Mapping The Unit Disc To The Hemisphere? 
Question: Can a disc drawn in the Euclidean plane be mapped to the surface of a hemisphere in Euclidean space ?  


If $U$ is the unit disc drawn in the Euclidean plane is there a map, $\pi$, which sends the points of $U$ to the surface of a hemisphere, $H,$ in Euclidean space ? 
Background and Motivation:
If $U$ is the unit disc centered at the origin consider $n$ chords drawn through the interior of $U$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $G$ induced by the discs and the chords has a vertex for each intersection point in the interior of $U$ and $2$ vertices for each chord incident to the boundary of $U.$ Naturally $G$ has an edge for each arc directly connecting two intersection points. $G$ is planar and $3-$connected. I know that $G$ has $n(n+3)/2, n(n+2)$ and $(n^2+n+4)/2$ vertices, edges and faces respectively. That $G$ is Hamiltonian-connected follows from R. Thomas and his work on Plummer's conjecture. I have conjectured that $G$ is $3-$colourable and $3-$choosable. 
Independently Felsner, Hurtado, Noy and Streinu :Hamiltonicity and Colorings of Arrangement Graphs, ask if the arrangement graph of great circles on the sphere is $3-$colourable. In addition they conjecture such an arrangement is $3-$choosable. 
Now I began to think the following


*

*Show my graph $G$ is $3-$colourable

*$\pi:G \to H$

*Glue $H$ to a copy of itself at the equator


then I could solve the conjecture by Felsner and his colleagues. Moreover if the map $\pi$ is bijective then any solution to Felsner's conjecture will also solve mine. The map $\pi$ does not need to preserve angles or surface areas. $\pi$ necessarily will have to map chords to great circles. See @JohnHughes excellent answer concerning the map $\pi$ and why vertical projection will not work. 
 
 A: I think you mean to say disk instead of circle.
If you mean disk then you can do it.
Just imagine of punching the disk from center, it will give you a hemisphere.
And if you want precisely the map then you can get it from Stereographic projection.
EDIT
It will look something like this when you punch a disk.

A: If by a "circle" you mean the set of all points inside a circle (e.g., points whose distance from some center $C$ is less than or equal to 1), then the answer is "yes" and one solution is called "stereographic projection;" another is "vertical projection". 
If you have a point $(x, y)$ in the unit disk (the "filled in circle"), the corresponding point, using vertical projection, is $(x, y, \sqrt{1 - x^2 - y^2})$. 
For stereographic projection, you send the point $P = (x, y)$ to a new point $Q$:
\begin{align}
h &= x^2 + y^2 \\
Q &= (\frac{x}{h+1}, \frac{y}{h+1}, \frac{h-1}{h+1})
\end{align}
The latter has the charm that it takes chords in your disk to great-circle arcs in the hemisphere, and preserves angles of intersection, although not lengths. 
A: Projecting the hemisphere onto the circle bounded by the equator from a point infinitely far above the North pole maps the hemisphere bijectively to the circle. Take the inverse of this map.
