Cross-Connections

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Introduction

The concept of cross-connection between partially ordered sets was originally introduced by Grillet (1974, 1974b, 1974c) in order to classify the class of fundamental regular semigroups. Grillet defines a cross-connection between two regular partial ordered sets $I$ and $\Lambda $ as a pair of order-preserving mappings

\begin{equation} \label{eq:1} \Gamma:I\to\Lambda^{*} \quad\text{and}\quad\Delta:\Lambda\to I^{*} \end{equation} satisfying certain axioms, where $I^{*}$ denote the dual of $I$. See Grillet (1974b) for definition of regular partially ordered sets, duals and axioms for cross-connections. Every regular semigroup $S$ induces a cross-connection between the partially ordered set $\Lambda (S)$ of all principal left ideals of $S$ under inclusion and the partially ordered set $I(S)$ of all principal right ideals. In Nambooripad (1994), this was generalized to cross-connections between normal categories. Again, given a regular semigroup $S$, category $\lcat $ of all principal left ideals is normal and similarly, the category $\rcat $ of all principal right ideals is also normal (see Propositions 4.3 and 4.4 for definitions of these categorioes). We can see that every regular semigroup $S$ induces a cross-connection between categories $\lcat $ and $\rcat $.

We observe that the configuration that occur in Grillet’s definition as well as its generalization in Nambooripad (1994) occures in many areas in mathematics. Our aim here is to describe cross-connections of a more general class of categories. We first identify an appropriate class of categories for which cross-section can be defined. These are categories with subobjects in which objects are sets, morhisms are mappings and inclusions are inclusions of sets. We refer to these categories as set-based categories ($\relax \mathcal {S}$-category for short). A cross-connection between to $\relax \mathcal {S}$-categorys $\cc $ and $\cd $ consists of two set-valued bifunctors \[ \Ga :\cc \times \cd \to \set \quad \text {and}\quad \De :\cc \times \cd \to \set \] and a natural isomorphism \[ \chi :\Ga \to \De \] between them. We can show that every cross-connection determines a semigroup. Also for any semigroup $S$ there is a cross-connection between $\lcat $ and $\rcat $ and there is a natural representation of $S$ by the cross-connection semigrpoup determined by the cross-connection between $\lcat $ and $\rcat $.

In the following discussion, we will follow Nambooripad (1994) for ideas regarding categories, subobject relations, etc. Moreover, to save repetition, we shall assume that categories under consideration are small, unless otherwise provided, so that they may be treated as partial algebras.

Categories with subobjects

We assume that the reader is familiar with the basic concepts such as categories, functors, natural transformations, etc. see for example (MacLane 1971). For the more specialized concepts like categories with subobjects and related ideas, we follow (Nambooripad 1994). We begin by reviewing briefly the definition of categories with subobjects.

Subobjects

Preorders

Recall that two morphisms $f,g\in \preord $ are parallel if $\dom f=\dom g$ and $\cod f=\cod g$. A preorder $\preord $ is a category such that for all $f,g\in \preord $, $f$ is parallel to $g$ if and only if $f=g$. Equivalently, $\preord $ is a preorder if and only if $\preord (p,q)$ contain atmost one element for all $p,q\in \vrt \preord $. It follows that the relation $\preceq $, defined for all $p,q\in \vrt \preord $, by \begin{equation*} p\preceq q\iff \preord(p,q)\ne\emptyset \end{equation*} is a quasi-order on the class $\vrt \preord $. In particular, if $\preord $ is small, then $(\vrt \preord ,\preceq )$ is a quasi-ordered set. Conversely, it is clear that any quasi-ordered set $\La $ uniquely determines a small preorder $\preord $ such that $\La =\vrt \preord $. The preorder $\preord $ is said to be strict if the quasi-order relation $\preceq $ above is a partial order; that is, $\preord $ has the property that for all $p,q\in \vrt \preord $, \[ \preord (p,q)\ne \emptyset \quad \text {and}\quad \preord (q,p)\ne \emptyset \implies p=q. \] The concepts of a small strict preorder and a partially ordered set are equivalent and will be used interchangeably in the sequel.

Choice of subobjects:

Recall (Krishnan 2000) that a choice of subobject in a category $\cc $ is sub-preorder $\preord \subseteq \cc $ satisfying the conditions:

1.$\preord $ is a strict preorder with $\vrt \preord =\vrt \cc $.

2.Every $f\in \preord $ is a monomorphism in $\cc $.

3.If $f,g\in \preord $ and if $f=hg$ for some $h\in \cc $, then $h\in \preord $.

Categories with subobjects

If $\preord $ is a choice of subobjects in $\cc $, the pair $(\cc ,\preord )$ is called a category with subobjects. For brevity, we shall say that a category $\cc $ has subobjects if a choice of subobjects in $\cc $ has been specified. Also, we shall use the notation $\vrt \cc $ to denote the preorder of subobjects as well as the partially ordered set of vertexes in $\cc $. No ambiguity will arise since both these are equivalent. We may then use the usual notation $\subseteq $ (see (Krishnan 2000)) to denote subobject relation in $\cc $. If $c\subseteq d$, the unique morphism in $\vrt \cc $, the inclusion of $c$ in $d$, is denoted by $\inc {c}{d}$. If $f\in \cc (c,d)$ and $c'\subseteq c$, then we write \[ f\mid c'=\inc {c'}{c}\circ f. \] As usual, $f\mid c'$ is called the restriction of $f$ to $c'$.

In the following $\cc , \cd $, etc. stands for catagories with subobjects in which $\vrt \cc , \vrt \cd $, etc. denote the corresponding preorder of subobjects. A functor $F:\cc \to \cd $ is said to be inclusion preserving if its vertex map $\vrt F$ is an order-preserving map of $\vrt \cc $ to $\vrt \cd $; that is, $\vrt F:\vrt \cc \to \vrt \cd $ is a functor of preorders. $F$ is an embedding if $F$ is faithful and $\vrt F$ is an order-embedding of $\vrt \cc $ into $\vrt \cd $. In particular, $\cc $ is a subcategory (with subobjects) of $\cd $ if $\cc \subseteq \cd $ as partial algebras. In this case, the inclusion $\subseteq $ is a category embedding of $\cc $ in $\cd $ whose vertex map is $\vrt \cc \subseteq \vrt \cd $.

The category $\set $ is clearly a category with subobjects in which the subobject relation coincides with usual set-theoretic inclusion. Similarly categories of groups $\Grp $, abelian groups $\abgrp $, etc., are categories with natural subobject relations.

It is clear that there is a category $\scat $ whose objects are small categories with subobjects and morphisms are inclusion preserving functors. Further, the assignments \[ \vrt {}:\cc \mapsto \vrt \cc ,\quad \text {and}\quad F\mapsto \vrt F \] is a functor of the category $\scat $ to the category of preorders (or the category of partially ordered sets).

Remark 2.1:

Notice that an abstract category $\cc $ may have more than one possible choices of subobjects. For example, if $\cc $ is a small category of sets, the the usual relation of inclusion among vertexes that are morphisms in $\cc $, gives a choice of subobjects in $\cc $. Also, identity on $\vrt \cc $ is also a choice of subobjects which may be different from the one given in the last sentence.

Categories with factorization

Equivalent monomorphisms

Recall that two monomorphisms $f,g\in \cc $ are equivalent if there exists $h,k\in \cc $ with $f=hg$ and $g=kf$ (see (Nambooripad 1994), Equation  I.4(15)). The fact that $f$ and $g$ are monomorphisms imply that the morphism $h$ is an isomorphism and $k=h^{-1}$. A monomorphism $f\in \cc (c,d)$ is called an embedding if $f$ is equivalent to an inclusion. Clearly, every inclusion is an embedding. It is important to notice that no two inclusions can be equivalent as mopnomorphisms (Krishnan 2000, see Lemma 1.10).

Equivalent epimorphisms

Eqivalence of epimorphisms are defined dually. As above, one can see that two epimorphisms $f,g\in \cc $ are equivalent if and only if there is an isomorphism $h\in \cc $ with $g=fh$.

Retractions

Recall also that a monomorphism $f:c\to d$ [epimorphism $h:c\to d$] splits if there exists $g:d\to c$ [$k:d\to c$] such that $fg=1_{c}$ [$kh=1_{d}$]. Notice that when $f$ is a split monomorphism, then $g$ is a split epimorphism and similarly if $h$ is a split epi, $k$ is a split mono. If an inclusion $\inc {c}{d}$ is said to be split if it is split as a monomorphism; that is, there exist $\e \in \cc $ with $\inc {c}{d}\e =1_{c}$. In this case, $\e $ is called a retraction which is clearly a split epimorphism.

Factorization property

A category $\cc $ with subobjects is said to have factorization property if every morphism $f\in \cc $ can be factorized as $f=kh$ where $k$ is an epimorphism and $h$ is an embedding. In this case we can choose $k$ and $h$ so that $h$ is an inclusion (see (Krishnan 2000), §  3.2.2). A factorization of the form $f=qj$ where $q$ is an epimorphism and $j$ is an inclusion, is called a canonical factorization. Thus $\cc $ has factorization property if and only if every morphism has canonical factorization. $\cc $ is said to have unique factorization if every morphism in $\cc $ has unique canonical factorization.

Image

Let $f$ be a morphism in the categopry $\cc $ with factorization. A canonical factorization $f=qj$ of $f$ is called an image factorization if $f=qj$ has the following universal property: if $f=q'j'$ is any canonical factorization of $f$, there is an inclusion $j$ with $q'=qj$ (see (Krishnan 2000), §  3.2.3). The factorization $f=qj$ with the universal property above, if it exists, is unique. The epimorphism $q=f^{\circ }$ is called the epimorphic component of $f$. If $j$ is the inclusion such that $f=f^{\circ }j$ then the vertex \begin{equation} \label{eq:2} \im f=\cod f^{\circ}=\dom(j) \end{equation} is called the image of $f$. Also the image facorizatrion of $f$ is the unique canonical factorzation \[ f=f^{\circ }\inc {\im f}{\cod f}. \] A morphism $f$ such that $f^{\circ }=f$ is said to be surjective. Note that any surjection is an epimorphism but not conversely. Also, the universal property of the image-factorization implies that, the epimorphic component $f^{\circ }$ of a morphism $f$ is surjective; that is, $(f^{\circ })^{\circ }=f^{\circ }$.

The category $\cc $ is said to have images if every morphism in $\cc $ has image.

Inverse image

If $\cc $ has images, and $f\in \cc $ then, as usual, we may write \begin{equation*} f(a)=\im(f\mid a)\quad\text{for all}\quad a\subseteq\dom f. \end{equation*} This defines a mapping \begin{equation*} \vmap[f]:a\mapsto f(a) \quad\text{of}\quad P(\dom f) \quad\text{to}\quad P(\cod f), \end{equation*} of the principal order ideal of the partially ordered set $P=\vrt \cc $ of subobjects of $\dom f$ to the partially ordered set of subobjects of $\cod f$. It is easy to see that the map $\vmap [f]$ is order-preserving. Moreover, for $f\in \cc $ and $x\subseteq \cod f$ we write \begin{equation} \label{eq:3} f^{-1}(x)=\max\{y\subseteq\dom f:f(y)\subseteq x\}. \end{equation} If the vertex $f^{-1}(x)$ exists it is called the inverse image or preimage of $x$. Observe that, for all $x\subseteq \cod f$, the set \begin{equation*} f^{-1}(P(x))=\{y\subseteq\dom f:f(y)\subseteq x\} \end{equation*} is an order ideal in the partially ordered set $P=\vrt \cc $ and the preimage of $x$ exists if and only if this ideal is non-empty and principal.

