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 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.