Equivalence of Definitions of Bijection

Theorem
The following definitions of bijection are equivalent:

Definition 1 iff Definition 2
From Injection iff Left Inverse, $f$ is an injection iff $f$ has a left inverse mapping.

From Surjection iff Right Inverse, $f$ is a surjection iff $f$ has a right inverse mapping.

Putting these together, it follows that $f$ is both an injection and a surjection iff $f$ has both a left inverse and a right inverse.

Definition 1 iff Definition 3
This is demonstrated in Bijection iff Inverse is Mapping.

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

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

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

Also by definition of mapping:


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

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

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

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

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 $\left({x, y}\right) \in f$.

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

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 $\left({x, y}\right) \in f$
 * $(2): \quad$ for each $y \in T$ there exists one and only one $x \in S$ such that $\left({x, y}\right) \in f$.

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