Set based categories

A category $\cc $ is concrete (see (Krishnan 2000), § 1̃.3) if there is a faithful functor $U:\cc \to \set $. Since, by Proposition(1.9) of (Krishnan 2000), small categories are concrete, the existence of such functors does not impose any additional restriction on the category under consideration. Here we shall consider a class $\setcat $ of small subcategories of $\set $ with images in which surjections are surjective maps (surjective morphisms in $\set $). In the sequel, $\set $ will be considered as a category with subobjects with respect to the set-theoretic inclusion. Unless otherwise stated, $\cc $, $\cd $, etc. will denote categories with images. Recall that, if $\cc $ is a subcategory of $\set $, then the inclusion $\inc {\cc }{\set }: \cc \subseteq \set $ is a category embedding of $\cc $ into $\set $.

Definition 3.1 (Definition of set-based categories).

Let $\cc $ be a small category and let $U:\cc \to \set $ be a functor. We say that $\cc $ is set-based ($\relax \mathcal {S}$-category for short) with respect to $U$ if the pair $(\cc ,U)$ satisfy the following:

(Sb:a)$U$ is an embedding.

(Sb:b)$U(\im f)=\im U(f)$ for all $f\in \cc $.

(Sb:c)The functor $U$ has the following property: for $c,c'\in \vrt \cc $ and $x\in U(c)\cap U(c')$ there is $d\in \vrt \cc $ such that

\[ d\subseteq c,\quad d\subseteq c'\quad \text {and}\quad x\in U(d). \]

Notice that the condition (Sb:b) implies that the functor $U$ preserves imagefactorizations:

Lemma 3.1.

Let $\cc $ be a category with images satisfying the condition (Sb:a). Then it satisfies (Sb:b) if and only if

(Sb:b)$^{*}$$U(f)^{\circ }=U(f^{\circ })$ for all $f\in \cc $.

If (Sb:b) (or (Sb:b)$^{*}$) hold, then

\begin{equation} \label{eq:4} U\left(f^{\circ}\inc{\im f}{\cod f}\right)=U(f)^{\circ}\inc{\im U(f)}{\cod U(f)} \end{equation} for all $f\in \cc $. Moreover, we have \begin{equation} \label{eq:5} U(f\mid c)=U(f)\mid U(c)\quad\text{and}\quad U\left(f(c)\right)=U(f)\left(U(c)\right) \end{equation}

for any $f\in \cc $ and $c\subseteq \dom f$.

Proof.

If (Sb:b)$^{\circ }$ holds for all $f\in \cc $ then

\[ \im U(f)=\cod U(f)^{\circ }=\cod U(f^{\circ })=U(\cod f^{\circ }) =U(\im f) \] and so, (Sb:b) holds. Conversely, assume that (Sb:b) is satisfied. Then, for any $f\in \cc $, \begin{align*} U(f)&=U\left(f^{\circ}\inc{\im f}{\cod f}\right)\\ &=U\left(f^{\circ}\right)\inc{U(\im f)}{U(\cod f)} &&\text{by (Sb:a)}\\ &=U\left(f^{\circ}\right)\inc{\im U(f)}{\cod U(f)} \end{align*}

using (Sb:b) and the fact that $U$ is a functor. The image factorization of $U(f)$ in $\set $ gives \begin{align*} U(f)&=U(f)^{\circ}\inc{\im U(f)}{\cod U(f)}. \end{align*}

Since $\inc {\im U(f)}{\cod U(f)}$ is a monomorphism, (Sb:b)$^{\circ }$ follows. The equality of \eqref{eq:4} also follows from the proof above.

By definition $f\mid c =\inc {c}{\dom f} f$ for any $c\subseteq \dom f$. Since $U$ is an inclusion preserving functor, we have \begin{align*} U(f\mid c) &= U(\inc{c}{\dom f})U(f) \\ & = \inc{U(c)}{U(\dom f)}U(f) \\ & = U(f)\mid U(c). \end{align*}

Since $f(c)=\im (f\mid c)$ by definition, we get \begin{align*} U\left(f(c)\right) & = U\left(\im(f\mid c)\right) = \im U(f\mid c) \\ & = \im\left(U(f)\mid U(c)\right) = U(f)\left(U(c)\right). \end{align*}

This proves \eqref{eq:5}.

It is clear that there is a (large) category $\SBC $ in which objects are $\relax \mathcal {S}$-categories and morphisms are inclusion-preserving functors that preserve images.

The following is an immediate consequence of the definition of $\relax \mathcal {S}$-category which will be useful in the sequal.

Corollary 3.2.

Let $\cc $ be an $\relax \mathcal {S}$-category with respect to the forgetful functor $U:\cc \to \set $. Assume that $\cd $ is a subcategory of $\cc $ having images such that the inclusion $\inc {\cd }{\cc }$ preserve images of morphisms in $\cd $. Then $\cd $ is an $\relax \mathcal {S}$-category withrespect to $U\vert \cd $.

Let $\cd $ be a subcategorty of the $\relax \mathcal {S}$-category satisfying the requtrements of the corollary above. Ihen $\cd $ will be called an $\relax \mathcal {S}$-subcategory of $\cc $.

Let $\cc $ be a category with images and $U:\cc \to \set $ be an embedding. Let \begin{equation} \label{eq:6} \im U=\left(\{U(c):c\in\vrt\cc\},\{U(f):f\in\cc\}\right) \end{equation} Let $f,g\in \cc $. Since $U$ is an embedding the composition $U(f)U(g)$ exists in $\set $ if and only if $fg$ exists in $\cc $ and \[ U(fg)=U(f)U(g). \] Consequently $\im U$ is a subcategory of $\set $. Since $U$ is an embedding $c\subseteq d$ in $\vrt \cc $ if and only if $U(c)\subseteq U(d)$ in $\set $. Hence $\im U$ is a category with subobjects in which the subobject relation is the set-theoretic inclusion. Notice that $\im U\subseteq \set $ is an embedding of $\im U$ into $\set $. It is routine to verify the following.

Proposition 3.3.

Let $U:\cc \to \set $ be an embedding of a category $\cc $ with subobjects to $\set $. Let $\im U$ denote the category with subobjects defined above. Then $\im U$ is a subcategory of $\set $ and the inclusion $\im U\subseteq \set $ is an embedding. Moreover $U:\cc \to \im U$ is an isomorphism of categories with subobjects and $\cc $ is a $\relax \mathcal {S}$-category with respect to $U$ if and only if $\im U$ is an $\relax \mathcal {S}$-category with respect to inclusion.

The result above shows that we may replace a $\relax \mathcal {S}$-category $\cc $ by $\im U$ in which objects are sets, morphisms are mappings of sets and the image-factorizations of morphisms in $\im U$ are those factorizations in $\set $. Consequently it may be possible to replace categorical manipulations in $\cc $ by elementary set-theoretic manipulations. For example, we have the following computations (cf. \eqref{eq:7}) of the image and the surjection of the composite of a pair of composable morphisms which is useful in the sequel. One can see, by elementary set-theoretic arguements, that \eqref{eq:7} holds in the category $\set $. In the next proof we use this fact and the properties of $\relax \mathcal {S}$-categories to extend the validity of \eqref{eq:7} to arbitrary $\relax \mathcal {S}$-categories. Recall that two morphisms $f,g\in \cc $ are parallel if they belong to the same home-set; that is, $\dom f=\dom g$ and $\cod f=\cod g$.

Proposition 3.4.

Suppose that $\cc $ is an $\relax \mathcal {S}$-category and that $f, g$ is a pair of composable morphisms in $\cc $. Then

\begin{equation} \label{eq:7} %%eq:22=Equation 9 \im(fg)=g\left(\im f\right)\quad\text{and}\quad (fg)^{\circ}= f^{\circ}\left(g\mid \im f\right)^{\circ}. \end{equation}

Moreover, if $g$ and $h$ are parallel and if $g^{\circ }=h^{\circ }$ then $(fg)^{\circ }=(fh)^{\circ }$ for all $f$ for which $fg$ exists and $(gk)^{\circ } =(hk)^{\circ }$ for all $k\in \cc $ for which $gk$ exists.

Proof.

The hypothesis implies that $f:c\to d$ and $g:d\to e$ for some $c,d,e\in \vrt \cc $. Then $\im f\subseteq d$ and

\begin{align*} U\left(\im(fg)\right) &=\im \left(U(f)U(g)\right) &&\quad\text{by Lemma 1}\\ &=U(g)\left(\im U(f)\right) &&\quad\text{since (7) hold in } \set\\ &=U\left(g(\im f)\right) &&\quad\text{by (5)} \end{align*}

Thus $U\left (\im (fg)\right )=U\left (g(\im f)\right )$. Since $U$ is an embedding, $\vrt U$ is injective and so $\im (fg) =g(\im f)$. \begin{align*} U\left((fg)^{\circ}\right)&=\left(U(fg)\right)^{\circ} &&\quad\text{by Lemma 1} \\ &=\left(U(f)U(g)\right)^{\circ} \\ &=U(f)^{\circ}\left(U(g)\mid \im U(f)\right)^{\circ} 7&&\quad\text{since (7) holds in } \set \\ &=U(f^{\circ})U\left((g\mid \im f)^{\circ}\right) &&\quad\text{by (5)}\\ &=U\left(f^{\circ}(g\mid \im f)^{\circ}\right). \end{align*}

Also by the first equality in (??), \[ \im (fg)=\cod (fg)^{\circ }=g(\im f)=\cod (g\mid \im f)^{\circ } \] and so $(fg)^{\circ }, f^{\circ }(g\mid \im f)^{\circ }\in \cc (\dom f, \im (fg))$. Since $U$ is faithful, we have $(fg)^{\circ }= f^{\circ }(g\mid \im f)^{\circ }$. This proves \eqref{eq:7} for all $\relax \mathcal {S}$-categories.

Now, to prove the remaining statements, let $g^{\circ }=h^{\circ }$. Suppose that $\dom g=\dom h=d$. We observe that for all $a\subseteq d$, by definition \begin{equation*} (g\mid a)^{\circ} =(\inc{a}{d}g^{\circ}\inc{\im g}{\cod g})^{\circ} =(\inc{a}{d}g^{\circ})^{\circ}=(\inc{a}{d}h^{\circ})^{\circ}=(h\mid a)^{\circ}. \end{equation*} Hence if $fg$ (or $fh$) exists, then \begin{equation*} (fg)^{\circ}=f^{\circ}(g\mid \im f)^{\circ} = f^{\circ}(h\mid \im f)^{\circ} =(fh)^{\circ}. \end{equation*} by (??). Since $g$ and $h$ are parallel, $gk$ exists if and only if $hk$ exists. If this holds, we have $\im g=\cod g^{\circ } =\cod h^{\circ }=\im h$ and again by \eqref{eq:7} \begin{equation*} (gk)^{\circ}=g^{\circ}(k\mid \im g)^{\circ}=h^{\circ}(k\mid \im h)^{\circ}=(hk)^{\circ}. \end{equation*}

This complets the proof.

