Inverse of Inverse of Bijection

Theorem

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

Then:

$\left({f^{-1}}\right)^{-1} = f$

Proof 1

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

From Bijection Composite with Inverse we have:

• $f^{-1} \circ f = I_S$, where $I_S$ is the identity mapping on $S$
• $f \circ f^{-1} = I_T$, where $I_T$ is the identity mapping on $T$

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

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 3.3$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 13$