Given function $f$, where $A⊆ \mathbb{R} $ is a symmetric domain with respect to 0,$ \;\; f:A \rightarrow\mathbb{R}$ and $f$ is an odd one-to-one function, I need to prove $f^{-1}$ is odd.
I was originally thinking of using the integral and proving that the integral over a domain symmetric to 0 would be 0, but now I'm thinking that doesn't imply that the function is odd.
How else could I go about proving $\;f^{-1}$ odd?
Let $y \in f(A)$. Then $y=f(x)$ for some $x \in A$. Since $f$ is odd, $f(-x) = -f(x) = -y$. Therefore $$f^{-1}(-y) = -x = -f^{-1}(y)$$
I would suggest looking at the function not as a formula but as a set of ordered pairs $(x,f(x))$. Then oddness would be a certain symmetry property for those ordered pairs. Now create $f^{-1}$ from those ordered pairs, and see whether it has the same property.
Let $f(x)$ be an odd, one-to-one continuous function, defined on some interval I.
If $f(x)$ is an odd function and if $f(x)=k|x\in I$, then:
$f(-x)=-f(x)=-k$
Now since:
$f(x)=k$ and $f(-x)=-k$
then:
$$f^{-1}(k)=x \tag{1}\label{eq:1} $$ and $$f^{-1}(-k)=-x \tag{2}\label{eq:2}$$
If we divide the equation \eqref{eq:1} by $-1$, we get:
$-f^{-1}(k)=-x$, and from equation \eqref{eq:2} $-x=f^{-1}(-k)$
Therefore:
$-f^{-1}(k)=-x=f^{-1}(-k)$
hence:
$f^{-1}(-k)=-f^{-1}(k)$
Take $x$ in the image of $f$, then $x=f(a)$ for some $a\in A$, since $f$ is odd $-x=f(-a)$, so $f^{-1}(-a)=-x$.
We just proved that if $f^{-1}(a)=x$ then $f^{-1}(-a)=(-x)$ in other words $f^{-1}(-a)=-f^{-1}(a) $