Remark 3.1:

Since epimorphisms are surjective in the categories $\Grp $, $\abgrp $, $\vct $, etc., they could regarded as (large) $\relax \mathcal {S}$-categories. Notice that these categories also have unique factorization property. Also any partial order is a $\relax \mathcal {S}$-category but don’t have unique factorization.

Example 3.1:

Let $\cc $ denote the category with vertex set $\vrt \cc =\{x,y\}$ and morphism set $\cc =\{1_{x},\al :x\to y,1_{y}\}$. Setting identity relation on $\vrt \cc $ as the subobject relation, $\cc $ becomes a category with factorization in which $\al $ is a surjection. Furthermore, $\cc $ has images. Assume that $X$ and $Y$ are disjoint sets and $f:X\to Y$ is a map such that $\im

f=Y' \subset Y$ ($Y'\ne Y$). Then the assignments \begin{equation*} U:x\mapsto X,\quad y\mapsto Y\quad\text{and}\quad \al\mapsto f \end{equation*} is an embedding of $\cc $ into $\set $. However, $\cc $ is not an $\relax \mathcal {S}$-category with respect to $U$. On the other hand, if $g:X\to Y$ is a surjection, then the assignments \begin{equation*} V(x)=X,\quad V(y)=Y\quad\text{and}\quad V(\al)=g \end{equation*}

is an embedding of $\cc $ into $\set $ such that $\cc $ is an $\relax \mathcal {S}$-category with respect to $V$. Similarly one can construct examples of distinct functors $U$ and $V$ such that $\cc $ is an $\relax \mathcal {S}$-category with respect to both $U$ and $V$.

It is clear from the remarks above that the structure of a $\relax \mathcal {S}$-category $(\cc ,U)$ depends both on the structure of $\cc $ as well as $U$. However, the following proposition discuss a class of categories $\cc $ (with images) that are $\relax \mathcal {S}$-categories with respect to functors $U$ where the functor $U$ is completely determined by the structure of $\cc $ itself. These are useful for discussing regular categories.

Proposition 3.5.

Let $\cc $ be a small category with images. Define assignments $\vrt U:\vrt \cc \to \vrt \set $ and $U:\cc \to \set $ as follows: for each $c\in \vrt \cc $

\begin{equation} \label{eq:7a} U(c)=\{f^{\circ}:f\in\cc,\; \im f\subseteq c\} \end{equation} and for each morphism $f:c\to d\in \cc $, let \begin{equation} \label{eq:7b} U(f):g^{\circ}\in U(c)\mapsto (gf)^{\circ}\in U(d). \end{equation}

Then $U:\cc \to \set $ is a functor satisfying the following:

(a)The functor $U:\cc \to \set $ is an embedding.

(b)For $a,b\in \vrt \cc $ and $h=h^{\circ }\in U(a)\cap U(b)$ there is $c'\in \vrt \cc $ such that $h\in U(c')\subseteq U(a)\cap U(b)$.

(c)$f\in \cc $ is a monomorphism if and only if $U(f)$ is injective.

(d)$f\in \cc $ is a split-epimorphism if and only if $U(f)$ is surjective.

Consequently $(\cc ,U)$ is an $\relax \mathcal {S}$-category if and only if every surjection in $\cc $ is a split-epimorphism.

Proof.

For each $c\in \vrt \cc $, $U(c)$ is a set and for each $f\in \cc (c,d)$, $U(f):g^{\circ }\in U(c)\mapsto (gf)^{\circ }$ is a well-defined map by Proposition 3.4. If $f:c\to d$ and $g:d\to e$ then for any $h^{\circ }\in U(c)$,

\begin{align*} && (h^{\circ})U(f)U(g)&=\left((hf)^{\circ}\right)U(g) && \\ && &=\left(h(fg)\right)^{\circ}=(h^{\circ})U(fg). \end{align*}

Hence $U:\cc \to \set $ is a functor. Taking $f=\inc {c}{d}$ in the above computation, we see that $U(\inc {c}{d})=\inc {U(c)}{U(d)}$. Hence $U:\cc \to \set $ is an inclusion-preserving functor.

(a) If $c\subseteq d$ and $f^{\circ }\in U(c)$ then $\im f \subseteq c \subseteq d$. Hence $f^{\circ }\in U(d)$ and so, $U(c)\subseteq U(d)$. Further, by Proposition 3.4 (see Equation (??)) \[ (f^{\circ })U(\inc {c}{d})=(f\inc {c}{d})^{\circ }=f^{\circ } =(f^{\circ })\inc {U(c)}{U(d)} \] Hence $U(\inc {c}{d})=\inc {U(c)}{U(d)}$. Conversely if $U(c) \subseteq U(d)$, then $1_{c}\in U(c)$ and so $1_{c}\in U(d)$. This gives $c\subseteq d$. Consequently, $\vrt U$ is an order-embedding and so, $U:\cc \to \set $ is an embedding if it can be shown that $U$ is faithful. To this end assume that $f,g\in \cc (c,d)$ and $U(f)= U(g)$. Then by \eqref{eq:7b}

\begin{equation*} 1_{c}U(f)=f^{\circ}=1_{c}U(g)=g^{\circ}. \end{equation*} Hence $f=g$. This proves that $U$ is faithful and so (a) is proved.

(b) Let $h=h^{\circ }\in U(c)\cap U(d)$. Then, clearly, $c'=\cod h \subseteq c$ and and $c'\subseteq d$. Hence $h\in U(c') \subseteq U(c)\cap U(d)$.

(c) Suppose that $f:c\to d$ is a monomorphism. If $(h^{\circ })U(f) =(g^{\circ })U(f)$ then $(hf)^{\circ }=(gf)^{\circ }$ and so, $hf=gf$. Since $f$ is a monomorphism, $h=g$ and so, $h^{\circ }=g^{\circ }$. Therefore $U(f)$ is injective. Conversely if $U(f)$ is injective and if $hf=gf$ then \[ (h^{\circ })U(f)=(hf)^{\circ }=(gf)^{\circ }=(g^{\circ })U(f). \] Since $U(f)$ is injective, $h^{\circ }=g^{\circ }$ and so, $h=g$. Hence $f$ is a monomorphism.

(d) Suppose that $f:c\to d\in \cc $ such that $U(f)$ is surjective. Since $1_{d}\in U(d)$, there exist $h=h^{\circ }:d\to c' \subseteq c$ such that $(h)U(f) = (hf)^{\circ }= 1_{d}$. Since $\cod hf = d$, we have $hf = 1_{d}$ and so, $f$ is a split epimorphism. On the other hand, if $f:c\to d$ is a split epimorphism then $hf=1_{d}$ for some $h:d\to c$. Then

$U(h)U(f)=1_{U(d)}$ and so, $U(f)$ is surjective. This complete the proof of (d).

If $\cc $ has images, statements (a) and (b) shows that $(\cc ,U)$ satisfies axioms (Sb:a) and (Sb:c) of Definition 3.1. Suppose that every surjection in $\cc $ split. Then by (d), $U(f^{\circ })$ is surjective for all $f\in \cc $. Hence $U(f^{\circ })=(U(f))^{\circ }$ and so by Lemma 3.1, $\cc $ satisfies axiom (Sb:b). Therefore $\cc $ is a

$\relax \mathcal {S}$-category. Conversely, if $\cc $ is an $\relax \mathcal {S}$-category then by Lemma 3.1, $U(f^{\circ }) = \left (U(f)\right )^{\circ }$ for all $f\in \cc $. Hence, by (d), every surjection split.

Given a category $\cc $ with images, we shall refer to the embedding $U:\cc \to \set $ constructed above as the natural embedding of $\cc $ into $\set $. $\cc $ is called a $\relax \mathcal {NS}$-category if it is a $\relax \mathcal {S}$-category with respect to its natural embedding.

The categories of left and right modules of a semigroup

Let $S$ be a semigroup. An action of $S$ on the left of a set $X$ is a map \begin{gather} a_{X}:S\times X\to X; \quad (s,x)\mapsto s.x\quad \text{such that }\quad s.(t.x)=(st).x \end{gather} for all $s,t\in S$. Often we shall write $sx$ for $s.x$ if there is no ambiguity. A left $S$-set (or a left $S$-module) is a pair $(X,a_{X})$ where $X$ is a set and $a_{X}$ is an action of $S$ on the left of $X$. A [left] $S$ map $\al :X\to Y$ from a left $S$-set $X$ to a left $S$-set $Y$ (or a left $S$-morphism) is a map $\al : X\to Y$ that satisfy the condition

\begin{equation} \label{eq:smap} (sx)\al=s(x\al) \quad\text{for all}\quad s\in S,\;\text{and}\; x\in X. \end{equation} We denote by $\Lmod [S]$ the category with $S$-modules as objects and $S$-maps as morphisms. There is a functor $U: \Lmod [S]\to \set $ sending every $X\in \vrt \Lmod $ to the underlying set $U(X)$ and every morphism $f:X\to Y\in \Lmod [S]$ to the map $U(f)$ determined by $f$. In the sequal, we refer to $U$ as the foregetful functor for left-modules. A subobject relation on $\vrt \Lmod $ is defined as follows: For $X\in \vrt \Lmod $ let $a_{X}:S\times U(X)\to U(X)$ denote the action of $S$ on $U(X)$ given by the module structure on $X$. For $X, Y\in \vrt \Lmod $ write \begin{equation} \label{eq:8} X\subseteq Y \iff U(X)\subseteq U(Y)\quad\text{and}\quad a_{X} =a_{Y}\mid (S\times U(X)). \end{equation} The relation on $\vrt \Lmod $ defined above is a subobject relation on $\Lmod [S]$. In fact, it is easy to check that the $J=\inc {X}{Y}$ is a morphism in $\Lmod [S]$ if and only if $X$ and $Y$ satisfy \eqref{eq:8}. The class of all these morphisms form a choice of subobjects in $\Lmod [S]$ (see 2.1.2). Notice that, if $X$ and $Y$ are submodules of $Z\in \Lmod $ then it follows from \eqref{eq:8} that $X\subseteq Y$ if and only if $U(X)\subseteq U(Y)$.

It is clear that the foregetful functor $U:\Lmod \to \set $ is faithful and preserve images. It follows that $(\Lmod , U)$ satisfies axiom (Sb:b) of Definition 3.1. If $X,Y\in \vrt \Lmod $ and $x\in U(X)\cap U(Y)$, then $D=\left \{sx:s\in S^{1}\right .$ is a lefmodule such that $x\in D\subseteq X$ and $x\in D\subseteq Y$. Therefore $(\Lmod , U)$ satisfies axiom (Sb:c) also. If $\cc \subseteq \Lmod $ is a small subcategory of $\Lmod $ with imagages and if $U_{0}=U\vert \cc $ then $(\cc ,U_{0})$ clearly satisfy axioms (Sb:b) and (Sb:c). Further, if the partially ordered set $\vrt \cc $ of leftmodules has a largest member $Z$, then it follows from \eqref{eq:8} that the functor $U_{0}$ is an embedding. In this case $\cc $ is an $\relax \mathcal {S}$-category with respect to $U_{0}$. For convenience of using these facts in the sequal we state as a propositionm below.

Proposition 4.1.

