Consider a generic equivalence relation $R$ on a set $S$. By definition, if we partition $S$ using the relation $R$ into $\pi_S$, whose members are the congruence classes $c_1, c_2…$ then
$aRb \text{ iff a and b are members of the same congruence class in } \pi_S$.
But what is the domain and codomain of $R$? Is it $S \rightarrow S $ or $S^2 \rightarrow \text{{true,false}}$?
The reason I ask is to have an idea of the members of $graph(R)$. Does it only contain ordered pairs which are equivalent, i.e. $(a,b)$; or all elements of $S \times S$ followed by whether they are equivalent, i.e. $((a,b), true)$?
Moreover, what would the image of $s \in S$ under $R$ look like? If one suggests that $R(s)$ would return a set of all the equivalent members, then the former definition is fitting.
I suppose the source of confusion is that we rarely think of equivalence relations as 'mapping' from an input to an output, instead it tells us if two objects are similar in some way.
Best Answer
One way to define a relation $R$ on a set $S$ is as a subset of $S \times S$. Using this definition, we can create a graph $G = (S, R)$ where $S$ is the vertex set and $R$ is the edge set. In other words, if $a, b \in S$ are vertices of the graph $G$, then $(a, b)$ is an edge of $G$ if and only if $(a, b) \in R$ (i.e. $aRb$).
Now for an equivalence relation, two vertices are equivalent if and only if they are in the same component. Equivalently, two vertices $a, b$ are equivalent if and only if there is a path from $a$ to $b$.