0
$\begingroup$

Show that $\{(x,y)\in\mathbb{R}^m\times \mathbb{R}^n ; |x|^2+|y|^2=1, y\neq 0\}$ is homeomorphic to $\mathbb{R}^m\times S^{n-1}$.

I tried to find an easy bijection (and then try to prove continuity and inverse's continuity) but I can't find this bijection.

$\endgroup$
2
$\begingroup$

Suppose that $m = 1$ and $n=2$, hence you are in $\mathbb{R}^3$. You should see that the space $X \subseteq \mathbb{R}^3$ is the unitary sphere $S^2 \subseteq \mathbb{R}^3$ minus the two antipodal points $(-1,0,0)$ and $(1,0,0)$. It should be clear that $X$ is homeomorphic to the meridian $\mathbb{R}$ times the equator $S^2 \cap \{ x = 0 \} \simeq S^1$, i.e. $\mathbb{R} \times S^1$ as wanted. This is a sort of Mercator projection of the earth planet without the two poles onto an infinite cylinder which is tangent to the equator. Can you write down formulas in this particular case?

Try to generalize the formulas found in the particular case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You mean sending $(x,y)$ to $\frac{1}{|y|}(x,y)\in\mathbb{R}^{m}\times \mathbb{S}^{n-1}$ right? In this case: Continuity and surjection are clear. Injection is not very easy but I think that I can prove it using $|x|^2+|y|^2=1$. But I can't see the form of inverse. Inverse's second componen it's obvious but inverse of first component I need to use that $|x|^2+|y|^2=1$ and is so algebraically difficult. $\endgroup$ – Gaston Burrull May 18 '12 at 23:20
  • 1
    $\begingroup$ Exactly! The inverse $\mathbb{R}^m \times S^{n-1} \to X$ is given by $(x,s) \mapsto (\vert x \vert^2 +1)^{-1/2} (x,s)$. You should be able to check that this is the inverse of $(x,y) \mapsto \vert y \vert^{-1} (x,y)$. $\endgroup$ – Andrea May 18 '12 at 23:42
  • $\begingroup$ Thank you. This kind of problems like "prove that there exist a function such that" are really hard for me. $\endgroup$ – Gaston Burrull May 19 '12 at 0:49
  • $\begingroup$ I think the best way to solve this kind of problems is to have a geometric picture in a particular case. $\endgroup$ – Andrea May 19 '12 at 7:54
  • $\begingroup$ I think so. Waiting that generalization be easy. $\endgroup$ – Gaston Burrull May 19 '12 at 8:43

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.