Let $\Lmod [S]$ be the category with subobjects in which objects are left $S$ modules, morphisms are left module morphisms and the subobject relation is defined by \eqref{eq:8}. The category $\Lmod [S]$ has images and the foregetful functor $U:\Lmod [S]\to \set $ is faithful and preserve subobjects. Moreover, the pair $(\Lmod , U)$

satisfy axioms (Sb:b) and (Sb:c) of Definition 3.1. If $\cc \subseteq \Lmod $ is a small subcategory of $\Lmod $ with images and if $U_{0}=U\vert \cc $ then $(\cc ,U_{0})$ satisfy axioms (Sb:b) and (Sb:c). Further, if there is a leftmodule $Z\in \vrt \Lmod $ which is an upperbound of partially ordered subset $\vrt \cc $ of $\vrt

\Lmod $ then, by \eqref{eq:8}, the functor $U_{0}$ is an embedding. In this case $\cc $ is an $\relax \mathcal {S}$-category with respect to $U_{0}$.

Proof.

The first part of the statement is routine to verify. \eqref{eq:8} shows that the natural forgetful functor preserves subobjects. To show that $\Lmod [S]$ has images, let $f:X\to Y\in \Lmod [S]$. Then $U(f)$ is a mapping from $U(X)$ to $U(Y)$ and $h=U(f)^{\circ }$ is a surjective map of $U(X)$ onto $Y'=\cod h=\im U(f)\subseteq U(Y)$. If $s\in S$ and $y\in Y'$ there is

$x\in X$ with \[ sy=s(h(x))=s(f(x))=f(sx)\in Y'. \] Hence there is a module $X'\subseteq Y$ with $U(X')=Y'$ and a morphism $f^{\circ }: X\to X'$ with $U(f^{\circ })=h$. Since $h$ is surjjecive, $f^{\circ }$ is an epimorphism. Also \begin{equation*} U(f) =h\inc{Y'}{U(Y)}=U(f^{\circ})\inc{U(X')}{U(Y)}=U(f^{\circ}\inc{X'}{Y}). \end{equation*} Since $U$ is faithful, $ f = f^{\circ }\inc {X'}{Y} $ . Thus $f=f^{\circ }\inc {X'}{Y}$ is a canonical factorization of $f$. Let $f=g\inc {X"}{Y}$ be any canonical factorization of $f$. Then \begin{equation*} U(f)^{\circ}\inc{U(X')}{U(Y)}=U(f)=U(g)\inc{U(X")}{U(Y)}. \end{equation*} Since the left-hand side of equation above is the image factoprization of $U(f)$ in $\set $, we have $U(X')\subseteq U(X")$. Since $X'$ and $X"$ are submodules of $Y$ we must have $X'\subseteq X"$. Therefore $X'=\im f$ by the definition of images (see 2.2). Moreover, we have \begin{equation*} U(\im f)=U(X')=\im U(f). \end{equation*} If $\cc \subseteq \Lmod $ is a small subcategory with images then it is clear that $\cc $ satisfy axioms (Sb:b) and (Sb:c) with respect to $U_{0 }=U\vert \cc $. Furthermore if there exists a largest module $Z$ for the partially ordered set $\vrt \cc $ then by \eqref{eq:8} $\vrt U_{0}$ is an orderembedding of $\vrt \cc $ into $\vrt \set $ and so, $(\cc , U_{0})$ satisfies

axiom (Sb:a). Therefore $\cc $ is ab $\relax \mathcal {S}$-category withrespect to $U_{0}$. This completes the proof.

Evidently $S^{1}$ (see (Clifford and Preston 1961), page 4), the semigroup obtained by adjoining identity to $S$ is a left [as well as right] $S$-module where the action is the multiplication by elements of $S$ on the left [right] of elements in $S^{1}$ . Similarly any left ideal $\La $ [righti deal $I$] of $S$ is a left [right] $S$-module and is a proper submodule of $S^{1}$. Let $\Lcat $ denote the full subcategory of $\Lmod [S]$ whose objects are left ideals of $S^{1}$; that is, if $X\in \vrt \Lcat $ then either $X=S^{1}$ or $X$ is a left ideal in $S$. Morphisms in this category are called right translations. It may be observed that, while $S^{1}$ is a faithful left (as well as right) module, a left [right] ideal may not be faithful as a module. For $a\in S$, the map $\rho _{a}:s\mapsto sa$ is a morphism in $\Lcat $ from $S^{1}$ onto the principal left ideal $L(a)=S^{1}a$. This map $\rho _{a}$ is called the principal right translation of $S$ determined by $a$.

Since, for every $X\in \vrt \Lcat $, the action of $S$ on $X$ is the restriction of the product in $S$ to $S\times U(X)$, $\vrt U$ is an order embedding of $\vrt \Lcat $ into $\vrt \set $. Thus the natural foregetful functor $U:\Lcat \to \set $ is an embedding of $\Lcat $ into $\set $. Therefore the pair $(\Lcat ,U)$ satisfies axiom (Sb:a) of Definition 3.1. Axioms (Sb:b) and (Sb:c) hold by Proposition 4.1. Thus we have:

Proposition 4.2.

Let $\Lcat $ denote the full subcategory of $\Lmod [S]$ whose objects are left ideals of $S^{1}$ and let $U:\Lcat \to \set $ be the natural forgetful functor. Then $\Lcat $ is an $\relax \mathcal {S}$-category with respect to $U$.

In the following, for brevity, we shall avoid explicit reference to the forgetful functor $U$ as far as possible. Instead, we shall assume that the given $\relax \mathcal {S}$-category has been identified with its image in $\set $ (see Proposition 3.3) so that $U$ reduces to the inclusion.

We can dualise the definitions and results above. Thus a right $S$ module is a set on which $S$ acts on the right in the usual sense. A morphism $f:X\to Y$ of right modules is a mapping that preserve rightr actions. There is a category $\Rmod $ in which objects are right $S$-modules and morphisms are morphisms of right modules. We can define a subobject relation in $\Rmod $ by dualising \eqref{eq:8}. The category $\Rmod $ with subobjects so defined has images and the foregetful functor $U:\Rmod \to \set $ (that forgets right action) preserve subobjects and images of morphisms.

Remark 4.1:

Notice that a left $S$-module $X$ uniquely determines a dual (contravariant) representation $\la :s\mapsto \la _{s}$ where $\la _{s}$ denote the map on the set $U(X)$ defined by

\[ \la _{s}(x)= s.x \quad \text {for all}\quad x\in X. \] Conversely every contravariant representation of $S$ by functions on the set $X$ uniquely determines a left $S$-module structure on $X$. Let $X$ and $Y$ be left $S$-modules giving representations $\rho $ and $\la $ respectively. A map $\al :X\to Y$ is a morphism of $S$-modules if and only if $\al $ commutes with $\rho $ and $\la $: \[ \la \circ \al = \al \circ \rho . \] Thus in the following we may replace modules by representations

and mprphisms by maps commuting with representations. We shall say a representation $\rho $ (by functions on a set $X$) or the associated module $M_{\rho }$ is faithful if the function $\rho :S\to \trns $ is injective. Dual remarks are valid for right modules of $S$ and representationys by functions on a set.

For each $a\in S$, $L(a)=S^{1}a$ and $R(a)=aS^{1}$ are principal left and right ideals of $S$ respectively. Notice that the natural left action of $S$ on $L(a)$ can be extended to an action of $S^{1}$ so that $L(a)$ is a left $S^{1}$-set which is generated by $a$ and thus $L(a)$ is a cyclic $S^{1}$-set. see (Krishnan 2000, S 2.5,2.6). A morphism $\rho :L(a)\to L(b)$ of left ideals is a mapping which satisfies the condition: there exists $s\in S^{1}$ such that $u\rho =us$ for all $u\in L(a)$. Thus $\rho $ is the restriction of the right translation $\rho _{s}:x\mapsto xs$ to $L(a)$ and is a morphism of the corresponding left $S^{1}$-set. We shall refer to a triplet $(a,s,b)\in S\times S^{1}\times S$ as left-admissible triple in $S$ if it satisfy the condition

\begin{equation} \label{eq:triple} as=tb\quad\text{for some}\quad t\in S^{1}. \end{equation} Given a left admissible triple $(a,s,b)$, it is clear that the map \begin{equation} \label{eq:9} \rho(a,s,b)=\rho_{s}\mid L(a):u\mapsto us \end{equation} is a morphism in $\Lcat $ induced by the principasl right translation $\rho _{s}$. Conversely every morphism $\rho :L(a)\to L(b)\in \Lcat $ between principal ideals in $S$ indused by a principal right translation $\rho _{s}$ determines a left admissible triple $(a,s,b)$ such that $\rho =\rho (a,s,b)$. Notice that these representations of certain morphisms in $\Lcat $ as left-admissible triples are not, in general, unique. In fact, we can see that \begin{equation} \label{eq:10} \rho(a,s,b)=\rho(a',s',b') \iff L(a)=L(a'),\; L(b)=L(b')\; \text{and}\; as=as' \; (\text{or} \; a's=a's'). \end{equation}

Proposition 4.3 (The category of principal left ideals).

To every semigroup $S$ there corresponds a category $\lcat $ specified as follows:

i)$\vrt \lcat =\{L(a):a\in S\}$.

ii)$\lcat (L(a),L(b))=\{\rho (a,s,b)=\rho _{s}\vert L(a):s\in S^{1} \quad \text {with}\quad as\in L(b)\}$.

iii)Composition in $\lcat $ is given by the rule \begin{equation*} \rho(a,s,b)\rho(c,t,d)=\begin{cases} \rho(a,st,d) &\text{if } L(b)=L(c);\\ \text{undefined}&\text{if } L(b)\neq L(c). \end{cases}

\end{equation*}

iv)$\rho (a,s,b)=\inc {L(a)}{L(b)}$ if and only if $\rho (a,s,b) =\rho (a,1,b)$. Moreover $\{\rho (a,1,b):a\in L(b)\}$ is a choice of subobjects in $\lcat $ and the foregetful functor $U:\lcat \to \set $ is an embedding category $\lcat $ (with choice of subobjects above) into $\set $.

v)If $\rho (a,s,b)$ is a morphism in $\lcat $, then \begin{equation} \label{eq:im-fac} \im\rho(a,s,b)=L(as) \quad\text{and}\quad \rho(a,s,b)= \rho(a,s,as)\rho(as,1,b) \end{equation}

gives the image-factorization of $\rho (a,s,b)$ in $\lcat $.

Furthermore, $\lcat $ is an $\relax \mathcal {S}$-category is an $\relax \mathcal {S}$-subcategory of $\Lcat $.

Proof.

By definition, for each left admissible triplet $(a,s,b)$, the map $\rho (a,s,b)=\rho _s\mid L(a)$ is a morphism in $\Lcat $. Hence

\begin{equation*} \lcat=\cup\{\lcat(L(a),L(b)):a,b\in S\} \end{equation*} is a subset of the set of morphisms of $\Lcat $. Also morphisms $\rho (a,s,b)$ and $\rho (c,t,d)$ are composable iff $L(b)=L(c)$. In this case, the triplet $(a,st,d)$ is left admissible and we haveith respect to \begin{align*} \rho(a,s,b)\rho(c,t,d) & =(\rho_{s}\vert L(a))(\rho_{t}\vert L(c))\\ & =\rho_{s}\rho_{t}\vert L(a) \\ & =\rho_{st}\vert L(a)\\ & =\rho(a,st,d). \end{align*}

Therefore $\lcat $ is closed with respect to composition of maps. It follows that the composition defined by the rule given in iii) coincides with composition of the corresponding maps. Now, for any $a\in S$, $(a,1,a)$ is left-admissible and $\rho (a,1,a)= 1_{L(a)}$. Hence $\lcat $ is the morphism set of a subcategory of $\Lcat $. Clearly, the vertex set of $\lcat $ is the set of all principal left ideals of $S$.

