# Semi-biproducts of monoids

Nelson Martins-Ferreira  Instituto Politécnico de Leiria, Leiria, Portugal
<EMAIL_ADDRESS>
###### Abstract.

It is shown that the category of _semi-biproducts_ of monoids is equivalent to
the category of _pseudo-actions_. A semi-biproduct of monoids is a new notion,
obtained through generalizing a biproduct of commutative monoids. By dropping
commutativity and requiring some of the homomorphisms in the biproduct diagram
to be merely identity-preserving maps, we obtain a semi-biproduct. A pseudo-
action is a new notion as well. It consists of three ingredients: a pre-
action, a factor system and a correction system. In the category of groups all
correction systems are trivial. This is perhaps the reason why this notion, to
the author’s best knowledge, has never been considered before.

###### Key words and phrases:

Semi-biproduct, biproduct, semi-direct product, groups, monoids, pointed map,
pseudo-action, correction factor, factor system, factor set, extension

##  1\. Introduction

In this paper, we introduce and study the notion of a semi-biproduct of
monoids. From a categorical point of view a biproduct is simultaneously a
product and a coproduct (see [13], p.194). Furthermore, when there is a
coincidence between products and coproducts in a category, each hom-set
carries the structure of a commutative monoid and parallel morphisms can be
added. In such a context a biproduct can be defined as a diagram of the form

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$
(1)

such that the following conditions hold:

$\displaystyle ps=1_{B},\quad pk=0_{X,B},$ (2) $\displaystyle qs=0_{B,X},\quad
qk=1_{X},$ (3) $\displaystyle kq+sp=1_{A}.$ (4)

In spite of the fact that biproducts are classically studied in categories
whose hom-sets are abelian groups (in which case the conditions $pk=0$ and
$qs=0$ are derived) [13], it is clear that the concept of a biproduct is
meaningful even when each hom-set is a commutative monoid. That seems to be
the most general case in which it makes sense — but we may ask the following
question: how to generalize the definition so that it can be applied to
groups, where semi-direct products are expected to appear? The present work
gives an answer to this question and it goes beyond groups. It provides a
generalization to the so called _Schreier extensions_ of monoids. However, we
answer the question at the expense of allowing consideration of functions
between underlying sets of groups/monoids, and hence stepping out from the
context of the category of groups/monoids. See [1] for a purely categorical
approach to semi-direct products of groups.

A Schreier extension of monoids (see [18]) can be seen as a sequence of monoid
homomorphisms

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{B,}$

in which $k=\ker(p)$, with the property that there exist two set-theoretical
functions, $s\colon{B\to A}$ and $q\colon{A\to X}$, satisfying conditions
$(\ref{eq: Sch-2})$–$(\ref{eq: Sch})$ below. Notice the similarity between
these conditions and the conditions $(\ref{eq: biproduct1})$–$(\ref{eq:
biproduct3})$ defining a biproduct. In particular, note that $(\ref{eq: Sch})$
is subsumed by $(\ref{eq: biproduct1})$ and $(\ref{eq: biproduct2})$ when $q$
is a homomorphism. Thus, the notion of semi-biproduct of monoids arrives
naturally as a diagram of the from $(\ref{diag: biproduct in additive cats})$
where $p$ and $k$ are monoid homomorphisms while $q$ and $s$ are zero-
preserving maps, satisfying conditions $(\ref{eq: biproduct1})$–$(\ref{eq:
biproduct3})$.

Let us see in more detail the long path that has given rise to this simple
definition. In the category of groups and group homomorphisms, every split
epimorphism

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ps=1_{B}$

induces an exact sequence

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{B,}$

with $X=\mathrm{Ker}(p)$. If fixing $X$ in additive notation (without assuming
that $X$ is an abelian group) and $B$ in multiplicative notation, then, up to
isomorphism, the group $A$ is recovered as a semi-direct product of the form
$X\rtimes_{\varphi}B$, with neutral element $(0,1)\in X\times B$, and group
operation

$(x,b)+(x^{\prime},b^{\prime})=(x+\varphi_{b}(x^{\prime}),bb^{\prime}).$ (5)

Recall that (see e.g. [14] and its references to previous work)

$\varphi_{b}\colon{X\to X}$

is defined, for every $b\in B$, as the map $\varphi_{b}(x)=q(s(b)+k(x))$, with
$q(a)=a-sp(a)\in X$, for every $a\in A$. Moreover, when $p$ is a surjective
homomorphism, but not necessarily a split epimorphism, it is still possible to
recover the group $A$ as a semi-direct product with a factor system,
$X\rtimes_{\varphi,\gamma}B$. A factor system, $\gamma\colon{B\times B\to X}$,
is any map
$\gamma(b,b^{\prime})=q(s(b)+s(b^{\prime}))=s(b)+s(b^{\prime})-s(bb^{\prime})$,
for some chosen set-theoretical section $s\colon{B\to A}$ of $p$, i.e., such
that $ps=1_{B}$. In this case, the group operation in
$X\rtimes_{\varphi,\gamma}B$ becomes

$(x,b)+(x^{\prime},b^{\prime})=(x+\varphi_{b}(x^{\prime})+\gamma(b,b^{\prime}),bb^{\prime}).$
(6)

When $A$ is abelian, the action $\varphi_{b}(x)=s(b)+k(x)-s(b)\in X$ is
trivial and so, the surjective homomorphisms with abelian domain are nothing
but (symmetric) factor systems.

The case of monoids is different (see [3]). It is no longer true that every
surjective monoid homomorphism $p\colon{A\to B}$ can be presented as a
sequence

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\langle
1,0\rangle}$$\textstyle{X\rtimes_{\varphi,\gamma}B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{B}}$$\textstyle{B}$

with $X=\mathrm{Ker}(p)$ and such that the underlying set of $A$ is in
bijection with the cartesian product $X\times B$. Thus, in monoids, the
transport of the structure from $A$ to $X\times B$ is not always possible. So,
there is no hope of having an isomorphism $A\cong X\rtimes_{\varphi,\gamma}B$.
A simple example to illustrate this is obtained as

$\textstyle{\\{1,a\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{\\{1,a,b\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{\\{1,b\\}}$
(7)

with $k$ the inclusion map, $p(1)=p(a)=1$, $p(b)=b$ and $a^{2}=a$,
$ab=ba=b^{2}=b$.

We may ask: which monoid homomorphisms exhibit a behaviour similar to the
behaviour of surjective homomorphisms of groups? The answer turns out to be
the so called Schreier extensions (see [10, 18] and references therein, see
also [3]). A Schreier extension $p\colon{A\to B}$ is nothing but a monoid
homomorphism with the property that there exists a set-theoretical section
$s\colon{B\to A}$ for which the function $X\times B\to A$, $(x,b)\mapsto
k(x)+s(b)$ is a bijection of the underlying sets, with $k=\ker(p)$.
Equivalently, a Schreier extension can be seen as a monoid homomorphism
$p\colon{A\to B}$, together with set-theoretical maps $s\colon{B\to A}$ and
$q\colon{A\to X}$, with $X=\mathrm{Ker}(p)$ and $k=\ker(p)\colon{X\to A}$,
such that

$\displaystyle ps(b)=b,$ $\displaystyle b\in B,$ (8) $\displaystyle
kq(a)+sp(a)=a,$ $\displaystyle a\in A,$ (9) $\displaystyle q(k(x)+s(b))=x,$
$\displaystyle x\in X,b\in B.$ (10)

The key feature of this work is the observation that, at the expense of having
to insert a correction system, condition $(\ref{eq: Sch})$ can be discarded.
The correction system controls the lack of condition $(\ref{eq: Sch})$ and
will be denoted by $x^{b}$. It is defined as $x^{b}=q(k(x)+s(b))$, for every
$x\in X$ and $b\in B$. While dropping condition $(\ref{eq: Sch})$, we lose the
bijection between the underlying set of $A$ and the cartesian product of sets
$X\times B$. However, if adding two simple conditions, namely $q(k(x))=x$ and
$q(s(b))=0$, then, $A$ is bijective to a subset of the cartesian product
$X\times B$. The subset consists of those pairs in $X\times B$ that are of the
form $(x^{b},b)$ for some $x\in X$ and $b\in B$ (see Theorem 1). The monoid
operation is

$(x,b)+(x^{\prime},b^{\prime})=((x+b\cdot x^{\prime}+(b\times
b^{\prime}))^{bb^{\prime}},bb^{\prime}).$ (11)

In this way we see that it is possible to consider monoid extensions in which
the middle object is not in bijection with the two ends, as it is the case in
diagram (7).

