Relationship between regular tessellations in general space Using the notation for regular tessellations $\{p,q\}$ denoting the tessellation consisting of p-gons , q of which meet at each vertex [e.g. $\{3,6\}$ in $\mathbb{R}^{2}$ is the equilateral triangle tessellation.]
My conjecture is that $\{p,q\}$ is a regular tessellation iff $\{q,p\}$ is a regular tessellation (both in the same space). (It is certainly true in $\mathbb{R}^{2}$) But I want to generalise to any two dimensional space i.e Spherical and Hyperbolic.
Any hints on how to prove this or a counter example would be very helpful.
Thank you
 A: This is true, and it is proved using the concept of the "dual tesselation", which works in each of the geometries you mention.
Suppose you have a regular tesselation by regular $p$-gons indexed as $\{T_i\}$ with $q$ meeting at each vertex. For each $p$-gon tile $T_i$ let $C_i$ be its center. If $T_i,T_j$ meet along a common edge, let $E_{ij}$ be the segment connecting $C_i$ to $C_j$, crossing the common edge of $T_i,T_j$ orthogonally at the midpoint. Given a vertex $V_j$ of the tesselation, enumerate the tiles meeting at $V$ in cyclic order as $T_1,…,T_q$. Let $S_j$ be the regular $q$-gon whose sides are $E_{12} E_{23} … E_{q-1,q} E_{q1}$. Then the $S_j$'s form a tesselation by regular $q$-gons of which $p$ meet at each vertex.
Try it out with some pictures, for example: the cube defines a $\{4,3\}$ tesselation of the 2-sphere, and its dual $\{3,4\}$ tesselation is defined by an octahedron. Also, a dodecahedron defines a $\{5,3\}$ tesselation of the 2-sphere, and its dual $\{3,5\}$ tessellation is defined by an icosahedron.
