Function is Odd Iff Inverse is Odd
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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:
| \(\ds y\) | \(=\) | \(\ds \map f x\) | |||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \map {f^{-1} } y\) | Image of Element under Inverse Mapping | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -x\) | \(=\) | \(\ds -\map {f^{-1} } y\) | Multiply both sides by $-1$ | |||||||||||
| Then: | |||||||||||||||
| \(\ds -y\) | \(=\) | \(\ds \map f {-x}\) | Definition of Odd Function | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {f^{-1} } {-y}\) | \(=\) | \(\ds \map {f^{-1} \circ f} {-x}\) | because $f^{-1}$ is a mapping | |||||||||||
| \(\ds \) | \(=\) | \(\ds -x\) | Composite of Bijection with Inverse is Identity Mapping | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\map {f^{-1} } y\) | from above | ||||||||||||
The result follows by definition of an odd function.
$\blacksquare$