The correction system $x^{b}$ has to satisfy some conditions (Definition 1).
In the case of groups it becomes trivial, that is $x^{b}=x$. To see it, simply
take $b^{\prime}=b^{-1}$ in equation $(\ref{eq: 53})$. The notion of a
correction system is inspired by the work of Leech and Wells on extending
groups by monoids ([12, 20]). The fact that the correction system
$\rho(x,b)=x^{b}$ is invisible in groups, hides the true nature of pseudo-
actions as a combination of structures $\rho$, $\varphi$, $\gamma$ (Definition
1) satisfying one major condition (29). Only the particular traces of (29) are
familiar in groups. This is better explained in Remark 1. A Schreier extension
of monoids is precisely a semi-biproduct with a trivial correction system,
which explains the similarity between Schreier extensions of monoids and all
extensions of groups. Examples of semi-biproducts of monoids that are not
Schreier extensions are presented in Sections 3 and 4. At this point we may
observe that the sequence (7), if completed with obvious section $s$ and
retraction $q$, is an example of a semi-biproduct of monoids. The celebrated
surjective homomorphism with the usual addition of natural numbers,

$\textstyle{\mathbb{N}\times\mathbb{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{+}$$\textstyle{\mathbb{N},}$

whose kernel is trivial, illustrates the fact that not every surjective monoid
homomorphism is part of a semi-biproduct diagram.

From a structural point of view, we are considering sequences of monoid
homomorphisms

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{B}$
(12)

together with set-theoretical maps, preserving the neutral element, but not
necessarily preserving the monoid operation,

$\textstyle{X}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$
(13)

and satisfying the conditions $(\ref{eq: biproduct1})$–$(\ref{eq:
biproduct3})$. When the maps $q$ and $s$ are monoid homomorphisms then this is
exactly the definition of a biproduct. It is then natural to use the term
_semi-biproduct_ when $q$ and $s$ are just zero-preserving maps. Particular
instances when just one of $q$ or $s$ is a morphism are worthwile studying,
but we shall no longer expect a product or a coproduct to appear. In general
it does not make sense to speak about the semi-biproduct of two objects,
unless some kind of pseudo-action is specified. This is no surprise, indeed,
in groups, semi-biproducts are the same as semi-direct products with a factor
system. Connections with homology and cohomology are expected via several
notions such as abstract kernel, obstruction, etc. (see for example [18] and
references there), and perhaps the notion of pseudo-action considered here
(Definition 1) can be used as a tool and clarify some classical
interpretations in group-cohomology.

This paper is organized as follows. In section 2 we give the definition of
semi-biproduct and pseudo-action and establish a categorical equivalence
between them (Theorem 4). In Section 3 we give examples of semi-biproducts of
monoids obtained from Schreier extensions and other types of more general
extensions. In Section 4 we explore the fact that a semi-biproduct may be
viewed as an extension of monoids with a specified section map and a specified
retraction map.

##  2\. Definitions and theorems

In this section, we introduce a new notion of _pseudo-action_ of monoids,
which enables a characterization of semi-biproducts of monoids, as defined in
the Introduction. Some previous works have been done in a similar direction
but the emphasis has always been on the side of (Schreier) extensions and
their classification (e.g. [4, 9, 10, 18]). Here we consider a semi-biproduct
as a mathematical object rather than something which arises from an extension
with appropriate choices for a section and a retraction. In a certain sense
this work goes in the direction of [11]. There, the Schreier condition
$(\ref{eq: Sch})$ is still being considered, but it shows that it is possible
to work in the wider context of unitary magmas. Another recent study on
(weakly) Schreier extensions can be found in [8]. There, up to some
differences concerning what is structure and what is property, the main focus
is on a certain quotient, while here we have put our emphasis on a certain
subset of the cartesian product between the two ends of a semi-biproduct
diagram.

###  2.1. Semi-biproducts of monoids

A _semi-biproduct_ of monoids is a diagram of the shape

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$
(14)

such that $p$, $k$, are monoid homomorphisms, while $q$ and $s$ are zero-
preserving maps, verifying the conditions

$\displaystyle ps$ $\displaystyle=$ $\displaystyle 1_{B}$ (15) $\displaystyle
qk$ $\displaystyle=$ $\displaystyle 1_{X}$ (16) $\displaystyle kq+sp$
$\displaystyle=$ $\displaystyle 1_{A}$ (17) $\displaystyle pk$
$\displaystyle=$ $\displaystyle 0_{X,B}$ (18) $\displaystyle qs$
$\displaystyle=$ $\displaystyle 0_{B,X}.$ (19)

We will work with additive notation on $X$ and multiplicative notation on $B$,
but neither is assumed to be commutative.

######  Theorem 1.

Let $(X,A,B,p,k,q,s)$ be a semi-biproduct of monoids. If we put for every
$b,b^{\prime}\in B$ and $x\in X$,

$\displaystyle b\cdot x=q(s(b)+k(x))$ (20) $\displaystyle x^{b}=q(k(x)+s(b))$
(21) $\displaystyle b\times b^{\prime}=q(s(b)+s(b^{\prime}))$ (22)

then, for every $a,a^{\prime}\in A$

$a+a^{\prime}=k(q(a)+p(a)\cdot q(a^{\prime})+(p(a)\times
p(a^{\prime})))+sp(a+a^{\prime})$ (23)

and

$a+a^{\prime}=k(u^{v})+s(v)$ (24)

with $u=q(a)+p(a)\cdot q(a^{\prime})+(p(a)\times p(a^{\prime}))\in X$ and
$v=p(a+a^{\prime})\in B$.

Moreover, the map $\beta\colon{A\to X\times B}$ with $\beta(a)=(q(a),p(a))$ is
always injective. An element $(y,b)\in X\times B$ is in the image of $\beta$
if and only if $y=x^{b}$ for some $x\in X$.

This means that $A$ is in bijection with the set $\\{(x^{b},b)\mid(x,b)\in
X\times B\\}$ with the inverse map for $\beta$ being
$\alpha(x^{b},b)=k(x^{b})+s(b)=k(x)+s(b)$.

Furthermore, for every $a\in A$, $q(a)=q(a)^{p(a)}$.

