Boolean Algebra Embeddings: Defining and Understanding the Role of Monomorphisms

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SUMMARY

The discussion centers on the definition of embeddings in Boolean algebra, specifically the injective map f: B → B' that satisfies conditions related to supremum and complement. Participants clarify that an embedding is indeed a monomorphism, and they debate the necessity of including the condition f(inf{x,y}) = inf'{f(x),f(y)}. The consensus is that this condition follows from the established properties due to de Morgan's laws, confirming the robustness of the definition.

PREREQUISITES
  • Understanding of Boolean algebra concepts, including supremum and infimum.
  • Familiarity with injective functions and monomorphisms in mathematical contexts.
  • Knowledge of de Morgan's laws and their application in algebraic structures.
  • Basic comprehension of mappings between algebraic systems.
NEXT STEPS
  • Study the properties of monomorphisms in category theory.
  • Explore the implications of de Morgan's laws in various algebraic structures.
  • Learn about the role of embeddings in lattice theory.
  • Investigate advanced topics in Boolean algebra, such as homomorphisms and isomorphisms.
USEFUL FOR

Mathematicians, computer scientists, and students studying algebraic structures, particularly those interested in Boolean algebra and its applications in theoretical computer science.

quasar987
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According to my notes, the definition of an embedding from a boolean algebra B in a boolean algebra B' is an injective map f:B-->B' such that for all x,y in B, f(sup{x,y}) = sup'{f(x),f(y)} and f(Cx)=C'(f(x)), where sup is the supremum in B and sup' is the complement in B', and where C is the complement in B and C' the complement in B'.

But I read on wiki that generally, an embedding is supposed to be a monomorphism. Aren't we missing the condition f(inf{x,y}) = inf'{f(x),f(y)}?
 
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quasar987 said:
But I read on wiki that generally, an embedding is supposed to be a monomorphism. Aren't we missing the condition f(inf{x,y}) = inf'{f(x),f(y)}?
Doesn't it follow from the other ones?
 
Right, because of the de Morgan laws !
 

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