\circ z+\left(-a\circ z+a\circ z-a\circ a^{-}+a\circ
a^{-}\circ b\right)\circ z$ $\displaystyle=-z+\left(a\circ a^{-}\circ z\circ
z^{-}+b\right)\circ z$ $\displaystyle=-z+a\circ a^{-}+b\circ z.$

Hence, by (Y1) it follows that

$\displaystyle-a\circ z+(a+b)\circ z=-z+a\circ a^{-}+b\circ z$ (5)

and so, by (1), we can write

$\displaystyle(a+b)\circ z=a\circ z-a\circ z+(a+b)\circ
z\underset{\eqref{key}}{=}a\circ z-z+a\circ a^{-}+b\circ z=a\circ z-z+b\circ
z,$

i.e., $z\in\mathcal{D}_{r}(S)$. Therefore, we get the claim. ∎

######  Corollary 2.10.

Let $(B,+,\circ)$ be a skew brace and $z\in B$. Then, $r_{z}$ is a solution by
$z$ if and only if $z\in\mathcal{D}_{r}(B)$. We call such a map $r_{z}$ the
_solution associated to $B$ deformed by $z$_.

The following theorem illustrates some properties of a deformed solution
$r_{z}$ on a dual weak brace $(S,+,\circ)$. In particular, $r_{Z}$ has a
behavior close to bijectivity and non-degeneracy since it is completely
regular in $\operatorname{Map}(S\times S)$.

######  Theorem 2.11.

Let $(S,+,\circ)$ be a dual weak brace, $z\in\mathcal{D}_{r}(S)$, and $r_{z}$
the solution associated to $S$ deformed by $z$. Then, considered the map
$\check{r}_{z^{-}}:S\times S\to S\times S$ given by

$\displaystyle\check{r}_{z^{-}}(a,b)=\left(a\circ b-a\circ z^{-}+z^{-},\
\left(a\circ b-a\circ z^{-}+z^{-}\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$, the following hold

$\displaystyle
r_{z}\,\check{r}_{z^{-}}\,r_{z}=r_{z},\qquad\check{r}_{z^{-}}\,r_{z}\,\check{r}_{z^{-}}=\check{r}_{z^{-}},\qquad\text{and}\qquad
r_{z}\check{r}_{z^{-}}=\check{r}_{z^{-}}r_{z}.$

Moreover, $\sigma^{z}_{a}$ and $\tau_{a}^{z}$ are completely regular elements
in $\operatorname{Map}(S)$, since

$\displaystyle\sigma_{a}^{z}\sigma_{a^{-}}^{z^{-}}\sigma_{a}^{z}=\sigma_{a}^{z},\qquad\sigma_{a^{-}}^{z^{-}}\sigma_{a}^{z}\sigma_{a^{-}}^{z^{-}}=\sigma_{a^{-}}^{z^{-}},\qquad$
$\displaystyle\&\qquad\sigma_{a}^{z}\sigma_{a^{-}}^{z^{-}}=\sigma_{a^{-}}^{z^{-}}\sigma_{a}^{z}$
$\displaystyle\tau_{a}^{z}\tau^{z}_{a^{-}}\tau_{a}^{z}=\tau_{a}^{z},\qquad\tau^{z}_{a^{-}}\tau_{a}^{z}\tau^{z}_{a^{-}}=\tau^{z}_{a^{-}},\qquad$
$\displaystyle\&\qquad\tau^{z}_{a}\tau_{a^{-}}^{z}=\tau_{a^{-}}^{z}\tau_{a}^{z},$

for every $a\in S$.

###### Proof.

In this proof, we will set for brevity $x^{0}:=x\circ x^{-}=x-x$, for every
$x\in S$.
Initially, if $a,b\in S$, the first component of $r_{z}\check{r}_{z^{-}}(a,b)$
is equal to

$\displaystyle X:=-\left(a\circ b-a\circ z^{-}+z^{-}\right)\circ
z+\left(a\circ b-a\circ z^{-}+z^{-}\right)^{0}\circ a\circ b\circ z$
$\displaystyle=-\left(a\circ b\circ z-z+z-a\circ z^{0}\right)+\left(a\circ
b-a\circ z^{-}+z^{-}\right)^{0}+a\circ b\circ z$ by Lemma 2.3
$\displaystyle=a+z^{0}-a\circ b\circ z+(a\circ b-a\circ z^{-}+z^{0}+a\circ
z^{-}-a\circ b)+a\circ b\circ z$ by (1) $\displaystyle=a-a\circ b\circ
z+(a\circ b)^{0}+(a\circ z)^{0}+a\circ b\circ z$ by (1)
$\displaystyle=a+b^{0}+z^{0}$ by (1)

and the first component of $\check{r}_{z^{-}}r_{z}(a,b)$ is equal to

$\displaystyle Y:=\left(-a\circ z+a\circ b\circ z\right)^{0}\circ a\circ
b-\left(-a\circ z+a\circ b\circ z\right)\circ z^{-}+z^{-}$
$\displaystyle=\left(-a\circ z+a\circ b\circ z\right)^{0}+a\circ
b-\left(z^{-}-a\circ z\circ z^{-}+a\circ b\right)\circ z^{0}+z^{-}$ by
($D$)-(4) $\displaystyle=(-a\circ z+a\circ b\circ z-a\circ b\circ z+a\circ
z)+(a\circ b)^{0}+a+z^{0}$ by (1) $\displaystyle=a+b^{0}+z^{0}$ by (1)

thus they are equal. Since the second components of
$r_{z}\check{r}_{z^{-}}(a,b)$ and $\check{r}_{z^{-}}r_{z}(a,b)$ are equal to
$X^{-}\circ a\circ b$ and $Y^{-}\circ a\circ b$, respectively, and $X=Y$, it
follows that $r_{z}\check{r}_{z^{-}}=\check{r}_{z^{-}}r_{z}$. Moreover, by the
previous paragraph and by (1), we compute

$\displaystyle r_{z}\,\check{r}_{z^{-}}\,r_{z}(a,b)$
$\displaystyle=r_{z}\left(Y,Y^{-}\circ a\circ b\right)=\left(-Y\circ z+Y\circ
Y^{-}\circ a\circ b\circ z\right)=r_{z}(a,b)$

and, by (1),

$\displaystyle\check{r}_{z^{-}}\,r_{z}\,\check{r}_{z^{-}}(a,b)=\check{r}_{z^{-}}\left(X,X^{-}\circ
a\circ b\right)$ $\displaystyle=\check{r}_{z^{-}}\left(X\circ X^{-}\circ
a\circ b-X\circ z^{-}+z^{-},\left(X\circ X^{-}\circ a\circ b-X\circ
z^{-}+z^{-}\right)^{-}\circ X\circ X^{-}\circ a\circ b\right)$
$\displaystyle=\check{r}_{z^{-}}(a,b).$

Furthermore, it is easy to check that
$\sigma_{a}^{z}\sigma_{a^{-}}^{z^{-}}(b)=\sigma_{a^{-}}^{z^{-}}\sigma_{a}^{z}(b)=a^{0}\circ
z^{0}\circ b$ and so they follow
$\sigma_{a}^{z}\sigma_{a^{-}}^{z^{-}}\sigma_{a}^{z}(b)=\sigma_{a}^{z}(b)$ and
$\sigma_{a^{-}}^{z^{-}}\sigma_{a}^{z}\sigma_{a^{-}}^{z^{-}}=\sigma_{a^{-}}^{z^{-}}(b)$.
The rest of the claim is a consequence of Lemma 2.7-2. ∎

Observe that if $(B,+,\circ)$ is a skew brace and $a\circ z=z+a$, for every
$a\in B$, then $\sigma^{z}_{a}=\lambda_{a}$, for every $a\in B$ (cf. [11,
Lemma 2.10]). This is equivalent to requiring that the map $\sigma^{z}$ is a
homomorphism, as we show next in the more general case.