###### Proof.

We observe:

$\displaystyle a+a^{\prime}$ $\displaystyle=$ $\displaystyle
kqa+(spa+kqa^{\prime})+spa^{\prime}\qquad(kq+sp=1)$ $\displaystyle=$
$\displaystyle kqa+kq(spa+kqa^{\prime})+sp(spa+kqa^{\prime})+spa^{\prime}$
$\displaystyle=$ $\displaystyle
kqa+kq(spa+kqa^{\prime})+spa+spa^{\prime}\qquad(ps=1,pk=0)$ $\displaystyle=$
$\displaystyle
kqa+kq(spa+kqa^{\prime})+kq(spa+spa^{\prime})+sp(spa+spa^{\prime})$
$\displaystyle=$ $\displaystyle
kqa+kq(spa+kqa^{\prime})+kq(spa+spa^{\prime})+s(pa+pa^{\prime})$
$\displaystyle=$ $\displaystyle
k(qa+q(spa+kqa^{\prime})+q(spa+spa^{\prime}))+sp(a+a^{\prime})$
$\displaystyle=$ $\displaystyle k(q(a)+p(a)\cdot q(a^{\prime})+(p(a)\times
p(a^{\prime})))+sp(a+a^{\prime}).$

We observe further that for every $a\in A$, $x\in X$, $b\in B$, if
$a=k(x)+s(b)$ then $a=k(x^{b})+s(b)$, which proves condition $(\ref{eq:
kuv+sv})$. This follows from the fact that for every $a\in A$, we have
$q(a)=q(a)^{p(a)}$. Indeed, $q(a)=q(kq(a)+sp(a))$ follows directly from
$a=kq(a)+sp(a)$.

If $\beta(a)=\beta(a^{\prime})$ then $q(a)=q(a^{\prime})$ and
$p(a)=p(a^{\prime})$, hence $a=a^{\prime}$. This proves $\beta$ is injective.

If $(x,b)\in X\times B$ is of the form $x=q(a)$, $b=p(a)$ for some $a\in A$,
then $x=x^{b}$ because $q(a)=q(a)^{p(a)}$ as already shown. If $(y,b)\in
X\times B$ is of the form $y=x^{b}$ for some $x\in X$, then there exists
$a=k(x)+s(b)\in A$ such that $\beta(a)=(x,b)$. In order to see that the map
$\alpha(x^{b},b)=k(x)+s(b)$ is the inverse of $\beta$ we observe:

$(x^{b},b)\mapsto k(x)+s(b)\mapsto(q(k(x)+s(b)),p(k(x)+s(b))=(x^{b},b)$

and

$a\mapsto(q(a),p(a))=(q(a)^{p(a)},p(a))\mapsto kq(a)+sp(a)=a.$

∎

As expected, a morphism between semi-biproducts is a diagram of the shape

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\scriptstyle{f_{1}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{3}}$$\scriptstyle{s}$$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k^{\prime}}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q^{\prime}}$$\scriptstyle{p^{\prime}}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s^{\prime}}$
(25)

in which the top and bottom rows are semi-biproducts, each $f_{i}$, $i=1,2,3$
is a homomorphism and the following compatibility conditions hold:
$f_{2}k=k^{\prime}f_{1}$, $p^{\prime}f_{2}=f_{3}p$, $f_{2}s=s^{\prime}f_{3}$,
$q^{\prime}f_{2}=f_{1}q$.

We will show that there is an equivalence of categories between the category
of semi-biproducts of monoids and a suitable category of pseudo-actions of
monoids.

###  2.2. Pseudo-actions of monoids

Let $X$ and $B$ be two monoids. For convenience, let us again use additive
notation for $X$ and multiplicative notation for $B$. Note that neither one is
assumed to be commutative.

######  Definition 1.

A pseudo-action of $B$ on $X$ consists in three different components:

  1. (1)

a correction system, that is, a map

$\displaystyle\rho\colon{X\times B\to X};\quad(x,b)\mapsto x^{b}$

such that for all $x\in X$ and $b\in B$

$\displaystyle x^{1}=x,\quad 0^{b}=0,$ (26)

  2. (2)

a pre-action, that is, a map

$\varphi\colon{B\times X\to X};\quad(b,x)\mapsto b\cdot x$

such that for all $b\in B$ and $x\in X$

$\displaystyle 1\cdot x=x,\quad b\cdot 0=0,$ (27)

  3. (3)

a factor system, that is, a map

$\displaystyle\gamma\colon{B\times B\to X};\quad(b,b^{\prime})\mapsto b\times
b^{\prime}$

such that for every $b\in B$

$\displaystyle 1\times b=0=b\times 1.$ (28)

The three components are related via one major condition, $(\ref{eq:
factor})$, which must hold for every $x,x^{\prime},x^{\prime\prime}\in X$ and
$b,b^{\prime},b^{\prime\prime}\in B$,

$\displaystyle(x+b\cdot((x^{\prime}+b^{\prime}\cdot
x^{\prime\prime}+(b^{\prime}\times
b^{\prime\prime}))^{b^{\prime}b^{\prime\prime}})+(b\times
b^{\prime}b^{\prime\prime}))^{bb^{\prime}b^{\prime\prime}}=$
$\displaystyle=((x+b\cdot x^{\prime}+(b\times
b^{\prime}))^{bb^{\prime}}+bb^{\prime}\cdot
x^{\prime\prime}+(bb^{\prime}\times
b^{\prime\prime}))^{bb^{\prime}b^{\prime\prime}}.$ (29)

Moreover, the following two compatibility conditions

$\displaystyle(b\cdot x)^{b}=b\cdot x$ (30) $\displaystyle(b\times
b^{\prime})^{bb^{\prime}}=b\times b^{\prime}$ (31)

are required for every $x\in X$ and $b,b^{\prime}\in B$.

The correction system $\rho$ is used to correct the fact that in general
$b\cdot(x+y)$ is not equal to $b\cdot x+b\cdot y$. Instead, we have the
equalities

$\displaystyle(b\cdot(x+y))^{b}=(b\cdot x+b\cdot y)^{b}$ (32)
$\displaystyle(x+y)^{b}=(x+y^{b})^{b}$ (33) $\displaystyle(x^{b}+b\cdot
y)^{b}=(x+(b\cdot y)^{b})^{b}$ (34)

which are obtained (Remark 1 — items 4, 5, 6 below) as particular cases of
$(\ref{eq: factor})$. The fact that the correction system is invisible in
groups (Remark 1 — item 2 below) is perhaps the reason why (to the author’s
best knowledge) it has not appeared before as an explicit structure. The idea
of having one major condition (29) accounting for associativity has been
implicitly used by Leech [12] and Wells [20] (see also [9]). The two
conditions (30) and (31) are needed in order to have an equivalence with semi-
biproducts (Theorem 4).

######  Remark 1.

The following particular cases of $(\ref{eq: factor})$ are of interest:

  1. (1)

Taking $x,x^{\prime},x^{\prime\prime}$ to be zero, we get

$(b\cdot(b^{\prime}\times b^{\prime\prime})^{bb^{\prime}}+(b\times
b^{\prime}b^{\prime\prime}))^{bb^{\prime}b^{\prime\prime}}=((b\times
b^{\prime})^{bb^{\prime}}+(bb^{\prime}\times
b^{\prime\prime}))^{bb^{\prime}b^{\prime\prime}}$ (35)

which, if we ignore the correction system, becomes the usual formula for a
factor system of monoids (see e.g. [10] and references therein).

  2. (2)

Taking $x=x^{\prime\prime}=0$ and $b=1$, we get (by putting again $x$ in the
place of $x^{\prime}$, $b$ in the place of $b^{\prime}$ and $b^{\prime}$ in
the place of $b^{\prime\prime}$ for readability)

$(x^{b}+b\times b^{\prime})^{bb^{\prime}}=(x+b\times
b^{\prime})^{bb^{\prime}}$ (36)

which explains how different the correction system is from being an action. In
particular, if $X$ is right cancellable and $B$ is a group then the correction
system is always trivial. Indeed, if we take $b^{\prime}=b^{-1}$ we obtain
$x^{b}+(b\times b^{-1})=x+(b\times b^{-1})$ and hence, cancelling out $b\times
b^{-1}$, we get $x^{b}=x$.

