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:


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

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 $$5.2$$
 * : $$\S 22.3$$