Let $X$ be a non-empty set, and let $R$ be an equivalence relation on $X$. Let $C$ be the set of all equivalence classes of $R$. So $C = \{A \subseteq X$ such that $A = [x]$ for some $x \in X\}$.
Now, define $f : X \rightarrow C$ by the rule $f(x) = [x], \forall x \in X$.
Prove that if $x \in X$, then there is one and only one equivalence class which contains $x$.
Suppose $X = \{1, 2, 3, 4, 5\}$ and that $R$ is an equivalence relation for which $1\,R\,3$, $2\,R\,4$ but not $1\,R\,2$, $1\,R\,5$, and $2\,R\,5$.
Write down the equivalence classes of $R$ and draw a diagram to represent the function $f$.
C = { [x] : x in X }.
If x in [a], [b], then x equivalent a and x equivalent b.
Thus a equivalent b, whereupon [a] = [b].