  3. (3)

Taking $x=x^{\prime}=0$ and $b^{\prime\prime}=1$, we get

$(b\cdot(b^{\prime}\cdot x^{\prime\prime})^{b^{\prime}}+(b\times
b^{\prime}))^{bb^{\prime}}=((b\times
b^{\prime})^{bb^{\prime}}+bb^{\prime}\cdot x^{\prime\prime})^{bb^{\prime}}$
(37)

which, in groups, becomes the familiar expression

$b\cdot(b^{\prime}\cdot x^{\prime\prime})=(b\times
b^{\prime})+bb^{\prime}\cdot x^{\prime\prime}-(b\times b^{\prime})$ (38)

stating that the factor system $b\times b^{\prime}$ measures, by conjugation,
the distance between a pre-action $\varphi$ and an ordinary action.

  4. (4)

Taking $b=b^{\prime}=1$ and $x^{\prime\prime}=0$, we get

$(x+x^{\prime})^{b^{\prime\prime}}=(x+x^{\prime
b^{\prime\prime}})^{b^{\prime\prime}}$ (39)

which is exactly the same as $(\ref{eq: correction2})$.

  5. (5)

Taking $b=b^{\prime\prime}=1$ and $x^{\prime}=0$, we get

$(x^{b^{\prime}}+b^{\prime}\cdot
x^{\prime\prime})^{b^{\prime}}=(x+(b^{\prime}\cdot
x^{\prime\prime})^{b^{\prime}})^{b^{\prime}}$ (40)

which is exactly the same as $(\ref{eq: correction3})$.

  6. (6)

Taking $b^{\prime}=b^{\prime\prime}=1$ and $x=0$, we get

$(b\cdot(x^{\prime}+x^{\prime\prime}))^{b}=((b\cdot x^{\prime})^{b}+b\cdot
x^{\prime\prime})^{b}$ (41)

which, if combined with $(\ref{eq: correction2})$ and $(\ref{eq:
correction3})$, gives $(\ref{eq: correction1})$.

Given a pseudo-action $(X,B,\rho,\varphi,\gamma)$ we construct (inspired by
[20]) a _synthetic diagram_

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\langle
1,0\rangle}$$\textstyle{R_{\rho,\varphi,\gamma}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{X}}$$\scriptstyle{\pi_{B}}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\langle
0,1\rangle}$ (42)

with $R_{\rho,\varphi,\gamma}=\\{(x,b)\in X\times B\mid x^{b}=x\\}$, equipped
with the restriction to the binary operation

$(x,b)+(x^{\prime},b^{\prime})=((x+b\cdot x^{\prime}+(b\times
b^{\prime}))^{bb^{\prime}},bb^{\prime})$ (43)

defined for every $x,x^{\prime}\in X$ and $b,b^{\prime}\in B$.

Let $(X,B,\rho,\varphi,\gamma)$ and
$(X^{\prime},B^{\prime},\rho^{\prime},\varphi^{\prime},\gamma^{\prime})$ be
two pseudo-actions of monoids. A morphism of pseudo-actions is a pair $(f,g)$
of monoid homomorphisms, $f\colon{X\to X^{\prime}}$, $g\colon{B\to
B^{\prime}}$ preserving each one of the three components of a pseudo-action,

$\displaystyle f(x^{b})$ $\displaystyle=$ $\displaystyle f(x)^{g(b)},$ (44)
$\displaystyle f(b\cdot x)$ $\displaystyle=$ $\displaystyle g(b)\cdot{f(x)},$
(45) $\displaystyle f(b\times b^{\prime})$ $\displaystyle=$ $\displaystyle
g(b)\times g(b^{\prime}).$ (46)

The equation

$f((x+b\cdot x^{\prime}+b\times b^{\prime})^{bb^{\prime}})=(f(x)+g(b)\cdot
f(x^{\prime})+g(b)\times g(b^{\prime}))^{g(bb^{\prime})}$ (47)

is a consequence of the previous ones and will be needed in proving that the
map ${R_{\rho,\varphi,\gamma}\to
R_{\rho^{\prime},\varphi^{\prime},\gamma^{\prime}}}$, defined for every pair
$(x,b)\in X\times B$, with $x^{b}=x$, as $(x,b)\mapsto(f(x),g(b))$, is a
monoid homomorphism.

###  2.3. The equivalence

In this subsection we will work towards an equivalence between pseudo-actions
and semi-biproducts of monoids. First we show that there is a functor from the
category of semi-biproducts into the category of pseudo-actions.

######  Theorem 2.

Let $(X,A,B,p,k,q,s)$ be a semi-biproduct of monoids. The system with three
components

$\displaystyle x^{b}=q(k(x)+s(b))$ (48) $\displaystyle b\cdot x=q(s(b)+k(x))$
(49) $\displaystyle b\times b^{\prime}=q(s(b)+s(b^{\prime}))$ (50)

is a pseudo-action from $B$ into $X$.

Moreover, if $(f_{1},f_{2},f_{3})$ is a morphism of semi-biproducts such as
the one displayed in (25) then the pair $(f_{1},f_{3})$ is a morphism of
pseudo-actions.

###### Proof.

Clearly, the formulas given for $x^{b}$, $b\cdot x$ and $b\times b^{\prime}$
are well defined maps $\rho$, $\varphi$ and $\gamma$ consistent with the three
components of a pseudo-action, respectively a correction system, a pre-action
and a factor system. The identities $x^{1}=x$, $0^{b}=0$, $1\cdot x=x$,
$b\cdot 0=0$, $b\times 1=0=1\times b$ are immediate consequences of
$qs=0_{X,B}$, $qk=1_{X}$, and the fact that the maps $s$ and $q$ preserve the
neutral element.

In order to prove condition (29) we observe that for every
$x,x^{\prime},x^{\prime\prime}\in X$ and $b,b^{\prime},b^{\prime\prime}\in B$,
the expression

$kx+sb+kx^{\prime}+sb^{\prime}+kx^{\prime\prime}+sb^{\prime\prime}\in A$

can be decomposed as

$\left(\left(kx+\left(\left(sb+kx^{\prime}\right)+sb^{\prime}\right)\right)+kx^{\prime\prime}\right)+sb^{\prime\prime}$
(51)

and as

$kx+\left(sb+\left(kx^{\prime}+\left(\left(sb^{\prime}+kx^{\prime\prime}\right)+sb^{\prime\prime}\right)\right)\right).$
(52)

Using the formulas

$\displaystyle kx+sb=k(x^{b})+s(b)$ (53) $\displaystyle sb+kx=k(b\cdot
x)+s(b)$ (54) $\displaystyle sb+sb^{\prime}=k(b\times
b^{\prime})+s(bb^{\prime})$ (55)

when appropriate and following the order of parenthesis indicated in (51) we
obtain the right hand side of (29). Following the order of parenthesis
indicated in (52) we obtain the left hand side of (29).

The condition $(x^{b})^{b}=x^{b}$ (which is also a consequence of (29)) as
well as the two conditions (30) and (31) are obtained by applying the map $q$
on both sides of equations (53)–(55).

