Equivalence of Definitions of Isometry of Metric Spaces

Definition 2
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces or pseudometric spaces.

Let $\phi: M_1 \to M_2$ be an isometry.

Then its inverse $\phi^{-1}: M_2 \to M_1$ is likewise an isometry.

$(1)$ implies $(2)$
Let $\phi: M_1 \to M_2$ be an isometry by definition 1.

Then by definition:
 * $\phi$ is a bijection such that:
 * $\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$

By Inverse of Bijection is Bijection, $\phi^{-1}$ is also a bijection.

Thus $\forall a, b \in A_2$ there exists $\phi^{-1} \left({a}\right)$ and $\phi^{-1} \left({b}\right)$ in $A_1$.

From Inverse of Inverse of Bijection:
 * $\forall a, b \in A_2: d_2 \left({\phi \left({\phi^{-1} \left({a}\right)}\right), \phi \left({\phi^{-1} \left({b}\right)}\right)}\right) = d_1 \left({\phi^{-1} \left({a}\right), \phi^{-1} \left({b}\right)}\right)$

and so:
 * $\forall a, b \in A_2: d_2 \left({a, b}\right) = d_1 \left({\phi^{-1} \left({a}\right), \phi^{-1} \left({b}\right)}\right)$

Thus $M_1$ and $M_2$ are isometric by definition 2.

$(2)$ implies $(1)$
Let $M_1$ and $M_2$ be isometric by definition 2.

Then by definition:
 * there exist inverse mappings $\phi: A_1 \to A_2$ and $\phi^{-1}: A_2 \to A_1$ such that:


 * $\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$
 * and:
 * $\forall u, v \in A_2: d_2 \left({u, v}\right) = d_1 \left({\phi^{-1} \left({u}\right), \phi^{-1} \left({v}\right)}\right)$

By Inverse of Bijection is Bijection, $\phi$ and $\phi^{-1}$ are both bijections.

Thus $\phi: M_1 \to M_2$ is a isometry by definition 1.