Inverse of Inverse of Bijection/Proof 1

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

From Bijection Composite with Inverse 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.