Prove that $g(x) = {x \over (1-|x|) } $ is a bijection.

Question

I'm struggling with the following homework assignment.

Let $X = \{x \in \mathbb{R}: -1 < x < 1\}$. We define $g: X \rightarrow \mathbb{R}$ by $g(x) = \frac{x}{1-|x|}$.

We must prove that $g$ is a bijection.

First, we prove $g$ is injective.

By definition of injection, $g$ is injective iff whenever $x_1,x_2 \in X$ and $g(x_1) = g(x_2)$, then $x_1 = x_2$.

Let $x_1,x_2 \in X$ and $g(x_1) = g(x_2)$. Want to show $x_1 = x_2$.

We have $\frac{x_1}{1-|x_1|} = \frac{x_2}{1-|x_2|} \Leftrightarrow x_1(1-|x_2|) = x_2(1-|x_1|)$

$\Leftrightarrow x_1-x_1|x_2|=x_2-|x_1|x_2$.

For cases $x_1,x_2 > 0$ and $x_1,x_2 < 0$ we get $x_1 = x_2$

But for the remaining two cases, how should I proceed?

Thanks in advance!

Answer 1

$\frac x{1-|x|}$ always has the same sign as $x$ when $x\in X$ (since the denominator is positive), so if $x_1$ and $x_2$ have different signs, then $g(x_1)$ and $g(x_2)$ also have different signs -- in particular they can't be equal.

Answer 2

you can do it like this you have

$$ x_1(1-|x_2|) = x_2(1-|x_1|)$$

there are two cases

case 1) $x_1=x_2$ and $1-|x_2|=1-|x_1|$ gives $x_1=x_2$

case 2) $x_1=1-|x_1|$ and $1-|x_2|=x_2$ now if $x_1 <0$ and $x_2<0$

it gives $x_1=1+x_1$ and $1+x_2=x_2$ $\to$ 1=0

which is not true

Answer 3

One can compute an explicit inverse:

Let $y = {x \over 1-|x|}$.

Note that $\operatorname{sgn} y = \operatorname{sgn} y$.

If $y = 0$ then $x=0$.

If $y>0$, then $(1-x)y = x$ or $x(1+y) = y$ which gives $x = {y \over 1+y}$.

If $y<0$, then $(1+x)y = x$ or $x(1-y) = y$ which gives $x = {y \over 1-y}$.

Hence we see that $y=x = {y \over 1+|y|}$.