Function is Odd Iff Inverse is Odd

Theorem

Let $f$ be an odd real function with an inverse $f^{-1}$.

Then $f^{-1}$ is also odd.

Proof

First note that we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle f \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle f^{-1} \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Inverse Mapping Image
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle -x\)	\(=\)	\(\displaystyle -f^{-1} \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Multiply both sides by $-1$
Then:
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle -y\)	\(=\)	\(\displaystyle f \left({-x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of odd function
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle f^{-1} \left({-y}\right)\)	\(=\)	\(\displaystyle f^{-1} \circ f \left({-x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because $f^{-1}$ is a mapping
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle -x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Bijection Composite with Inverse
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle -f^{-1} \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above

The result follows by definition of an odd function.

$\blacksquare$