What's the correct term for it?

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Discussion Overview

The discussion revolves around the terminology used to describe a relationship between mappings f:A->A and g:B->B, particularly in the context of a one-to-one mapping h:A->B that satisfies a specific condition involving these functions. The scope includes theoretical aspects of mathematics, particularly related to function mappings and their properties.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant asks for the correct term for the relationship between the mappings f and g given the existence of a mapping h that relates them.
  • Another participant points out that for the equality f(a) = g(h(a)) to hold, the codomain of f must equal the codomain of g.
  • A different participant notes that since f maps from A to A and g maps from B to B, g(h(x)) will be in B, not A, raising a concern about the initial formulation.
  • A subsequent reply corrects the earlier statement, suggesting it should be h(f(a)) = g(h(a)).
  • One participant mentions that the relationship described resembles a "commutative diagram," a concept from Algebraic Topology and Category Theory, although they do not provide a specific name for the function h.

Areas of Agreement / Disagreement

Participants express differing views on the formulation of the relationships between the functions and the conditions under which they hold. There is no consensus on a specific term for the function h or the relationship itself.

Contextual Notes

There are limitations regarding the assumptions about the codomains of the functions and the specific conditions under which the relationships are defined. The discussion does not resolve these issues.

zef
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How do we call such relation between mappings f:A->A and g:B->B when there exists one-to-one mapping h:A->B such that for any a from A f(a)=g(h(a))?

Example.

f: {T, F) -> {T, F}
where f(x) = not x

g : {0, 1} -> {0, 1}
where f(x)=1-x

h: {T, F} -> {0, 1}
where h(T)=1, h(F)=0
 
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Hmm, technically f(a) cannot be equal to g(h(a)) unless codomain(f) = codomain(g).
 
You just said f:A->A and g:B->B. For any function h:A->B, g(h(x)) is in B, not A.
 
Sorry, it should be h(f(a))=g(h(a))
 
I don't know any specific name for the function h, but what you have is a "commutative diagram" used in Algebraic Topology and Category Theory.
 

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