Image of Element under Inverse Mapping

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

Then:
 * $\forall x \in S, y \in T: f \left({x}\right) = y \iff f^{-1} \left({y}\right) = x$

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

From the definition of inverse mapping:
 * $f^{-1} = \left\{{\left({y, x}\right): \left({x, y}\right) \in f}\right\}$

Let $y = f \left({x}\right)$.

From the definition of the preimage of an element:
 * $f^{-1} \left({y}\right) = \left\{{s \in S: \left({y, x}\right) \in f}\right\}$

Thus:
 * $x \in f^{-1} \left({y}\right)$

However, $f^{-1}$ is a mapping. Therefore, from the definition of a mapping:


 * $\forall y \in T: \left({y, x_1}\right) \in f^{-1} \land \left({y, x_2}\right) \in f^{-1} \implies x_1 = x_2$

Thus:
 * $\forall s \in f^{-1} \left({y}\right): s = x$

Thus:
 * $f^{-1} \left({y}\right) = \left\{{x}\right\}$

That is:
 * $x = f^{-1} \left({y}\right)$

Source

 * : $\S 5$: Theorem $5.2$
 * : $\S 22.3$