Parity of Inverse of Permutation

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Then:
 * $\forall \pi \in S_n: \map \sgn \pi = \map \sgn {\pi^{-1} }$

Proof
From Parity Function is Homomorphism:
 * $\map \sgn {I_{S_n} } = 1$

Thus:
 * $\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$

The result follows immediately.