# 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 $f^{-1} \left({y}\right)$ has exactly one element.

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

## Proof

### 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.

$\Box$

### 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.

$\blacksquare$

## Sources

{SourceReview|username = Prime.mover|complicated}}

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 1.9$: Inverse Functions, Extensions, and Restrictions