Proof similarity transformations are bijective. I'm trying to prove that similarity transformations from $\mathbb{R}^2$ to $\mathbb{R}^2$ are bijective, I've already proven its injectivity by contradiction, so I'm really just stuck on the surjectivity.
Let's define similarity transformations, we call a function $\psi : \mathbb{R}^2 \longrightarrow  \mathbb{R}^2$ a similarity transformation if it is not constant and if there exists an $\alpha \in \mathbb{R}_{>0}$ such that for all $x, y \in \mathbb{R}^2$ so that $x \neq y$  we have $\alpha \cdot d(x,y) = d(\psi(x), \psi(y))$.
I proved that a similarity transformation is injective by contradiction, a contradiction follows from the definition of non-injectivity and the fact that $\alpha$ is bigger than zero.
Any help would be greatly appreciated!
 A: My solution would show that $\psi$ is a affine map, i.e. $\psi(v)=b+Av$ for some vector $b$ and a matrix $A$. If you have this, then you would be done, since injectivity will give you surjectivity (dimension argument). 
To see that $\psi$ is indeed affine you can proceed in different ways. Do you know that isometries are affine? If yes just use that $\sqrt{\alpha^{-1}}\psi$ is an isometry and use this to argue that $\psi$ is affine.
Another way would be direct calculation: Define $\phi(v):=\psi(v)-\psi(0)$. It will suffice to show that $\phi$ is linear. First observe that $\phi$ is also a similar transformation since:
$$ d(\phi(x),\phi(y))=\Vert \phi(x)-\phi(y) \Vert=\Vert \psi(x)-\psi(y) \Vert=\alpha\cdot d(x,y).  $$
The next step would be showing, that $\phi$ preserves the scalarproduct, i.e. $<\phi(x),\phi(y)>=\alpha^2<x,y>$. First observe that $\phi(0)=\psi(0)-\psi(0)=0$ then you have:
$$<\phi(x),\phi(x)>=\Vert \phi(x)-\phi(0)\Vert^2=  \alpha^2\cdot\Vert x-0\Vert^2= \alpha^2 <x,x> $$ 
Now use that 
$$ <x,y>=-1/2(<x-y,x-y>-<x,x>-<y,y>) $$
for the general case. Note that this allready shows that $\phi$ preserves orthogonality. Take your favorit orthonormalbasis, $e_1,e_2$, then $f_1:=\phi(\alpha^{-1}e_1),f_2:=\phi(\alpha^{-1}e_2)$ is also a orthonormalbasis.
Now assume that $\phi$ is not linear, i.e. there exists $x,y\in \mathbb{R}^2$ and $a,b \in \mathbb{R}$, s.t. $\phi(ax+by)\neq a\phi(x)+b\phi(y)$ then you have for either $f_1$ or $f_2$:
$$
\alpha^2\cdot <ax+by,\alpha^{-1}e_i>=<\phi(ax+by),f_i>\neq <a\phi(x)+b\phi(y),f_i>\\
=a\cdot <\phi(x),f_i>+b\cdot <\phi(y),f_i>
= \alpha^2\cdot <ax+by,\alpha^{-1}e_i>.
$$
The $\neq$ now follows because otherwise we would have $\phi(ax+by)= a\phi(x)+b\phi(y)$, since the $f_i$ form a ONB. But now we arrive at a contradiction. 
A: Hint:
Prove $\psi$ is an affine map, and the associated linear map is injective. Hence, as it is an endomorphism in finite dimension, this linear map is bijective.
A: So what remains to prove is that for any $y$ there is an $x$ such that $\psi(x) = y$.
Well lets assume this is false. Then there exists a $y$ for which no $x$ satisfies $\psi(x) = y$
Now take any point $x_0$ and let $y_0 = \psi(x_0)$ be it's image. Define $R = d(y_0,y) / \alpha$ 
Next define the circle $C = \{z \in \mathbb R^2 \, : \, d(x_0, z) = R\}$. The idea is that this circle is mapped in such a way that $y_0$ "should" lie on the image
$$\psi(C) = \{z' \in \mathbb R^2 \, : \, \exists z \in C \, ,z' = \psi(z) \}$$
The intuition is that a similarity transform only rotate and enlarge so this image should also be a circle, now centered at $y_0$ with radius $d(y_0, y)$, the set $\{z \in \mathbb R^2\, : \, d(z, x_0) = d(x_0,y)\}$ and this one contains  $y_0$. Then we'd have $x$ by going backwards. Lets prove it. 
Every point $z'$ on $\psi(C)$ satisfies $d(y_0,z') = d(y_0, y)$ because by definitions there exists some $z \in C$ such that $d(y_0,z') = d(\psi(x_0),\psi(z)) = \alpha d(x_0, z) = \alpha R = d(y_0,y)$
Okay but this doesn't mean $\psi(C) = \{z \in \mathbb R^2\, : \, d(z, x_0) = d(x_0,y)\}$ (that it is a circle) only $\psi(C) \subseteq\{z \in \mathbb R^2\, : \, d(z, x_0) = d(x_0,y)\}$ (that it is contained in one). 
Unfortunately here I see no recourse but to invoke continuity and topology. A similarity transform is clearly continuous meaning $\psi(C)$ must be connected like $C$ but if $y \not \in \psi(C)$ then it wouldn't be connected since it's be punctured.**
Therefore $y \in \psi(C)$ and by definition there then exists some $x \in C \subset \mathbb R^2$ such that $y = \psi(x)$
**I'm not happy with this part as it's specific intuition about $\mathbb R^2$ and the standard metric but it's too in the evening late for me to come up with a better alternative. 
