Composite of Mapping with Inverse

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

Then:
 * $\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\mathcal R_f}$

where:
 * $\mathcal R_f$ is the equivalence induced by $f$
 * $\eqclass x {\mathcal R_f}$ is the $\mathcal R_f$-equivalence class of $x$.

Proof
Let $y = \map f x$.

Then by the definition of induced equivalence:
 * $x \in \eqclass x {\mathcal R_f}$

By the definition of the inverse of a mapping:
 * $f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$

Thus:
 * $\eqclass x {\mathcal R_f} = \set {s \in \Dom f: \map f s = \map f x}$

By definition:
 * $\map {f^{-1} } y = \eqclass x {\mathcal R_f}$

Hence the result.

Also see

 * Composite of Inverse of Mapping with Mapping