Bijection has Inverse Mapping

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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 $f$.


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 $s \in S$.


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


\(\displaystyle f \left({s'}\right)\) \(=\) \(\displaystyle f \left({g \left({f \left({s}\right)}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f \left({s}\right)\) 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.


Also see