What surface is represented by the \(6\dash\)gon with edges identified according to the symbol \(xyzx\inverseof{y} \inverseof{z}\) ?
Solution
1. Label the six polygon vertices cyclically \(v_0,\dots,v_5\).The edge pairings identify all six vertices.
The two \(x\)-edges have the same orientation, giving \(v_0\sim v_3\) and \(v_1\sim v_4\). The \(y,y^{-1}\) pairing gives \(v_1\sim v_5\) and \(v_2\sim v_4\). The \(z,z^{-1}\) pairing gives \(v_2\sim v_0\) and \(v_3\sim v_5\). These relations put every vertex in one equivalence class.
2. The quotient therefore has one vertex, three edges, and one \(2\)-cell, so\[\chi=1-3+1=-1.\]
There is one quotient edge for each of the labels \(x,y,z\).
3. The quotient surface is nonorientable.
The label \(x\) occurs twice with the same boundary orientation. In an orientable polygon presentation, each edge label must occur once in each orientation; a same-direction pairing produces a crosscap.
4. A closed connected nonorientable surface \(N_h=\#^h\mathbb{RP}^2\) has Euler characteristic \(2-h\). Hence \(2-h=-1\), so \(h=3\).
Apply the classification theorem for closed connected surfaces.
5. Thus the polygon represents\[\boxed{\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2.}\]
This is the unique closed connected nonorientable surface of Euler characteristic \(-1\).