# Distance-Preserving Image Isometric to Domain for 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 distance-preserving mapping, then:

- $\phi: M_1 \to \Img \phi$

is an isometry.

## Proof

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

Let $\phi$ be a distance-preserving mapping from $M_1$ to $M_2$.

Let $A = \Img \phi$ be the image of $\phi$.

By Subspace of Metric Space is Metric Space, $\struct {A, d_2}$ is a metric space.

As $\phi$ is a distance-preserving mapping, by Distance-Preserving Mapping is Injection of Metric Spaces it is injective.

From Restriction of Injection is Injection, $\phi: M_1 \to \Img \phi$ is an injection.

From Surjection iff Image equals Codomain, $\phi: M_1 \to \Img \phi$ is a surjection.

Thus $\phi \to \Img \phi$ is by definition a bijection.

Thus $\phi: M_1 \to \Img \phi$ is a bijective distance-preserving mapping.

Hence, by definition, $\phi: M_1 \to \Img \phi$ is an isometry.

$\blacksquare$