Composite of Quotient Mappings

Theorem
Let $\mathcal R_1$ be an equivalence on $S$, and $\mathcal R_2$ be an equivalence on the quotient set $S / \mathcal R_1$.

We can find an equivalence $\mathcal R_3$ on $S$ such that $\left({S / \mathcal R_1}\right) / \mathcal R_2$ is in one-to-one correspondence with $S / \mathcal R_3$ under the mapping:


 * $\phi: \left({S / \mathcal R_1}\right) / \mathcal R_2 \to S / \mathcal R_3: \left[\!\left[{\left[\!\left[{x}\right]\!\right]_{\mathcal R_1}}\right]\!\right]_{\mathcal R_2} \mapsto \left[\!\left[{x}\right]\!\right]_{\mathcal R_3}$.

Proof
Define $\mathcal R_3$ to be the equivalence induced by:


 * $x \mapsto \left[\!\left[{\left[\!\left[{x}\right]\!\right]_{\mathcal R_1}}\right]\!\right]_{\mathcal R_2}$

By definition of $\mathcal R_3$:


 * $\left[\!\left[{\left[\!\left[{x}\right]\!\right]_{\mathcal R_1}}\right]\!\right]_{\mathcal R_2} = \left[\!\left[{\left[\!\left[{y}\right]\!\right]_{\mathcal R_1}}\right]\!\right]_{\mathcal R_2} \implies \left[\!\left[{x}\right]\!\right]_{\mathcal R_3} = \left[\!\left[{y}\right]\!\right]_{\mathcal R_3}$

Therefore, $\phi$ is well-defined.

Again by definition of $\mathcal R_3$:


 * $\left[\!\left[{x}\right]\!\right]_{\mathcal R_3} = \left[\!\left[{y}\right]\!\right]_{\mathcal R_3} \implies \left[\!\left[{\left[\!\left[{x}\right]\!\right]_{\mathcal R_1}}\right]\!\right]_{\mathcal R_2} = \left[\!\left[{\left[\!\left[{y}\right]\!\right]_{\mathcal R_1}}\right]\!\right]_{\mathcal R_2}$

which means $\phi$ is an injection.

Lastly, note that every element of $S / \mathcal R_3$ is of the form $\left[\!\left[{x}\right]\!\right]_{\mathcal R_3}$ for some $x \in S$.

It is now immediate from the definition of $\phi$ that it is surjective.

Hence $\phi$ is a bijection, as desired.