Show $f$ is a surjective isometry. Give $f:(\mathbb{R}^2,d^{1}) \rightarrow (\mathbb{R}^2,d^{\infty})$ let $f$ be defined $$f(x_1,x_2)=(x_1+x_2,x_1-x_2)$$ My ideal is to try to show that $f$ preserves the origin and that $f$ is continuous and injective, thereby implying $f$ is surjective. I wonder if it is possible to show that $$d^{1}(x,y) = d^{\infty}(f(x),f(y))$$ to show that $f$ is an isometry, but I do not know how to manipulate the terms, for example what does $d^{\infty}(f(x_1,x_2),f(y_1,y_2))$ even look like? At any rate, here is my work thus far. 

I can show that $f$ preserves the origin.$ f(0,0)=
  \left[ {\begin{array}{cc}
   1 &amp; 1 \\
   1 &amp; -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}0\\0\end{array} } \right]=(0,0).$

Moreover, $f$ is invertible so that,$ f^{-1}(0,0)= \frac{1}{2}
  \left[ {\begin{array}{cc}
   1 &amp; 1 \\
   1 &amp; -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}0\\0\end{array} } \right]=(0,0). $
For any $a,b \in R^{2}$ we have 
$ f(a,b)=
  \left[ {\begin{array}{cc}
   1 &amp; 1 \\
   1 &amp; -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}a\\b\end{array} } \right]=(a+b,a-b).$

So that, $ f^{-1}(a+b,a-b)= \frac{1}{2}
  \left[ {\begin{array}{cc}
   1 &amp; 1 \\
   1 &amp; -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}a+b\\a-b\end{array} } \right]= \left(   \frac{(a+b)-(a-b)}{2},\frac{(a+b)+(b-a)}{2} \right)= \left( \frac{2a-b+b}{2}, \frac{a+2b-a}{2}\right) = (a,b)$

I am pretty certain that this shows that $f$ is a surjective isometry.I believe I have shown that $f$ and $f^{-1}$ are injective with the origin preserved, also both spaces have the same dimension and the kernel of $f$ contains only the origin, therfore $f$ is a surjective isometry. But, something feels off.
So I have been able to find a solution. 

I must show that $$d^{1}(x,y) = d^{\infty}(f(x),f(y)).$$ I have that $$f(x) = f(x_1,x_2)= (x_1+x_2,x_1-x_2)$$ On the L.H.S. of the first equation we have $$d^{1}(x,y)= |(x_1-y_1)|+|(x_2-y_2)|$$ which is the Taxi Cab metric on $\mathbb{R}^2.$ On the R.H.S. we have $$ d^{\infty}(f(x),f(y)) =  d^{\infty}(f(x_1,x_1),f(y_1,y_2))= d^{\infty}((x_1+x_2,x_1-x_2),(y_1+y_2,y_1-y_2))$$ So that we get $$max:\{|(x_1+x_2)-(y_1+y_2)|,|(x_1-x_2)-(y_1-y_2)|\}$$ 
and this is $$max:\{|(x_1-y_1)+(x_2-y_2)|,|(x_1-y_1)-(x_2-y_2)|\}$$ and it is clear that  $$d^{1}(x,y) = d^{\infty}(f(x),f(y)).$$ The rest is shown above by showing that $f$ is invertible.
So I have been able to find a solution. 

I must show that $$d^{1}(x,y) = d^{\infty}(f(x),f(y)).$$ I have that $$f(x) = f(x_1,x_2)= (x_1+x_2,x_1-x_2)$$ On the L.H.S. of the first equation we have $$d^{1}(x,y)= |(x_1-y_1)|+|(x_2-y_2)|$$ which is the Taxi Cab metric on $\mathbb{R}^2.$ On the R.H.S. we have $$ d^{\infty}(f(x),f(y)) =  d^{\infty}(f(x_1,x_1),f(y_1,y_2))= d^{\infty}((x_1+x_2,x_1-x_2),(y_1+y_2,y_1-y_2))$$ So that we get $$max:\{|(x_1+x_2)-(y_1+y_2)|,|(x_1-x_2)-(y_1-y_2)|\}$$ 
and this is $$max:\{|(x_1-y_1)+(x_2-y_2)|,|(x_1-y_1)-(x_2-y_2)|\}$$ and it is clear that  $$d^{1}(x,y) = d^{\infty}(f(x),f(y)).$$ The rest is shown above by showing that $f$ is invertible.