# Bijection has Inverse Mapping

 It has been suggested that this page or section be merged into Mapping is Injection and Surjection iff Inverse is Mapping/Necessary Condition/Proof 2. (Discuss)

## Theorem

Let $S$ and $T$ be sets.

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

Then there exists a mapping $g: T \to S$ which is the inverse mapping of $f$.

## Proof

Let $t \in T$.

Then as $f$ is a surjection:

$\exists s \in S: t = f \left({s}\right)$

As $f$ is an injection, there is only one $s \in S$ such that $t = f \left({s}\right)$.

Define $g \left({t}\right) = s$.

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

$\forall t \in T: \exists s \in S: g \left({t}\right) = s$

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

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

By the definition of $g$:

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

Let $s \in S$.

Let:

$(2): \quad s' = g \left({f \left({s}\right)}\right)$.

Then:

 $\displaystyle f \left({s'}\right)$ $=$ $\displaystyle f \left({g \left({f \left({s}\right)}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle f \left({s}\right)$ from $(1)$ $\displaystyle \implies \ \$ $\displaystyle s$ $=$ $\displaystyle s'$ as $f$ is an injection $\displaystyle$ $=$ $\displaystyle g \left({f \left({s}\right)}\right)$ from $(2)$

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

$\blacksquare$