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.
Bijection between the plane and the surface of a sphere
-
3You should speci$f$y i$f$ 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). – 2011-06-19
3 Answers
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.
-
1This is just the usual trick for bijecting $[0,1]$ with $[0,1)$ :-), nonetheless very nice. – 2011-06-19
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.)
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.
-
1It's not clear that the OP was searching for a *homeomorphism*, per se; the OP specifies only a bijection. – 2011-06-19