up | next | prev | ptail | tail |
Another class’s of categories that will be of interest in the sequel are defined here [see Nambooripad, 1994].
To avoid repetition, we assume throughout this section that , , etc., stand for small categories with factorization property.
A normal factorization of a morphism is a factorization of the form where is a retraction, is an isomorphism and . A morphism is said to be regular if there is such that and . We proceed to investigate the relation between regularity and normal factorization.
Proposition 6.1. Let be a split inclusion and be a retraction in .
Proof.
Proposition 6.2. Suppose that every inclusion in splits. Then every morphism in has a unique canonical factorization. In particular, has images.
Proof. Let be canonical factorizations of the morphism . Since and are split there exist morphisms and in with and . Then
and since is an epimorphism we have . Similarly . Hence and are equivalent as monomorphisms. Since these are inclusions, we have . Then and since is a monomorphism, we have .
Since canonical factorizations are unique, the existence of follows from its definition 2.2.5.
Theorem 6.3. A morphism in a category (with factorization) is regular if and only if has normal factorization where is a split inclusion. Thus the category is regula if and only if
In particular, every regular category is a -category. (see § 3)
Proof. Suppose that be regular so that there is with and . Let and be canonical factorizations. Then we have
Since , epimorphisms and , are monomorphisms, we have
where and . Then is a retraction which is a right inverse of and is a retraction which is a right inverse of . Moreover, is an isomorphism with as its inverse. Therefore,
is a normal factorization of in which is a split inclusion.
Suppose that be a normal factorization of in which splits. Let be the retraction which is the right inverse of and let be the left inverse of the retraction . Let . Then simple calculations show that and .
It now follows that is regular if and only if it satisfies conditions a) and b). It follows from Proposition 6.2 that has images. By b), every surjection in splits and hence by Proposition 3.5, is an -category with respect to the natural embedding and so, is an -category. (see §3).
Let be a regular semigroup. Since every principal left ideal in has at least one idempotent generator, we may write objects (vertexes) in as for and morphisms as where and is left compatible (see Proposition 4.3). Since and we may replace with if necessary and assume that . Conversely if then is left admissible. Thus
(28) |
When is regular we shall use these representations of objects and morphism of with out further comment.
We proceed to discuss some important properties of the category [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 E. Krishnan, 2000, Chapter 2 and 3].
Proposition 6.4. Let be a regular semigroup. Then
is a normal factorization of . Every normal factorization of has this form.
Consequently when is a regular semigroup, is a regular category.
Proof.
(a) If , by Equation (13) we have , , , and . Since , . Similarly . Conversely, if the given conditions hold, it is clear that conditions of Equation (13) are satisfied and hence we have .
(b) If then is left admissible and so is a morphism from to in by definition. Hence is a mapping of to which is one-to-one by (a). If is any morphism in then and so is left-admissible. Also, since we have . This proves (b).
(c) Suppose that be a monomorphism. Since , by [Nambooripad, 1979, Definition 1.1 and Theorem 1.1] . Let . Dually let Then there is a unique with and [see E. Krishnan, 2000]. If then and since is a monomorphism, we have . This shows that is injective. If is injective and if , then which gives . Hence . If this holds and if with then by Proposition 4.3 (iii) where . This also shows that is a split monomorphism.
If is an epimorphism we can choose such that and satisfies . Since is an epimorphism, . So is surjective as a map which is also a split epimorphism. If is surjective, there is some with . Since this gives . Again, if this holds, as in the previous paragraph, we can find so that which shows that is a split epimorphism.
(d) If then there is . By (c) is both an epimorphism and a monomorphism which split and so, is an isomorphism. Conversely, if is an isomorphism, by (c), and so, .
(e) Clearly if and only if so that . If this holds, for any ,
Hence . Let be a retraction that is a right inverse of . Then and since , we have
Also, and hence . Thus . Conversely if then
Thus is a retraction that is a right inverse of .
(f) Let . As in the proof of (b), we can pick and . Then by (e), is a retraction which is a right inverse of the inclusion , is an isomorphism by (d) and . Moreover by Proposition 4.3 (iii), we have
This gives a normal factorization of . Assume that be a normal factorization of . Then is a retraction and by (e) we can be represent as with . Since and is an isomorphism, by (e), we may represent in the form where and . Since by (e) we have
Now using Proposition 4.3 (iii), we have
Therefore the normal factorization has the desired form.
The definition of shows that it is a category with factorization. Statement (e) above says that every inclusion in splits and by (f) every morphism has a normal factorization. By Proposition 3.5, is regular.
Remark 6.1: If denote the opposite semigroup of with multiplication given by where the right hand side is the product in , then it is easy to see that and . Using these it is possible to translate any statement about the category of right ideals of a semigroup as a statement regarding the category of left ideals of the opposite semigroup and vice versa.
up | next | prev | ptail | top |