Equivalence of Definitions of Isometry of Metric Spaces

Theorem
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.

Proof
By definition of isometry, $\phi$ is a bijection such that:
 * $\forall a, b \in M_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$

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

Thus $\forall a, b \in M_2$ there exists $\phi^{-1} \left({a}\right)$ and $\phi^{-1} \left({b}\right)$ in $M_1$.

From Inverse of Inverse of Bijection:
 * $\forall a, b \in M_2: d_1 \left({\phi^{-1} \left({a}\right), \phi^{-1} \left({b}\right)}\right) = d_2 \left({\phi \left({\phi^{-1} \left({a}\right)}\right), \phi \left({\phi^{-1} \left({b}\right)}\right)}\right)$

and so:
 * $\forall a, b \in M_2: d_1 \left({\phi^{-1} \left({a}\right), \phi^{-1} \left({b}\right)}\right) = d_2 \left({a, b}\right)$

Hence the result.