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So I know in order to prove a function is bijective, you need to prove that it is both injective and surjective. I know that to prove it is an injection, I need to make $f(x) = f(y)$, and try to get $x=y$ from that, but I can't seem to manipulate the equations to do so.

Also, how would I prove that this is surjective?

Michael Hardy
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goodguy1234
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    If $x|x|=y|y|$ then show $x^2=y^2$, which means that $x=y$ or $x=-y$. Look at what happens when $x=-y$. – Thomas Andrews Jul 26 '17 at 20:06
  • observe that $$x|x|=\begin{cases}x^2,&x\ge 0\-x^2,&x<0\end{cases}$$ – Masacroso Jul 26 '17 at 20:22
  • I would usually write x(y+1) rather than x\left(y+1\right) to get $x\left( y+1\right),$ reserving "left" and "right" for occasions when making the sizes match is an issue, as in $\displaystyle x\left( 1+ \int_0^\infty 1,dx \right).$ However, notice this difference: $$ \begin{align} \text{without “left'' and “right'': } & \quad x|x| \ \text{with “left'' and “right'': } & \quad x \left| x\right| \end{align} $$ Apparently the fact that the same symbol is used on both sides means this case is different. – Michael Hardy Jul 26 '17 at 20:26
  • $\ldots,$somewhat like this what happens when French notation is used for open intervals and \left] and \right[ are appropriate: $$ \begin{align} \text{without: } & \qquad x\in ]3,5[ \qquad \text{This is obviously wrong.} \ \text{with: } & \qquad x\in\left]3,5\right[ \qquad \text{$\ldots$ and now it's right.} \end{align} $$ – Michael Hardy Jul 26 '17 at 20:30

3 Answers3

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Note that $|f(x)|=x^2$. Also note that $f(-x)=-f(x)$.

Assume $f(x)=f(y)$. Then $|f(x)|=|f(y)|$, so $x^2=y^2$ and $x=y$ or $x=-y$. In the latter case, $f(y)=-f(x)$, so $f(x)=f(y)=0$, which means $x=y=0$.

Let $y\in\Bbb R$ be given. If $y\ge 0$ then $y=f(\sqrt y)$. If $y<0$ then $y=f(-\sqrt{-y})$.

Possibly simpler alternative: Show that $f$ is a bijection $[0,\infty)\to [0,\infty)$, and also a bijection $(-\infty,0)\to(-\infty,0)$.

Hagen von Eitzen
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$$\text{Suppose } x\left| x \right| = y\left| y\right|. \tag 1$$

Either $x\ge0$ or $x<0.$

If $x\ge0$ then we cannot have $y<0$ since then one side of line $(1)$ would be positive and the other negative. So we would have $y\ge0.$ But if $x\ge0$ and $y\ge 0$ then $x\left|x\right| = x^2,$ and similarly we have $y\left|y\right|=y^2.$ So we have $x^2 = y^2.$ That means $x=\pm y,$ but they're both $\ge 0.$ So $x=y.$

That's if $x\ge0.$ Now figure out what happens if $x<0.$

Michael Hardy
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It has the inverse function $$ g(x) = \begin{cases} x/\lvert x \rvert^{1/2} & x \neq 0 \\ 0 & x=0 \end{cases}. $$

Chappers
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