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

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## 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.

- the inverse $f^{-1}$ of $f$ is such that:
- for each $y \in T$, the preimage $\map {f^{-1} } y$ 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 mapping such that:

- $(1): \quad f$ is an injection
- $(2): \quad f$ is a surjection.

Let $t \in T$.

Then as $f$ is a surjection:

- $\exists s \in S: t = \map f s$

As $f$ is an injection, there is only one $s \in S$ such that $t = \map f s$.

Define $\map g t = s$.

As $t \in T$ is arbitrary, it follows that:

- $\forall t \in T: \exists s \in S: \map g t = s$

such that $s$ is unique for a given $t$.

That is, $g: T \to S$ is a mapping.

By the definition of $g$:

- $(1): \quad \forall t \in T: \map f {\map g t} = t$

Let $s \in S$.

Let:

- $(2): \quad s' = \map g {\map f s}$

Then:

\(\displaystyle \map f {s'}\) | \(=\) | \(\displaystyle \map f {\map g {\map f s} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map f s\) | from $(1)$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s\) | \(=\) | \(\displaystyle s'\) | as $f$ is an injection | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map g {\map f s}\) | from $(2)$ |

Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other.

$\blacksquare$

$\Box$

### 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.

$\blacksquare$