So I was attempting to solve a topology exercise when the following question came to me.
The objective was trying to fine a bijection $f$ between the following disks in $\Bbb R^2$:
$$D_1 :=\{(x,y)\in \Bbb R^2: x^2 + y^2 \leq1\}$$
$$D_2 :=\{(x,y)\in \Bbb R^2: x^2 + y^2 \leq 4\}$$
So $D_1$ is a disk of radius $1$ centered at the origin, and $D_2$ is a disk of radius $2$ centered at the origin.
The first thing that came to my mind was to use polar coordinates, so let's redefine both disks as:
$$D_1 :=\{(r,\varphi): r\in [0,1] \wedge \varphi \in [0,2\pi]\}$$
$$D_2 :=\{(r,\varphi): r\in [0,2] \wedge \varphi \in [0,2\pi]\}$$
Now we can just simply scale disk 1 into disk 2:
$$f:D_1\to D_2$$
$$f(r,\varphi)=(2r,\varphi)$$
My question about the injectivity of this function, more concretely in the center of the disks.
Let $\varphi_1,\varphi_2 \in [0,2\pi]$, with $\varphi_1 \neq \varphi_2$. Then how do we treat points like $(0,\varphi_1)$ and $(0,\varphi_2)$. In the disk they represent the same point: the center of the disk. But, when learning about double integrals with polar coordinates, my teacher taught us that when we use polar coordinates to describe a disk we are just defining an rectangle in the $rO\varphi$ plane, instead of in the $xOy$ plane:
And all the points of the form $(0,\varphi)$ are in that line in $\varphi-$axis and are indeed different points.
So how do we treat the points with this form? Are they considered all the same point and thus $(0,\varphi_1) = (0,\varphi_2)$? Or are they considered different points as seen in the $rO\varphi$ plane?
Best Answer
Your function is simply $(x,y)\mapsto2(x,y)$, and therefore it is injective (indeed, bijective).
Concerning your final questions, all the pairs $(0,\varphi_1)$, with $\varphi\in[0,2\pi]$, describe the same point (the origin), and therefore there is no problem there.