Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I try solving the following excercise:

Show by stereoscopic projection that the $S^2$ sphere is locally homeomorphic to $R^2$.

I tried to solve this by using the cotangens function on the two angles defining the surface of the ball by the projection:

$\phi : S^2 \rightarrow \mathbb{R}^2, \ (\alpha, \beta) \mapsto \left(\cot \alpha, \cot \frac{\beta}{2}\right)$ for $\alpha \in (0, \pi), \beta \in (0, 2\pi)$

However the poles and one of the connecting lines ($\beta = 0$) is not defined by this map. Is there a neat trick to get this out of the way?

share|cite|improve this question
It seems like you're already labeling points on the sphere with two real numbers, and so at least implicitly already are viewing the sphere as locally homeomorphic to $\mathbb{R}^2$. – wckronholm Jul 25 '11 at 15:55
@wckronholm: you can label all points of any subset of $\mathbb R^2$ with two real numbers but this surely doesn't assume anything about local homeomorphisms... – Marek Jul 25 '11 at 19:28
@Marek: Of course, but points on $S^2$ aren't points in a subset of $\mathbb{R}^2$. Presumably, Helium is thinking of the sphere as the units in $\mathbb{R}^3$ and so the $\alpha$ and $\beta$ need to be explained somehow. If these come from spherical coordinates, then what work is there really to do? – wckronholm Jul 25 '11 at 20:57
@wckronholm: well, if we are to prove anything we need to have some definition of $S^2$. As this is obviously an elementary question it's clear from the context that $S^2$ is understood as a subset of $\mathbb R^3$ (not to mention that OP talks about the stereographic projection) and not as a CW-complex with 1 0-cell and 1 2-cell (say); which requires some definition of spheres anyway... – Marek Jul 25 '11 at 21:24

Considering $S^2$ as the points $(x, y, z)$ in $\mathbb{R}^3$ with $x^2 + y^2 + z^2 = 1$, then stereographic projection $S^2 - \{(0,0,1)\} \to \mathbb{R}^2$ is a homeomorphism. (You can check directly that the standard stereographic projection formula defines a continuous bijection with a continuous inverse.) This shows that all points except $\{(0,0,1)\}$ have neighborhoods homeomorphic to an open subset of $\mathbb{R}^2$. Interchanging the roles of $x$ and $z$ gives a stereographic projection $S^2 - \{(1,0,0)\} \to \mathbb{R}^2$ which is also a homeomorphism.

(This is one way of getting the result with just two "patches" in your atlas.)

share|cite|improve this answer
this works too. – ncmathsadist Jul 25 '11 at 23:06

Now rotate the sphere 90 degrees and repeat with two new poles. You have 4 patches that show $\mathbb{R}^2$ is locally homeomorphic to $S^2$.

share|cite|improve this answer
Is it possible to show that that 4 patches are a maximal atlas or do I need to use more than those? – Helium Jul 25 '11 at 13:10
I think it has you covered. – ncmathsadist Jul 25 '11 at 13:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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