Is $f(x) = x^2$ an isometry?

Question

I am wondering if $f(x) = x^2$ is an isometry? My definition of an isometry is the following:

Let $(X, d)$ and $(\dot{X}, \dot{d})$ be metric spaces. $f: X \rightarrow \dot{X}$ is an isometry if:

$\dot{d}(f(x), f(y)) = d(x,y), \forall x,y \in X$

For my question I would choose:

• $f(x) = x^2$
• $X = \mathbb{R}$
• $d(x, y) = |x-y|$
• $\dot{X} = \mathbb{R}_{\geq0}$
• $\dot{d}(x,y) = ||\sqrt{x}|-|\sqrt{y}||$

It is also true that every isometry is injective. However, $f(x)=x^2$ is not injective which is clear.
Could you tell me what I am missing here? Thanks in advance.

Answer 1

$d(1,-1) = 2$

$\dot{d}(f(1),f(-1)) = 0$