######  Proposition 2.12.

Let $(S,+,\circ)$ be a dual weak brace and $z\in S$. Then, the function
$\sigma^{z}:\left(S,\circ\right)\to\operatorname{Map}(S)$ is a homomorphism if
and only if $a\circ z=z+a$, for every $a\in S$.

###### Proof.

If $a\in S$ and $\sigma^{z}$ is a homomorphism, by
$\sigma^{z}_{z^{-}}\sigma^{z}_{z^{-}\circ z}(a)=\sigma^{z}_{z^{-}\circ
z^{-}\circ z}(a)$, using (1), we get $z^{-}\circ(-z+a\circ z)\circ
z=z^{-}\circ a\circ z$. Thus, by the last identity and (1), we obtain

$\displaystyle z+a$ $\displaystyle=z+z\circ\left(z^{-}\circ a\circ
z\right)\circ z^{-}=z+z\circ z^{-}\circ(-z+a\circ z)\circ z\circ z^{-}$
$\displaystyle=z-z+a\circ z=a\circ z.$

Conversely, if $a,b\in S$, we have

$\displaystyle\sigma_{a}^{z}\sigma_{b}^{z}(c)$
$\displaystyle=-a-z+z+a\circ\left(-b-z+z+b\circ c\right)$
$\displaystyle=-a-z+z+a-a\circ b+a-a+a\circ b\circ c$ by (1)
$\displaystyle=-a\circ b\circ z+a\circ b\circ c\circ z$
$\displaystyle=\sigma_{a\circ b}^{z}(c).$

Therefore, the claim follows. ∎

In the case of a dual weak brace $S$, even if
$\sigma^{z}:\left(S,\circ\right)\to\operatorname{Map}(S)$ is a homomorphism,
in general, $\sigma_{a}^{z}$ does not coincide with $\lambda_{a}$, since
$\sigma_{a}^{z}(b)=\lambda_{a}(b)+z\circ z^{-}$. In the study of deformed
solutions, the following question arises.

######  Question 1.

Let $(S,+,\circ)$ be a dual weak brace. For which parameters $z,w\in S$, are
the deformed solutions $r_{z}$ and $r_{w}$ equivalent?

We recall that two solutions $r$ and $s$ on two sets $S$ and $T$,
respectively, are said to be _equivalent_ if there exists a bijective map
$\varphi:S\to T$ such that $(\varphi\times\varphi)r=s(\varphi\times\varphi)$
(see [13]). In direction of 1, in the context of skew braces, in [11, Example
2.14] and [11, Example 2.15] one can find instances of different parameters
which give rise to non-equivalent deformed solutions. Here, the following
example shows that in the case of dual weak braces, even deformed solutions by
idempotents are not equivalent in general.

######  Example 5.

Let $X=\\{e,x,y\\}$ and $\left(S,\circ\right)$ the commutative inverse monoid
on $X$ with identity $e$ satisfying the relations $x\circ x=y\circ y=x$ and
$x\circ y=y$. Note that $a^{-}=a$, for every $a\in S$. Consider the trivial
weak braces on $S$, namely $a+b=a\circ b$, for all $a,b\in S$. Then, by 2.9,
we have two solutions $r_{e}=r$ and $r_{x}$ related to the two idempotents $e$
and $x$, respectively, for which the maps $\sigma^{e}$ and $\sigma^{x}$ are
explicitly given by
$\sigma^{e}_{a}\left(b\right)=\lambda_{a}\left(b\right)+e=\lambda_{a}(b)=a\circ
a\circ b$ and
$\sigma^{x}_{a}\left(b\right)=\lambda_{a}\left(b\right)+x=-a+a\circ b+x=a\circ
a\circ b\circ x.$ If the two solutions $r_{x}$ and $r_{e}$ were equivalent via
a bijection $\varphi:S\to S$, then, in particular, we would have that
$\varphi\left(a\circ a\circ b\circ
x\right)=\varphi\left(a\right)\circ\varphi\left(a\right)\circ\varphi\left(b\right)$,
for all $a,b\in S$. Thus, if $a=b=e$ we have that
$\varphi\left(x\right)=\varphi\left(e\right)\circ\varphi\left(e\right)\circ\varphi\left(e\right)=\varphi\left(e\right)$,
a contradiction.

Observe that if $z,w\in S$ give rise to two deformed solutions $r_{z}$ and
$r_{w}$, respectively, and there exists
$\varphi\in\operatorname{Aut}\left(S,+,\circ\right)$ such that
$\varphi\left(z\right)=w$, then $r_{z}$ and $r_{w}$ are trivially equivalent
via $\varphi$. In the special case of a two-sided skew brace, such a map
$\varphi$ exists when $z$ and $w$ are in the same conjugacy class, as we show
in the next result.

######  Proposition 2.13.

Let $\left(B,+,\circ\right)$ be a two-sided skew brace and $z,w\in B$
belonging to the same conjugacy class in $\left(B,\circ\right)$. Then, the
deformed solutions $r_{z}$ and $r_{w}$ are equivalent.

###### Proof.

As a consequence of 2.6, $r_{z}$ and $r_{w}$ are deformed solutions on $B$. By
[23, Proposition 2.3] and [18, Lemma 4.1], all the inner automorphisms of
$(B,\circ)$ are skew brace automorphisms of $B$. By the assumption, there
exist $c\in B$ such that $w=c^{-}\circ z\circ c$, thus $r_{z}$ and $r_{w}$ are
equivalent via the inner automorphism $\varphi_{c}$ given by
$\varphi_{c}\left(a\right)=c^{-}\circ a\circ c$, for any $a\in B$. In
particular,
$\left(\varphi_{c}\times\varphi_{c}\right)r_{z}=r_{w}\left(\varphi_{c}\times\varphi_{c}\right)$.
∎

Note that 2.13 could is also true in the context of dual weak braces whenever
the map $\varphi_{c}$ is bijective.

######  Remark 2.14.

Note that the converse of 2.13 is not true. To show this, it is enough to
consider the trivial brace $(B,+,+)$ on the cyclic group
$\mathbb{Z}/2\mathbb{Z}$. Then, the solution $r_{0}$ coincides with the
solution $r_{1}$ but $0$ and $1$ trivially belong to different conjugacy
classes.

Motivated by 1, in the next section, we deepen the right distributor
$\mathcal{D}_{r}(S)$ of any dual weak brace $S$, focusing mainly on the case
of skew braces.

##  3 The right distributor of a skew brace

In this section, we study the right distributor of any skew brace
$(B,+,\circ)$, i.e., the set $\mathcal{D}_{r}(B)=\\{z\in
B\,\mid\,\forall\,a,b\in B\quad(a+b)\circ z=a\circ z-z+b\circ z\\}$, that is a
subgroup of $(B,\circ)$.

According to 2.6, $\mathcal{D}_{r}(B)=B$ if and only if $B$ is a two-sided
skew brace. The other limit case is when there exists only the trivial
deformation, in other words, $\mathcal{D}_{r}(B)=\\{0\\}$. We give some
examples below.

######  Examples 6.

  1. 1.

