# need help showing a particular function is a bijection ((URGENT))

• Nov 19th 2009, 04:47 PM
rgriss1
need help showing a particular function is a bijection ((URGENT))
I need help with the following problem:

Show the function

f(x) = x / (1 + |x|) is a bijection
f: R -> (-1, 1)

I know that f(b) = f(a) => b = a... and if you can show this then the function is a bijection. I know I'm forgetting something obvious, but the absolute value in the function is throwing me for a loop.

• Nov 19th 2009, 04:56 PM
Jose27
Let $g:(-1,1)\rightarrow \mathbb{R}$ such that $g(y)=\frac{y}{1-\vert y \vert }$. Prove $g\circ f=id_{\mathbb{R} }$ and $f\circ g= id_{(-1,1)}$
• Nov 19th 2009, 11:04 PM
Drexel28
Quote:

Originally Posted by rgriss1
I know that f(b) = f(a) => b = a... and if you can show this then the function is a bijection.

Be careful friend. This shows that $f$ is injective. To show that $f$ is bijective it must be, as you said, injective but surjective as well. In other words, we need the codomain of the mapping to be the same as the image of $f$ under the domain. In more understandable terms we need there to be no "excess" stuff in what is mapped to. So for example it is pretty clear that if we append $\pi$ to this mapping (i.e. rewrite it as $f:\mathbb{R}\mapsto(-1,1)\cup\{\pi\}$) that this mapping has "excess" stuff in the codomain. Nothing in $(-1,1)$ can "reach" (is mapped to) $\pi$. So, in closing, to show that $f:X\mapsto Y$ is bijective you need to show

1. That $f:X\mapsto Y$ is injective-via how you said $f(x)=f(y)\implies x=y$

2. That $f:X\mapsto Y$ is surjective- there is no excess stuff in the codomain. To do this you need to show that for any element of the codomain something in the domain is mapped to it. In other words, given any $y\in Y$ you must find a $x\in X$ such that $f(x)=y$

In your particular example, injectivity is a tad tricky. To do it, let us assume that

injectivity: $f(x)=\frac{x}{1+|x|}=\frac{y}{1+|y|}=f(y)$. Note if either $x,y=0$ then the result is trivial. So let us assume that $x,y\ne0$. Then $\frac{x}{y}=\frac{1+|y|}{1+|x|}$. Notice though that the RHS is positive always, so that means that the LHS is as well. This implies that $x,y$ have the same sign.

Now let us go back to $\frac{x}{1+|x|}=\frac{y}{1+|y|}$. Doing a little alg. manip. we can see that $x+x|y|=y+y|x|$. Note though that since $x,y$ have the same sign that $x|y|=y|x|$ and the result follows.

Surjectivity: So now we need to show that given any $\kappa\in(-1,1)$ we can find some $\tau\in\mathbb{R}$ such that $f(\tau)=\kappa$. Once again if $\kappa=0$ we see that $\tau=0$. So assume that $\kappa\ne0$. Then $\frac{\tau}{1+|\tau|}=\kappa\implies \kappa \tau=1+|\tau|$. Once again we see that since the RHS is positive that $\kappa,\tau$ share signs. So let us split this into two cases. If $\kappa<0\implies \tau<0$ the above becomes $\kappa\tau=1-\tau\implies \tau=\frac{1}{1+\kappa}$. If $\kappa>0\implies\tau>0$ then the equation is reduced to $\kappa\tau=1+\tau\implies \tau=\frac{1}{1-\kappa}$. Therefore $\tau=\begin{cases} \frac{1}{1+\kappa} & \mbox{if}\quad \kappa<0 \\ 0 & \mbox{if}\quad \kappa=0 \\ \frac{1}{1-\kappa} & \mbox{if} \quad \kappa>0 \end{cases}$. It is easily verifiable that for $\kappa\in(-1,1)$ the above implies that $\tau\in\mathbb{R}$. This is what we wanted to do.

Putting these two together shows that the mapping $f:\mathbb{R}\to(-1,1)$ given by $f(x)=\frac{x}{1+|x|}$ is bijective.

Note: There was nothing wrong with Jose27's approach. He merely noted that if $g:Y\mapsto X$ and $f\circ g=\iota_Y$ and $g\circ f=\iota_X$ where $\iota$ is the identity mapping that $f$ is invertible, which is logically equivalent bijective.