Equivalence of Definitions of Bijection

Necessary Condition
Suppose $f^{-1}: T \to S$ is a mapping.

Then by definition:
 * $\forall y \in T: \exists x \in S: \tuple {y, x} \in f^{-1}$

Thus for all $y \in T$ there exists at least one $x \in S$ such that $\tuple {y, x} \in f^{-1}$.

Also by definition of mapping:


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

Thus for all $y \in T$ there exists at most one $x \in S$ such that $\tuple {y, x} \in f^{-1}$.

Thus for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.

Sufficient Condition
Suppose that for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.

Then by definition $f^{-1}$ is a mapping.

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

Then as $f$ is a mapping:
 * for all $x \in S$ there exists a unique $y \in T$ such that $\tuple {x, y} \in f$.

Similarly, as $f^{-1}$ is also a mapping:
 * for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.

Sufficient Condition
Let $f \subseteq S \times T$ be a relation such that:
 * $(1): \quad$ for each $x \in S$ there exists one and only one $y \in T$ such that $\tuple {x, y} \in f$
 * $(2): \quad$ for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.

Then by definition:
 * from $(1)$, $f$ is a mapping.
 * from $(2)$, $f^{-1}$ is a mapping.