# Equivalence relations, surjective maps, partitions and fibers.

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.