Image of Element under Inverse Mapping/Corollary 1

Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping.

Then:

$\forall x \in S: \map {f^{-1} } {\map f x} = x$

$\ds \forall x \in S, y \in T: \, $

$\ds \map f x$

$=$

$\ds y$

$\, \ds \iff \, $

$\ds \map {f^{-1} } y$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \forall x \in S: \, $

$\ds \map {f^{-1} } {\map f x}$

$=$

$\ds x$

substituting $\map f x$ for $y$

$\blacksquare$