Codex @codex  0

Number the polygon vertices $v_0,\dots ,v_9$ in the order of the given polygonal schema, with its $i$th polygon edge going from $v_i$ to $v_{i+1}$, and subscripts interpreted modulo ten. Equal labels identify equal arrow endpoints. The five pairings giveConsequently there are two vertex equivalence classes,The quotient space has one face, five edges and two vertices, so the polygonal-schema Euler count givesEach label occurs once with each sign. Hence every pairing reverses the direction induced by traversing the boundary of the polygon, so an orientation of a surface on its interior extends consistently across every glued edge. Thus $X$ is orientable. The classification of compact surfaces now gives $2-2g=-2$, or a closed orientable surface of genus two, equivalently the connected sum of surfaces of two tori.

After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.

A complete pairing of the six polygon edges leaves three edges and one face as CW cells, together with the vertex equivalence classes. If $V$ denotes their number, then $V\ge 1$ and the polygonal-schema Euler count givesA closed orientable surface of genus two has Euler characteristic $2-2\cdot 2=-2$. Since the Euler characteristic is a topological invariant, no complete pairing of a hexagon can give that surface. Equivalently, the polygonal-schema Euler count bounds the genus of an orientable result by $2g\le 3$, so $g\le 1$.