Mapping is Injection and Surjection iff Inverse is Mapping/Proof 2

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

Let $f: S \to T$ be a mapping.

Then:
 * $f: S \to T$ can be defined as a bijection in the sense that:
 * $(1): \quad f$ is an injection
 * $(2): \quad f$ is a surjection.

iff:


 * the inverse $f^{-1}$ of $f$ is such that:
 * for each $y \in T$, the preimage $f^{-1} \left({\left\{{y}\right\}}\right)$ has exactly one element.


 * That is, such that $f^{-1} \subseteq T \times S$ is itself a mapping.

Proof
Let $f: S \to T$ be a mapping.

Necessary Condition
Let $f: S \to T$ be a bijection in the sense that:
 * $(1): \quad f$ is an injection
 * $(2): \quad f$ is a surjection.

By Bijection has Inverse Mapping‎, $f^{-1}$ is a mapping.

Sufficient Condition
Let $f^{-1}: T \to S$ be a mapping.

By Inverse Mapping is Bijection, both $f$ and $f^{-1}$ are bijections.

Hence, in particular, $f$ is a bijection.