Let us now suppose $(f_{1},f_{2},f_{3})$ is a morphism of semi-biproducts such
as the one displayed in (25). We have to show that the pair $(f_{1},f_{3})$ is
a morphism of pseudo-actions. To simplify notation let us consider $f=f_{1}$,
$g=f_{3}$ and $h=f_{2}$. We observe,

$\displaystyle f(x)^{g(b)}$ $\displaystyle=$ $\displaystyle
q^{\prime}(k^{\prime}f(x)+s^{\prime}g(b))$ $\displaystyle=$ $\displaystyle
q^{\prime}(hk(x)+hs(b))$ $\displaystyle=$ $\displaystyle
q^{\prime}h(k(x)+s(b))$ $\displaystyle=$ $\displaystyle fq(k(x)+s(b))$
$\displaystyle=$ $\displaystyle f(x^{b}).$

Similarly, we prove

$f(b\cdot
x)=fq(sb+kx)=q^{\prime}h(sb+kx)=q^{\prime}(sg(b)+k^{\prime}f(x))=g(b)\cdot
f(x)$

and

$f(b\times
b^{\prime})=fq(sb+sb^{\prime})=q^{\prime}h(sb+sb^{\prime})=q^{\prime}(s^{\prime}g(b)+s^{\prime}g(b^{\prime}))=g(b)\times
g(b^{\prime}).$

∎

The previous result gives a functor from the category of semi-biproducts into
the category of pseudo-actions. The synthetic construction (42) produces a
functor in the other direction.

######  Theorem 3.

Let $(X,B,\rho,\varphi,\gamma)$ be a pseudo-action of monoids, the synthetic
diagram (42) is a semi-biproduct of monoids. Moreover, if

$(f,g)\colon{(X,B,\rho,\varphi,\gamma)\to(X^{\prime},B^{\prime},\rho^{\prime},\varphi^{\prime},\gamma^{\prime})}$

is a morphism of pseudo-actions then the triple $(f,h,g)$, with

$h\colon{R_{\rho,\varphi,\gamma}\to
R_{\rho^{\prime},\varphi^{\prime},\gamma^{\prime}}},$

defined as $h(x,b)=(f(x),g(b))$, is a morphism of semi-biproducts.

###### Proof.

Given a pseudo-action $(X,B,\rho,\varphi,\gamma)$ we consider the synthetic
diagram (42). The operation (43) is well defined on the set

$R_{\rho,\varphi,\gamma}=\left\\{\left(x^{b},b\right)\mid x\in X,b\in
B\right\\}$

and it is associative by (29). With neutral element $(1,0)$ we see that the
set $R=R_{\rho,\varphi,\gamma}$ has a monoid structure. Clearly, the maps
$\langle 1,0\rangle\colon{X\to R}$ and $\pi_{B}\colon{R\to B}$ are monoid
homomorphisms while $\langle 0,1\rangle\colon{B\to R}$ and $\pi_{X}\colon{R\to
X}$ are zero-preserving maps. Furthermore, we have $\pi_{B}\langle
0,1\rangle=1_{B}$, $\pi_{X}\langle 1,0\rangle=1_{X}$, $\pi_{B}\langle
1,0\rangle=0$ and $\pi_{X}\langle 0,1\rangle=0$. Finally, we observe that for
every $(x,b)\in X\times B$, from (43) we have

$(x,b)=(x,1)+(0,b)=(x^{b},b)$

which means that if $(x,b)\in R$ then $(x,b)=(x,1)+(0,b)$. This shows that the
synthetic construction is a semi-biproduct.

From equation (47) we see that the map $h(x,b)=(f(x),g(b))$ is a monoid
homomorphism. Since it is compatible with the respective semi-biproduct
synthetic structure maps, we conclude that it is a morphism of semi-
biproducts. ∎

In order to prove an equivalence of categories between the category of semi-
biproducts and the category of pseudo-actions, we observe that if starting
with a structure $(x^{b},b\cdot x,b\times b^{\prime})$ for a pseudo-action of
$X$ on $B$, constructing its associated synthetic semi-biproduct and then
extracting a new pseudo-action out of it, we obtain the structure

$(x^{b},(b\cdot x)^{b},(b\times b^{\prime})^{bb^{\prime}}).$

The role of equations (30) and (31) is to ensure that the new structure is in
fact equal to the original one. On the other hand, if starting with a semi-
biproduct of monoids, extracting its associated pseudo-action, and then
reconstructing the synthetic semi-biproduct, we get a natural isomorphism of
semi-biproducts.

######  Theorem 4.

There is an equivalence between the category of semi-biproducts of monoids and
the category of pseudo-actions.

###### Proof.

Let us start with a semi-biproduct $(X,A,B,p,k,q,s)$ and let
$(X,B,\rho,\varphi,\gamma)$ be its associated pseudo-action. The synthetic
semi-biproduct constructed from the pseudo-action is isomorphic to the
original semi-biproduct as illustrated

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\langle
1,0\rangle}$$\textstyle{R_{\rho,\varphi,\gamma}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\pi_{X}}$$\scriptstyle{\pi_{B}}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\langle
0,1\rangle}}$ (56)

with the morphisms $\alpha$ and $\beta$ defined as in Theorem 1. We have to
show that $\alpha$ and $\beta$ are monoid homomorphisms. It is immediate to
see that they are compatible with the semi-biproduct structuring maps, that is

$\displaystyle p\alpha(x,b)=p(kx+sb)=b,$ $\displaystyle
q\alpha(x,b)=q(kx+sb)=x^{b}=x\quad,(x,b)\in R$
$\displaystyle\alpha(0,b)=s(b),\quad\alpha(x,1)=k(x)$
$\displaystyle\pi_{B}\beta(a)=p(a),\quad\pi_{X}\beta(a)=q(a)$
$\displaystyle\beta s(b)=(0,b),\quad\beta k(x)=(x,1).$

In order to prove that $\alpha$ is a homomorphism we observe

$\alpha((x,b)+(x^{\prime},b^{\prime}))=k((x+b\cdot x^{\prime}+b\times
b^{\prime})^{bb^{\prime}})+s(bb^{\prime})$

and

$\displaystyle\alpha(x,b)+\alpha(x^{\prime},b^{\prime})$ $\displaystyle=$
$\displaystyle kx+sb+kx^{\prime}+sb^{\prime}$ $\displaystyle=$ $\displaystyle
kx+k(b\cdot x^{\prime})+sb+sb^{\prime}$ $\displaystyle=$ $\displaystyle
kx+k(b\cdot x^{\prime})+k(b\times b^{\prime})+s(bb^{\prime})$ $\displaystyle=$
$\displaystyle k(x+b\cdot x^{\prime}+b\times b^{\prime})+s(bb^{\prime})$
$\displaystyle=$ $\displaystyle k((x+b\cdot x^{\prime}+b\times
b^{\prime})^{bb^{\prime}})+s(bb^{\prime}).$

In order to prove that $\beta$ is a homomorphism we observe that
$\beta(a+a^{\prime})=(q(a+a^{\prime}),p(a+a^{\prime}))$ while
$\beta(a)+\beta(a^{\prime})=(q(a),p(a))+(q(a^{\prime}),p(a^{\prime}))$. This
means that $\beta$ is a homomorphism as soon as

$q(a+a^{\prime})=(q(a)+p(a)\cdot q(a^{\prime})+(p(a)\times
p(a^{\prime})))^{p(a+a^{\prime})}$ (57)

which follows from (23).

