Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that maps $(x,y)$ to $(\arctan{x}, \arctan{y})$. It maps the $\mathbb{R}^2$ plane to the interior of a square.

Define $g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that maps $(x,y)$ to $\displaystyle \left(\frac{x}{\sqrt{x^2+y^2+1}}, \frac{y}{\sqrt{x^2+y^2+1}} \right)$. It maps the $\mathbb{R}^2$ plane to an open (unit) disk.

I realize that $f$ and $g$ are different in the following sense:

  1. Add the boundaries to their images to make them a square with the boundary and a closed disk. (It's like compactification or "closure".)
  2. The square with the boundary is a manifold with corners, and the closed disk is a manifold with boundary, and they are not diffeomorphic to each other.

I suspect the difference between $f$ and $g$ as I noted above is a manifestation of something more fundamental. Hence my somewhat vague-wording question: How should I understand this difference between the two functions? What makes them have different "boundary structures"?

[I'd like to add some words on the background of what I am working on in case it helps people understand where I come from.

As a consequence of the different "boundary structures" between the square and the closed disk, $f$ has four "limiting functions" corresponding to the four sides of the square. For example, the upper side of the square corresponds $\lim_{y \rightarrow \infty} f = \tilde{f}$, where $\tilde{f}(x) = \left(\arctan{x}, \frac{\pi}{2}\right)$. In contrast, $g$ has only one "limiting function", the one corresponding to the circle.

Why am I interested in this kind of "limiting functions"? Because I am studying how to approximate certain functions. If we add a scaling factor $N$ to the definition of $f$ and have it now send $(x,y)$ to $(\arctan{x}, \frac{1}{N}\arctan{y})$, where $N \gg 1$, the square will now be squeezed along the y direction and become a thin rectangle. Then I can approximate $f$ with, say, $\tilde{f}$ as defined above, which, in a certain sense, is simpler but has little loss of fidelity. For $g$, similarly I can have it map $(x,y)$ to $\left(\frac{x}{\sqrt{x^2+y^2+1}}, \frac{1}{N}\frac{y}{\sqrt{x^2+y^2+1}} \right)$ and have the disk squeezed to an ellipse shape, but in this case I have only one "limiting" or approximating function.]

Any input (perspectives, references, correcting my terminologies or notations, etc.) on this will be received gratefully.

  • 2
    $\begingroup$ To be clear, the closed disk and the closed square are diffeomorphic to one another. One way to see this is to inscribe the bounding circle of the disk in the bounding square and consider a radial map that takes the boundary of the circle to the boundary of the square. $\endgroup$
    – 211792
    Sep 14, 2016 at 0:19
  • $\begingroup$ For what it's worth, you presumably want $\sqrt{x^{2} + y^{2} + 1}$ (i.e., the "${} + 1$" inside the radical) in the denominators defining $g$. ;) $\endgroup$ Sep 14, 2016 at 0:57
  • $\begingroup$ @AndrewD.Hwang It seems both work in this case. No? $\endgroup$
    – Lei
    Sep 14, 2016 at 1:48
  • 4
    $\begingroup$ @AustinC: Not by my definition of diffeomorphic they aren't. Homeomorphic, sure. $\endgroup$ Sep 14, 2016 at 5:45
  • $\begingroup$ @LeiHuang: $(x, y) \mapsto \sqrt{x^{2} + y^{2}}$ isn't differentiable at the origin. $\endgroup$ Sep 14, 2016 at 10:30


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