Equivalence of Definitions of Inverse Mapping

Theorem
Let $S$ and $T$ be sets.

Definition 1 implies Definition 2
Let $f^{-1}: T \to S$ be an inverse mapping of $f: S \to T$ by definition 1.

From Bijection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.

By Bijection Composite with Inverse:
 * $f^{-1} \circ f = I_S$
 * $f \circ f^{-1} = I_T$

Thus $f^{-1}: T \to S$ is an inverse mapping of $f: S \to T$ by definition 2.

Definition 2 implies Definition 1
Let $g: T \to S$ be an inverse mapping of $f: S \to T$ by definition 2.

From Left and Right Inverse Mappings Implies Bijection it follows that both $f$ and $g$ are bijections.

From Bijection iff Inverse is Mapping it follows that $g = f^{-1}$.

Thus $g: T \to S$ is an inverse mapping of $f: S \to T$ by definition 1.