Prove inverse of odd function is odd

Question

I need to prove this statement:

The bijective function $f: X\to Y$ is odd. Prove $f^{-1}$ is odd.

And i want to know if my proof is fine.

My attempt was:

Let $x \in X, y \in Y$.

We have $f$ as $f(x) = y$, and $f^{-1}(y) = x$ as the inverse mapping.

Now, the key steps:

$f^{-1}(-y) = f^{-1}(-y)$

$f^{-1}(-y) = f^{-1} \circ -f(x)$, but since $f$ is odd $-f(x) = f(-x)$

Getting $f^{-1}(-y)= f^{-1} \circ f(-x)$ and since the composition between a function and its inverse is always the identity, we have

$f^{-1}(-y) = -x \iff f^{-1}(y) = -f^{-1}(y)$. Hence $f^{-1}$ is odd.

Thanks for reading.

Answer 1

For all $ x \in X$,

$$f^{-1}(f(x))=x \implies$$

$$f^{-1}(f(-x))=-x \implies$$

$$f^{-1}(-f(x))=-x \implies$$

$$f^{-1}(-f(x))=-f^{-1}(f(x))$$

So, for each $ y\in Y$ we have $y=f(x)$.

Your proof is correct.