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