# Show $f(x)=\frac{x}{1-|x|}$ is a homeomorphism from (-1,1) to $\mathbb{R}$ [duplicate]

I have to Show $$f:(-1,1) \to \mathbb{R} ;\quad f(x)=\frac{x}{1-|x|}$$ is a homeomorphism from (-1,1) to $$\mathbb{R}$$

I am considering a function to be a homeomorphism if it is continuous, has an inverse and this inverse is also continuous.

So firstly I argued $$f$$ is continuous at $$(-1,1)$$ since both $$(x) \quad \& \quad \ (1-|x|)$$ are continuous there. Also, I should add that $$(1-|x|)$$ is never 0 at $$(-1,1)$$.

So here comes my first problem: I wanted to show that $$f$$ is an injection because than I could use a theorem which states:

if $$f$$ is a continuous injection from any interval to $$\mathbb{R}$$, then its inverse is also continuous.

The thing is, I can't manage to prove $$f$$ is an injection. I've tried to show that $$\frac{x}{1-|x|}=\frac{y}{1-|y|} \Rightarrow x=y$$ but I couldn't isolate things properly, I guess.

Any help would be appreciated. Also, this was the only way I could think to solve the problem. If you have any other method, feel free to share :)

• Another one: math.stackexchange.com/q/1138344/42969 – both found with Approach0 Oct 19, 2019 at 17:29
• Multiplying your equation by $(1-|x|)(1- |y|)$ results in $x(1-|y|) = y(1-|x|)$. Oct 19, 2019 at 17:31
• you can split the function to two intervals according to its behaviour, then differentiate to see its strictly increasing on each interval and hence a bijection.
– omer
Oct 19, 2019 at 18:23

You want to show that the function $$f : (-1,1)\to\mathbb R$$ is bijective.

First we prove injectivity, as you suggested. For this, let $$f(x) = f(y)$$, that is, $$\frac x{1-|x|}=\frac y{1-|y|}$$. This can only be true if $$x$$ and $$y$$ have the same signs (or one of them is zero, but then the other one also must be zero). Multiplying by $$(1-|x|)(1-|y|)$$ gives $$x(1-|y|) = y(1-|x|)$$, i.e., $$x-x|y| = y-y|x|$$. But as $$x$$ and $$y$$ have the same sign, we have $$x|y| = y|x|$$. So, it follows that $$x=y$$.

Now, surjectivity. For this, let $$z\in\mathbb R$$ be arbitrary. We have to find some $$x$$ such that $$f(x) = z$$, i.e., $$\frac x{1-|x|} = z$$. This is equivalent to $$x+z|x| = z$$. If $$z\ge 0$$, then we try $$x\ge 0$$ for which the equation is $$(1+z)x = z$$ and so $$x = \frac z{1+z}$$, which is in $$[0,1)$$ and therefore indeed a solution. Let $$z<0$$. Here, we try $$x<0$$, for which the equation is $$(1-z)x = z$$ with solution $$x = \frac z{1-z}$$. This is indeed $$<0$$ and $$>-1$$ (i.e., $$x\in (-1,0)$$). Hence, for each $$z\in\mathbb R$$ we have found a solution of $$f(x) = z$$.

So, $$f:(-1,1)\to\mathbb R$$ is bijective and thus has an inverse function. As you said already, this inverse is then continuous and we're done.

$$f(00, f'(-10.$$ The function is increasing on $$x \in (1,1)$$ So the function $$f(x):(-1,1) \rightarrow R$$ is bijection and $$f^{-1}(x)$$ exists and it is given as $$f^{-1}(x<0)=\frac{x}{1-x},~ f^{-1}(x>0)=\frac{x}{1+x},~f^{-1}(0)=0.$$ See the fig. belpw $$f(x)$$ is blu and $$f^{-1}(x)$$ is the red one.

Hint:

You can check that the map \;\left\{\begin{aligned}g:\mathbf R&\longrightarrow (-1,1),\\ x&\longmapsto \dfrac x{1+|x|}\end{aligned}\right. is continuous and satisfies $$f\circ g=\operatorname{id}_{\mathbf R},\qquad g\circ f=\operatorname{id}_{(-1,1)}.$$