Epicompletion in frames with skeletal maps. II: Compact normal joinfit frames
In an earlier paper [``Epicompletion in frames with skeletal maps. I: Compact regular frames'', Appl. Categ. Struct. 16, No. 4, 521--533 (2008; Zbl 1156.06004)], to which the current paper is a sequel, the authors proved that in the category of compact regular frames with skeletal maps (a frame homomorphism is skeletal if it maps dense elements to dense elements), the subcategory consisting of extremally disconnected frames (which the authors call ``strongly projectable'') coincides with the epicomplete objects. In the paper under review, a frame \(L\) is said to be joinfit if, for each \(0<a\leq b\in L\), there exists \(c\in L\) with \(c<b\) such that \(b=a\vee c\). Let \(\mathbf{A}\) denote the category of compact normal joinfit frames with skeletal maps, and \(\mathbf{B}\subseteq \mathbf{A}\) the subcategory consisting of strongly projectable objects of \(\mathbf{A}\) with the same morphisms. The main achievement of the paper is the construction (without invoking the Axiom of Choice, no less) of an epireflection \(\psi:\mathbf{A}\to\mathbf{B}\). This is achieved via a series of subsidiary results of independent interest such as the preservation of normality by coproducts if each summand is compact, normal and joinfit. An indispensable tool in the construction of the epireflection is the notion of the absolute of a compact frame. This is the frame of ideals of the Booleanization (or, as the authors prefer to say, the frame of polars) of the given compact frame. The paper ends with a number of open questions and workpoints, premier among which is whether \(\psi\) is a monoreflection. This latter question has however since been resolved (in the affirmative) by the first-named author. The proof appears in the forthcoming paper [``Epicompletion in frames with skeletal maps. III: When maps are closed'', Appl. Categ. Struct., DOI: 10.1007/s10485-009-9194-3]. Reviewer's remark: The reader should note that in the definition of skeletal maps (Definition 1.4) there is an omission of the polar symbol \(\perp\). What appears as \(h(x)^{\perp}=1\) should be \(h(x)^{\perp\perp}=1\).
scientific article; zbMATH DE number 5633348
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