# Image of Element under Inverse Mapping

## 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:

$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$

## Proof

### Sufficient Condition

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

From the definition of inverse mapping:

$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$

Let $y = f \paren x$.

From the definition of the preimage of an element:

$\map {f^{-1} } y = \set {s \in S: \tuple {y, x} \in f}$

Thus:

$x \in \map {f^{-1} } y$

However, $f^{-1}$ is a mapping.

Therefore, by definition:

$\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$

Thus:

$\forall s \in f^{-1} \paren y: s = x$

Thus:

$\map {f^{-1} } y = \set x$

That is:

$x = \map {f^{-1} } y$

$\Box$

### Necessary Condition

Let $\map {f^{-1} } y = x$.

By definition of inverse mapping:

$\map f x = y$

$\blacksquare$