Let $\left(B,+,\circ\right)$ be the brace on $\left(\mathbb{Z},+\right)$ with
$a\circ b=a+\left(-1\right)^{a}b$, for all $a,b\in\mathbb{Z}$, cf. [22,
Proposition 6]. Then, $\mathcal{D}_{r}\left(B\right)=\\{0\\}$ (it is enough to
choose $a,b\in\mathbb{Z}$ both odd).

  2. 2.

Let $\left(B,+,\circ\right)$ be the brace on
$\left(\mathbb{Z}/6\mathbb{Z},+\right)$ with $a\circ
b=a+\left(-1\right)^{a}b$, for all $a,b\in\mathbb{Z}/6\mathbb{Z}$, see [21,
Example 3]. Similarly to $1.$, we obtain
$\mathcal{D}_{r}\left(B\right)=\\{0\\}$.

  3. 3.

Let $\left(B,+,\circ\right)$ be the brace on
$\left(\mathbb{Z},\circ\right)=\left<\,g\,\right>$ with
$g^{k}+g^{l}=g^{k+\left(-1\right)^{k}l}$, for all $k,l\in\mathbb{Z}$. Then, if
$k,l,z\in\mathbb{Z}$, since

$\displaystyle\left(g^{k}+g^{l}\right)\circ
g^{z}=g^{k+z+\left(-1\right)^{k}l}\quad\ \&\quad g^{k}\circ
g^{z}-g^{z}+g^{l}\circ g^{z}=g^{k+z+\left(-1\right)^{k}\left(l+2z\right)}$

it follows that $\mathcal{D}_{r}\left(B\right)$ is trivial.

  4. 4.

If $\left(B,\circ\right)=\left<\,g\,\right>$ is the cyclic group of order $m$,
with $m$ even, then by defining the operation of $+$ as in $3.$, we obtain a
skew brace such that $D_{r}\left(B\right)$ is trivial.

The following is an example of a skew brace in which $\mathcal{D}_{r}(B)$ is
not trivial.

######  Example 7.

Let us consider the skew brace $(B,+,\circ)$ from 6-1 and the brace
$U_{9}:=\left(U\left(\mathbb{Z}/2^{9}\mathbb{Z}\right),+_{1},\circ\right)$,
with $a+_{1}b=a-1+b$ and $a\circ b=ab$, for all $a,b\in
U(\mathbb{Z}/2^{9}\mathbb{Z})$, with the usual operations in the ring modulo
$2^{9}$. Then, $U_{9}\times B$ is a left skew brace such that
$\mathcal{D}_{r}\left(U_{9}\times B\right)=U_{9}\times\\{0\\}$.

More generally, one can prove the following results related to the right
distributor of any dual weak brace. Let us first recall that an inverse
subsemigroup $I$ of an inverse semigroup $S$ is _full_ if
$\operatorname{E}(S)\subseteq I$ (see [17, p. 19]).

######  Proposition 3.1.

Let $(S,+,\circ)$ be a dual weak brace. Then, $\mathcal{D}_{r}(S)$ is a full
inverse subsemigroup of the Clifford semigroup $(S,\circ)$ containing the
center $\zeta(S,\circ)$ of $(S,\circ)$.

###### Proof.

Initially, if $e\in E\left(S\right)$, by (1) we have that
$\left(a+b\right)\circ e=a+e+b+e=a\circ e-e+b\circ e,$ for all $a,b\in S$, and
so $e\in\mathcal{D}_{r}(S)$. Moreover, if $a,b\in S$ and
$z_{1},z_{2}\in\mathcal{D}_{r}(S)$, by ($D^{\prime}$), we get

$\displaystyle(a+b)\circ(z_{1}\circ z_{2})$ $\displaystyle=\left(a\circ
z_{1}+\left(z_{1}^{-}+b\right)\circ z_{1}\right)\circ z_{2}$
$\displaystyle=a\circ z_{1}\circ
z_{2}+\left(z_{2}^{-}+\left(z_{1}^{-}+b\right)\circ z_{1}\right)\circ z_{2}$
$\displaystyle=a\circ z_{1}\circ z_{2}+\left(z_{2}^{-}\circ z_{1}^{-}\circ
z_{1}+\left(z_{1}^{-}+b\right)\circ z_{1}\right)\circ z_{2}$ by (1)
$\displaystyle=a\circ z_{1}\circ z_{2}+\left(\left(z_{1}\circ
z_{2}\right)^{-}+b\right)\circ z_{1}\circ z_{2},$

namely $z_{1}\circ z_{2}\in\mathcal{D}_{r}(S)$. Besides, by (1),

$\displaystyle(a+b)\circ z_{1}^{-}$ $\displaystyle=\left(a\circ z_{1}^{-}\circ
z_{1}+\left(z_{1}^{-}-z_{1}^{-}+b\circ z_{1}^{-}\right)\circ z_{1}\right)\circ
z_{1}^{-}$ $\displaystyle\underset{\eqref{cond_zmeno}}{=}\left(a\circ
z_{1}^{-}-z_{1}^{-}+b\circ z_{1}^{-}\right)\circ z_{1}\circ z_{1}^{-}$
$\displaystyle=a\circ z_{1}^{-}-z_{1}^{-}+b\circ z_{1}^{-},$

i.e., $z_{1}^{-}\in\mathcal{D}_{r}(S)$. Besides, if $z\in\zeta(S,\circ)$, then
$(a+b)\circ z=z\circ a-z+z\circ b=a\circ z-z+b\circ z,$ for all $a,b\in S$,
i.e., $z\in\mathcal{D}_{r}(S)$. Therefore, the claim follows. ∎

######  Corollary 3.2.

Let $(B,+,\circ)$ be a skew brace. Then, $\mathcal{D}_{r}(B)$ is a subgroup of
$(B,\circ)$ containing the center $\zeta(B,\circ)$ of the group $(B,\circ)$.

In general, $\mathcal{D}_{r}(S)$ is not an inverse subsemigroup of the
additive semigroup, unless we get into particular cases.

######  Proposition 3.3.

Let $(S,+,\circ)$ be a dual weak brace in which the Clifford semigroup $(S,+)$
is commutative, then $\mathcal{D}_{r}(S)$ is a two-sided dual weak subbrace of
$S$.

###### Proof.

By 3.1, it is enough to show that $\mathcal{D}_{r}(S)$ is an inverse
subsemigroup of $(S,+)$. Clearly,
$\operatorname{E}(S,+)\subseteq\mathcal{D}_{r}(S)$ because
$\operatorname{E}(S,\circ)=\operatorname{E}(S,+)$. Moreover, if
$x,y\in\mathcal{D}_{r}(S)$ and $a,b\in S$, we have

$\displaystyle(a+b)\circ(x+y)$ $\displaystyle=a\circ x-x+b\circ x-b-a+a\circ
y-y+b\circ y$ $\displaystyle=a\circ x-a+a\circ y-(x+y)+b\circ x-b+b\circ y$
$\displaystyle=a\circ(x+y)-(x+y)+b\circ(x+y),$

and thus $x+y\in\mathcal{D}_{r}(S)$. Now, by Lemma 1.3-4,

$\displaystyle(a+b)\circ(-x)$ $\displaystyle=a+b-(a+b)\circ x+a+b=a+b-a\circ
x+x-b\circ x+a+b$ $\displaystyle=a-a\circ x+a+x+b-b\circ
x+b=a\circ(-x)-(-x)+b\circ(-x),$

and so $-x\in\mathcal{D}_{r}(S)$. Therefore, the claim follows. ∎

######  Corollary 3.4.

Let $(B,+,\circ)$ be a brace, then $\mathcal{D}_{r}(B)$ is a two-sided
subbrace of $B$.

