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 ı f . It remains to show that s is a bijection. In the same way that we deﬁned s , we can also deﬁne a map s W Y ! Y such that f D s ı f . I claim that s and s are inverse maps, so both are bijective. In fact, we have f D s ı f D s ı s ı 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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 Fall '08
 EVERAGE
 Logic, Algebra, Equivalence relation, Binary relation, Isomorphism

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