A set of all open circles (circles without elements lying on the border) in $\mathbb{R}^2$ is called E. A function $f∶ E → \mathbb{R}^3$ is defined as $f(\mathcal{K}) = (x, y, r)$, where $(x, y)$ is the center of the circle $\mathcal{K}$, and $r$ is the radius of the circle $\mathcal{K}$.
- Is $f$ "onto" $\mathbb{R}^3$ (is that an surjection)? If not, determine its set of values $R_f$.
- Is $f$ an injection? If not, point out a counter-example. If yes, determine $f^{-1}$
Now, I think that $f$ is "onto" $\mathbb{R}^3$. I can't think of any point in $\mathbb{R}^3$ that could be excluded from being covered by those circles.
As for the second one, I am sure that this is an injection. The only possibility to get the same circle is to enter the same information into the function.
However, I don't know how to express that more formally. Am I right in my way of thinking? Could somebody please help me structure my thoughts?
$f$ ia not onto. The radius of a ball must be positive, so in particular $(0,0,-1)$ is not in the image. The image of $f$ will be $\mathbb R^2 \times (0, \infty)$. It is, however, injective. Indeed, let $A, B \in E$ such that $f(A)=f(B)$. Let $p$ denote their common center and $r$ their common radius. Then $q \in A$ iff $|p-q| < r$ iff $q \in B$. Hence, $A=B$. Because the map is not onto, the inverse can only be defined from the image $\mathbb R^2 \times (0, \infty)$. For $(p,r) \in \mathbb R^2 \times (0, \infty)$, we define $f^{-1}(p,r)=\{q\in \mathbb R^2 : |p-q| < r\}$, i.e. the open ball of radius $r$ centered at $p$.