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 ...
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