Example 2

\* The equivalence relation "$x \in \mathbf{Z}$ and $y \in \mathbf{Z}$ and $x - y$ is divisible by 4" is finer than the equivalence relation "$x \in \mathbf{Z}$ and $y \in \mathbf{Z}$ and $x - y$ is divisible by 2". \*

Let R and S be two equivalence relations on the same set E, such that S is finer than R. Let $f$ and $g$ be the canonical mappings of E onto E/R and E onto E/S respectively; then the function $f$ is compatible with S. Let $h$ be the function induced by $f$ on passing to the quotient with respect to S; then $h$ is a mapping of E/S onto E/R. The equivalence relation associated with $h$ on E/S is called the quotient of R by S and is denoted by R/S. The relation $x \equiv y \pmod{R}$ is equivalent to $g(x) \equiv g(y) \pmod{R/S}$, and the equivalence classes with respect to R/S are the images under $g$ of the equivalence classes with respect to R. Let $h = j \circ h_2 \circ h_1$ be the canonical decomposition (no. 5) of the mapping $h$. Then $h_1$ is the canonical mapping of E/S onto (E/S)/(R/S), $j$ is the identity mapping of E/R, and $h_2$ is a one-to-one mapping of (E/S)/(R/S) onto E/R. The mapping $h_2$ and its inverse are said to be canonical.

¶ Conversely, consider an arbitrary equivalence relation T on the set E/S, and let R be the equivalence relation on E which is the inverse image under $g$ of the relation T (no. 6). Since the relation $x \equiv y \pmod{R}$ is equivalent to $g(x) \equiv g(y) \pmod{T}$, it follows that T is equivalent to R/S.