My question deals with the following exercise from Hartshorne's Euclid: Geometry and Beyond:
Following our general principles, we say that two models $M,M'$ of our geometry are isomorphic if there exists a 1-to-1 mapping $\phi:M\to M'$ of the set of points of $M$ onto the set of points $M'$, written $\phi(A) = A'$, that sends lines to lines, preserves betweenness, i.e., $A\ast B\ast C$ in M $\iff$ $A'\ast B'\ast C'$ in M', and preserves congruence of line segments, i.e., $AB\cong CD$ in M$ \iff$ $A'B'\cong C'D'$ in $M'$
If $A=(a_1,a_2),B=(b_1,b_2)\in\mathbb{R}^2$, the exercise asks you to first show that $\mathbb{R}^2$ endowed with the taxicab metric: $$d(A,B)=|a_1-b_1|+|a_2-b_2|$$ Is isomorphic to the model of $\mathbb{R}^2$ endowed with the infinity distance: $$d(A,B)=sup\{|a_1-b_1|,|a_2-b_2|\}$$ How can we proceed in order to prove these models are isomorphic? How should we describe our isomorphism $\phi:\mathbb{R}^2\to \mathbb{R}^2$? I'm definitely having trouble with sending congruent lines to congruent lines.
The second part of the exercise claims that the taxicab metric model is not isomorphic to that of the standard model, i.e., the cartesian plane $\mathbb{R}^2$ equipped with the usual Euclidean distance. Hartshorne leaves a hint to this exercise: in order to prove two models are not isomorphic, one just has to find a statement which is true in one model but not in the other. I'm thinking, for example, that the line: $$y=x+1$$ Cuts the circle of center the origin $(0,0)$, and radius $1$ in infinitely many points in the taxicab metric model, a property that never holds in the standard model, which should make it clear that both models are not isomorphic. However, the first part still puzzles me, and I can't imagine how to proceed.
Thanks in advance for your time, any comments will be appreciated.