Composite of Mapping with Inverse

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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.

$\blacksquare$


Also see


Sources