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Is there a simple bijective function between the surface of the unit sphere in $\mathbb{R}^3$ and the real plane? I am aware of stereographic projection and Riemann sphere but these seem to map the north pole to infinity.

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there is no continuous bijection from $S^2$ to $\mathbb{R}^2$ – yoyo Jun 19 '11 at 20:25
@yoy How can I prove it? Is there a simple principle? – TROLLKILLER Jun 19 '11 at 20:27
A continuous bijection between $S^2$ -compact, and $R^2$ Hausdorff.... – gary Jun 19 '11 at 20:31
$S^2$ is compact. A continuous image of a compact set is compact. $\mathbb{R}^2$ is not compact, so $\mathbb{R}^2$ cannot be a continuous image (let alone a continuous bijection) of $S^2$. – Arturo Magidin Jun 19 '11 at 20:33
You should specify if you want any bijection (just set-theoretic), or if you want some property attached to it; there are no continuous bijections, but there are certainly lots of purely set-theoretic bijections (since the two sets have the same cardinality). – Arturo Magidin Jun 19 '11 at 20:40
up vote 7 down vote accepted

Start from the stereographic bijection $B$ between the sphere minus its North pole $p$ and the plane. One needs a point of the plane to be the image of $p$ hence we will make some room for it. Choose any injective sequence $(x_k)_{k\ge0}$ of points in the plane and define a shift $S$ on the plane by $S(x_k)=x_{k+1}$ for every $k\ge0$ and $S(x)=x$ for every other point $x$ of the plane. Then $S\circ B$ is a bijection between the sphere minus $p$ and the plane minus $x_0$. Extend it to a bijection between the sphere and the plane by sending $p$ to $x_0$. You are done.

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Yes, a nice use for Hilbert's infinite hotel. – André Nicolas Jun 19 '11 at 20:39
Exactly. $ $ $ $ – Did Jun 19 '11 at 20:40
This is just the usual trick for bijecting $[0,1]$ with $[0,1)$ :-), nonetheless very nice. – Asaf Karagila Jun 19 '11 at 21:44

It is not possible if you want a continuous bijection: Assume there is such a bijection h. Then $h^-1$ is also a bijection between $S^2$ and the plane. But $S^2$ is compact, and the plane is Hausdorff; a continuous bijection between compact and Hausdorff is a homeomorphism. But $S^2$ is compact, and the plane is not, and compactness is a topological property, i.e., it is preserved by homeomorphisms (basically since compactness is defined in terms of open sets.)

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The unit sphere is compact. Any image of it under a continuous function is compact. The plane is not bounded; hence it is not compact. Therefore no such homeomorphism exists.

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It's not clear that the OP was searching for a homeomorphism, per se; the OP specifies only a bijection. – amWhy Jun 19 '11 at 21:24

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