![]() See excellent answer concerning the map $\pi$ and why vertical projection will not work. ![]() $\pi$ necessarily will have to map chords to great circles. The map $\pi$ does not need to preserve angles or surface areas. Moreover if the map $\pi$ is bijective then any solution to Felsner's conjecture will also solve mine. In the sample above, this person has similar Graphs I and II and this is reflected in Graph III. In that case, Graph III serves to confirm the information in Graphs I and II. You are then using behaviors that are not as comfortable or natural for you. Then I could solve the conjecture by Felsner and his colleagues. Style (Graph II), this may cause stress if done over a long period of time. ![]() Glue $H$ to a copy of itself at the equator.In addition they conjecture such an arrangement is $3-$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. I have conjectured that $G$ is $3-$colourable and $3-$choosable. Thomas and his work on Plummer's conjecture. That $G$ is Hamiltonian-connected follows from R. I know that $G$ has $n(n+3)/2, n(n+2)$ and $(n^2+n+4)/2$ vertices, edges and faces respectively. 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. 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. 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 ? Question: Can a disc drawn in the Euclidean plane be mapped to the surface of a hemisphere in Euclidean space ?
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