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 $$\left[\left[{x}\right]\right]_{\mathcal{R}_f} = \left\{{s \in \mathrm{Dom} \left({f}\right): f \left({s}\right) = f \left({x}\right)}\right\}$$

By definition: $$f^{-1} \left({y}\right) = \left[\left[{x}\right]\right]_{\mathcal{R}_f}$$ hence the result.