Proving a function is a one to one correspondence

Question

I understand that to show a function is a one to one correspondence, you have to show that the function is both one to one and onto. Proving a function is one to one seems simple enough. However for many of the examples I have seen, when they prove a function is onto, they have a really simple linear equation like $f(x)=3x+6$.

So for a more complicated function like $f(x)=\frac{x}{1-|x|}$, I'm having difficulty proving that it has one to one correspondence between $(-1,1)$ and $(\infty, -\infty)$. It's a pretty obvious conclusion looking at a graph, but how would I prove it more rigorously? More specifically, how do I show that it is onto?

Answer 1

To show that $f:(-1,1)\to\Bbb R$ is onto means to show that

for every $y\in\Bbb R$, the equation $y=f(x)$ has a solution $x\in(-1,1)$.

In this case we have $$y=\frac{x}{1-|x|}\ .$$ For $x\in(-1,1)$ the denominator is always positive, so $y\ge0$ if and only if $x\ge0$. Therefore $$\eqalign{y\ge0\quad &\Leftrightarrow\quad x\ge0\cr &\Leftrightarrow\quad y=\frac{x}{1-x}\cr &\Leftrightarrow\quad x=\frac{y}{y+1}\ .\cr}$$ The value of $x$ we have just found is in $(-1,1)$ because $y\ge0$ and $y<y+1$. Therefore $y=f(x)$ has a solution if $y\ge0$; similarly, it has a solution if $y<0$. So $f$ is onto.

Answer 2

$f(x)=\frac{1}{1-|x|}$ is not injective on $(-1,1)$ -- for example, $f(\frac12)=f(-\frac12)=2$.

Answer 3

To show onto you need to take an arbitrary member of the range and find a member of the domain that maps to it.

For $f(x) = \frac{x}{1-|x|}$ with the member of the range being called $y$ you get $y = \frac{x}{1-|x|} \to y(1-|x|) = x \to y-y|x|=x \to y = x + y|x|$ From this you have two cases

$x \geq 0 \to y = x + yx \to \frac{y}{1+y} = x$ and since we know $x$ is positive we have $y \in (-\infty, -1) \cup [0,\infty)$

$x < 0 \to y = x - yx \to \frac{y}{1-y} = x$ and since we know $x$ is negative we have $y \in (-\infty, 0) \cup (1, \infty)$

The union of the possible values for $y$ is $(-\infty, \infty)$