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I got this question in an exam recently. I haven't been able to solve it. The question goes like this :

Does there exist a continuous surjective map from $\mathbb{R}^3\setminus \mathbb{S}^2$ to $\mathbb{R }^2\setminus \{0\}$?

I was proceeding in kind of a naïve fashion by trying to construct a suitable map from $\mathbb{R}^3$ to $\mathbb{R }^2$ which vanishes exactly on the unit sphere. But I doubt of that will help. Even though I am not quite sure I think the claim is false and will require some nontrivial use of tricky techniques on topology. Can someone help me with some hints?

shadow10
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1 Answers1

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We only need to map the open unit ball $B=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2<1\}$ to $\mathbb{R}^2\setminus\{0\}$. The part outside the unit ball $B$ can be mapped to any constant point except the origin of $\mathbb{R}^2$.

First, map the 3D open ball $B$ to the 2D open disk $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$ by projection: $f(x,y,z)=(x,y)$.

Next, map the open disk $D$ to the right half-plane $\mathbb{R}^2_{x>0}$ by $$g(x,y)=\left(\frac{1-x^2-y^2}{(1-x)^2+y^2},\frac{2y}{(1-x)^2+y^2}\right)$$ which is inspired by the complex function $z\mapsto \dfrac{1+z}{1-z}$

Finally, map $\mathbb{R}^2_{x>0}$ to $\mathbb{R}^2\setminus\{0\}$ by $$h(x,y)=(x^3-3xy^2, 3x^2y-y^3)$$ which is inspired by the complex function $z\mapsto z^3$.

Mathis
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    Another map that works: For $(x,y,z)$ in the open unit ball of $\mathbb R^3,$ define

    $$f(x,y,z) = \exp \left ( \frac{x+iy}{1-(x^2+y^2)}\right),$$ where $\exp w$ is the complex exponential function.

     – zhw. Dec 13 '17 at 18:19
  • @Mathis How $z\to z^3$ takes upper Half plane to C/0 Can you explain please – Curious student Nov 27 '18 at 10:23