6. Regular categories

Another class’s of categories that will be of interest in the sequel are defined here [see Nambooripad1994].

6.1. Normal factorization and regularity

To avoid repetition, we assume throughout this section that 𝒞, 𝒟, etc., stand for small categories with factorization property.

A normal factorization of a morphism f 𝒞(c,d) is a factorization of the form f = euj where e : c c is a retraction, u : c d d is an isomorphism and j = ȷdd. A morphism f : c d is said to be regular if there is f : d c such that fff = f and fff = f. We proceed to investigate the relation between regularity and normal factorization.

Proposition 6.1. Let j = ȷdc be a split inclusion and 𝜖 : d c be a retraction in 𝒞.

(a)
If j is an epimorphism in 𝒞 then c = d and j = 1c.
(b)
If 𝜖 is a monomorphism in 𝒞 then c = d and 𝜖 = 1d.

Proof.

(a) Since j : c d is a split inclusion there is 𝜖 : d c with j𝜖 = 1c. Then j(𝜖j) = j = j1d. Since j is an epimorphism, we have 𝜖j = 1d. By axiom (3) of 2.1.2, 𝜖 = ȷcd. Hence c d c. Since is a partial order on 𝖛𝒞, we have c = d and then j = 1c.

(b) Let j = ȷdc so that 𝜖j𝜖 = 𝜖. Since 𝜖 is a monomorphism, we have 𝜖j = 1d and by axiom (3) of 2.1.2, 𝜖 = ȷcd. Consequently c = d and so 𝜖 = ȷdc = 1d.

Proposition 6.2. Suppose that every inclusion in 𝒞 splits. Then every morphism in 𝒞 has a unique canonical factorization. In particular, 𝒞 has images.

Proof. Let f = xj = yj be canonical factorizations of the morphism f 𝒞. Since j = ȷdc and j = ȷdc are split there exist morphisms u and v in 𝒞 with ju = 1c and jv = 1c. Then

yjuj = xjuj = xj = yj

and since y is an epimorphism we have (ju)j = juj = 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 imf follows from its definition 2.2.5.

Theorem 6.3. A morphism f in a category 𝒞 (with factorization) is regular if and only if f has normal factorization f = 𝜖uj where j is a split inclusion. Thus the category 𝒞 is regula if and only if

a)
every inclusion in 𝒞 splits; and
b)
every morphism in 𝒞 has normal factorization.

In particular, every regular category is a 𝒩𝒮-category. (see § 3)

Proof. Suppose that f : c d be regular so that there is f : d c with fff = f and fff = f. Let f = xj and f = yj be canonical factorizations. Then we have

xjyjxj = xj andyjxjyj = yj

Since x, y epimorphisms and j, j are monomorphisms, we have

jyjx = 1d andjxjy = 1c

where d = codx and c = cody. Then 𝜖 = xjy : c c is a retraction which is a right inverse of j = cc and 𝜖 = yjx : d d is a retraction which is a right inverse of j = ȷdd. Moreover, jx = u is an isomorphism with jy = v as its inverse. Therefore,

f = xj = xjyjxj = 𝜖uj

is a normal factorization of f in which j is a split inclusion.

Suppose that f = 𝜖uj be a normal factorization of f in which j splits. Let 𝜖 be the retraction which is the right inverse of j and let j be the left inverse of the retraction 𝜖. Let f = 𝜖u1j. Then simple calculations show that f = fff and f = fff.

It now follows that 𝒞 is regular if and only if it satisfies conditions a) and b). It follows from Proposition 6.2 that 𝒞 has images. By b), every surjection in 𝒞 splits and hence by Proposition 3.5, 𝒞 is an 𝒮-category with respect to the natural embedding and so, 𝒞 is an 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 𝕃0(S) as L(e) = Se for e E(S) and morphisms ρ : Se Sf as ρ = ρ(e,s,f) where e,f E(S) and (e,s,f) is left compatible (see Proposition 4.3). Since es Sf and ρ(e,s,f) = ρ(e,es,f) we may replace s with es if necessary and assume that s eSf. Conversely if esf eSf then (e,s,f) is left admissible. Thus

𝖛𝕃0(S) = {L(e) = Se : e E(S)} and𝕃0(S) = {ρ(e,s,f) : e,f E(S),s eSf}. (28)

When S is regular we shall use these representations of objects and morphism of 𝕃0(S) with out further comment.

We proceed to discuss some important properties of the category 𝕃0(S) [see Nambooripad1994, 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 E. Krishnan2000, Chapter 2 and 3].

Proposition 6.4. Let S be a regular semigroup. Then

(a)
ρ(e,u,f) = ρ(e,v,f) if and only if 𝜖 e, f f, u eSf, v eSf and v = eu.
(b)
The map ρ(e,s,f)s is a bijection of 𝕃0(S)(Se,Sf) onto eSf for all e,f E(S).
(c)
A morphism ρ = ρ(e,s,f) is a monomorphism [epimorphism] if and only if ρ is injective [surjective]; this is true if and only if e s [s f]. Every monomorphism [epimorphism] in 𝕃0(S) split.
(d)
Se and Sf are isomorphic if and only if e 𝒟 f; if this is the case, there is a bijection of the set of isomorphisms of Se onto Sf and the -class Le Rf.
(e)
A morphism ρ is the inclusion Se Sf if and only if ef = e and ρ = ρ(e,e,f). Then ρ : Sf Se is the right inverse of ρ if and only if ρ = ρ(f,g,e) where g Le ω(f).
(f)
Let ρ = ρ(e,u,f) be a morphism in 𝕃0(S). For any g Ru ω(e) and h E(Lu)
ρ = ρ(e,g,g)ρ(g,u,h)ρ(h,h,f)

is a normal factorization of ρ. Every normal factorization of ρ has this form.

