TITLE: Isometry of two metric spaces
QUESTION [7 upvotes]: How to prove that $(\Bbb R^2, d_1)$ and $(\Bbb R^2, d_\infty)$ are isometric?
My approach
Let $f:(\Bbb R^2, d_1)\to(\Bbb R^2, d_\infty)$ be a function defined by $f(x, y) = (x+y,x-y)$. I can easily prove that $f$ is bijective, but I can not prove the isometry of $f$.

REPLY [3 votes]: The map $f$ maps the square $Q_1$ with vertices $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$ onto the square $Q_\infty\subset({\mathbb R}^2,d_\infty)$ with vertices $(1,1)$, $(1,-1)$, $(-1,-1)$, $(-1,1)$.
Since $Q_1$ is the closed unit ball in $({\mathbb R}^2,d_1)$, and $Q_\infty$ is the closed unit ball in $({\mathbb R}^2,d_\infty)$, it follows from general principles about norms and linear maps that $f$ is an isometry.