Unformatted text preview: f x 00 if f.x / D f.x 00 / . We might as well assume that f is surjective, as we can replace Y by the range of f without changing the equivalence relation. The equivalence classes of ± f are the ﬁbers f ± 1 .y/ for y 2 Y . See Exercise 2.6.1 . On the other hand, given an equivalence relation ± on X , deﬁne X= ± to be the set of equivalence classes of ± and deﬁne a surjection ± of X onto X= ± by ±.x/ D ŒxŁ . If we now build the equivalence relation ± ± associated with this surjective map, we just recover the original equivalence relation. In fact, for x ;x 00 2 X , we have x ± x 00 , Œx Ł D Œx 00 Ł , ±.x / D ±.x 00 / , x ± ± x 00 . We have proved the following result: Proposition 2.6.11. Let ± be an equivalence relation on a set X . Then there exists a set Y and a surjective map ± W X ! Y such that ± is equal to the equivalence relation ± ± ....
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra, Sets

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