functions - Proving a bijection with absolute value

let

$g(x) = \frac{x}{1-|x|}$ for all $x \in (-1,1)$

a) show that $g$ is a bijection from $(-1,1)$ to $\mathbb{R}$

Here is what I have done.

Say $g(x)=y$ so $$y = \frac{x}{1-|x|}$$
iff $$y(1-|x|)=x$$

Therefore, for each y, there exists an x such that $f(x)=y). So$f$ is a surjection.

Suppose $g(x)= g(z)$ for some $x,z \in (-1,1)$, then
$$\frac{x}{1-|x|}=\frac{z}{1-|z|}$$
iff
$$x-z=x|z|-z|x|$$

Then there are four cases, one $x,z>0$ and $x,z<0$ where obviously $x=z$, but for the other two cases, I am only concerned with figuring out one, because obviously the other one follows, but I have gotten to the point where
$$x-z=x(2z)$$ for $x>0$, $y<0$. Am I correct thus far? Where do I go next? Thank you for any help.

Answer

For the injection part, is $\dfrac{x}{1-|x|}=\dfrac{z}{1-|z|}$ really possible if $x$ and $z$ have different signs ? One of the two terms is $\geq0$ and the other $\leq 0$.

For the surjection part, I would advise you to separate the cases where $y$ is positive, and $y$ is negative.

If $y$ is positive then $x$ must be positive, so $|x| = x$. You have $y = \dfrac{x}{1-x}$, so $y(1-x)-x = 0$ thus $-x(1+y)+y = 0$ and $x = \dfrac{y}{1+y}$. You have know found a $x$ that works. Note that the final step is legit because as $y$ is positive, $(1+y)$ can't be $0$.

Don't forget the case $y < 0$ now !