= Solution
{c}
Label the <hexagon> <polygon vertices> $v_0,\ldots,v_5$ in the direction in which its boundary word is $abca^{-1}b^{-1}c^{-1}$, with the first side running from $v_0$ to $v_1$. Matching the endpoints according to the arrows gives
$$
\begin{aligned}
a:&\quad v_0\sim v_4,\quad v_1\sim v_3,\\
b:&\quad v_1\sim v_5,\quad v_2\sim v_4,\\
c:&\quad v_2\sim v_0,\quad v_3\sim v_5.
\end{aligned}
$$
Thus the two vertex <equivalence classes> are $\{v_0,v_2,v_4\}$ and $\{v_1,v_3,v_5\}$.
The <polygonal schema> gives a <compact> <connected> <surface> with empty <manifold boundary>. Interior points have <disc> <neighbourhoods>; a paired <polygon edge> glues two half-<disc> <neighbourhoods> into a <disc>. At each of the two vertex <equivalence classes>, its three corner sectors join cyclically, so the <vertex link in a polygonal schema> is a <circle>, again giving a <disc> <neighbourhood>. Pairing every label once positively and once negatively makes these gluings consistent with an <orientation>, so the resulting <surface> is <orientable>.
The resulting <CW complex> has <CW cells> comprising two vertices, three edges and one face. Its <Euler characteristic> is
$$
\chi=V-E+F=2-3+1=0.
$$
By the <classification of compact surfaces>, a <connected> <closed orientable surface> with $\chi=0$ has <genus> $g$ satisfying $2-2g=0$, hence $g=1$. \b[The <quotient space> is a <torus>.]
Back to article page