######  Remark 3.5.

Observe that if $(S,+,\circ)$ is a dual weak brace and
$\lambda_{z}\left(\mathcal{D}_{r}(S)\right)\subseteq\mathcal{D}_{r}(S)$, for
every $z\in\mathcal{D}_{r}(S)$, then $\mathcal{D}_{r}(S)$ is an inverse
subsemigroup of the additive Clifford semigroup $(S,+)$. Indeed, it is enough
to observe that if $z,w\in\mathcal{D}_{r}(S)$, then
$z+w=z\circ\lambda_{z^{-}}\left(w\right)$ and that
$-z=\lambda_{z}\left(z^{-}\right)$.

It becomes natural to wonder when $\mathcal{D}_{r}(B)$ is an ideal of a skew
brace $(B,+,\circ)$. Let us recall that a subset $I$ of $B$ is an _ideal_ of
$B$ if it is both a normal subgroup of $(B,+)$ and $(B,\circ)$ and $I$ is
_$\lambda$ -invariant_, namely $\lambda_{a}(I)\subseteq I$, for every $a\in B$
(see [14, Definition 2.1]). According to [4, Lemma 3.16], $I$ is an ideal of
$B$ if and only if $(I,+)$ is a normal subgroup of $(B,+)$, $I$ is
$\lambda$-invariant and $\gamma$-invariant, where $\gamma:B\to B$ is a map
such that $\gamma_{a}(b)=b\circ a-a$, for all $a,b\in B$. Consequently, we
have the following.

######  Proposition 3.6.

Let $(B,+,\circ)$ be a skew brace. Then, $I$ is an ideal of $B$ if and only if

  1. 1.

$(I,+)$ is normal subgroup,

  2. 2.

$a\circ I=a+I$ and $I\circ a=a+I,$ for any $a\in B$.

The next general result provides a sufficient condition for obtaining an ideal
starting from a normal subsemigroup of the multiplicative structure
$(B,\circ)$.

######  Proposition 3.7.

Let $(B,+,\circ)$ be a skew brace and $(I,\circ)$ a $\lambda$-invariant and
$\lambda^{op}$-invariant normal subgroup of $(B,\circ)$. Then, $(I,+)$ is a
normal subgroup of $(B,+)$.

###### Proof.

Initially, if $i,j\in I$, then, by Lemma 1.3, $i-j=i\circ\lambda_{i\circ
j}\left(j^{-}\right)\in I$. Moreover, if $a\in B$ and $i\in I$, then
$a+i-a=a\circ\lambda_{a^{-}}(i)-a=\lambda^{op}_{a}\left(\lambda_{a^{-}}(i)\right)\in
I.$ Thus $a+I=I+a$, for any $a\in B$, and $(I,+)$ is a normal subgroup of
$(B,+)$. ∎

######  Remark 3.8.

Let $(B,+,\circ)$ be a skew brace and $(I,+)$ a normal subgroup of $(B,+)$.
Then, for all $a\in B$ and $i\in I$, $\lambda^{op}_{a}(i)\in I$ if and only if
$\lambda_{a}(i)\in I$.

######  Example 8.

Let $U_{9}\times B$ the skew brace in 7. Then, by 3.7,
$\mathcal{D}_{r}(U_{9}\times B)$ is an ideal.

In $(B,+,\circ)$ is a skew brace, the group $\mathcal{D}_{r}(B)$ can be
related to $\operatorname{Fix}(B)$ and to its annihilator
$\operatorname{Ann}\left(B\right)$. According to [9, Definition 1.5],

$\displaystyle\operatorname{Fix}\left(B\right)=\\{a\in B\ |\ \forall\ x\in
B\quad\lambda_{x}\left(a\right)=a\\}$

and it is a left ideal of $B$, i.e., $(\operatorname{Fix}\left(B\right),+)$ is
a subgroup of $(B,+)$ and it is $\lambda$-invariant. Moreover, according to
[5, Definition 7], the _annihilator_ of $B$ is an ideal of $B$ and it is
defined by

$\displaystyle\operatorname{Ann}\left(B\right)=\operatorname{Soc}\left(B\right)\cap\zeta\left(B,\circ\right)$

where $\operatorname{Soc}\left(B\right)=\\{a\in B\,\mid\,\forall\ b\in B\quad
a+b=a\circ b\quad\&\quad\ a+b=b+a\\}$.

######  Proposition 3.9.

Let $(B,+,\circ)$ be a skew brace. Then, the following hold:

  1. 1.

$\operatorname{Fix}(B)\subseteq\mathcal{D}_{r}(B)$;

  2. 2.

$\operatorname{Ann}(B)\subseteq\mathcal{D}_{r}(B)$.

###### Proof.

$1.$ Let $z\in\operatorname{Fix}(B)$, then $x\circ z=x+z$, for every $x\in B$.
Hence, for all $a,b\in B$,

$a\circ z-z+b\circ z=a+z-z+b+z=a+b+z=(a+b)\circ z$

and so $z\in\mathcal{D}_{r}(B)$, thus
$\operatorname{Fix}(B)\subseteq\mathcal{D}_{r}(B).$
$2.$ Since $\operatorname{Ann}(B)\subseteq\zeta(B,\circ)$, by 3.2,
$\operatorname{Ann}(B)\subseteq\mathcal{D}_{r}(B)$. ∎

##  4 Deformed solutions and unital near-trusses

In this section, we will focus on a new way to obtain solutions by extending
solutions from skew braces to unital near-trusses. To achieve it we will focus
on the sigma map which coincides with the approach contained in [11].

First let us fix an arbitrary near-truss $T$, then we can define the map
$\check{\sigma}:T\times T\times T\to T$ given by

$\check{\sigma}_{a}^{z}(b):=[ab,\,az,\,z],$

for all $a,b,z\in T$. Observe that in the case of skew braces consideration of
particular $\check{\sigma}^{z}$ maps leads to new solutions. A similar
approach cannot be taken here as the image of $\check{\sigma}^{z}$ is not
necessarily an invertible element. Although the map still has some interesting
properties which we list below since they are essential for the main theorem
of this section.

######  Lemma 4.1.

Let $T$ be a near-truss and $z\in T$. Then, the following properties of
$\check{\sigma}^{z}$ hold, for all $a,b,c,d\in T$:

  1. 1.

$a\,b=[\check{\sigma}^{z}_{a}(b),z,az],$

  2. 2.

$\check{\sigma}^{z}_{a}(z)=z,$

  3. 3.

if $T$ is unital, then $\check{\sigma}^{z}_{1_{T}}(a)=a,$

  4. 4.

$\check{\sigma}^{z}_{a}\left([b,c,d]\right)=[\check{\sigma}^{z}_{a}(b),\check{\sigma}^{z}_{a}(c),\check{\sigma}^{z}_{a}(d)]$,

  5. 5.

