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Thread: Proof trouble

  1. #1
    Aug 2010

    Proof trouble

    Hi, I'm working out of Dummit and Foote and I need to know if I'm proceeding correctly in a proof. Problem: for the surjective map from A to B f, prove that the relation

    a ~ b iff f(a)=f(b)

    is an equivalence relation whose equivalence classes are the fibers of f.

    So it makes sense but I don't know if I'm necessarily 'showing it'. I say:

    if f(a) = f (b)
    then a= b
    so (a ~ b ) -> a=b

    for any f(x) in y there exists a set containing only f(x). the preimage of any of these sets defines the fiber of f over their respective elements...

    Long story short, I'm running in circles and would appreciate a helping hand. Thanks!
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  2. #2
    Super Member
    Aug 2009
    You defined a relation ~ by $\displaystyle a \sim b \Leftrightarrow f(a)=f(b)$ where $\displaystyle f:A \to B$ is a surjection and $\displaystyle a,b \in A$.

    1) Reflexivity:
    $\displaystyle a \sim a \Leftrightarrow f(a)=f(a)$

    $\displaystyle a \sim b \Rightarrow f(a)=f(b) \Rightarrow f(b)=f(a)$

    3) Transitivity:
    $\displaystyle a \sim b, b \sim c \Rightarrow f(a)=f(b)=f(c)$

    Also, f is not neccessarily injective - so $\displaystyle f(a)=f(b) \Rightarrow a=b$ is not always true.
    The second part is straightforward - take a look at the definition of a fiber and you should see why every equivalence class of ~ is a fiber.
    Last edited by Defunkt; Aug 25th 2010 at 05:50 PM. Reason: latex
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