Distance-Preserving Surjection is 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.

Let $\phi: M_1 \to M_2$ be a surjection such that:
 * $\forall a, b \in M_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$

Then $\phi$ is an isometry.

Proof
The premises satisfy all elements of the definition of isometry except for bijectivity.

As we presume $\phi$ to be surjective we need only show that it is injective.

Let $a, b \in A_1$ and suppose that $\phi \left({a}\right) = \phi \left({b}\right)$.

Then by the definition of a metric space:


 * $d_2 \left({ \phi \left({a}\right), \phi \left({b}\right) }\right) = 0$

By our other premise, this means:


 * $d_1 \left({ a, b }\right) = 0$

Thus by the definition of a metric space, $a = b$.

Hence $\phi$ is injective, and the result follows.