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Math Help - Equivalence class

  1. #1
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    Equivalence class

    Problem: Let  f: A \to B be a surjective map of sets. Prove that the relation  a \sim b \Leftrightarrow  f(a) = f(b) is an equivalence relation whose equivalence classes are the fibers of  f .

    My Work/Thoughts: What does a surjective map of sets mean? I think it means that for all  b \in B there is some  a \in A such that  f(a) = b . An equivalence relation has to be symmetric, reflexive, and transitive. I know that  a \sim b really means that  (a,b) \in R which is some subset of a set  A \times A . So we have to show that these 3 properties hold.

    (i) reflexive:  a \sim a for all  a \in A . So  f(a) = f(a) ?

    (ii) symmetric:  a \sim b \Rightarrow b \sim a for all  a,b \in A . So  f(a) = f(b) and  f(b) = f(a) .

    (iii) transitive:  a \sim b and  b \sim c \Rightarrow a \sim c for all  a,b,c \in A . So  f(a) = f(b) \ \text{and} \ f(b) = f(c) \Rightarrow f(a) = f(c) for some  a,b,c \in A ? So the '  = ' sign acts like the '  \sim '?

    Now I have to show that the equivalence classes are the fibers of  f . So by definition, an equivalence class is  \{x \in A|x \sim a \} . This is where I become stuck. A fiber is the inverse image of one element of the codomain mapped to the domain right? So is it  \{x \in A| f(a) = f(b) \} ?

    Thanks
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  2. #2
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    Quote Originally Posted by shilz222 View Post
    [B] Now I have to show that the equivalence classes are the fibers of  f . So by definition, an equivalence class is  \{x \in A|x \sim a \} . This is where I become stuck. A fiber is the inverse image of one element of the codomain mapped to the domain right?
    You work is just fine. However, this particular problem requires the function to be a surjection.
    In general, the fibers of a function define an equivalence relation on its domain. But the that may vary with your particular definitions.
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