Is an isometry always a bijection?

[3029298]

A function from the plane to itself which preserves the distance between any two points is called an isometry. Prove that an isometry must be a bijection.

To prove that an isometry is injective is easy:
For an isometry: $$||f(x)-f(y)||=||x-y||$$
If $$x\neq y$$ then $$||x-y||>0$$ and therefore $$||f(x)-f(y)||>0$$ and $$f(x)\neq f(y)$$.
But to prove that an isometry is surjective... how should I do that?

 

[Hurkyl]

Have you thought about what all this means in terms of high-school Euclidean geometry?

 

[3029298]

Well, I tried to make some drawings in the plane, but they do not seem to lead to a solution...

 

[Hurkyl]

Well, at least have you found other things that must be preserved by an isometry?

 

[3029298]

Well, if you have three points with $$||f(x)-f(y)||=||f(y)-f(z)||=||f(z)-f(x)||$$ this is also true for $$||x-y||=||y-z||=||z-x||$$. This means that a triangle remains a triangle. But I do not know what first step I have to see to do the proof...

 

[3029298]

Is there anyone who can help me on this?

 

[g_edgar]

Consider the tiling of the plane by regular triangles with all edges 1. What does your isometry do to its vertices? Once you know that, consider a general point (it is in one of those triangles) and see what you can say about what the isometry does to it.