# Open unit disc is homeomorphic to sphere minus a point

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.

• The usual buzzphrase is "stereographic projection". Jul 6, 2017 at 21:24
• Compose the two homeomorphisms you mentioned. Jul 6, 2017 at 21:24
• i am looking for some other answer Jul 6, 2017 at 21:43
• If you're considering conformal structures (as suggested by the use of hyperbolic surface), the sphere minus one point is the plane, not a disk. The plane and disk are homeomorphic, as you say, but not conformally isomorphic. Jul 6, 2017 at 22:18
• When you write the formulas for the two homeomophisms you mentioned, and compose those formulas as suggested by @JoaoCaminada, you will get one formula for the homeomorphism you want. You can then pretend you didn't know where it came from, and it will be "a direct map". Generally speaking, just because a function is a composition of two other functions does not make it any less of a function. Jul 7, 2017 at 0:15

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.

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?

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.

• what is latitude and longitude. Jul 6, 2017 at 21:46
• Well, they're how we put coordinates on a sphere for geography. If that doesn't work, I'll edit for something else. Jul 6, 2017 at 21:53
• It would help if you tell us how you identify specific points on the sphere. Jul 6, 2017 at 21:54
• i visualize sphere sitting in $R^3$ Jul 6, 2017 at 22:10
• @Tony jacobs how did you get this map Jul 13, 2017 at 17:03