Prove that every isometry on $\mathbb{R}^2$ is bijective

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.

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.

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.