$\check{\sigma}^{z}_{a}\,\check{\sigma}^{z}_{b}(c)=\check{\sigma}^{z}_{a\,b}(c)$,

  6. 6.

for any homomorphism of near-trusses $f:T\to S$, it holds
$f\check{\sigma}^{z}_{a}(b)=\check{\sigma}^{f(z)}_{f(a)}\,f(b).$

###### Proof.

Initially, note that 1., 2. and 3. easily follows from the definition of the
map $\check{\sigma}^{z}$.
4. If $a,b,c,d\in T$, then we get

$\displaystyle\check{\sigma}_{a}^{z}\left([b,c,d]\right)$
$\displaystyle=\left[\left[[ab,az,[z,z,az]],ac,ad\right],az,z\right]$
$\displaystyle=[[[ab,az,z],z,az],ac,[ad,az,z]]$
$\displaystyle=[[ab,az,z],[ac,az,z],[ad,az,z]]$ by (2)
$\displaystyle=[\check{\sigma}^{z}_{a}(b),\check{\sigma}^{z}_{a}(c),\check{\sigma}^{z}_{a}(d)].$

5. Moreover, if $a,b,c\in T$, $\check{\sigma}^{z}_{a}\check{\sigma}^{z}_{b}(c)=[[abc,abz,az],az,z]=[abc,abz,z]=\check{\sigma}^{z}_{a\,b}(c)$.
6. If $f:T\to S$ is a near-truss homomorphism and $a,b\in T$, then

$f\check{\sigma}^{z}_{a}(b)=[f(a)f(b),f(a)f(z),f(z)]=\check{\sigma}^{f(z)}_{f(a)}\,f(b),$

which completes the proof. ∎

######  Proposition 4.2.

Let $T$ be a near-truss and $z\in T$. Then $a\in T$ is left cancellative if
and only if $\check{\sigma}^{z}_{a}$ is injective. Moreover, if $T$ is unital
and $a\in T$ is invertible, then $\check{\sigma}^{z}_{a}$ is bijective.

###### Proof.

Let us assume that there exists $a\in T$ such that $\check{\sigma}^{z}_{a}$ is
injective. Then, for all $b,c\in T$ such that $ab=ac$, we get that

$[ab,az,z]=[ac,az,z]\implies\check{\sigma}^{z}_{a}(b)=\check{\sigma}^{z}_{a}(c)\implies
b=c,$

and thus $a$ is left cancellative. Conversely, if $a\in T$ is left
cancellative, then for all $a,b\in T$, we get that

$\check{\sigma}^{z}_{a}(b)=\check{\sigma}_{a}^{z}(c)\implies[ab,az,z]=[ac,az,z]\implies
ab=ac\implies b=c,$

thus $\check{\sigma}^{z}_{a}$ is injective. For the second statement, if $a\in
T$ is invertible, for any $c\in T$, we get
$\check{\sigma}^{z}_{a}\left[a^{-1}c,a^{-1}z,z\right]=[[c,z,az],az,z]=c.$
Therefore, $\check{\sigma}^{z}_{a}$ is a surjection, and, by the previous
considerations, it is a bijection. ∎

To introduce the solutions associated to near-trusses, let us consider the
following setup. Let $T$ be a near-truss and $\mathrm{T}(B)$ the near-truss
associated to a skew brace $(B,+,\circ)$ by taking the ternary operation
$[a,b,c]:=a-b+c$, for all $a,b,c\in B$ and $\cdot=\circ$. In this section, we
will denote by $1$ the identity of $B$. Moreover, let us consider the
following exact diagram of unital near-truss homomorphisms:

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
6.61632pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
25.58478pt\raise 7.5208pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 46.61632pt\raise
3.01385pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
46.61632pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathrm{T}(B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
25.7678pt\raise-8.20134pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\gamma}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
6.61633pt\raise-3.01385pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
91.43925pt\raise 5.15279pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.15279pt\hbox{$\scriptstyle{t}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 115.70314pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
115.70314pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\\{1\\}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$ (6)

In this case, by exactness we understand $t^{-1}(1)=\mathrm{Im}(\pi)$ and
$\pi\gamma=\operatorname{id}_{\mathrm{T}(B)}$, that is $\pi$ is surjective and
$\gamma$ is injective. Having this data we can consider the following map
$\check{\eta}:=\gamma\pi\check{\sigma}:T\times T\times T\to T$ given by

$\check{\eta}_{a}^{z}(b):=[\gamma\pi(a)\gamma\pi(b),\gamma\pi(a)\gamma\pi(z),\gamma\pi(z)],$

which will allow us to extend the solution of the skew brace $B$ into $T$.
Before we proceed further, let us present in the following lemma the basic
properties of $\check{\eta}$, the proof of which is omitted since it is an
easy consequence of Lemma 4.1 and the fact that $\gamma$ and $\pi$ are
homomorphisms of near-trusses.

######  Lemma 4.3.

Let $T$ be a near-truss and $z\in T$. Then, the map $\check{\eta}$ has the
following properties for all $a,b,c,d\in T$:

  1. 1.

if $\check{\eta}^{z}_{a}(z)=\gamma\pi(z)$, that is if $\gamma\pi(z)=z$, then
$\check{\eta}^{z}_{a}(z)=z,$

  2. 2.

if $T$ is unital, then $\check{\eta}^{z}_{1_{T}}(a)=\gamma\,\pi(a)$,

  3. 3.

$\check{\eta}^{z}_{a}{([b,c,d])}=[\check{\eta}^{z}_{a}(b),\check{\eta}^{z}_{a}(c),\check{\eta}^{z}_{a}(d)],$

  4. 4.

$\check{\eta}^{z}_{a}\,\check{\eta}^{z}_{b}(c)=\check{\eta}^{z}_{ab}(c),$

  5. 5.

if $T$ is unital, then
$\check{\eta}^{z}_{a}(b)\,\check{\eta}^{z}_{a}(b)^{-1}=1_{T}$, where
$\check{\eta}^{z}_{a}(b)^{-1}=\gamma\left(\left(\pi\,\check{\eta}^{z}_{a}(b)\right)^{-1}\right)$
and $\left(\pi\,\check{\eta}^{z}_{a}(b)\right)^{-1}$ is the inverse of the
element $\pi\,\check{\eta}^{z}(b)$ in $(B,\,\cdot)$.

######  Lemma 4.4.

Let $T$ be a unital near-truss and $z\in T$. Then, the following property is
satisfied by the map $\check{\eta}^{z}$

$\displaystyle
T_{1}(a,b,c):=\check{\eta}^{z}_{ab}(c)\check{\eta}^{z}_{\check{\eta}^{z}_{ab}(c)^{-1}a\check{\eta}^{z}_{b}(c)}(\check{\eta}^{z}_{b}(c)^{-1}bc)=\check{\eta}^{z}_{a}(b)\check{\eta}^{z}_{\check{\eta}^{z}_{a}(b)^{-1}ab}(c)=:T_{2}(a,b,c),$

for all $a,b,c\in T$.

