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Thread: Fibers

  1. #1
    Senior Member tukeywilliams's Avatar
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    Fibers

    7. $\displaystyle f: A \rightarrow B $ is a surjective map of sets. We want to prove that $\displaystyle a \sim b $ if and only if $\displaystyle f(a) = f(b) $ is an equivalence relation whose equivalence classes are fibers of $\displaystyle f $.
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  2. #2
    MHF Contributor

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    Here it is: $\displaystyle
    \begin{array}{l}
    \left( {\forall x \in A} \right)\left[ {f(x) = f(x)} \right] \\
    \left[ {f(x) = f(y)\quad \Rightarrow \quad f(y) = f(x)} \right] \\
    \left[ {f(x) = f(y)\quad \& \quad f(y) = f(z)\quad \Rightarrow \quad f(x) = f(z)} \right] \\
    \end{array}$
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