Open unit disc is homeomorphic to sphere minus a point

Question

How to show that sphere minus a point is homeomorphic to open unit disc in $\mathcal R^2$.By using this i will be able to show that sphere minus a point can be thought as a hyperbolic surface . I know that open unit disc is homeomorphic to complex plane which is inturn homeomorphic to sphere minus a point by stereographic projection and composition of these two will work. I am looking for a direct map between open unit disc and sphere minus a point. i have no knowledge of algebraic topology so a simple answer without using fancy terms will be appriciated.

Answer 1

Can you just send the south pole to the center, leave longitude alone, and map latitude to radius via a linear $(-90^\circ,90^\circ)\to (0,1)$?

Also, if you already know how to send the sphere to the plane, and the plane to the disc, then the composition of those two maps will be a direct map doing what you want.

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If the geography terminology isn't comfortable, let's try some more explicit algebra. For the open disc, take the set of points $D=\{x,y:x^2+y^2<1\}$. For the sphere with one point missing, use $S=\{(x,y,z):x^2+y^2+z^2=1, z\neq1\}$. We construct a map $f:S\to D$.

Let $f(0,0,-1)=(0,0)$, and for $z>-1$, take $f(x,y,z)=(xt,yt)$, where $t=\frac{z+1}{2\sqrt{x^2+y^2}}$.

What does this map do? It sends the bottom of the sphere to the center of the disc: $(0,0)$. Points near the bottom of the sphere have $z$ close to $-1$, so they get sent to points close to $(0,0)$. Points near the missing point on top get sent very close to the edge of the disc. The direction of a point from the origin, i.e., the $\theta$ of polar coordinates, is preserved.

To see that it is one-to-one, and onto, we can write down its inverse: $f^{-1}(x,y)=\left(\frac{2x\sqrt{1-u^2}}{u+1},\frac{2y\sqrt{1-u^2}}{u+1},u\right)$, where $u=-1+2\sqrt{x^2+y^2}$. You can verify that these functions are inverses by calculating $f(f^{-1}(x,y))$ and $f^{-1}(f(x,y,z))$. You can also verify that the ranges of $f$ and $f^{-1}$ really are in $D$ and $S$, respectively.

Does this help?

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Edit: To answer the question, How did I cook up this map?

First, I visualized a sphere sitting on top of a disc, and I imagined it opening up at the top and becoming flat. There's no need for any rotation in this scenario, so the $(x,y)$ coordinates of the point just need to maybe move in or out from the origin, but stay pointing in the same $xy$-direction.

In order to keep the bottom of the sphere in the center of the circle and get the top of the sphere out to the circumference, I thought, why not just map $z$ to $r$. Since the $z$-coordinates all come from $[-1,1)$, and we want $r$-values on $[0,1)$, I thought of the linear map $r=\frac{z+1}{2}$. If we start with the point $(x,y,z)$, its purely $xy$-distance from the origin is $\sqrt{x^2+y^2}$, so if we want it to be at distance $r$ instead, we need to multiply both $x$ and $y$ by the factor $\frac{r}{\sqrt{x^2+y^2}}$.

That's how I cooked up $f$, and to find the formula for $f^{-1}$, I just worked out how to reverse the above process. That part was more algebraic, and less about visualizing a melting sphere.