###### Proof.

For brevity, for any $x\in T$, let us denote by $\overline{x}:=\gamma\pi(x)$.
So, we can write, for all $a,b\in T$,
$\check{\eta}^{z}_{a}(b)=\left[\overline{a}\overline{b},\
\overline{a}\overline{z},\ \overline{z}\right]$ and, since
$\pi\gamma=\operatorname{id}_{\mathrm{T}(B)}$,
$\overline{\check{\eta}^{z}_{a}(b)}=\check{\eta}^{z}_{a}(b)$. If $a,b,c\in T$,
we get

$\displaystyle T_{1}(a,b,c)$
$\displaystyle=\check{\eta}^{z}_{ab}(c)\left[\check{\eta}^{z}_{ab}(c)^{-1}\overline{a}\check{\eta}^{z}_{b}(c)\check{\eta}^{z}_{b}(c)^{-1}\overline{b}\overline{c},\
\check{\eta}^{z}_{ab}(c)^{-1}\overline{a}\check{\eta}^{z}_{b}(c)\overline{z},\
\overline{z}\right]$
$\displaystyle=\left[\overline{a}\overline{b}\overline{c},\
\overline{a}\check{\eta}^{z}_{b}(c)\overline{z},\
\check{\eta}^{z}_{ab}(c)\overline{z}\right]$
$\displaystyle=\left[\overline{a}\overline{b}\overline{c},\
\left[\overline{a}\overline{b}\overline{c}\overline{z},\
\overline{a}\overline{b}\overline{zz},\ \overline{a}\overline{zz}\right],\
\left[\overline{a}\overline{b}\overline{c}\overline{z},\overline{a}\overline{b}\overline{zz},\overline{zz}\right]\right]$
$\displaystyle=\left[\overline{a}\overline{b}\overline{c},\
\overline{a}\overline{zz},\overline{zz}\right]$ by (2)
$\displaystyle=\left[\overline{a}\overline{b}\overline{c},\
\overline{a}\overline{b}\overline{z},\
\left[\overline{a}\overline{b}\overline{z},\overline{a}\overline{z}\overline{z},\overline{z}\overline{z}\right]\right]$
$\displaystyle=\check{\eta}^{z}_{a}(b)\left[\check{\eta}^{z}_{a}(b)^{-1}\overline{a}\overline{b}\overline{c},\
\check{\eta}^{z}_{a}(b)^{-1}\overline{a}\overline{b}\overline{z},\
\overline{z}\right]$ $\displaystyle=T_{2}(a,b,c),$

and so the claim follows. ∎

######  Remark 4.5.

Let $T$ and $S$ be near-trusses and $f:T\to S$ a surjective homomorphism. If
there exists $z\in T$ such that $[a,b,c]z=[az,bz,cz]$, for all $a,b,c\in T$,
then

$[s,p,q]f(z)=[sf(z),pf(z),qf(z)],$

for all $s,p,q\in S$.

######  Theorem 4.6.

Let $T$ be a unital near-truss and $B$ a skew brace. Let $z\in T$ such that
$\pi(z)\in\mathcal{D}_{r}(B)$ and suppose there exist two homomorphisms of
unital near-trusses $\pi:T\to\mathrm{T}(B)$ and $\gamma:\mathrm{T}(B)\to T$
such that $\pi\gamma=\mathrm{id}_{B}$. Then, the map
$\check{\eta}:=\gamma\pi\check{\sigma}$ gives rise to the solution
$\check{r}^{z}:T\times T\to T\times T$ defined by

$\check{r}^{z}(a,b):=\left(\check{\eta}^{z}_{a}(b),\
\left(\check{\eta}^{z}_{a}(b)\right)^{-1}ab\right),$

for all $a,b\in T$.

###### Proof.

To denote the second component of $\check{r}^{z}$, we introduce the map
$\check{\varepsilon}:T\times T\times T\to T$ given by
$\check{\varepsilon}^{z}_{b}\left(a\right):=\left(\check{\eta}^{z}_{a}(b)\right)^{-1}ab$,
for all $a,b\in S$. To get the claim, we need to show the equalities
(Y1)-(Y3). Observe that (Y1) easily follows from the Lemma 4.3-4&5 and that
$\gamma\left(1_{B}\right)=1_{T}$. If $a,b,c\in T$, for (Y2), we have that

$\displaystyle\check{\varepsilon}^{z}_{\check{\eta}^{z}_{\check{\varepsilon}^{z}_{b}\left(a\right)}\left(c\right)}\check{\eta}^{z}_{a}\left(b\right)$
$\displaystyle=\left(\check{\eta}^{z}_{ab}(c)\right)^{-1}\check{\eta}^{z}_{a}(b)\
\check{\eta}^{z}_{\check{\varepsilon}^{z}_{b}(a)}(c)$ by Lemma 4.3-4
$\displaystyle=(\check{\eta}^{z}_{ab}(c))^{-1}T_{2}(a,b,c)=(\check{\eta}^{z}_{ab}(c))^{-1}T_{1}(a,b,c)$
by Lemma 4.4
$\displaystyle=(\check{\eta}^{z}_{ab}(c))^{-1}\check{\eta}^{z}_{ab}(c)\
\check{\eta}^{z}_{\check{\varepsilon}^{z}_{\check{\eta}^{z}_{b}(c)}(a)}(\check{\varepsilon}^{z}_{c}(b))$
$\displaystyle=\check{\eta}^{z}_{\check{\varepsilon}^{z}_{\check{\eta}^{z}_{b}(c)}(a)}\check{\varepsilon}^{z}_{c}(b)$

and, for (Y3), observe that

$\displaystyle\check{\varepsilon}^{z}_{\check{\varepsilon}^{z}_{c}(b)}\check{\varepsilon}^{z}_{\check{\eta}^{z}_{b}(c)}(a)$
$\displaystyle=\left(\check{\eta}^{z}_{\check{\varepsilon}^{z}_{\check{\eta}^{z}_{b}(c)}(a)}(\check{\varepsilon}^{z}_{c}(b))\right)^{-1}\check{\eta}^{z}_{ab}(c)^{-1}\
a\ \check{\eta}^{z}_{b}(c)\check{\eta}^{z}_{b}(c)^{-1}bc$
$\displaystyle=\left(\check{\eta}^{z}_{ab}(c)\eta^{z}_{\check{\varepsilon}^{z}_{\check{\eta}^{z}_{b}(c)}(a)}(\check{\varepsilon}^{z}_{c}(b))\right)^{-1}abc$
$\displaystyle=T_{1}(a,b,c)^{-1}abc=T_{2}(a,b,c)^{-1}abc$ by Lemma 4.4
$\displaystyle=\left(\check{\eta}^{z}_{a}(b)\check{\eta}^{z}_{\check{\varepsilon}^{z}_{b}(a)}(c)\right)^{-1}abc=\varepsilon^{z}_{c}\varepsilon^{z}_{b}(a).$

Therefore, $r^{z}$ is a solution on $T$. ∎

The construction from 4.6 gives new, non-necessarily bijective, solutions.

######  Corollary 4.7.

For every unital near-truss $T$, there exists the solution
$\check{r}^{1_{T}}(a,b)=\left(1_{T},ab\right),$ given by $B=\\{1\\}$,
$a\overset{\pi}{\mapsto}1$ and $\gamma(1)=1_{T}$.

######  Corollary 4.8.

The restriction of $\check{r}^{z}:T\times T\to T\times T$ to
$\check{r}^{z}|_{\gamma(B)\times\gamma(B)}$ gives an equivalent solution to
the solution $\check{r}^{\pi(z)}$ on $B$ via the bijection $\pi|_{\gamma(B)}.$

The following corollary states when the solution $\check{r}^{z}$ in 4.6 is
left non-degenerate.

######  Corollary 4.9.

Let $T$ be a unital near-truss and $a,z\in T$. The map
$\check{\eta}^{z}_{a}:T\to T$ is injective if and only if
$\pi:T\to\mathrm{T}(B)$ is a bijection.

