Inverse of Inverse of Bijection

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Theorem

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


Then:

$\paren {f^{-1} }^{-1} = f$

where $f^{-1}$ is the inverse of $f$.


Proof 1

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

From Composite of Bijection with Inverse is Identity Mapping we have:

$f^{-1} \circ f = I_S$
$f \circ f^{-1} = I_T$

where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.

The result follows from Left and Right Inverses of Mapping are Inverse Mapping.

$\blacksquare$


Proof 2

A mapping is a relation.

Thus it follows that Inverse of Inverse Relation can be applied directly.

$\blacksquare$


Sources

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