I'm trying to answer this question(how to show the map $f(x_1,x_2) = (x_1+x_2, x_1-x_2)$ is a surjective isometry for specified metrics on $\Bbb R^2$) as an exercise, but I think I'm stuck and this is probably not an isometry. Also, I don't understand why the OP is using $d^{\infty}(f(x),f(y)):=\max:\{|(x_1+x_2)-(y_1+y_2)|,|(x_1-x_2)-(y_1-y_2)|\}$ . It doesn't seem right since $d^{\infty}:=\sqrt{\sum_{i=1}^2(x_i-y_i)^2}$. I have shown bijection so I won't post it here, for the sake of brevity.
Show $f:\Bbb{R}^2\rightarrow\Bbb{R^2}$ such that $f(x)=f((x_1,x_2)):=(x_1+x_2,x_1-x_2)$ where $d'(x,y):=\sum_{i=1}^2\vert x_i-y_i\vert$ and $d^{\infty}:=\sqrt{\sum_{i=1}^2(x_i-y_i)^2}$ is a surjective isometry.
(Show Isometry)
Let $x=(x_1,x_2)$ and $y=(y_1,y_2)$. So we have $d'(x,y)=\sum_{i=1}^2\vert x_i-y_i\vert=\vert x_1-y_1\vert+\vert x_2-y_2\vert$ and
\begin{align}
d^\infty (f(x),f(y))&=d^\infty((x_1 +x_2),(x_1-x_2))\\
&=\sqrt{[(x_1+x_2)-(y_1+y_2)]^2+[(x_1-x_2)-(y_1-y_2)]^2}\\
&=\sqrt{[(x_1+x_2)^2-2(x_1+x_2)(y_1+y_2)+(y_1+y_2)^2]+[(x_1-x_2)^2-2(x_1-x_2)(y_1-y_2)+(y_1-y_2)^2]}\\
&=\sqrt{[(x_1+x_2)^2+(x_1-x_2)^2]+[(y_1+y_2)^2+(y_1-y_2)^2]-2(x_1+x_2)(y_1+y_2)-2(x_1-x_2)(y_1-y_2)}\\
&=\sqrt{[2x_{1}^2+2x_{2}^2]+[2y_{1}^2+2y_{2}^2]-2[2x_1y_1+2x_2y_2]}\\
&=\sqrt{2x_{1}^2-2(2x_1y_1)+2y_{1}^2+2x_{2}^2-2(2x_2y_2)+2y_{2}^2}\\
&=\sqrt{2(x_1-y_1)^2+2(x_2-y_2)^2}
\end{align}
Finally, I want to ask if I still need to show that $f$ is a homeomorphism.
Best Answer
Firstly, clearly it's not an isometry between the $1$-norm and the $2$-norm. This can be most intuitively seen by looking at the spheres of the two metrics. The transformation $T$ is linear, and the sphere of the $1$-norm is a polygon. Its image, therefore, is also a polygon. But if it were an isometry, we'd also expect it to be the $2$-norm's circular sphere, which is absurd.
Secondly, you don't need to show $f$ is a homeomorphism. Showing a map is an isometry shows continuity automatically (it's super-easy: pick $\delta = \varepsilon$). Isometries are also injective, and their inverses (when considered mapping onto their range) are also isometries. So, if you can show $f$ is surjective, then $f^{-1}$ is an isometry, and also continuous. Thus, $f$ is a homomorphism (when surjective).