A map $f:([a,b], |\cdot|) \to ([c,d], |\cdot|)$ is an isometry if and only if $d-c = b-a$.

Question

I was asked to prove the following problem: A map $f:([a,b], |\cdot|) \to ([c,d], |\cdot|)$ is an isometry if and only if $d-c = b-a$.

But I think this is not correct, specifically the sufficient part. Can anyone please suggest the correct version of the question and a hint to solve it?

Answer 1

The correct definition of an isometry under the $|\cdot |$ metric is $|f(x)-f(y)|=|x-y|$ for all $x,y\in [a,b]$.

Answer 2

Certainly, $f:[0,1]\to[0,2]$ defined by $f(x)=x$ is an isometry but $1\neq 2$. So the correct formulation will be: There exists a surjective isometry $f:[a,b]\to[c,d]$ iff $b-a=d-c$

Answer 3

The map $f(x)=x^2$, from $([0,1],|\cdot|)$ to $([0,1],|\cdot|)$ is not an isometry, despite the fact that $1-0=1-0$.