Let $B_{a}$ be the open ball centered at 0 of radius $a$ in $\Bbb {R}^{n}$, that is $B_{a}=\{x\in \Bbb {R}^{n}:\left\lVert x\right\rVert\lt a\}$. Define a map $\phi:B_{a}\to \Bbb {R}^{n}$ by $\phi (x)=\frac{ax}{\sqrt{a^{2}-{\left\lVert x\right\rVert}^{2}}}$. Prove that $\phi$ is a homeomorphism.

So I have to find the contiuous inverse function of $\phi$?

  • 3
    $\begingroup$ My guess is that you re supposed to prove that $\phi$ is a homeomorphism, not a homomorphism. $\endgroup$ – José Carlos Santos Mar 11 at 14:47
  • $\begingroup$ oh...sorry! I edited it. So I have got it wrong... $\endgroup$ – Maggie Mar 11 at 14:50
  • $\begingroup$ But now your question makes no sense. Being a homeomorphism has nothing to do with being a linear map. $\endgroup$ – José Carlos Santos Mar 11 at 14:52
  • 3
    $\begingroup$ You may want to start by finding the inverse $\endgroup$ – Scott Mar 11 at 14:53
  • $\begingroup$ For the curious, here's a picture of the mapping. It takes a ball and "stretches it out" to infinity. The "edges" of the ball get stretched the most, and the boundary ends up at infinity. In the image, the top is the original with arbitrary coloring, and the colors of corresponding points remain the same in the mapped ball below (for reference). Includes $x$ up to radius=0.999. $\endgroup$ – Scott Mar 11 at 15:10

In fact the best approach is to find a continuous inverse function of $\phi$. This is not hard. Write $$y = \frac{ax}{\sqrt{a^2-{\left\lVert x\right\rVert}^{2}}} .$$ Take the norm on both sides, square and resolve for $\lVert x\rVert$. This gives $$\lVert x\rVert = \frac{a\lVert y\rVert}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}}.$$ Since $y = \lambda x$ with $\lambda > 0$, we get for $x \ne 0$ $$\frac{y}{\lVert y\rVert} = \frac{x}{\lVert x\rVert}$$ and therefore $$x = \lVert x\rVert \frac{x}{\lVert x\rVert} = \lVert x\rVert \frac{y}{\lVert y\rVert} = \frac{a\lVert y\rVert}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}} \frac{y}{\lVert y\rVert} = \frac{a y}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}} .$$ The map $$ \psi : \mathbb R^n \to B_a, \psi(y) = \frac{a y}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}} $$ is now easily verified to be the desired inverse of $\phi$. Note that $\psi(0) = 0$ and, for $y \ne 0$, $\lVert \psi(y) \rVert = \frac{a \lVert y \rVert}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}} < \frac{a \lVert y \rVert}{\sqrt{ {\left\lVert y\right\rVert}^{2}}} = a$, i.e. $\psi(\mathbb R^n) \subset B_a$.

| cite | improve this answer | |
  • $\begingroup$ That's awesome. I was really confused how to find the inverse. But I have some questions...why can't we just take off the norm of the equation of $\left\lVert x\right\rVert$. $\endgroup$ – Maggie Mar 13 at 14:59
  • $\begingroup$ And does $\psi(\mathbb R^n) \subset B_a$ mean that $\forall \epsilon \gt 0, p\in \Bbb {R}^{n}, \exists \delta \gt 0$ such that $\psi(B_{\delta}(p)\cap \Bbb {R}^{n})\subset B_{\epsilon}(\psi (p))\cap \psi (\Bbb {R}^{n})\subset B_{\epsilon}(\psi (p))\cap B_{a}$ and imply continuous? $\endgroup$ – Maggie Mar 13 at 15:12
  • 1
    $\begingroup$ Question 1: In the first step we only get an equation for $\lVert x \rVert$. This implies of course the desired equation for $x$ since $y$ is a positive multiple of $x$. But we need a formal argument as in my answer. Question 2: We want to define a function $\psi : \mathbb R^n \to B_a$. The definition $\psi(y) = \frac{a y}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}}$ trivially produces an element of $\mathbb R^n$, but we still must show $\psi(y) \in B_a$, i.e. $\lVert \psi(y) \rVert < a$. $\endgroup$ – Paul Frost Mar 13 at 15:41
  • 1
    $\begingroup$ This has nothing to do with $\epsilon$-$\delta$-arguments. The continuity of $\psi$ follows from the fact that all elementary algebraic operations are continuous. $\endgroup$ – Paul Frost Mar 13 at 15:41
  • $\begingroup$ thanks so much! Your Explanation is clear and helpful $\endgroup$ – Maggie Mar 13 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.