Embedding of $\mathbb R^2$ into the unit sphere $S^2$? How to find an embedding of $\mathbb R^2$ into the unit sphere $S^2$?
I know there is such an embedding because $S^2$ is a one-point compactification of $\mathbb R^2$, but I don't know how to write this embedding explicitly.
Thanks!
 A: Consider the usual 2 charts of the sphere: they give homeomorphisms from $\mathbb{R}^2$ and open disks in $S^2$. They are examples of the embedding you are looking for. 
A: A well known construction is the so called "stereographic projection". Think of $\def\R{\mathbb R}\R^2$ embedded into $\R^3$ as $\R^2\times\{0\}$ and the sphere embedded as the set of euclidian unit vectors. Now for each $x \in \R^2$, think of the line joining it with $e_3 = (0,0,1)$. It hits the sphere at exactly one other point besides $e_3$. Call this point $\phi(x)$. Then $\phi\colon \R^2 \to S^2$ is an embedding.
To give $\phi$ exactly, note that points on the line are of the form 
$$ tx + (1-t)e_3 = (tx_1, tx_2, 1-t) $$
which is on the sphere iff 
$$ t^2x_1^2 + t^2x_2^2 + (1-t)^2 = 1 \iff (1 + x_1^2+ x_2^2)t^2 - 2t = 0 $$
Hence $t=0$ (which corresponds to $e_3$) or 
$$ t = \frac{2}{1+x_1^2 + x_2^2} \leadsto \phi(x) = \left(\frac{2x_1}{1+x_1^2+x_2^2}, \frac{2x_2}{1+x_1^2+x_2^2}, \frac{-1+x_1^2+x_2^2}{1+x_1^2+x_2^2}\right) $$
A: A couple of ways to do this. The classic way is to use stereographic projection. This gives you pretty much the nicest map you could hope for from real $n$-space to the $n$-sphere (which misses a single point) and explicitly gives you the one-point compactification.
Maybe the easier way to just show that it embeds though is the note that the southern hemisphere of a sphere (without its boundary circle on the equator) is homeomorphic to an open disc and an open disc is homemorphic to the plane, so composing these two homeomorphisms gives you your embedding.