To see that $\alpha$ and $\beta$ define a natural isomorphism between a semi-
biproduct and its associated synthetic one, let us consider the diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\scriptstyle{\beta}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta^{\prime}}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{h}$$\textstyle{R^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha^{\prime}}$
(58)

in which $f_{2}$ is part of a morphism $(f_{1},f_{2},f_{3})$ between semi-
biproducts and $h$ is the induced morphism between respective associated
synthetic semi-biproducts, that is, $h(x,b)=(f_{1}(x),f_{3}(b))$. We observe

$h\beta(a)=(f_{1}q(a),f_{3}p(a))=(q^{\prime}f_{2}(a),p^{\prime}f_{2}(a))=\beta^{\prime}f_{2}(a)$

and

$\displaystyle f_{2}\alpha(x,b)$ $\displaystyle=$ $\displaystyle
f_{2}(kx+sb)=f_{2}k(x)+f_{2}s(b)$ $\displaystyle=$ $\displaystyle
k^{\prime}f_{1}(s)+s^{\prime}f_{3}(b)=\alpha^{\prime}h(x,b).$

If we start with a pseudo-action $(X,B,\varphi,\rho,\gamma)$, build its
associated synthetic semi-biproduct, and extract the pseudo-action associated
to it, then a new pseudo-action is obtained, let us call it
$(X,B,\rho^{\prime},\varphi^{\prime},\gamma^{\prime})$. The new pseudo-action
is $\rho^{\prime}(x,b)=x^{b}$, $\varphi^{\prime}(b,x)=(b\cdot x)^{b}$ and
$\gamma^{\prime}(b,b^{\prime})=(b\times b^{\prime})^{bb^{\prime}}$. The two
conditions (30) and (31) ensure that the new pseudo-action and the original
one are the same. ∎

##  3\. Examples

Every Schreier split extension ([4]) of a monoid $X$ by a monoid $B$ is a
semi-biproduct of monoids. In this case $b\times b^{\prime}=0$ and $x^{b}=x$
for all $x\in X$, $b,b^{\prime}\in B$. More generally, every Schreier
extension of monoids is a semi-biproduct in which $x^{b}=x$. In fact, a semi-
biproduct of monoids $(X,A,B,p,k,q,s)$ is a Schreier extension if and only if
$x^{b}=x$ for all $x\in X$ and $b\in B$.

Here is a general procedure for the construction of semi-biproducts which are
not necessarily Schreier extensions. The procedure is obtained from a pseudo-
action $(X,B,\rho,\varphi,\gamma)$ by isolating the subset

$\\{(\rho(x,b),b)\mid x\in X,b\in B\\}\subseteq X\times B$

and then working backwards in order to recover the rest of the structure while
assuming that the map $s$ is of the form $s(b)=(u(b),b)$ for some map $u$ from
$B$ to $X$.

Let $X$ (written additively) and $B$ (written multiplicatively) be two monoids
and let us suppose the existence of a subset $R\subseteq X\times B$,
considered as a binary relation, so that we write $xRb$ in the place of
$(x,b)\in R$, together with two maps $u\colon{B\to X}$ and $q\colon{R\to X}$
satisfying the following conditions:

$\displaystyle\text{ for all $x\in X$, $xR1$ and $q(x,1)=x$},$ (59)
$\displaystyle\text{ for all $b\in B$, $u(b)Rb$ and $q(u(b),b)=0$},$ (60) if
$xRb$, $yRb$ and $q(x,b)=q(y,b)$ then $x=y$. (61)

For every monoid structure on the set $R$, for which $(0,1)$ is the neutral
element and the projection map $(x,b)\mapsto b$ is a homomorphism, we put:

$\displaystyle x\oplus x^{\prime}$ $\displaystyle=$ $\displaystyle
q((x,1)+(x^{\prime},1))$ $\displaystyle b\times b^{\prime}$ $\displaystyle=$
$\displaystyle q((u(b),b)+(u(b^{\prime}),b^{\prime}))$ $\displaystyle b\cdot
x$ $\displaystyle=$ $\displaystyle q((u(b),b)+(x,1))$ $\displaystyle x^{b}$
$\displaystyle=$ $\displaystyle q((x,1)+(u(b),b)).$

If $x\oplus x^{\prime}=x+x^{\prime}$ for all $x,x^{\prime}\in X$ (where $+$ is
the monoid operation on $X$) and if $q(x,b)^{b}=q(x,b)$, whenever $xRb$, then
we have a semi-biproduct

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$
(62)

with $p(x,b)=b$, $k(x)=(x,1)$, $s(b)=(u(b),b)$. The monoid operation on $R$ is
given by the formula

$(x,b)+(x^{\prime},b^{\prime})=((x+b\cdot x^{\prime}+b\times
b^{\prime})^{bb^{\prime}},bb^{\prime}).$ (63)

Indeed, $k$ is a homomorphism because $\oplus$ is the monoid operation defined
on $X$, $p$ is a homomorphism by assumption, $pk=0$, $ps=1$, $qk=1$, $qs=0$ by
construction and we have, for every $xRb$,

$(x,b)=(q(x,b),1)+(u(b),b)$ (64)

which means that $kq+sp=1_{R}$. It is not difficult to see that
$(\ref{eq:xRb})$ follows from the fact that $q(x,b)^{b}=q(x,b)$, together with
the fact that by condition $(\ref{item:q})$ the map $\langle
q,p\rangle\colon{R\to X\times B}$ is injective. In particular, when we force
$q(x,b)=x$ and $u(b)=0$ then we are reduced to the analysis of monoid
structures on $R$. In that case, the condition $q(x,b)^{b}=q(x,b)$ becomes
$x^{b}=x$ whenever $xRb$, and if we are looking for semi-biproducts which are
not Schreier extensions then we have to consider relations $R$ which are not
in bijection with the set $X\times B$. Let us work a concrete example.

Take $X=\\{0,s\\}$ with $s+s=s$ and $B=\\{1,t\\}$ with $t^{2}=t$. Clearly,
$R=\\{(0,1),(s,1),(0,t)\\}$ is the only possible proper subset of $X\times B$
with $xR1$ and $0Rb$ for every $x\in X$ and $b\in B$. There are two possible
solutions to turn $R$ into a monoid with $0R1$ as neutral element and such
that the map $p(x,b)=b$ is a homomorphism.

  1. (1)

One possibility is to map $0R1\mapsto 1$, $sR1\mapsto-1$, $0Rt\mapsto 0$ and
take the usual multiplication on the set $\\{-1,0,1\\}$. However, in this case
we find that $s\oplus s=0$ while $s+s=s$. So, it does not give a semi-
biproduct.

  2. (2)

Another possibility is to consider the chain semilattice structure on $R$ as
follows

+ | $0R1$ | $sR1$ | $0Rt$
---|---|---|---
$0R1$ | $0R1$ | $sR1$ | $0Rt$
$sR1$ | $sR1$ | $sR1$ | $0Rt$
$0Rt$ | $0Rt$ | $0Rt$ | $0Rt$

In this case we have

$x$ | $b$ | $x^{\prime}$ | $b^{\prime}$ | $x\oplus x^{\prime}$ | $b\times b^{\prime}$ | $b\cdot x$ | $x^{b}$
---|---|---|---|---|---|---|---
$0$ | $1$ | $0$ | $1$ | $0$ | $0$ | $0$ | $0$
$0$ | $t$ | $s$ | $1$ | $s$ | $0$ | $0$ | $0$
$s$ | $1$ | $0$ | $t$ | $s$ | $0$ | $s$ | $s$
$s$ | $t$ | $s$ | $t$ | $s$ | $0$ | $0$ | $0$

and since $x\oplus x^{\prime}=x+x^{\prime}$ and $x^{b}=x$ whenever $xRb$ (note
that $s^{t}=0$ but $(s,t)\notin R$) we obtain a semi-biproduct of monoids

$\textstyle{\\{0,s\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{1}}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{1}}$$\scriptstyle{\pi_{2}}$$\textstyle{\\{1,t\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{2}}$
(65)

with $\pi_{2}(x,b)=b$, $\iota_{2}(b)=(0,b)$, $\pi_{1}(x,b)=x$,
$\iota_{1}(x)=(x,1)$. It cannot be a Schreier extension because $R$ has three
elements while $X\times B$ has four elements. Note that this example is the
same as (7), from Introduction.