It is obvious that $L(a)\subseteq L(b)$ if and only if $(a,1,b)$ is an admissible triple and $\rho (a,1,b)=\inc {L(a)}{L(b)}$. Therefore inclusions between vertices of $\lcat $ are morphisms of the form $\rho (a,1,b)$ with $a,b \in S$. It is easy to verify that the set $\{\rho (a,1,b):a\in L(b)\}$ is a choice of subobjects for $\lcat $. Further, for any $\rho =\rho (a,s,b)$, we can see that the function $\rho $ maps the set $L(a)$ onto the set $L(as)$. Therefore $L(as)$ is the image of $\rho $ in $\lcat $ as well as in $\Lcat $. Since $\Lcat $ is an $\relax \mathcal {S}$-category, by Corollary 3.2 $\lcat $ is an $\relax \mathcal {S}$-subcategory of $\Lcat

$.

Let $\phi :S\to T$ be a homomorphism of semigroups. If $(a,s,b)$ is a left admissible triple in $S$, by \eqref{eq:triple} $as=tb$ for some $t\in S^{1}$. Then $a\phi , b\phi \in T$, $s\phi ,t\phi \in T^{1}$ and $(a\phi )(s\phi )=(t\phi )(b\phi )$. Therefore $(a\phi ,s\phi ,b\phi )$ is an admissible triple in $T$. We see that the equations

\begin{equation} \label{eq:18} \begin{aligned} \lcat[\phi]\left(L(a)\right)&=L(a\phi)\\ \lcat[\phi]\left(\rho(a,s,b)\right)&=\rho(a\phi,s\phi,b\phi) \end{aligned} \end{equation} for each $a\in S$ and admissible triple $(a,s,b)$ in $S$ is an inclusion-preserving functor $\lcat [\phi ]:\lcat [S]\to \lcat [T]$. Furthermore the assignments \[ \lcat : S\mapsto \lcat [S],\quad \phi \mapsto \lcat [\phi ] \] is a functor $\mathbb {L}_{0}$ from the category $\sg $ of semigroups to the category $\SBC $ of small $\relax \mathcal {S}$-categories.

Dually there is a full subcategory $\Rcat \subseteq \Rmod $ of all right ideals of $S^{1}$ in which morphisms are called left-translations. The morphism \[ \la _{a}: t\mapsto at;\quad S^{1}\to R(a) \] in $\Rcat $ is called the principal left translation of $S$ determined by $a$.

Dually we can define the category $\rcat $ of principal right ideals. A morphism in $\rcat $ from $R(a)$ to $R(b)$, denoted by $\la (a,s,b): R(a)\to R(b)$ is a left translation $\la _{s}\vert R(a)$ with $s\in S^{1}$ and $sa\in R(b)$. We shall say that a triplet $(a,s,b)\in S\times S^{1}\times S$ satisfying the condition $sa\in R(b)$ is right admissible. Notice that $(a,s,b)$ is right admissible if and only if $sa=bt$ for some $t\in S^{1}$. Moreover all definition and results for $\lcat $ holds for $\rcat $ with appropriate modifications. Proof of the following statement (the dual of Proposition 4.3) is obtained by dualising (that is, replacing left ideals by right ideals, left translations by right translations, etc.) the proof of Proposition 4.3.

Proposition 4.4 (The category of principal right ideals).

To every semigroup $S$ there corresponds a category $\rcat $ specified as follows:

i)$\vrt \rcat =\{R(a):a\in S\}$.

ii)$\rcat (R(a),R(b))=\{\la (a,s,b)=\la _{s}\vert R(a):s\in S^{1} \quad \text {with}\quad sa\in R(b)\}$.

iii)Composition in $\rcat $ is given by the rule \begin{equation*} \la(a,s,b)\la(c,t,d)=\begin{cases} \la(a,ts,d) &\text{if } R(b)=R (c);\\ \text{undefined}&\text{if } R(b)\neq R(c). \end{cases}

\end{equation*}

iv)$\la (a,s,b)=\inc {R(a)}{R(b)}$ if and only if $\la (a,s,b) =\la (a,1,b)$. Moreover $\{\la (a,1,b):a\in R(b)\}$ is a choice of subjects in $\rcat $ and $\rcat $ is a category with subobjects in which the set-theoretic inclusions among vertexes in $\rcat $ is the subobject relation.

v)If $\la (a,s,b)$ is a morphism in $\rcat $, then

\begin{equation*} \im\la(a,s,b)=R(sa) \quad\text{and}\quad \la(a,s,b)=\la(a,s,sa)\la(sa,1,b) \end{equation*}

gives the image-factorization of $\la (a,s,b)$ in $\rcat $. Furthermore, $\rcat $ is an $\relax \mathcal {S}$-category and is a subcategory of $\Rcat $.

The proposition Proposition 4.3 shows that there are close relations between the structure of the category $\lcat $ and the semigroup $S$. For example, the following observations are useful:

Corollary 4.5.

Let $S$ be a semoigroup and let $\rho =\rho (a,s,b) :L(a)\to L(b)$ be a morphism in $\lcat $. Then we have

(a)$\rho $ is surjective if and only if $as$ $\lr $ $b$;

(b)$\rho $ is a split injection if and only if there exists $s'\in S^{1} $ such that $(b,s',a)$ is left admissible and $ass'=a$.

(c)$\rho $ is a split surjection if and only if if there exists $s'\in S^{1} $ such that $(b,s',a)$ is left admissible and $bs's=b$.

(d)$\rho $ is an isomorphism if only if $a$ $\rr $ $as$ $\lr $ $b$.

Statements dual to the above are valid for morphisms in $\rcat $.

Proof.

The statement (a) is an immediate consequence of the definition of $\rho (a,s,b)$. To prove (b), assume that an element $s'\in S^{1}$ exists satisfying the requirements. Then $\si :\rho (b,s',a)$ is a morphism in $\lcat $ such that

\begin{equation*} \rho\si = \rho(a,s,b)\rho(b,s',a) = \rho(a,ss',a) \end{equation*} by Proposition 4.3. Since $ass'=a$, by (??) we have $\rho (a,ss',a)=\rho (a,1,a)=1_{L(a)}$. Therefore $\rho $ is a split monomorphism. Conversely if $\rho =\rho (a,s,b)$ is a split monomorphism, then there is $\si =\rho (b,s',a):L(b)\to L(a)$ such that $\rho \si =1_{L(a)}$. $(b,s',a)$ is left admissible and the fact that $\rho $ is a split monomorphism gives \begin{equation*} \rho\si = \rho(a,s,b)\rho(b,s',a) = \rho( a,ss',a)=\rho(b,1,b) \end{equation*} and so, $ass'=a$. Consequently, the element $s'$ above satisfies the requirements.

To prove (c), assume that $\rho =\rho (a,s,b):L(a)\to L(b)$ is a split epimorphoism so that $\si \rho =1_{L(b)}$ for some morphism $\si =\rho (b,s',a)$. Then by the definition of composition in $\lcat $ $\si \rho =\rho (b,s's,b)$ and by (??), we have \[ \si \rho =\rho (b,s's,b)=\rho (b,1,b) \] This gives $bs's=b$. On the other hand, if $s'$ exist satisfying the requirements in (c), then $\si =\rho (b,s',a)$ is a morphism in $\lcat $ from $L(b)$ to $L(a)$ and since $bs's=b$, by (??), \[ \si \rho = \rho (b,s's,b) \rho (b,1,b)=1_{L(b)}. \] Hence $\rho $ is a split epimorphism. The statement (d) follows from (b) and

(c).

Remark 4.2:

Notice that $\lcat $ is a proper subcategory of the category $\Lmod $ of all left $S^{1}$-systems. In general, $\lcat $ may not be a full subcategory of $\Lmod $. Another category which may be of interest in this connection is the full subcategory $\Lcat $ of $\Lmod $ whose objects are left ideals of $S$. We have

\[ \lcat \subseteq \Lcat \subseteq \Lmod . \] Suppose that $\rho :L(a)\to L(b)$ is a morphism in $\Lcat $. If $a$ is a regular element in $S$ there is an idempotent $e\lr a$. Let $s=e\rho $. Then $s=es\in L(b)$ and $(e,s,b)$ is anadmissible triple. By \eqref{eq:9} and \eqref{eq:10} $\rho =\rho (e,s,b)=\rho (a,s,b)$. It follows that the inclusion $\lcat

\subseteq \Lcat $ is fully-faithful if $S$ is regular. Dual remarks hold for the category $\rcat $ of all principal right ideals, the category $\Rcat $ of all right ideals and the category $\Rmod $ of all right $S^{1}$-systems.

The semigroup of cones

The following definition is basic in what follows in this section. Here, unless otherwise provided, we shall assume that $\cc $, $\cd $, etc., are small $\relax \mathcal {S}$-categories.

Definition 5.1 (Cones).

Let $d\in \vrt \cc $. A map $\ga :\vrt \cc \to \cc $ is called a cone from the base $\vrt \cc $ to the vertex $d$ (or simply a cone in $\cc $ to $d$) if $\ga $ satisfies the following:

Con-1$\ga (c)\in \cc (c,d)$ for all $c\in \vrt \cc $.

Con-2If $c'\subseteq c$ then $\inc {c'}{c}\ga (c)=\ga (c')$.

Given the cone $\ga $ we denote by $c_{\ga }$ the the vertex of $\ga $ and for each $c\in \vrt \cc $, the morphism $\ga (c):c\to c_{\ga }$ is called the component of $\ga $ at $c$. The cone $\ga $ is said to be proper [normal] if $\ga $ satisfies the condition Con-3 [respectively Con-4] below:

Con-3$\bigvee _{c\in \vrt \cc }\im (\ga (c))=c_{\ga }$.

Con-4There exists $c\in \vrt \cc $ such that $\ga (c):c\to c_{\ga }$ is an isomorphism.

Since $U$ is an (order) embedding of $\vrt \cc $ into $\set $, it is clear that the statement Con-3 is equivalent to \begin{equation} \label{eq:Con-3} \bigcup_{c\in\vrt\cc}U(\im\ga(c))=U(c_{\ga}). \end{equation} It is useful to observe that a sufficient condition for the cone $\ga $ to be proper is that there exists $c\in \vrt \cc $ such that $\ga (c):c\to c_{\ga }$ is surjective (that is, $\ga (c)=\epi [\ga (c)]$). It follows that any normal cone is proper.

Let $\con $ [respectively $\pcon $, and $\ncon $] denote the set of all cones [respectively proper and normal] cones in $\cc $. Then \[ \ncon \subseteq \pcon \subseteq \con . \] When $\ga \in \con $ is normal we write \begin{equation} \label{eq:11} \mset[\ga] =\{c\in\vrt\cc: \ga(c)\;\;\text{is an isomorphism}\}. \end{equation} Therefore the cone $\ga $ is normal if and only if $\mset [\ga ] \ne \emptyset $ .

