Why is $x^2+y^2=1$ in $\mathbb{C}$ a sphere missing two points? and the relationship with solution in $\mathbb{R}$?

Question

The first question was asked and had an anwser here, but I can't understand why the cylinder is homeomorphic to a sphere missing two points.

Also, I was wondering where the two points are. Are they the two polar points of the sphere? And is the solution in $\mathbb{R}$ the equator?

Answer 1

Let $z=x+\mathrm{i}y$ and $w=x-\mathrm{i}y$—note that these linear relations are invertible.

Then $x^2+y^2=1$ iff $zw=1$, so $\{(x,y):\mathbb{C}^2|x^2+y^2=1\}$ and $\{(z,w):\mathbb{C}^2|zw=1\}$ are isomorphic over $\mathbb{C}$. The latter, in turn, is isomorphic over $\mathbb{C}$ to $\{z:\mathbb{C} | z \neq 0\}$.

Now we switch to working over $\mathbb{R}$. Since a complex number vanishes iff its real and imaginary parts do, $\{z:\mathbb{C}| z\neq 0\}$ becomes $\{(u,v):\mathbb{R}^2|(u,v)\neq 0 \}$ where $u$ and $v$ are the real and imaginary parts of $z$, respectively—a plane (over $\mathbb{R}$) punctured at the origin.

Using stereographic projection

$$\begin{align}(u,v)&=\frac{(u',v')}{1-w'}\\ (u',v',w')&=\frac{(2u,2v,u^2+v^2-1)}{u^2+v^2+1}\end{align}$$ we find that $\{(u,v):\mathbb{R}^2|(u,v)\neq 0 \}$ is isomorphic over $\mathbb{R}$ to $$\{ (u',v',w'):\mathbb{R}^3 | (u')^2 + (v')^2 + (w')^2=1\text{ and } (u',v',w')\neq (0,0,\pm 1) \}\text{.}$$

Finally, following the real solutions $(x,y)$ of $x^2+y^2=1$ to the end of the calculation, we successively find that $u=x$, $v=y$, $u'=u$, $v'=v$, and $w'=\frac{u^2+v^2-1}{u^2+v^2+1}=0$. Therefore the real solutions are indeed mapped to the equator of the sphere.

Answer 2

Think of the sphere $x^2+y^2+z^2=1$ and the cylinder $x^2+y^2=1.$ A ray from the origin through any point on the sphere other than a pole intersects the cylinder in exactly one point, and it's pretty clear that this is a homeomorphism.