###### Proof.

Assume that $\check{\eta}^{z}_{a}$ is an injection. Then, for all $b,c\in T$,
we get

$\gamma\pi(b)=\gamma\pi(c)\iff\check{\eta}^{z}_{a}({b})=\check{\eta}^{z}_{a}({c})\iff
b=c,$

where equivalence follows by the 4.2 and the fact that
$\check{\eta}^{z}_{a}=\check{\sigma}_{\gamma\pi(a)}^{\gamma\pi(z)}$. Hence,
$\gamma\pi$ is injective, and thus $\pi$ is injective. Indeed, let us assume
$\pi$ is not injective, then there exist $x\not=y\in T$ such that
$\pi(x)=\pi(y)$ and so $\gamma\pi(x)=\gamma\pi(y)$, that is a contradiction.
Since $\pi$ is surjective by definition, it is a bijection. Conversely, if
$\pi$ is injective, it is bijective with inverse $\gamma$ and so
$\check{\eta}^{z}_{a}=\check{\sigma}^{z}_{a}$. By 4.2, $\check{\eta}_{a}^{z}$
is a bijection. ∎

######  Corollary 4.10.

Let $T$ be a unital near-truss and $B$ a skew brace. Consider the unital near-
truss $\mathcal{T}:=\mathrm{T}(B)\times T$, $\pi:\mathcal{T}\to\mathrm{T}(B)$,
$(b,t)\mapsto b$, $\gamma:\mathrm{T}(B)\to\mathcal{T}$, $b\mapsto(b,1)$, and
$z=(p,s)\in\mathcal{T}$ such that $\pi(z)=p\in\mathcal{D}_{r}(B)$. Then, the
map $r^{z}:\mathcal{T}\times\mathcal{T}\to\mathcal{T}\times\mathcal{T}$ given
by

$r^{z}\left((b,t),(b^{\prime},t^{\prime})\right):=\left(\left(\left[bb^{\prime},bp,p\right],1\right),\left([bb^{\prime},bp,p]^{-1}bb^{\prime},tt^{\prime}\right)\right)$

is a solution on $\mathcal{T}$.

######  Example 9.

Let us consider the brace
$\left(U\left(\mathbb{Z}/8\mathbb{Z}\right),+_{1},\circ\right)$ on the units
of the ring of integers modulo $8$, $U\left(\mathbb{Z}/8\mathbb{Z}\right)$,
endowed with sandwich addition $a+_{1}b:=a-1+b$, for all $a,b\in
U(\mathbb{Z}/8\mathbb{Z})$, and with the modulo multiplication. Let us
consider the unital near-truss
$\mathbb{Z}^{8}:=\mathrm{T}\left(U(\mathbb{Z}/8\mathbb{Z})\right)\times\mathbb{Z}$,
$\pi:\mathbb{Z}^{8}\to U(\mathbb{Z}/8\mathbb{Z}),(k,n)\mapsto k$,
$\gamma:U(\mathbb{Z}/8\mathbb{Z})\to\mathbb{Z}^{8},k\mapsto(k,1)$, and
$z=(3,p)$ for some fixed $p\in\mathbb{Z}$. Then, the map
$r^{(3,p)}:\mathbb{Z}^{8}\times\mathbb{Z}^{8}\to\mathbb{Z}^{8}\times\mathbb{Z}^{8}$
defined by

$r^{(3,p)}((k,n),(l,m))=\left((kl-3k+3,1),((kl-3k+3)^{-1}kl,nm)\right)$

is a deformed solution on $\mathbb{Z}^{8}$.

To conclude, in the last part of this section, we discuss when a unital near-
truss can be decomposed into the product of a skew brace and a unital near-
truss. Firstly, observe that the diagram (6) can be extended to

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
10.50002pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\\{1\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
50.50002pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
50.50002pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathrm{ker}_{1}(\pi)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
96.33177pt\raise 5.30833pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.30833pt\hbox{$\scriptstyle{i}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 126.1392pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
126.1392pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
158.3403pt\raise 7.5208pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 179.37184pt\raise
3.01385pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
179.37184pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathrm{T}(B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
158.52333pt\raise-8.20134pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\gamma}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
139.37186pt\raise-3.01385pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
224.19478pt\raise 5.15279pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.15279pt\hbox{$\scriptstyle{t}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 248.45866pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
248.45866pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\\{1\\}}$}}}}}}}\ignorespaces}}}}\ignorespaces,$

where $i:\mathrm{ker}_{1}(\pi)\to T$ is the canonical embedding. This
extension allows us to define the following map

$\varphi:\mathrm{ker}_{1}(\pi)\times\mathrm{T}(B)\to
T,\quad(s,b)\mapsto[i(s),1,\gamma(b)],$

where the structure of the near-truss
$\mathrm{ker}_{1}(\pi)\times\mathrm{T}(B)$ is given by componentwise
operations.

######  Remark 4.11.

Note that $\varphi$ is a bijection. Indeed, for any $t\in T$,
$[t,\gamma\pi(t),1]\in\mathrm{ker}_{1}(\pi)$ and
$t=[[t,\gamma\pi(t),1],1,\gamma\pi(t)]=\varphi\left([t,\gamma\pi(t),1],\pi(t)\right),$
so $\varphi$ is surjective. For the injectivity, observe that if
$\varphi(s,b)=\varphi(s^{\prime},b^{\prime})$, then
$\pi\varphi(s,b)=\pi\varphi(s^{\prime},b^{\prime}),$ and
$\gamma(b)=\gamma(b^{\prime}).$ By injectivity of $\gamma$, we get that
$b=b^{\prime}$ and $\varphi(s,b)=\varphi(s^{\prime},b)$, this further implies
that $i(s)=i(s^{\prime})$, and by bijectivity of $i$, we get that
$s=s^{\prime}$. Hence, $\varphi$ is a bijection.

Given $T$ a near-truss, to identify when
$T\cong\mathrm{ker}_{1}(\pi)\times\mathrm{T}(B)$, if $S\subset T$, let us
introduce the set

$\mathrm{Fix}_{T}(S):=\\{t\in T\ |\ \forall{s\in S}\quad[1,s,st]=t\ \\}.$

Having this background, we are ready to specify when $\varphi$ is a
homomorphism of near-trusses.

######  Theorem 4.12.

Let $T$ be a unital near truss and $B$ a skew brace. Then, the map
$\varphi:\mathrm{ker}_{1}(\pi)\times B\to T$ is an isomorphism of near-trusses
if and only if
$\gamma(B)\subseteq\mathrm{Fix}_{T}\left(\mathrm{ker}_{1}(\pi)\right)$ and,
for all $s\in\mathrm{ker}_{1}(\pi)$ and $b\in B$,
$i(s)\gamma(b)=\gamma(b)i(s)$ and

$\displaystyle[i(s),1,\gamma(b)]=[\gamma(b),1,i(s)].$ (7)

###### Proof.

First, let us assume that $\varphi$ is an isomorphism of near-trusses, then,
if $s,s^{\prime}\in\mathrm{ker}_{1}(\pi)$ and $b\in B$,

$[i(ss^{\prime}),1,\gamma(b)]=\varphi(ss^{\prime},b)=\varphi(s,1)\varphi(s^{\prime},b)=i(s)[i(s^{\prime}),1,\gamma(b)]=[i(ss^{\prime}),i(s),i(s)\gamma(b)].$

Hence, $\gamma(b)=[1,i(s),i(s)\gamma(b)]$ and so
$\gamma(B)\subseteq\mathrm{Fix}_{T}\left(\mathrm{ker}_{1}(\pi)\right)$.
Moreover, observe that

$i(s)\gamma(b)=\varphi(s,1)\varphi(1,b)=\varphi(s,b)=\varphi(1,b)\varphi(s,1)=\gamma(b)i(s),$

and

$[i(s),1,\gamma(b)]=\varphi(s,b)=\varphi\left([1,1,s],[b,1,1]\right)=[\varphi(1,b),\varphi(1,1),\varphi(s,1)]=[\gamma(b),1,i(s)].$

