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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 ?

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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
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.

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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.

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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.

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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 $y$ ?

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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

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