Recall that the preorder $\vrt \cc $ is a subcategory of $\cc $ so that the inclusion $J=\inc {\vrt \cc }{\cc }$ is a functor of $\vrt \cc $ to $\cc $. Any cone in $\cc $ in the sense of the definition above, is a natural transformation from the base $J$ to the constant functor of $\vrt \cc $ at vertex $c_{\ga }$; that is, a cone in the sense of classical category theory (see for example, (MacLane 1971)).

Given a cone $\ga $ in $\cc $ and a morphism $f\in \cc (c_{\ga },d)$ the map defined, for all $c\in \vrt \cc $, by: \begin{align} \left(\ga\ast f\right)(c) &=\ga(c)f \label{eq:12} \end{align}

is clearly a cone with vertex $d$. Furthermore, if $f,g\in \cc $ is a pair of composable morphisms with $\dom f =c_{\ga }$,we have: \begin{align} \ga\ast (fg) &=\left(\ga\ast f\right)\ast(g\mid \im f). \label{eq:13} \end{align}

The construction given in \eqref{eq:12} is especially useful when $\ga $ is proper.

Proposition 5.1.

Let $\ga $ be a proper cone in $\cc $ and $f\in \cc (c_{\ga },d)$. Then the cone $\ga \ast f$ is prooper if and only if $f$ is surjective.

Proof.

Suppose that $\eta =\ga \ast f$ where $f$ is surjective. Since $\ga $ is proper, Con-3 or equivalently, \eqref{eq:Con-3} hold. Thus

\begin{equation} U(c_{\ga}) =\bigcup_{c\in\vrt\cc}\im(U(\ga(c))) \notag \end{equation} Since by \eqref{eq:12}, $\ga \ast f$ is the map sending $c\in \vrt \cc $ to $\ga (c)f$, we have \begin{equation*} U(\ga\ast f)(c) = U(\ga(c))U(f) = U(\ga(c))\circ U(f). \end{equation*} Let $y\in U(d)$. Since the map $U(f):U(c_{\ga })\to U(d)$ is surjective there is $x\in U(c_{\ga })$ such that $xU(f)=y$. Since the cone $\ga $ is proper, by Cone-3, there is $c\in \vrt \cc $ such that $x\in \im U(\ga (c))$. This implies that $x=uU(\ga (c))$ for some $u\in U(c)$. Then $y = \left (uU(\ga (c))\right )U(f)$. Therefore $y \in U\im \left (\ga (c)f\right )$. \begin{align*} \text{It follows that } &U(d) \subseteq\bigcup_{c\in\vrt\cc}\im U\left(\ga(c)f\right).\\ \text{The inclusion } &\bigcup_{c\in\vrt\cc}\im U\left(\ga(c)f\right) \subseteq U(d) \quad \text{is obvious.} \\ \text{Hence } &U(d) =\bigcup_{c\in\vrt\cc}\im U \left(\ga(c)f\right). \end{align*}

Therefore by Cone-3, $\ga \ast f$ is proper.

Suppose that the cone $\ga \ast f$ is proper. By Cone-3 we have \begin{equation*} U(d) = \bigcup_{c\in\vrt\cc}U\left(\im (\ga(c)f)\right). \end{equation*} Hence, for $y\in U(d)$ there is $c\in \vrt \cc $ such that $y\in U\left (\im (\ga (c)f)\right )$. Then there exist $x\in U(c)$ such that $y=xU(\ga (c)f) =uU(f)$ where $u=xU(\ga (c))$. Therefore $U(f):U(c_{\ga })\to U(d)$ is surjective. The fact that $\cc $ is an $\relax \mathcal {S}$-category now shows that $f:c_{\ga }\to d$ is

surjective.

The semigroup of cones

Recall that $\pcon $ denote the set of all proper cones in $\cc $. For $\ga ,\eta \in \pcon $, let \begin{equation} \label{eq:14} \ga\eta=\ga\ast\epi[\eta(c_{\ga})]. \end{equation} By Proposition 5.1 $\ga \eta $ is a unique proper cone with vertex $\im \eta (c_{\ga }) \subseteq c_{\eta }$. Therefore this defines a binary operation in $\pcon $.

Proposition 5.2.

Let $\cc $ be a small $\relax \mathcal {S}$-category. Then $\pcon $ is a semigroup with respect to the binary composition defined by \eqref{eq:14}.

Proof.

It is sufficient to show that the binary composition defined by \eqref{eq:14} is associative. Let $\al ,\be ,\ga \in \pcon $ and $c\in \vrt \cc $. Then

\begin{align*} \left(\al(\be\ga)\right)(c)&=\al(c)\left((\be\ga)(c_{\al})\right)^{\circ}\\ &=\al(c)\left((\be(c_{\al}))\left((\ga(c_{\be}))\right)^{\circ} \right)^{\circ}\\ &=\al(c)\epi[(\be(c_{\al})) (\ga(c_{\be}))] \\ &=\al(c)\epi[(\be(c_{\al}))]\epi[(\ga(c_{\al\be}))] \end{align*}

Thus $\al (\be \ga )=(\al \be )\ga $. If $\ga , \eta \in \pcon $ then by Proposition 5.1 $\ga \eta \in \pcon $. Thus $\pcon $ is a semigroup.

Representations of $\pcon $ by transformations

Let $\cc $ be an $\relax \mathcal {S}$-categorywith forgetful functor $U$. Since $\vrt \cc $ is a subcategory of the small category $\cc $ and $\set $ is (small) complete, the functor $\vrt U:\vrt \cc \to \set $ has the direct limit see (MacLane 1971, page 105). The direct-limit $\dlim (\vrt U)$ of $\vrt U$ consists of a set, which is denoted by $\De $, and a cone (the universal or limiting cone) from the base $\vrt U$ to the vertex $\De $ see (Nambooripad 1994, page 11 for details). If $U$ is the inclusion functor $\cc \subseteq \cd $, then we write $\dlim \cc $ for the limit $\dlim (\cc \subseteq \cd )$ if the category $\cd $ in which the limit is evaluvated is clear from the context. In particular, unless otherwise stated explicitly, $\dlim \cc $ will stand for the direct limit of the inclusion functor $\cc \subseteq \set $ (evaluvated in $\set $). We use these notations in the following lemma which provides a convenient representation for $\dlim (\vrt U)$.

Proposition 5.3.

Suppose that $(\cc ,U)$ be an $\relax \mathcal {S}$-category. Define

\begin{align} \De &=\bigcup_{c\in\vrt\cc} U(c), \label{eq:12a} \end{align}

and let $\jmath $ be the function defined, for all $c\in \vrt \cc $, by \begin{align} \jmath(c) &=\inc{U(c)}{\De} \label{eq:12b} \end{align}

Then $\jmath $ is a proper cone from the base $\vrt U$ to the vertex $\De $ and \begin{align} \dlim(\vrt U) & = (\De,\jmath). \label{eq:12c} \end{align}

If $\eta \in \pcon [\cc ]$, then there is a unique map $\tilde {\eta }:\De \to U(c_{\eta })$ such that \begin{equation} \label{eq:13a} \eta(c)=\inc{U(c)}{\De}\circ\tilde{\eta} \quad\text{for all}\quad c\in \vrt\cc. \end{equation}

Furthermore, the cone $\eta $ is proper if and only if $\tilde {\eta }$ is surjective.

Proof.

Since $\vrt \cc \subseteq \set $ is an order embedding we see that the map $\jmath :c\mapsto \inc {c}{\De }$ is a cone from the base $\vrt \cc $ to the vertex $\De $. We prove that $\jmath $ is universal. Suppose that $\al :c\mapsto \al (c)$ is a cone from the base $\vrt \cc $ to the vertex $X$. For each $x\in \De $ define

\[ xf=x\al (c)\quad \text {if}\quad x\in c. \] Since, by \eqref{eq:12a}, $\De $ is the union of sets $c\in \vrt \cc $, $xf$ is defined for all $x\in \De $. If $x\in U(c_{i})$, $i=1,2$, then by the axiom (S:c) of Definition 3.1 there is $d\in \vrt \cc $ such that $d\subseteq c_{i}$, $i=1,2$ and $x\in d$. Since, as noted above, $\vrt \cc \subseteq \set $ is inclusion preserving and $\al $ is a cone on $\vrt \cc $ we have \begin{equation*} x\al(c_{1})=x\al(d)=x\al(c_{2}). \end{equation*} This shows that $f:\De \to X$ is a map. The definition of $f$ shows that, for all $c\in \vrt \cc $, \[ f\mid c=\inc {c}{\De }\circ f=\jmath (c)\circ f=\al (c). \] Thus the following diagram commutes \begin{equation} \label{dig:1} XY[1] \end{equation} for all $c\in \vrt \cc $. To prove the uniqueness of $f$, suppose that $g:\De \to X$ is another map such that \begin{gather*} \jmath(c)\circ g=\al(c)\quad\text{for all}\quad c\in\vrt\cc. \end{gather*} Then we have \begin{gather*} g\mid c =\al(c)=f\mid c \end{gather*} for all $c\in \vrt \cc $ which implies that $f=g$. Therefore the cone $\jmath $ is universal from the base $\vrt \cc $ to the vertex $\De $.

Let $\eta \in \pcon $. Since $\jmath $ is universal from $\vrt U$ to $\De $ there is a unique map $\tilde {\eta }:\De \to U(c_{\eta })$ making the digram ?? commute or equivalently, \eqref{eq:13a} holds.

\begin{equation} \label{dig:2} XY[2] \end{equation}

Suppose that $S$ is a smigroup and let $\cc $ be a subcategory of $\Lmod $ with images. Let $U:\Lmod \to \set $ be the natural forgetful functor. By Propositions 4.1 and 4.2, $\cc $ is an $\relax \mathcal {S}$-category with respect to the foregetful functor $U\vert \cc =U_{0}:\cc \to \set $. Vertices of $\cc $ are $S$-modules and so $\De $ defined by \eqref{eq:12a} carries a unique module structure such that $\inc {c}{\De }$ is a module morphism for each $c\in \vrt \cc $. Furthermore, as in Proposition 5.3, the map $\jmath :c\mapsto \jmath (c)= \inc {c}{\De }$ is a cone from the base $\vrt \cc $ to the vertex $\De $ which is universal. It follows that $(\De ,\jmath )$ is the direct limit $\dlim \vrt \cc $ evaluvated in $\Lmod $. If $\cc \subseteq \Lcat $ then we can show similarly that $(\De ,\jmath )$ is the direct limit $\dlim \vrt \cc $ evaluvated in $\Lcat $. Thus we have:

Proposition 5.4.

Let $S$ be a smigroup. Suppose that $\cc \subseteq \Lmod $ [or $\cc \subseteq \Lcat $] and that $\De $ and $\jmath $ are defined by \eqref{eq:12a} and \eqref{eq:12b}. Then \eqref{eq:12c} holds in which the direct limit $\dlim (\vrt U)$ is evaluvated in $\Lmod $ if $\cc \subseteq \Lmod $ [or in $\Lcat $ if $\cc \subseteq \Lcat $].

