S-偏序系的1-纯性质研究与绝对1-纯分类

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for every a, a

∈ A. Let H ⊆ A × A is an subset. Deﬁne θ(H) by

aθ(H)a

⇔ a ≤

ρ(H)

≤

ρ(H)

a, a

∈ A, where ρ = ρ(H) is the S-act congruence on A generated by H. Then θ(H) is an S-act

congruence. Moreover θ(H) is an S-poset congruence on A, where the factor S-poset A/θ(H)

is equipped with the order

[a]

θ(H)

≤ [a

]

θ(H)

⇔ a ≤

ρ(H)

We call θ(H) the S-poset congruence generated by H.

Let α be any cardinal greater than 1, we call a right ideal I of S a right α-ideal, if I has

a generating set G ⊆ S of fewer than α elements. Specially, right 2-ideals are principally right

ideals and right ℵ

-ideals are ﬁnitely generated right ideals.

We say that an S-poset A is regularly α-injective if every right α-ideal I of S, every

regular monomorphism ι: I → S and every S-poset homomorphism f: I → A there exists

an S-poset homomorphism g: S → A such that gι = f. So regularly 2-injective S-posets are

regularly principally weakly injective S-posets and regularly ℵ

-injective S-posets are regularly

fg-weakly weakly injective S-posets.

Clearly, a regularly weakly injective S-poset is regularly α-injective for which α is an

arbitrary cardinal. If α ≥ β > 1, then any regularly α-injective must be regularly β-injective.

If all right ideals are α-ideals, then regularly α-injective is regularly weakly injective.

2 Properties of regularly α-injective S-posets

In this section, we will give necessary and suﬃcient conditions for an S-posets to be regular

α-injective.

Lemma 1. ([11])An S-poset A is regular injective if and only if for any S-poset C, any

S-subposet B ⊆ C, and any S-poset homomorphism f: B → A, there exists an S-poset homo-

morphism g: C → A such that g|

= f.

Corollary 2. An S-poset A is regular α-injective if and only if for any α-ideal I, I ⊆ S, and

any S-poset homomorphism f: I → A, there exists an S-poset homomorphism g: S → A such

that g|

= f.

Proposition 3. An S-poset A is regularly α-injective if and only if for any S-poset homomor-

phism f: K → A where K ⊆ S is an α-ideal, there exists an element a ∈ A such that f(k) = ak

for each k ∈ K.

Proof. Necessity. Let ι: K → S be an inclusion mapping. Then there exists an S-poset

homomorphism g: S → A such that gι = f because A is regularly α-injective. Let g(1) = a.

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S-偏序系的1-纯性质研究与绝对1-纯分类

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