Conversely, assume that
$\gamma(B)\subseteq\mathrm{Fix}_{T}\left(\mathrm{ker}_{1}(\pi)\right)$,
$i(s)\gamma(b)=\gamma(b)i(s)$, and (7) holds, for all
$s\in\mathrm{ker}_{1}(\pi)$ and $b\in B$. This implies that
$\gamma(b)=[1,i(s),i(s)\gamma(b)]$ and thus

$\varphi(s,b)=[i(s),1,\gamma(b)]=i(s)\gamma(b).$

Then, $\varphi(1,1)=1$ and, for all $s,s^{\prime}\in\mathrm{ker}_{1}(\pi)$ and
$b,b^{\prime}\in B$,

$\displaystyle\varphi(s,b)\varphi(s^{\prime},b^{\prime})=i(s)\gamma(b)i(s^{\prime})\gamma(b^{\prime})=i(s)i(s^{\prime})\gamma(b)\gamma(b^{\prime})=i(ss^{\prime})\gamma(bb^{\prime})=\varphi(ss^{\prime},bb^{\prime}),$

that is $\varphi$ is a homomorphism of monoids. To show it is also a
homomorphism of heaps, let
$s,s^{\prime},s^{\prime\prime}\in\mathrm{ker}_{1}(\pi)$ and
$b,b^{\prime},b^{\prime\prime}\in B$, then

$\displaystyle\varphi\left(\left[s,s^{\prime},s^{\prime\prime}\right],\left[b,b^{\prime},b^{\prime\prime}\right]\right)$
$\displaystyle=[i([s,s^{\prime},s^{\prime\prime}]),\gamma(1),\gamma([b,b^{\prime},b^{\prime\prime}])]$
by $\gamma(1)=1$
$\displaystyle=[i([s,s^{\prime},s^{\prime\prime}]),[\gamma(b^{\prime}),\gamma(b),\gamma(1)],\gamma(b^{\prime\prime})]$
by (2)
$\displaystyle=[[i(s),i([s^{\prime},s^{\prime\prime},1]),1],\gamma([b^{\prime},b,1]),\gamma(b^{\prime\prime})]$
$\displaystyle=[i(s),[i([s^{\prime},s^{\prime\prime},1]),1,\gamma([b^{\prime},b,1])],\gamma(b^{\prime\prime})]$
$\displaystyle=[i(s),\gamma([b^{\prime},b,1]),[1,i([s^{\prime},s^{\prime\prime},1]),\gamma(b^{\prime\prime})]]$
by (7)
$\displaystyle=[i(s),\gamma([b^{\prime},b,1]),[1,i(1),[i([1,s^{\prime},s^{\prime\prime}]),i(1),\gamma(b^{\prime\prime})]]]$
by $i(1)=1$ and (7)
$\displaystyle=[i(s),[i(1),1,\gamma([b^{\prime},b,1])],[i([1,s^{\prime},s^{\prime\prime}]),i(1),\gamma(b^{\prime\prime})]]$
by (2)
$\displaystyle=[i(s),[\gamma([b^{\prime},b,1]),1,i(1)],[i([1,s^{\prime},s^{\prime\prime}]),i(1),\gamma(b^{\prime\prime})]]$
by (7)
$\displaystyle=[[i(s),1,1],\gamma([b^{\prime},b,1]),[i([1,s^{\prime},s^{\prime\prime}]),1,\gamma(b^{\prime\prime})]]$
by $i(1)=1$ and (2)
$\displaystyle=[i(s),\gamma([b^{\prime},b,1]),[1,i(s^{\prime}),[i(s^{\prime\prime}),1,\gamma(b^{\prime\prime})]]]$
$\displaystyle=[[i(s),\gamma([b^{\prime},b,1]),1],i(s^{\prime}),[i(s^{\prime\prime}),1,\gamma(b^{\prime\prime})]]$
$\displaystyle=[[[i(s),1,\gamma(b)],\gamma(b^{\prime}),1],i(s^{\prime}),[i(s^{\prime\prime}),1,\gamma(b^{\prime\prime})]]$
by (2)
$\displaystyle=[[i(s),1,\gamma(b)],[i(s^{\prime}),1,\gamma(b^{\prime})],[i(s^{\prime\prime}),1,\gamma(b^{\prime\prime})]]$
by (7)
$\displaystyle=[\varphi(s,b),\varphi(s^{\prime},b^{\prime}),\varphi(s^{\prime\prime},b^{\prime\prime})].$

Therefore, $\varphi$ is a homomorphism of near-trusses. ∎

In the end, let us show the following example of a solution that is not coming
from the Cartesian product of a near-truss and a skew brace.

######  Example 10.

Let us consider the near-truss associated to the ring $\mathbb{R}$ and the
trivial brace $(\mathrm{Odd}(\mathbb{Q}),\cdot,\cdot)$, with
$\mathrm{Odd}(\mathbb{Q}):=\displaystyle\left\\{\dfrac{2n+1}{2m+1}\ |\
n,m\in\mathbb{Z}\right\\}$ and $\cdot$ the multiplication of rational numbers.
Then, the map
$\check{\sigma}^{z}:\mathrm{Odd}(\mathbb{Q})\mapsto\mathrm{Aut}(\mathbb{R},+_{1})$
$a\mapsto\check{\sigma}_{a}^{1}$ is a group homomorphism, where
$r+_{1}s:=r-1+s$, for all $r,s\in\mathbb{R}$. Thus, we can consider
$\mathrm{Odd}(\mathbb{Q})\rtimes_{\sigma^{1}}(\mathbb{R},+_{1})$ with
operations given by

$\displaystyle(q,r)+(q^{\prime},r^{\prime})$
$\displaystyle:=\left(qq^{\prime},r+_{1}\check{\sigma}^{1}_{q}\left(r^{\prime}\right)\right)\qquad\&\qquad(q,r)\cdot(q^{\prime},r^{\prime}):=\left(qq^{\prime},rr^{\prime}\right),$

where $q,q^{\prime}\in\mathrm{Odd}(\mathbb{Q})$ and
$r,r^{\prime}\in\mathbb{R}$. One can easily check that
$\mathrm{Odd}(\mathbb{Q})\rtimes_{\sigma^{1}}(\mathbb{R},+_{1})$ with

$[(q,r),\left(q^{\prime},r^{\prime}\right),\left(q^{\prime\prime},r^{\prime\prime}\right)]:=(q,r)-\left(q^{\prime},r^{\prime}\right)+\left(q^{\prime\prime},r^{\prime\prime}\right),$

for all $q,q^{\prime}\in\mathrm{Odd}(\mathbb{Q})$ and
$r,r^{\prime}\in\mathbb{R}$, and $\cdot$ is a unital near-truss. Then, by the
4.6, by taking $z=1$, the maps $(q,r)\overset{\pi}{\mapsto}q$ and
$q\overset{\gamma}{\mapsto}(q,1)$, the map

$r^{1}\left(\left(q,r\right),\left(q^{\prime},r^{\prime}\right)\right)=\left(\left(qq^{\prime}-_{1}q,1\right),\left(qq^{\prime}-_{1}q)^{-1}qq^{\prime},rr^{\prime}\right)\right)$

is a solution on
$\mathrm{Odd}(\mathbb{Q})\rtimes_{\sigma^{1}}(\mathbb{R},+_{1})$.

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