Image of Element under Inverse Mapping

Theorem
Let $$f$$ be a mapping such that its inverse $$f^{-1}$$ is also a mapping.

Then $$f \left({x}\right) = y \iff f^{-1} \left({y}\right) = x$$.

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:

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

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

Thus $$f^{-1} \left({y}\right) = \left\{{x}\right\}$$, or $$x = f^{-1} \left({y}\right)$$.