Exercise 2.6

Proof. First, we will show that $B$ can even be a range for $g$, i.e. we must show that $g(x)\in B$ for every $x\in A$, as this is not necessarily obvious. So for any $x\in A$ we have that $x\ge 1$, and thus $x^2\ge 1$ as well. Then we have $x^2-1\ge 0$ so that the product $x(x^2-1)\ge 0$ since of course $x\ge 1>0$. Therefore $g(x)=f(x)=x^3-x=x(x^2-1)\ge 0$ so that $g(x)\in B$.

Next, we show that $g$ is monotonically increasing, from which injectivity follows. Suppose that $x,y\in A$ where $x<y$. Since we have $x,y\ge 1>0$, it follows that $x^2<y^2$, and therefore $x^2-1<y^2-1$. Thus we have

since both $x,y\ge 1>0$ and $x^2-1,y^2-1\ge 0$. This shows that $g$ is monotonically increasing. It follows that $g$ is injective because, if we consider $x,y\in A$ where $x\ne y$, then it has to be that either $x<y$ or $x>y$. In the former case we have $g(x)<g(y)$ and in the latter $g(x)>g(y)$ so that $g(x)\ne g(y)$ either way.

We show that $g$ is surjective in a roundabout way that depends on calculus since $g$ is cubic and so does not yield a simple algebraic inverse function. Consider any $y\in B$ so that of course $y\ge 0$. If $y=0$ then clearly $g(1)=0=y$, so assume that $y>0$. Let $a_0=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{2,y\right\}$ so that of course there is a real $a$ such that $a>a_0$ since the reals are unbounded. Then we of course have $a>a_0\ge 2\ge 0$ so that $a\in A$. We also have

Then we have that $g(1)=0<y<g(a)$, and that of course $1<2\le a_0<a$. It then follows from the intermediate value theorem that there is an $x\in (1,a)$ such that $f(x)=y$ since clearly $g$ is continuous by elementary calculus. We note that of course $1\le x$ so that $x\in A$. This shows that $g$ is surjective since $y$ was arbitrary, which in turn completes the proof that $g$ is a bijection. □