March 7, 2017

Algebras and Coalgebras by Yde Venema

By Yde Venema

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1. 8 Let f : A → B be a map between Boolean algebras. Then (i) if f is monotone then so is f σ ; (ii) if f is an operator then f σ is a complete operator; (iii) if f is additive or multiplicative then f is smooth. Proof. 14). For the remaining part, we need to show that if f is normal and additive in each coordinate, then f σ is normal and completely additive in each coordinate. Leaving the easy proof for normality as an exercise for the reader, concerning additivity, we will prove that if f : A0 × A1 → B is additive in its first coordinate and monotone in its second, then f σ preserves all non-empty joins in its first coordinate.

4 that for an arbitrary element b of a Boolean algebra B, the sets b↑ and b↓ are defined as b↑ = {a ∈ B | b ≤ a} and b↓ = {a ∈ B | a ≤ b}. 11 For any Boolean algebra B, the sets σ ↑ := {p↑ | p ∈ K} and σ ↓ := {u↓ | u ∈ O} both form a topology on Aσ ; and so does the set σ := {p↑ ∩ u↓ | K p ≤ u ∈ O}, which is in fact identical to the join σ ↑ ∨ σ ↓ in the lattice of topologies on Aσ . In the sequel, we will write [p, u] for the interval between p and u, that is, [p, u] = p↑ ∩ u↓. 41 Proof. 3(ii).

While it may not be the case that the lower and the upper extension agree in all cases, both kinds of extensions generally display good behavior; often they even improve on the original map. 1. 8 Let f : A → B be a map between Boolean algebras. Then (i) if f is monotone then so is f σ ; (ii) if f is an operator then f σ is a complete operator; (iii) if f is additive or multiplicative then f is smooth. Proof. 14). For the remaining part, we need to show that if f is normal and additive in each coordinate, then f σ is normal and completely additive in each coordinate.

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