Consequently when S is a regular semigroup, 𝕃0(S) is a regular category.

Proof.

(a) If ρ(e,u,f) = ρ(e,v,f), by Equation (13) we have 𝜖 e, f f, eu Sf, ev Sf and eu = ev = v. Since u = eu Sf, u eSf. Similarly v eSf. Conversely, if the given conditions hold, it is clear that conditions of Equation (13) are satisfied and hence we have ρ(e,u,f) = ρ(e,v,f).

(b) If u eSf then (e,u,f) is left admissible and so ρ(e,u,f) = ρuSe is a morphism from L(e) = Se to L(f) in 𝕃0(S) by definition. Hence uρ(e,u,f) is a mapping of eSf to 𝕃0(S)(L(e),L(f)) which is one-to-one by (a). If ρ : L(e) L(f) is any morphism in 𝕃0(S) then u = eρ eSf and so (e,u,f) is left-admissible. Also, since eρ = u = eρ(e,u,f) we have ρ = ρ(e,u,f). This proves (b).

(c) Suppose that ρ = ρ(e,s,f) be a monomorphism. Since s eSf, by [Nambooripad1979, Definition 1.1 and Theorem 1.1] Rs ω(e). Let g Rs ω(e). Dually let h Ls ω(e) Then there is a unique s𝒱(s) with ss = g and ss = h [see E. Krishnan2000]. If ρ = ρ(f,s,e) then ρρρ = ρ and since ρ is a monomorphism, we have ρρ = 1Se. This shows that ρ is injective. If ρ is injective and if g Rs ω(e), then eρ = es = s = gs = gρ which gives g = e. Hence e s. If this holds and if s𝒱(s) with ss = e then by Proposition 4.3 (iii) ρρ = ρ(e,e,e) where ρ = ρ(f,s,e). This also shows that ρ is a split monomorphism.

If ρ is an epimorphism we can choose s𝒱(s) such that s ssωf and ρ(f,s,e) = ρ satisfies ρρρ = ρ. Since ρ is an epimorphism, ρρ = 1Sf. So ρ is surjective as a map which is also a split epimorphism. If ρ is surjective, there is some x S1 with xs = f. Since sf = s this gives s f. Again, if this holds, as in the previous paragraph, we can find ρ = ρ(f,s,e) so that ρρ = ρ(f,f,f) = 1Sf which shows that ρ is a split epimorphism.

(d) If e 𝒟 f then there is s Le Rf. By (c) ρ(e,s,f) is both an epimorphism and a monomorphism which split and so, ρ(e,s,f) is an isomorphism. Conversely, if ρ(e,s,f) : Se Sf is an isomorphism, by (c), e s f and so, e 𝒟 f.

(e) Clearly Se Sf if and only if e ωl f so that ef = e. If this holds, for any x Se,

xρ(e,e,f) = xe = x = xȷSf Se.

Hence ρ(e,e,f) = ȷSf Se. Let ρ = ρ(f,u,e) be a retraction that is a right inverse of ρ. Then ρρ = 1Se and since u fSe, we have

u = uρ(e,e,f)ρ(f,u,e) = uρ(e,eu,e) = u(eu) = (ue)u = u2.

Also, eu = eρ(f,u,e) = e and hence u e ωl f. Thus u Le ω(f). Conversely if g Le ω(f) then

ρ(e,e,f)ρ(f,g,e) = ρ(e,e,e) = 1Se

Thus ρ(f,g,e) is a retraction that is a right inverse of ρ.

(f) Let ρ = ρ(e,s,f). As in the proof of (b), we can pick g Rs ω(e) and h E(Ls). Then by (e), ρ(e,g,g) : Se Sg is a retraction which is a right inverse of the inclusion ρ(g,g,e) : Sg Se, ρ(g,s,h) is an isomorphism by (d) and ρ(h,h,f) : Sh Sf. Moreover by Proposition 4.3 (iii), we have

ρ(e,g,g)ρ(g,s,h)ρ(h,h,f) = ρ(e,gsh,f) = ρ(e,s,f).

This gives a normal factorization of ρ(e,s,f). Assume that ρ = 𝜖σj be a normal factorization of ρ = ρ(e,s,f). Then 𝜖 : Se Sg is a retraction and by (e) we can be represent 𝜖 as 𝜖 = ρ(e,g,g) with g g ω e. Since ρ(e,g,g) = ρ(e,g,g) and σ is an isomorphism, by (e), we may represent σ in the form ρ(g,t,h) where g t h and Sh Sf. Since j = ȷSf Sh = ρ(h,h,f) by (e) we have

ρ(e,s,f) = ρ(e,g,g)ρ(g,t,h)ρ(h,h,f).

Now using Proposition 4.3 (iii), we have

s = eρ(e,s,f) = eρ(e,g,g)ρ(g,t,h)ρ(h,h,f) = egth = t.

Therefore the normal factorization ρ = 𝜖σj has the desired form.

The definition of 𝕃0(S) shows that it is a category with factorization. Statement (e) above says that every inclusion in 𝕃0(S) splits and by (f) every morphism has a normal factorization. By Proposition 3.5, 𝕃0(S) is regular.

Remark 6.1: If Sop denote the opposite semigroup of S with multiplication given by a b = b.a where the right hand side is the product in S, then it is easy to see that 𝕃0(Sop) = 0(S) and 0(Sop) = 𝕃0(S). Using these it is possible to translate any statement about the category 0(S) of right ideals of a semigroup S as a statement regarding the category 𝕃0(Sop) of left ideals of the opposite semigroup Sop and vice versa.