Inverse of Isometric Isomorphism

Theorem

Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be normed division rings.

Let $\phi:R \to S$ be a mapping.

Then $\phi:R \to S$ is an isometric isomorphism if and only if $\phi^{-1}: S \to R$ is also an isometric isomorphism.

Proof

By Inverse of Algebraic Structure Isomorphism is Isomorphism then:

$\phi: R \to S$ is an ring isomorphism if and only if $\phi^{-1}: S \to R$ is also an ring isomorphism.

Let $d_R$ and $d_S$ be the metric induced by the norms $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively.

By Inverse of Isometry of Metric Spaces is Isometry then:

$\phi: \struct {R, d_R} \to \struct {S, d_S}$ is an isometry if and only if $\phi^{-1}: \struct {S, d_S} \to \struct {R, d_R}$ is also an isometry.

The result follows.

$\blacksquare$