As above, let $(\cc ,U)$ be a $\relax \mathcal {S}$-category with forgetful functor $U$. If $\cc \subseteq \Lmod $ [or $\cc \subseteq \Lcat $], $\eta (c)$ and $\jmath (c)=\inc {c}{\De }$ are module morphisms [translations of ideals] for every $c\in \vrt \cc $. The equation above shows that the map $\tilde {\eta }:\De \to c_{\eta }$ is a module morphism [translations of ideals]. Define $\phi =\phi _{\cc }$ on $\pcon $ as follows: \begin{equation} \phi(\ga)=\tilde{\ga}\circ\jmath(c_{\ga})\quad\text{for all}\quad \ga\in\pcon\label{eq:13b} \end{equation} It is clear that, for each $\ga \in \pcon $, $\phi (\ga )\in \trns [\De ]$ is a transformation of $\De $. We have

Proposition 5.5.

Let $\cc $ be a $\relax \mathcal {S}$-category. For each $\ga \in \pcon $, let $\phi (\ga )$ be defined by \eqref{eq:13b}. Then

\begin{equation*} \phi_{\cc}=\phi:\pcon\to\trns[\De],\quad\ga\mapsto\phi(\ga) \end{equation*}

is a faithful representation of the semigroup $\pcon $ by transformations on $\De $.

Proof.

We shall show that $\phi :\pcon \to \trns [\De ]$ is a homomorphism. Let $\al ,\be \in \pcon $. By \eqref{eq:14},

\[ (\al \be )(c)=\al (c)\be (c_{\al })\quad \text {for}\quad c\in \vrt \cc . \] Hence, for all $x\in \De $, we have \begin{align*} x\phi(\al\be)&=x\tilde{\al\be}\jmath(c_{\al\be}) \\ &=x(\al\be)(c)\jmath(c_{\al\be}) \\ &=x\al(c)(\be(c_{\al}))^{\circ} \jmath(c_{\al\be}) \end{align*}

by (??). Since $\vrt \cc \subseteq \set $ is an order embedding this gives \begin{align*} x\phi(\al\be)&=x \al(c) \be(c_{\al})\jmath(c_{\be})\\ &=x\tilde{\al}\tilde{\be}\jmath(c_{\be})=x\phi(\al)\phi(\be). \end{align*}

by \eqref{eq:13b}. Therefore $\phi (\al \be )=\phi (\al )\phi (\be )$.

Assume that $\al ,\be \in \pcon $ and $\phi (\al )=\phi (\be )$.

For $c\in \vrt \cc $ let \[ \pcon (c)=\{\ga \in \pcon :c_{\ga }\subseteq c\}. \] The result above shows that each $f\in \cc (c,d)$ induces a map \begin{equation} \label{eq:15} \rho _{f}:\ga\mapsto\ga\ast\epi[(f\mid c_{\ga})] \end{equation} of the set $\pcon (c)$ to the set $\pcon (\im f)$. The partial transformation $\rho _{f}$ of $\pcon $ will be called the translation induced by $f$. Notice that $\rho _{f} =\rho _{\epi [f]}$. Hence, in the following, we may restrict our consideration to translations induced by surjections.

Let $\ga \in \pcon $. For each $c\in \vrt \cc $ and for $f\in \cc (c,d)$, define

\begin{align} \hfn{\ga}{c}&=\{\ga\ast\epi[f]:f\in\cc(c_{\ga},c)\}, \label{eq:16}\\ (\ga\ast\epi[g])\hfn{\ga}f&=\ga\ast\epi[(gf)] \quad\text{for all } g\in\cc(c_{\ga},c).\label{eq:17} \end{align}

Clearly, $\hfn {\ga }c$ is a set for all $c\in \vrt \cc $ and by the uniqueness of epimorphic component, $\hfn {\ga }f$ is a map of the set $\hfn {\ga }c$ to $\hfn {\ga }d$ for $f\in \cc (c,d)$.

In the following we use some standard constructions and terminologies of category theory; see MacLane (1971, Sections III.2 and III.3) and Nambooripad (1994, Sections I.2 and I.3). In particular, given two functors $F,G:\cc \to \cd $ we write $F\subseteq G$ if and only if \begin{equation*} F(c) \subseteq G(c) \quad\text{for all } c\in\vrt\cc \text{ and the map} \quad c\mapsto \inc{F(c)}{G(c)} \end{equation*} is a natural transformation.

Proposition 5.6.

For each $\ga \in \pcon $, \eqref{eq:16} and \eqref{eq:17} defines an inclusion preserving covariant functor $\hfn {\ga }-:\cc \to \set $. Moreover, there is a unique natural isomorphism

\[ \eta _{\ga }:\hfn {\ga }-\nattr \cc (c_{\ga },-)\quad \text {such that}\quad \eta _{\ga }(c_{\ga })(\ga )=1_{\ga }. \]

Consequently, $\hfn {\ga }-$ is a representable functor and $c_{\ga }$ represents it.

Proof.

Let $g\in \cc (c,d)$ and $h\in \cc (d,u)$. Then by \eqref{eq:16} and \eqref{eq:17},

\begin{align*} (\ga\ast\epi[f])\hfn{\ga}{gh}&=\ga\ast\epi[(fgh)] \\ &=\left((\ga\ast\epi[f])\hfn{\ga}g\right)\hfn{\ga}h\quad &&\text{for all }\quad \ga\ast\epi[f]\in\hfn[c]{\ga}.\\ \text{Therefore }\quad \hfn{\ga}{gh}&=\left(\hfn{\ga}g\right) \left(\hfn{\ga}h\right). \end{align*}

If $c\subseteq d$ then it follows readily from \eqref{eq:17} that \[ \hfn {\ga }{\inc {c}{d} } = \inc {\hfn{\ga}{c} } {\hfn{\ga}{d} }. \] In particular, $\hfn {\ga }{1_{c} }=1_{\hfn {\ga }c}$. In view of the remarks following \eqref{eq:17}, this shows that $\hfn {\ga }-:\cc \to \set $ is an inclusion preserving covariant functor.

To prove the remaining statements, by Yoneda lemma see (MacLane 1971, Section III.2) and Nambooripad (1994, section I.2), it is sufficient to show that $\ga $ is universal element for $\hfn {\ga}-$ in $\hfn {\ga}{c_{\ga} }$. Thus we must show that given any $\ga \ast \epi [f]\in \hfn {\ga }c$, there is a unique $h:c_{\ga }\to c$ such that \[ (\ga )\hfn {\ga }h=\ga \ast \epi [f]. \] By \eqref{eq:17}, the above equation holds for the choice $h=f$. To see that this choice is unique, assum that $k:c_{\ga }\to c$ also satisfies this equation. Then, again by \eqref{eq:17}, we have $\ga \ast \epi [f]=\ga \ast \epi [k]$. Since $\ga $ is proper, $\dset \ne \emptyset $. Choose $c\in \dset $. By \eqref{eq:12}, we have \[ \ga (c)\epi [f]=\ga (c)\epi [k]\quad \text {which implies}\quad \epi [f] =\epi [k] \] since $\ga (c)$ is surjective (see \eqref{eq:11}). Hence $\im f= \im g$ and so, $f=\epi [f]\inc {\im f}{c}=\epi [k]\inc {\im k}{c}=k$. This completes the

proof.

The functor $\hfn {\ga }:\cc \to \set $ is called the covariant $\hfnt $ etermined by the cone $\ga \in \pcon $.

Regular categories

Another class’s of categories that will be of interest in the sequel are defined here see (Nambooripad 1994).

Normal factorization and regularity

To avoid repetition, we assume throughout this section that $\cc $, $\cd $, etc., stand for small categories with factorization property.

A normal factorization of a morphism $f\in \cc (c,d)$ is a factorization of the form $f=euj$ where $e:c\to c'$ is a retraction, $u:c'\to d' \subseteq d$ is an isomorphism and $j=\inc {d'}{d}$. A morphism $f:c\to d$ is said to be regular if there is $f':d\to c$ such that $ff'f=f$ and $f'ff'=f'$. We proceed to investigate the relation between regularity and normal factorization.

Proposition 6.1.

Let $j=\inc {c}{d}$ be a split inclusion and $\e :d\to c$ be a retraction in $\cc $.

(a)If $j$ is an epimorphism in $\cc $ then $c=d$ and $j=1_{c}$.

(b)If $\e $ is a monomorphism in $\cc $ then $c=d$ and $\e =1_{d}$.

Proof.

(a) Since $j:c\subseteq d$ is a split inclusion there is $\e :d\to c$ with $j\e =1_{c}$. Then $j(\e j)=j=j1_{d}$. Since $j$ is an epimorphism, we have $\e j=1_{d}$. By axiom (3) of Section 2.1.2, $\e =\inc {d}{c}$. Hence $c\subseteq d\subseteq c$. Since $\subseteq $ is a partial order on $\vrt \cc $, we have $c=d$ and then $j=1_{c}$.

(b) Let $j=\inc {c}{d}$ so that $\e j\e =\e $. Since $\e $ is a monomorphism, we have $\e j=1_{d}$ and by axiom (3) of Section 2.1.2, $\e =\inc {d}{c}$. Consequently $c=d$ and so $\e =\inc {c}{d}=1_{d}$.

Proposition 6.2.

Suppose that every inclusion in $\cc $ splits. Then every morphism in $\cc $ has a unique canonical factorization. In particular, $\cc $ has images.

Proof.

Let $f=xj=yj'$ be canonical factorizations of the morphism $f\in \cc $. Since $j=\inc {c}{d}$ and $j'=\inc {c'}{d}$ are split there exist morphisms $u$ and $v$ in $\cc $ with $ju=1_{c}$ and $j'v =1_{c'}$. Then

\[ yj'uj=xjuj=xj=yj' \] and since $y$ is an epimorphism we have $(j'u)j=j'uj=j'$. Similarly $(jv)j'=j$. Hence $j$ and $j'$ are equivalent as monomorphisms. Since these are inclusions, we have $j=j'$. Then $xj=yj$ and since $j$ is a monomorphism, we have $x=y$.

Since canonical factorizations are unique, the existence of $\im f$ follows from its definition 2.2.5.

Theorem 6.3.

A morphism $f$ in a category $\cc $ (with factorization) is regular if and only if $f$ has normal factorization $f=\e uj$ where $j$ is a split inclusion. Thus the category $\cc $ is regula if and only if

a)every inclusion in $\cc $ splits; and

b)every morphism in $\cc $ has normal factorization.

In particular, every regular category is a $\relax \mathcal {NS}$-category. (see § Section 3)

Proof.

Suppose that $f:c\to d$ be regular so that there is $f':d\to c$ with $ff'f=f$ and $f'ff'=f'$. Let $f=xj$ and $f'=yj'$ be canonical factorizations. Then we have

\[ xjyj'xj=xj \quad \text {and}\quad yj'xjyj'=yj' \] Since $x$, $y$ epimorphisms and $j$, $j'$ are monomorphisms, we have \[ jyj'x=1_{d'} \quad \text {and}\quad j'xjy=1_{c'} \] where $d'=\cod x$ and $c'=\cod y$. Then $\e '=xjy:c\to c'$ is a retraction which is a right inverse of $j'=\in {c'}{c}$ and $\e =yj'x:d\to d'$ is a retraction which is a right inverse of $j=\inc {d'}{d}$. Moreover, $j'x=u$ is an isomorphism with $jy=v$ as its inverse. Therefore, \[ f=xj=xjyj'xj=\e u j \] is a normal factorization of $f$ in which $j$ is a split inclusion.

