Proving an equivalence relation using inverse functions

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SUMMARY

The discussion centers on proving that the set (f × f)-1(Γ) is an equivalence relation on A, given a function f: A → B and an equivalence relation Γ on B. The proof establishes that if (f(a), f(a′)) ∈ Γ, then (a, a′) must also hold an equivalence relation, confirming that (f × f)-1(Γ) ⊂ A × A is indeed an equivalence relation. The conclusion is that the properties of equivalence relations are preserved under the inverse image of functions.

PREREQUISITES
  • Understanding of equivalence relations in set theory
  • Familiarity with functions and their properties
  • Knowledge of inverse functions and their implications
  • Basic concepts of set notation and operations
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  • Study the properties of equivalence relations in detail
  • Learn about inverse functions and their applications in set theory
  • Explore examples of functions and equivalence relations in mathematical proofs
  • Investigate the relationship between functions and their inverse images
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Mathematics students, educators, and anyone studying abstract algebra or set theory, particularly those focusing on functions and equivalence relations.

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Homework Statement


Let f : A → B be a function and let Γ ⊂ B × B be an equivalence relation on B. Prove that the set (f × f)^-1 (Γ) ⊂ A × A (this can be described as {(a, a′) ∈ A × A|(f(a), f(a′)) ∈ Γ}) is an equivalence relation on A.

Homework Equations


The Attempt at a Solution


Let (f(a),f(a’)) ⊂ Γ. Since f(a) and f(a’) hold an equivalence relation with each other, it follows that a and a’ hold an equivalence relation with each other. Since f(a) and f(a’) are arbitrary elements of Γ, it follows that (fxf)-1Γ ⊂ A x A is an equivalence relation on A.

I'm not sure if thi is the right approach. In particular I am not sure that i can say that f(a) and f(a') holding an equivalence relation means that a and a' hold one too.
 
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Does anybody know if I am correct? I'm not sure if I'm skipping steps.
 

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