Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $d(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ for $x=(x_1,x_2), y=(y_1,y_2)$.
A isometry of $\mathbb{R^2}$ is an image $f:\mathbb{R^2}\to\mathbb{R^2}:d(x,y)=d(f(x),f(y))$. Show that every isometry is bijective. I don't know where to start, any hints ?

share|cite|improve this question
Showing that $f$ is injective is trivial. For surjectivity, see this post. – David Mitra Feb 18 '13 at 16:41
every isometry from $\mathbb R^2 \to \mathbb R^2$? Because in general ismometry is only injective. – Mathematician Feb 18 '13 at 16:43

4 Answers 4

up vote 4 down vote accepted

Let $ABC$ be the triangle with vertices $(1,0)$, $(0,1)$, and $(0,0)$. Let $f$ be an isometry as defined in the post. Suppose $f$ takes $A$, $B$, and $C$ to $A'$, $B'$, and $C'$.

There is a combination $\phi$ of rotation and/or reflection and/or translation that takes $ABC$ to $A'B'C'$. Then $\phi^{-1}\circ f$ is an isometry as defined in the post. Note that it leaves $A$, $B$, and $C$ fixed.

Given an unknown point $P=(x,y)$, if we know the distances from $P$ to $A$, $B$, and $C$, we know $x$ and $y$. Since $\phi^{-1}\circ f$ fixes $A$, $B$, and $C$, it is the identity. Thus $f=\phi$.

Rotations, reflections, and translations are surjective, and therefore $f$ is.

share|cite|improve this answer
This is beautifully geometric. – k.stm Feb 18 '13 at 17:41
There are many proofs, with many flavours. For pure geometry, the start should have been a general triangle, but I thought the one with the familiar corners would make the visualization easier – André Nicolas Feb 18 '13 at 17:48
It took me some time to understand, but it was worth it :) Thanks for this enlightening post :) – Kasper Feb 18 '13 at 18:44

You can show, that every isometry on a finite dimensional hilbertspace is affine linear. So with the fact, that every isometry is injective you get the bijection.

share|cite|improve this answer
Generally affine, linear in case $f(0)=0$. – Jonas Meyer Feb 18 '13 at 16:50
@JonasMeyer: apparently in some parts what you and I call "affine" are called "affine linear". For example: or even, gasp!, nLab. – Willie Wong Feb 18 '13 at 16:57

If you assume $f(0) = 0$, in the $2$-dimensional case you can show surjectivity by observing that circles are mapped to themselves:

For any $x ∈ ℝ^2$ with $\lVert x \rVert = r$, since $d(f(x)),0) \overset{f(0) = 0}{=} d(x,0) = r$, you have $f(rS_1) \subseteq rS_1$.

Now you can show that an injective continuous map $f \colon S_1 → S_1$ is bijective:

Since $S_1$ is compact and $\operatorname{img}(f) \subseteq S_1$ is hausdorff, $f$ maps homeomorphically to $\operatorname{img}(f)$. If $\operatorname{img}(f) ≠ S_1$, say $a \notin \operatorname{img} (f)$, then – as a connected space – it’s homeo-morphic to an interval since $\operatorname{img}(f) \subseteq (S_1\setminus\{a\}) \underset{ae^{2πit}}{\cong} (0..1)$, which is a contradiction.

share|cite|improve this answer

To prove it is injective, you want to show that if $f(x)=f(y)$, then $x=y$.

If $f(x)=f(y)$, then what is $d(f(x),f(y))$ ? What is $d(x,y)$ equal to ? What can you then say about $x$ and $y$ ?

share|cite|improve this answer
Aaaah, I got. Thanks. $f(x)=f(y) \implies d(f(x),f(y))=0 \implies d(x,y)=0 \implies x=y$. – Kasper Feb 18 '13 at 16:47
Yes that's it ! – user62705 Feb 18 '13 at 16:50
Any hint for showing it is surjective ? – Kasper Feb 18 '13 at 16:51
as i told you every isometry is affine linear, if you have 2 times 2 matrix with rank 2, (because of the injective) you get the bijection – Dominic Michaelis Feb 18 '13 at 16:57
If someone has another proof, I am interested in that. – user62705 Feb 18 '13 at 17:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.