# Prove that an isometry is a bijection

attempt of solution:

In order to show a bijection, we require two steps: 1) show that it is injective 2) show that it is surjective

I have already proved the part to show injectivity but i am not quite sure how to show surjectivity

The idea I have is to pick a point y in the codomain of the isometry function, f, and show that there exists a point x in the domain such that f(x) = y

I don't know how to start on this, any help would be highly appreciated

Thanks

• What is your definition of isometry? (Some consider the bijective condition to be part of the definition) – Dan Rust Sep 19 '13 at 21:02
• This is not true in general. The identity map from $\mathbb Q$ to $\mathbb R$ (with the usual metrics) is an isometry, but it is not a bijection. – John Gowers Sep 19 '13 at 21:03
• If isometry means distance-preserving map then $[0,\infty)\to[0,\infty)$, $x\mapsto 1+x$, is a non-bijective isometry. Something is missing in your problem (isometry of what?) – user8268 Sep 19 '13 at 21:06
• – John Gowers Sep 19 '13 at 21:15

We will assume that surjectivity is not built into the definition. The ambient spaces have not been specified. We will assume that $\phi$ is a distance-preserving map from $S=\mathbb{R}^n$ to $\mathbb{R^n}$.
Let $p$ be in the target space, and let $q_1,q_2, \dots, q_{n+1}$ be points in the image of $S$ under $\phi$.
Let $d_i$ be the distance from $p$ to $q_i$. Let $b_i$ be such that $\phi(b_i)=q_i$.
Then there is a unique $a$ such that the distance from $a$ to $b_i$ is $d_i$ for all $i$. It follows that $\phi(a)=p$.
Remark: The assumption that the domain and codomain are each $\mathbb{R}^n$ is a strong one. But without some assumptions, one cannot expect to be able to prove the result, since there are straightforward counterexamples.