If we now consider the case when $X=\\{0,s\\}$ with $s+s=0$, while keeping
$B=\\{1,t\\}$ with $t^{2}=t$, then we observe that the same mapping
$0R1\mapsto 1$, $sR1\mapsto-1$, $0Rt\mapsto 0$ as before, with the usual
multiplication on the set $\\{-1,0,1\\}$, gives a semi-biproduct

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{\\{1,-1,0\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$
(66)

if we let $X\cong\\{1,-1\\}$ and $B\cong\\{1,0\\}$ have the structure of the
usual multiplication while $p,k,q,s$ are obvious inclusions and retractions.
Thus we see that even when $X$ is a group (but not $B$), we can have an
extension which is not a Schreier extension.

In the case when $B$ is a group, say $B=\\{1,b\\}$ with $b^{2}=1$, we may
still produce one example of a semi-biproduct as soon as we consider
$X=\\{1,a\\}$ and $A=\\{1,a,b\\}$ (compare with diagram (7)) with
$a^{2}=b^{2}=a$ and $ab=ba=b$ (note that in this case $X$ does not admit
cancellation, otherwise every semi-biproduct would be a Schreier extension as
observed in Remark 1, item 2.

A simple example that illustrates the case when $R$ is in bijection with
$X\times B$, that is when $x^{b}=x$, is the following. Put $X=B=\mathbb{N}$
the additive monoid of natural numbers and consider the order relation
$R=\\{(x,b)\in\mathbb{N}^{2}\mid x\geq b\\}$ together with the two maps
$q(x\geq b)=x-b$ and $u(b)=b$. The usual component-wise addition on $R$
verifies all the conditions specified by the construction scheme outlined
above and hence it gives rise to a semi-biproduct. Namely,
$(\mathbb{N},R,\mathbb{N},p,k,q,s)$ with $p(x\geq b)=b$, $k(x)=(x\geq 0)$,
$q(x\geq b)=x-b$ and $s(b)=(b\geq b)$. In this case $q$ and $s$ are both
homomorphisms and hence $x^{b}=x$ for every
$(x,y)\in\mathbb{N}\times\mathbb{N}$ so that we obtain an isomorphism

$\textstyle{\mathbb{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\scriptstyle{q}$$\scriptstyle{p}$$\textstyle{\mathbb{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{\mathbb{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\langle
1,0\rangle}$$\textstyle{\mathbb{N}\times\mathbb{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\pi_{1}}$$\scriptstyle{\pi_{2}}$$\textstyle{\mathbb{N}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\langle
0,1\rangle}$

with $\beta(x\geq b)=(x-b,b)$ and $\alpha(x,b)=(x+b\geq b)$. So we see that
the usual order relation on the natural numbers is part of a semi-biproduct of
monoids whose associated synthetic semi-biproduct is the familiar biproduct on
the natural numbers.

##  4\. The underlying extension

In this section we observe some slight adaptations of well-known results
suggesting that semi-biproducts can be studied in the light of
$S$-protomodular categories [5]. A semi-biproduct $(X,A,B,p,k,q,s)$ can be
seen as an extension of monoids $X\to A\to B$ equipped with a section $s$ and
a retraction $q$. Moreover, such extensions are pullback stable and the pair
$(k,s)$ is jointly epimorphic in the sense that every monoid homomorphism
$g\colon{A\to Y}$, from $A$ into a monoid $Y$, is completely determined by the
homomorphism $gk\colon{X\to Y}$ and the map $gs\colon{B\to Y}$, as
$g(a)=gkq(a)+gsp(a)$, for every $a\in A$. Similarly, the pair $(q,p)$ is
jointly monomorphic in the sense that a homomorphism $f\colon{Z\to A}$, from a
monoid $Z$ into the monoid $A$, is completely determined by the homomorphism
$pf\colon{Z\to B}$ and the map $qf\colon{Z\to X}$, as $f(z)=kqf(z)+spf(z)$,
for every $z\in Z$.

######  Proposition 1.

Let $(X,A,B,p,k,q,s)$ be a semi-biproduct of monoids. The morphism $k$ is the
kernel of the morphism $p$.

###### Proof.

Let $f\colon{Z\to A}$ be a morphism such that $pf=0$. Then the map
$\bar{f}=qf$ is a homomorphism

$\displaystyle qf(z+z^{\prime})$ $\displaystyle=$ $\displaystyle
q(fz+fz^{\prime})=q(kqf(z)+spf(z)+kqf(z^{\prime})+spf(z^{\prime}))$
$\displaystyle=$ $\displaystyle q(kqf(z)+0+kqf(z^{\prime})+0)$
$\displaystyle=$ $\displaystyle qk(qf(z)+qf(z^{\prime}))=qf(z)+qf(z^{\prime})$

and it is unique with the property $k\bar{f}=f$. Indeed, if $k\bar{f}=f$ then
$qk\bar{f}=qf$ and hence $\bar{f}=qf$. ∎

######  Proposition 2.

Let $(X,A,B,p,k,q,s)$ be a semi-biproduct of monoids. The morphism $p$ is the
cokernel of the morphism $k$.

###### Proof.

Let $g\colon{A\to Y}$ be a morphism and suppose that $gk=0$. It follows that
$g=gsp$,

$g=g1_{A}=g(kq+sp)=gkq+gsp=0+gsp=gsp,$

and consequently the map $\bar{g}=gs$ is a homomorphism,

$\displaystyle
gs(bb^{\prime})=gs(ps(b)ps(b^{\prime}))=gsp(sb+sb^{\prime})=g(sb+sb^{\prime})=gs(b)+gs(b^{\prime}).$

The fact that $\bar{g}=gs$ is the unique morphism with the property
$\bar{g}p=g$ follows from $\bar{g}ps=gs$ which is the same as $\bar{g}=gs$. ∎

######  Proposition 3.

Semi-biproducts of monoids are stable under pullback.

###### Proof.

Let $(X,A,B,p,k,q,s)$ be a semi-biproduct of monoids displayed as the bottom
row in the following diagram which is obtained by taking the pullback of $p$
along an arbitrary morphism $h\colon{C\to B}$, with induced morphism $\langle
k,0\rangle$ and map $\langle sh,1\rangle$,

(71)

We have to show that the top row is a semi-biproduct of monois. By
construction we have $\pi_{2}\langle sh,1\rangle=1_{C}$, $\pi_{2}\langle
k,0\rangle=0$, $q\pi_{1}\langle sh,1\rangle=qsh=0$, $q\pi_{1}\langle
k,0\rangle=qk=1_{X}$. It remains to prove the identity

$(a,c)=(kq(a),0)+(sh(c),c)=(kq(a)+sh(c),c)$

for every $a\in A$ and $c\in C$ with $p(a)=h(c)$, which follows from
$a=kq(a)+sp(a)=kq(a)+sh(c)$. ∎

The following list of examples illustrates some aspects of semi-biproducts
when considered as an extension with specified section and retraction. Let us
consider eight different diagrams displayed as

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
7.53471pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
14.6924pt\raise-7.58331pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{k}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
31.53471pt\raise-2.15277pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
31.53471pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
14.93666pt\raise 7.34027pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{q}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 7.53471pt\raise
2.15277pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
53.86743pt\raise 7.34027pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{p}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 70.964pt\raise 2.15277pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
70.964pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
53.98776pt\raise-6.65971pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{s}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 46.964pt\raise-2.15277pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces}}}}\ignorespaces,\quad
i=1,\ldots,8.$ (72)

