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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- Jul 16th 2007, 11:29 AMtukeywilliamsFibers
**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 $. - Jul 16th 2007, 11:40 AMPlato
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}$