Composite of Mapping with Inverse

Theorem
Let $f: S \to T$ be a mapping.

Then:
 * $\forall x \in S: f^{-1} \circ f \left({x}\right) = \left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$

where $\mathcal R_f$ is the equivalence induced by $f$, and $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$ is the $\mathcal R_f$-equivalence class of $x$.

Proof
Let $y = f \left({x}\right)$.

Then by the definition of Induced Equivalence:
 * $x \in \left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$

By the definition of the inverse of a mapping:
 * $f^{-1} = \left\{{\left({y, x}\right): \left({x, y}\right) \in f}\right\}$

Thus:
 * $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f} = \left\{{s \in \operatorname{Dom} \left({f}\right): f \left({s}\right) = f \left({x}\right)}\right\}$

By definition:
 * $f^{-1} \left({y}\right) = \left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$

Hence the result.