Can you give me a hint for how the inverse function of $f\colon (-1,1)\to \mathbb{R}\colon f(x)=\frac{x}{1-x^2}$ looks?
I need to show a homeomorphism!
$\endgroup$ 22 Answers
$\begingroup$You need to show that $f$ is continuous, injective, surjective, and has a continuous inverse. The first two are rather simple and I think you can show that. Showing that $f$ is surjective amounts to using the intermediate value theorem (hint, $f$ is monotonic increasing on $(-1,1)$ and unbounded). To show that $f$ has a continuous inverse, you just need to show that $f$ is an open map - that is $f$ maps open sets to open sets. You should first show that $f$ maps open intervals to open intervals - use the fact, again, that $f$ is monotonic increasing here.
$\endgroup$ 2 $\begingroup$$\begin{align} f(x)&=\frac{x}{1-x^2} \\ f(x)(1-x^2)&=x \\ f(x)-f(x)x^2&=x \\ 0&=f(x)x^2+x-f(x) \\ x&=\frac{-1\pm\sqrt{1+4f(x)^2}}{2f(x)} \end{align}$
Note, as mentioned in the comments below, that this does not give an inverse function, as a function cannot map two values to one point. Rather, you need to choose the correct inverse and demonstrate that it is indeed an inverse over the interval you are considering.
$\endgroup$ 5
