Is there a simple bijective function between the surface of the unit sphere in $\mathbb{R}^3$ and the real plane? I am aware of stereographic projection and Riemann sphere but these seem to map the north pole to infinity.
[Math] Bijection between the plane and the surface of a sphere
general-topology
Related Solutions
For the Riemann sphere, it is the one point compactification of the plane. So yes, rather by definition it is compact.
Generally, a non-compact topological space can be compactified by suitable addition of "points at infinity". Particularly useful ones (besides the one-point compactification above) include the Stone-Cech compactification, which in some sense adds the "most points", and whose existence for any topological space follows from axiom of choice.
Another compactification that often arises is the Penrose/conformal compactification of space-time in general relativity.
A propos your question on hyperbolic plane. Topologically the hyperbolic plane is the same as the normal plane. By considering the Poincare disk model and "adding" the boundary of the disk, you compactify the hyperbolic plane. As mentioned above, any topological space admits some compactification. What's more useful is to consider special compactifications: for example the one-point compactification of the complex plane into the Riemann sphere is conformal relative to the additional geometrical structure on the two surfaces. Similarly the Penrose compactifications respect the geometrical structure on space-times.
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.
Best Answer
Start from the stereographic bijection $B$ between the sphere minus its North pole $p$ and the plane. One needs a point of the plane to be the image of $p$ hence we will make some room for it. Choose any injective sequence $(x_k)_{k\ge0}$ of points in the plane and define a shift $S$ on the plane by $S(x_k)=x_{k+1}$ for every $k\ge0$ and $S(x)=x$ for every other point $x$ of the plane. Then $S\circ B$ is a bijection between the sphere minus $p$ and the plane minus $x_0$. Extend it to a bijection between the sphere and the plane by sending $p$ to $x_0$. You are done.