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    Equivalence relations, surjective maps, partitions and fibers.

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    Consider any surjective map f from a set X onto another set Y. We can define a relation on X by x_1 ~ x_2 if f(x_1) = f(x_2). Check that this is an equivalence relation. Show that the associated partition of X is the partition into "fibers" f^(-1) (y) for y in Y.

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    Problem:

    Consider any surjective map f from a set X onto another set Y. We can define a relation on X by:
    x_1 ~ x_2 if f(x_1) = f(x_2).
    Check that this is an equivalence relation. Show that the associated partition of X is the partition into "fibers" f^(-1) (y) for y in Y.

    Solution:

    In order to see if the above relation is an equivalence one, we have to check out if the equivalence axioms are satisfied:

    a) Reflexivity:
    This is almost obvious, since
    b) Symmetry:
    This is again obvious for this case ( f(x1) = f(x1) ).

    (one example of relation reflexive but not symmetric is the perpendicularity: two lines or vectors can be perpendicular one on each other, but never a line or vector can be perpendicular on itself!)

    c) Transitivity: this is to prove that
    According to the definition of our relation, we have:

    But

    so that ...

    Solution Summary

    Equivalence relations, surjective maps, partitions and fibers are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

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