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

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

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

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

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

Now I have to show that the equivalence classes are the fibers of $\displaystyle f $. So by definition, an equivalence class is $\displaystyle \{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 $\displaystyle \{x \in A| f(a) = f(b) \} $?

Thanks