Each monoid $A_{i}$ has underlying set $\\{0,a,b,c,d\\}$ and respective
operation defined by the table

$\displaystyle A_{1}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&c&d\\\
b&b&b&c&d\\\ c&c&d&d&d\\\ d&d&d&d&d\end{array}\right),$ $\displaystyle
A_{2}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&c&d\\\ b&b&b&c&d\\\
c&c&c&d&c\\\ d&d&d&c&d\end{array}\right),$ $\displaystyle
A_{3}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&c&c\\\ b&b&b&c&c\\\
c&c&c&c&c\\\ d&d&c&c&c\end{array}\right),$ $\displaystyle
A_{4}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&c&c\\\ b&b&b&c&c\\\
c&c&c&c&c\\\ d&c&c&c&c\end{array}\right)$

and

$\displaystyle A_{5}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&d&d\\\
b&b&b&d&d\\\ c&c&c&d&d\\\ d&d&d&d&d\end{array}\right),$ $\displaystyle
A_{6}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&d&d\\\ b&b&b&d&d\\\
c&d&d&d&d\\\ d&d&d&d&d\end{array}\right),$ $\displaystyle
A_{7}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&d&d\\\ b&b&b&d&d\\\
c&c&c&c&c\\\ d&d&d&d&d\end{array}\right),$ $\displaystyle
A_{8}=\left(\begin{array}[]{ccccc}0&a&b&c&d\\\ a&a&b&d&d\\\ b&b&b&d&d\\\
c&d&d&c&d\\\ d&d&d&d&d\end{array}\right).$

The monoid $X$ is the three element linear chain (sup-semi-lattice) whereas
the monoid $B$ is the two element linear chain. More specifically we consider
$X$ as the set $\\{0,a,b\\}$ and $B$ as the set $\\{0,c\\}$ with operation
tables

$\displaystyle X=\left(\begin{array}[]{ccc}0&a&b\\\ a&a&b\\\
b&b&b\end{array}\right),$ $\displaystyle B=\left(\begin{array}[]{cc}0&c\\\
c&c\end{array}\right).$

The given presentation for the monois $X$, $B$ and $A_{i}$ permits us to
consider $k$ as the inclusion homomorphism, $s$ as the inclusion map, $p$ as
the homomorphism defined by $p^{-1}(0)=\\{0,a,b\\}$, $p^{-1}(c)=\\{c,d\\}$ and
$q$ as the map defined by $q^{-1}(0)=\\{0,c\\}$, $q^{-1}(a)=\\{a,d\\}$,
$q^{-1}(b)=\\{b\\}$, in all the eight cases.

For each structure $(X,A_{i},B,p,k,q,s)$, $i=1,\ldots,8$, we observe:

  1. (1)

The cases 1 to 4 are not semi-biproducts because the element $d\in A$ cannot
be obtained as $d=kq(d)+sp(d)$, indeed $kq(d)+sp(d)=a+c=c\neq d$.

  2. (2)

The cases 5 to 8 are semi-biproducts of monoids. Indeed, we have $pk=0$,
$ps=1_{B}$, $qk=1_{X}$, $qs=0$ and we have $0=kq(0)+sp(0)=0+0$,
$a=kq(a)+sp(a)=a+0$, $b=kq(b)+sp(b)=b+0$, $c=kq(c)+sp(c)=0+c$ in all eight
cases. However, the condition $d=kq(d)+sp(d)=a+c$ is satisfied in $A_{5}$ to
$A_{8}$ but not in $A_{1}$ to $A_{4}$ as explained in the previous item.

  3. (3)

The inclusion map $s$ is a homomorphism in the cases 3, 4, 7 and 8 but not in
the cases 1, 2, 5, 6.

  4. (4)

The map $q$ is never a homomorphism. Indeed, in the cases 1 to 4 we have
$q(b+c)=q(c)=0$ whereas $q(b)+q(c)=b+0=b$. In the cases 5 to 8 we have
$q(b+c)=q(d)=a$ whereas $q(b)+q(c)=b+0=b$.

  5. (5)

In the cases 2, 4, 6 and 8 the monoid $A_{i}$ is commutative whereas in the
other cases it is not.

  6. (6)

In all cases $k$ is the kernel of $p$ but for example in the cases 1 and 2 the
pair $(q,p)$ is not jointly monomorphic in the sense above. Indeed, the
inclusion homomorphism $f\colon{Z\to A_{i}}$ with $Z=\\{0,d\\}$ is not
determined by $qf$ and $pf$ since

$kqf(d)+spf(d)=kq(d)+sp(d)=a+c=c\neq d.$

  7. (7)

In all cases except 2 the morphism $p$ is the cokernel of $k$. In the cases 5
to 8 by Proposition 2, in the cases 1, 3 and 4 because $p$ is a split
epimorphism (in case 1, $s$ is not a homomorphism but $s^{\prime}(c)=d$ is).
In the case 2 we see that the pair $(k,s)$ is not jointly epimorphic in the
sense above. For example, the homomorphism $g\colon{A_{2}\to Y}$, with
$Y=\\{0,c,d\\}$ obtained as $Y=A_{2}([1,4,5],[1,4,5])$, and defined by
$g^{-1}(0)=\\{0,a,b\\}$, $g^{-1}(c)=\\{c\\}$, $g^{-1}(d)=\\{d\\}$ is not
determined by $gk$ and $gs$. Indeed, $g(d)=d$ cannot be obtained as $gkq(d)=0$
and $gsp(d)=c$. In particular we observe that there is no $\bar{g}\colon{B\to
Y}$ such that $g=\bar{g}p$.

The previous propositions are simple adaptations of well-known results,
especially when $s$ is a homomorphism. Moreover, since monoid monomorphisms
are injective maps we observe further that each pair $(k,s)$ in a semi-
biproduct is jointly extremal-epimorphic. Indeed, given any monomorphism
$m\colon M\to A$ for which there exists a homomorphism $l\colon X\to M$ and a
map $t\colon B\to M$ such that $ml=k$ and $mt=s$, we have
$a=kq(a)+sp(a)=m(lq(a)+tp(a))$, and so $m$ is an isomorphism. Note that this
argument is not dual because not every epimorphism is a surjective map.

##  5\. Conclusion

We have introduced the notion of semi-biproduct of monoids. Considerations on
categorical aspects related to this notion can be found in [16]. Several
directions using different techniques are pointed out there. The main issue
involves dealing with morphisms and maps on the same ground. One possibility
is to consider maps as an extra structure in higher dimensions [6, 15].
Another one is to consider maps as imaginary morphisms [2, 19]. It seems to be
relevant the investigation on both directions and to determine whether they
lead to different settings or if rather there is a more general setting which
unifies both directions. Developing a general categorical framework in which
to study semi-biproducts is desirable due to several interesting cases that
occur in different contexts. For example, semi-biproducts can be studied in
the context of preordered monoids (see [17], see also [7] for preordered
groups) where the maps $q$ and $s$ can be required to be monotone maps rather
than merely zero-preserving maps. The context of topological monoids [10] is
also worthwhile studying since we can take the maps $q$ and $s$ to be
continuous or not and possible variations on that.

## Acknowledgements

The author would like to thank the referee for a very careful reading of the
text, which led to the current improved version; and the Editor for helpful
suggestions on Section 4. This work was supported by Fundação para a Ciência e
a Tecnologia (FCT UID-Multi-04044-2019), Centro2020 (PAMI – ROTEIRO/0328/2013
– 022158) and by the Polytechnic of Leiria through the projects
CENTRO-01-0247: FEDER-069665, FEDER-069603, FEDER-039958, FEDER-039969,
FEDER-039863, FEDER-024533.

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