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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

user84324
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    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
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    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
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    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
  • Possible duplicate: http://math.stackexchange.com/questions/170989/a-isometric-map-in-metric-space-is-surjective and http://math.stackexchange.com/questions/36502/isometries-of-mathbbrn – John Gowers Sep 19 '13 at 21:15

2 Answers2

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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.

André Nicolas
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Suppose $\sigma \in $ ENDV, ker($\sigma$)=0, since $\sigma$ is isometry.

Then with rank-nullity theory, dim(Im($\sigma$))= dim(V), i.e. Im($\sigma$) = V, now we can conclude that $\sigma$ is a bijection.

Gansheng
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