College Algebra Exam Review 121

College Algebra Exam Review 121 - y and not on the choice...

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2.6. EQUIVALENCE RELATIONS AND SET PARTITIONS 131 When do two surjective maps f W X ! Y and f 0 W X ! Y 0 deter- mine the same equivalence relation on X ? The condition for this to happen turns out to be the following: Definition 2.6.12. Two surjective maps f W X ! Y and f 0 W X ! Y 0 are similar if there exists a bijection s W Y ! Y 0 such that f 0 D s ı f . X f 0 q q q q q Y 0 f q q q q q ± ± ± ± ± ± q q q q q q q = s Y Proposition 2.6.13. Two surjective maps f W X ! Y and f 0 W X ! Y 0 determine the same equivalence relation X if, and only if, f and f 0 are similar. Proof. It is easy to see that if f and f 0 are similar surjections, then they determine the same equivalence relation on X . Suppose, on the other hand, that f W X ! Y and f 0 W X ! Y 0 are surjective maps that define the same equivalence relation ± on X . We want to define a map s W Y ! Y 0 such that f 0 D s ı f . Let y 2 Y , and choose any x 2 f ± 1 .y/ . We wish to define s.y/ D f 0 .x/ , for then s.f.x// D s.y/ D f 0 .x/ , as desired. However, we have to be careful to check that this makes sense; that is, we have to check that s.y/ really depends only on
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Unformatted text preview: y and not on the choice of x 2 f 1 .y/ ! But in fact, if N x is another element of f 1 .y/ , then x f N x , so x f N x , by hypothesis, so f .x/ D f . N x/ . We now have our map s W Y ! Y such that f D s f . It remains to show that s is a bijection. In the same way that we dened s , we can also dene a map s W Y ! Y such that f D s f . I claim that s and s are inverse maps, so both are bijective. In fact, we have f D s f D s s f , so f.x/ D s .s.f.x/// for all x 2 X . Let y 2 Y ; choose x 2 X such that y D f.x/ . Substituting y for f.x/ gives y D s .s.y// . Similarly, s.s .y // D y for all y 2 Y . This completes the proof that s is bijective. n...
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