Suppose that $f=\e uj$ be a normal factorization of $f$ in which $j$ splits. Let $\e '$ be the retraction which is the right inverse of $j$ and let $j'$ be the left inverse of the retraction $\e $. Let $f'=\e 'u^{-1}j'$. Then simple calculations show that $f=ff'f$ and $f'=f'ff'$.

It now follows that $\cc $ is regular if and only if it satisfies conditions a) and b). It follows from Proposition 6.2 that $\cc $ has images. By b), every surjection in $\cc $ splits and hence by Proposition 3.5, $\cc $ is an $\relax \mathcal {S}$-category with respect to the natural embedding and so, $\cc $ is an $\relax \mathcal

{NS}$-category. (see § 3).

Let $S$ be a regular semigroup. Since every principal left ideal in $S$ has at least one idempotent generator, we may write objects (vertexes) in $\lcat $ as $L(e)=Se$ for $e\in \bset (S)$ and morphisms $\rho :Se\to Sf$ as $\rho =\rho (e,s,f)$ where $e,f\in \bset (S)$ and $(e,s,f)$ is left compatible (see Proposition 4.3). Since $es\in Sf$ and $\rho (e,s,f)=\rho (e,es,f)$ we may replace $s$ with $es$ if necessary and assume that $s\in eSf$. Conversely if $esf\in eSf$ then $(e,s,f)$ is left admissible. Thus \begin{equation} \label{eq:19} \vrt\lcat=\{L(e)=Se:e\in\bset(S)\}\quad\text{and}\quad \lcat =\{\rho(e,s,f):e,f\in\bset(S),\; s\in eSf\}. \end{equation} When $S$ is regular we shall use these representations of objects and morphism of $\lcat $ with out further comment.

We proceed to discuss some important properties of the category $\lcat $ see (Nambooripad 1994, Definition III.3, Lemma III.12 and Lemma III.13). Here we shall follow (Nambooripad 1979) for notations and terminiology related to biordered set, regular semigroups, etc. see also (Krishnan 2000, Chapter 2 and 3).

Proposition 6.4.

Let $S$ be a regular semigroup. Then

(a)$\rho (e,u,f)=\rho (e',v,f')$ if and only if $\e $ $\lr $ $e'$, $f$ $\lr $ $f'$, $u\in eSf$, $v\in e'Sf'$ and $v=e'u$.

(b)The map $\rho (e,s,f)\mapsto s$ is a bijection of $\lcat (Se,Sf)$ onto $eSf$ for all $e,f\in \bset (S)$.

(c)A morphism $\rho =\rho (e,s,f)$ is a monomorphism [epimorphism] if and only if $\rho $ is injective [surjective]; this is true if and only if $e$ $\rr $ $s$ [$s$ $\lr $ $f$]. Every monomorphism [epimorphism] in $\lcat $ split.

(d)$Se$ and $Sf$ are isomorphic if and only if $e$ $\dr $ $f$; if this is the case, there is a bijection of the set of isomorphisms of $Se$ onto $Sf$ and the $\hrs $-class $L_{e}\cap R_{f}$.

(e)A morphism $\rho $ is the inclusion $Se\subseteq Sf$ if and only if $ef=e$ and $\rho =\rho (e,e,f)$. Then $\rho ':Sf\to Se$ is the right inverse of $\rho $ if and only if $\rho '=\rho (f,g,e)$ where $g\in L_{e}\cap \om (f)$.

(f)Let $\rho =\rho (e,u,f)$ be a morphism in $\lcat $. For any $g\in R_{u}\cap \om (e)$ and $h\in \bset (L_{u})$ \[\rho =\rho (e,g,g)\rho (g,u,h)\rho (h,h,f)\]

is a normal factorization of $\rho $. Every normal factorization of $\rho $ has this form.

Consequently when $S$ is a regular semigroup, $\lcat $ is a regular category.

Proof.

(a) If $\rho (e,u,f)=\rho (e',v,f')$, by \eqref{eq:10} we have $\e $ $\lr $ $e'$, $f$ $\lr $ $f'$, $eu\in Sf$, $e'v\in Sf'$ and $e'u=e'v=v$. Since $u=eu\in Sf$, $u\in eSf$. Similarly $v\in e'Sf'$. Conversely, if the given conditions hold, it is clear that conditions of \eqref{eq:10} are satisfied and hence we have $\rho (e,u,f)=\rho (e',v,f')$.

(b) If $u\in eSf$ then $(e,u,f)$ is left admissible and so $\rho (e,u,f) = \rho _{u}\mid Se$ is a morphism from $L(e)=Se$ to $L(f)$ in $\lcat $ by definition. Hence $u\mapsto \rho (e,u,f)$ is a mapping of $eSf$ to $\lcat (L(e),L(f))$ which is one-to-one by (a). If $\rho :L(e)\to L(f)$ is any morphism in $\lcat $ then $u = e\rho \in eSf$ and so $(e,u,f)$ is left-admissible. Also, since $ e\rho =u=e\rho (e,u,f) $ we have $\rho =\rho (e,u,f)$. This proves (b).

(c) Suppose that $\rho =\rho (e,s,f)$ be a monomorphism. Since $s\in eSf$, by (Nambooripad 1979, Definition 1.1 and Theorem 1.1) $R_{s}\cap \om (e)\ne \emptyset $. Let $g\in R_{s}\cap \om (e)$. Dually let $h\in L_{s}\cap \om (e)$ Then there is a unique $s'\in \sinv (s)$ with $ss'=g$ and $s's=h$ see (Krishnan 2000). If $\rho ' =\rho (f,s',e)$ then $\rho \rho '\rho =\rho $ and since $\rho $ is a monomorphism, we have $\rho \rho '=1_{Se}$. This shows that $\rho $ is injective. If $\rho $ is injective and if $g\in R_{s}\cap \om (e)$, then $e\rho =es=s=gs=g\rho $ which gives $g=e$. Hence $e$ $\rr $ $s$. If this holds and if $s'\in \sinv (s)$ with $ss'=e$ then by Proposition 4.3 (iii) $\rho \rho ' =\rho (e,e,e)$ where $\rho '= \rho (f,s',e)$. This also shows that $\rho $ is a split monomorphism.

If $\rho $ is an epimorphism we can choose $s'\in \sinv (s)$ such that $s$ $\lr $ $s's\om f$ and $\rho (f,s',e) =\rho '$ satisfies $\rho \rho '\rho =\rho $. Since $\rho $ is an epimorphism, $\rho '\rho =1_{Sf}$. So $\rho $ is surjective as a map which is also a split epimorphism. If $\rho $ is surjective, there is some $x\in S^{1}$ with $xs=f$. Since $sf=s$ this gives $s$ $\lr $ $f$. Again, if this holds, as in the previous paragraph, we can find $\rho ' =\rho (f,s',e)$ so that $\rho '\rho =\rho (f,f,f)=1_{Sf}$ which shows that $\rho $ is a split epimorphism.

(d) If $e\dr f$ then there is $s\in L_{e}\cap R_{f}$. By (c) $\rho (e,s,f)$ is both an epimorphism and a monomorphism which split and so, $\rho (e,s,f)$ is an isomorphism. Conversely, if $\rho (e,s,f):Se\to Sf$ is an isomorphism, by (c), $e$ $\rr $ $s$ $\lr $ $f$ and so, $e$ $\dr $ $f$.

(e) Clearly $Se\subseteq f$ if and only if $e$ $\omr [l]$ $f$ so that $ef=e$. If this holds, for any $x\in Se$, \[ x\rho (e,e,f)=xe=x=x\inc {Se}{Sf}.\] Hence $\rho (e,e,f)=\inc {Se}{Sf}$. Let $\rho '=\rho (f,u,e)$ be a retraction that is a right inverse of $\rho $. Then $\rho \rho ' =1_{Se}$ and since $u\in fSe$, we have \[ u=u\rho (e,e,f)\rho (f,u,e)=u\rho (e,eu,e)=u(eu)=(ue)u=u^{2}.\] Also, $eu=e\rho (f,u,e)=e$ and hence $u$ $\lr $ $e\omr [l]f$. Thus $u\in L_{e}\cap \om (f)$. Conversely if $g\in L_{e}\cap \om (f)$ then \[\rho (e,e,f)\rho (f,g,e)=\rho (e,e,e)=1_{Se} \] Thus $\rho (f,g,e)$ is a retraction that is a right inverse of $\rho $.

(f) Let $\rho =\rho (e,s,f)$. As in the proof of (b), we can pick $g\in R_{s}\cap \om (e)$ and $h\in \bset (L_{s})$. Then by (e), $\rho (e,g,g):Se\to Sg$ is a retraction which is a right inverse of the inclusion $\rho (g,g,e):Sg\subseteq Se$, $\rho (g,s,h)$ is an isomorphism by (d) and $\rho (h,h,f):Sh\subseteq Sf$. Moreover by Proposition 4.3 (iii), we have \begin{equation*} \rho(e,g,g)\rho(g,s,h)\rho(h,h,f)=\rho(e,gsh,f)=\rho(e,s,f). \end{equation*} This gives a normal factorization of $\rho (e,s,f)$. Assume that $\rho =\e \si j$ be a normal factorization of $\rho =\rho (e,s,f)$. Then $\e :Se\to Sg$ is a retraction and by (e) we can be represent $\e $ as $\e =\rho (e,g',g)$ with $g$ $\lr $ $g'\omr e$. Since $\rho (e,g',g)=\rho (e,g',g')$ and $\si $ is an isomorphism, by (e), we may represent $\si $ in the form $\rho (g',t,h)$ where $g'$ $\rr $ $t$ $\lr $ $h$ and $Sh\subseteq Sf$. Since $j=\inc {Sh}{Sf}=\rho (h,h,f)$ by (e) we have \[\rho (e,s,f)=\rho (e,g',g')\rho (g',t,h)\rho (h,h,f). \] Now using Proposition 4.3 (iii), we have

\begin{equation*} s=e\rho(e,s,f)=e\rho(e,g',g')\rho(g',t,h)\rho(h,h,f)=eg'th=t. \end{equation*}

Therefore the normal factorization $\rho =\e \si j$ has the desired form.

The definition of $\lcat $ shows that it is a category with factorization. Statement (e) above says that every inclusion in $\lcat $ splits and by (f) every morphism has a normal factorization. By Proposition 3.5, $\lcat $ is regular.

Remark 6.1:

If $\opsg $ denote the opposite semigroup of $S$ with multiplication given by $a\circ b=b.a$ where the right hand side is the product in $S$, then it is easy to see that $\lcat [\opsg ]=\rcat $ and $\rcat [\opsg ]=\lcat $. Using these it is possible to translate any statement about the category $\rcat $ of right ideals of a semigroup $S$ as a statement regarding the category $\lcat [\opsg ]$ of left ideals of the opposite semigroup $\opsg $ and vice versa.

References

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Grillet, P.A. (1974a), "Structure of regular semigroups — I: a representation", Semigroup Forum 8: 177–183 

Grillet, P.A. (1974b), "Structure of regular semigroups — II: cross-connections", Semigroup Forum 8: 254–259 

Grillet, P.A. (1974c), "Structure of regular semigroups — III: the reduced case", Semigroup Forum 8: 260–265 

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