1. The problem statement, all variables and given/known data suppose [itex]f~:~A \rightarrow B[/itex] be a surjective map of sets. Prove that the relation [itex]a Rb \iff f(a)=f(b)[/itex] is a equivalence relation whose equivalence classes are the fibers of [itex]f[/itex]. 2. Relevant equations 3. The attempt at a solution I was able to easily prove that the relation satisfied the reflexive and symmetric part, but there are a few details of the transitive case that I am uncertain of. Here is some of my work: Suppose that [itex](a,b) \in R[/itex] and [itex](b,c) \in R[/itex], meaning that the function [itex]f[/itex] maps the elements a and b to the same element in the codomain, call it [itex]\alpha[/itex], and the function maps b and c to the same element; that is, [itex]f(a) = f(b) [/itex] and f(b) = f(c). Does this immediately imply that [itex]f(a) = f(c)[/itex], or do I have to do some further reasoning, such as reiterating that [itex]f[/itex] is indeed a function, which means that [itex]f[/itex] can only associate one element in its codomain to each element in its domain, suggesting that [itex]f(b) = \alpha = \beta[/itex]? For instance, suppose in the problem that [itex]f[/itex] was not specified as a function. Then if [itex]f[/itex] was [itex]f(x) = \pm \sqrt(x)[/itex], then it wouldn't be transitive, right?