Inverse Mapping is Bijection

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Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ and $g: T \to S$ be inverse mappings of each other.


Then $f$ and $g$ are bijections.


Proof

\(\displaystyle f \paren x\) \(=\) \(\displaystyle f \paren y\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \paren {f \paren x}\) \(=\) \(\displaystyle g \paren {f \paren y}\) $\quad$ as $g$ is a mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle y\) $\quad$ Definition 1 of Inverse Mapping $\quad$



Thus $f$ is by definition an injection.


\(\displaystyle t\) \(\in\) \(\displaystyle T\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \paren t\) \(\in\) \(\displaystyle S\) $\quad$ as $g$ is a mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle f \paren {g \paren t}\) \(=\) \(\displaystyle t\) $\quad$ Definition 1 of Inverse Mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\, \displaystyle \exists s \in S: \, \) \(\displaystyle f \paren s\) \(=\) \(\displaystyle t\) $\quad$ setting $s = g \paren t$ $\quad$

Thus $f$ is by definition a surjection.


Similarly:

\(\displaystyle g \paren x\) \(=\) \(\displaystyle g \paren y\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle f \paren {g \paren x}\) \(=\) \(\displaystyle f \paren {g \paren y}\) $\quad$ as $f$ is a mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle y\) $\quad$ Definition 1 of Inverse Mapping $\quad$

Thus $g$ is by definition an injection.


\(\displaystyle s\) \(\in\) \(\displaystyle S\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle f \paren s\) \(\in\) \(\displaystyle T\) $\quad$ as $f$ is a mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \paren {f \paren s}\) \(=\) \(\displaystyle s\) $\quad$ Definition 1 of Inverse Mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\, \displaystyle \exists t \in T: \, \) \(\displaystyle g \paren t\) \(=\) \(\displaystyle s\) $\quad$ setting $t = f \paren s$ $\quad$

Thus $g$ is by definition a surjection.


So $f$ and $g$ are both injections and surjections.


The result follows by definition of bijection.

$\blacksquare$


Also see


Sources