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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:

enter image description here

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?

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    $\begingroup$ They are different names for the same point in $\Bbb R^2$, just as $\langle 1,0\rangle$ and $\langle 1,2\pi\rangle$ denote the same point. (You might want to take $\varphi\in[0,2\pi)$ rather than in $[0,2\pi]$.) $\endgroup$ Aug 22, 2020 at 17:06

2 Answers 2

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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.

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The correct way to define polar co-ordidinates is using the domain $r > 0$ , $\varphi$ in $(0, 2\pi)$ so that the transformation between cartesian and polar co-ordinates is a diffeomorphism (has a differentiable inverse).

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  • $\begingroup$ This way there is no representation for $(0,0)$ or any point on the positive $x$ axis, so we cannot actually transform one disk to the other. $\endgroup$
    – David K
    Aug 22, 2020 at 17:34
  • $\begingroup$ Correct, that's one of the drawbacks of using polar co-ordinates. $\endgroup$
    – PeteBabe
    Aug 22, 2020 at 17:38

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