Lipschitz Equivalence yields Lipschitz Equivalent Metric

Theorem
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $f: M_1 \to M_2$ be a Lipschitz equivalence.

Let $d_3: A^2 \to \R$ be a mapping defined as:


 * $\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$

Then $d_3$ yields a metric on $M_1$ which is Lipschitz equivalent to $d_1$.