TITLE: Showing a function bijective
QUESTION [6 upvotes]: Given that $f:\mathbb R\to \mathbb R$ is continuous and $\vert x-y\vert \leq \vert f(x)-f(y)\vert$. We need to show that $f$ is bijective.
It is easy to show that $f$ is one one. But how to show that $f$ is onto?
I wanted to use intermediate value theorem. For that if $z\in \mathbb R$, I need to find $x,y\in \mathbb R$ such that $f(x)<z<f(y)$. But how to do so? Any hint will be appreciated.

REPLY [4 votes]: If $f: \mathbb{R} \to\mathbb{R}$ is continuous and injective then it is strictly monotone.
WLOG let us assume $f$ is increasing. Then $$|x| < |f(x)-f(0)| \implies f(0)-f(x)< |x| < f(x)-f(0)$$
$\displaystyle f(x) > f(0) + |x| \implies \lim_{x \to\infty} f(x) \to\infty$
$\displaystyle f(x) > f(0) - |x| \implies \lim_{x \to -\infty} f(x) \to -\infty$.

Added. To see that $f:\mathbb{R} \to\mathbb{R}$ is continuous and injective implies monotone observe that if $f$ isn't monotone then you have $x_{1},x_{2},x_{3} \in \mathbb{R}$ with $x_{1}<x_{2}<x_{3}$ but $f(x_{1}) > f(x_{2})$ and $f(x_{2}) <f(x_{3})$. Now try applying the Intermediate Value Theorem.