= Solution
{c}
\b[False: the result is again a <torus> with one <open disc> removed.] Label the <hexagon> <polygon vertices> clockwise from the upper left as
$$
v_0=(0,2),\quad v_1=(1,2),\quad v_2=(2,1),\quad v_3=(2,0),\quad v_4=(1,0),\quad v_5=(0,1).
$$
With the same arrow labels as in part (ii),
$$
\begin{aligned}
a:&\quad v_5\sim v_3,\quad v_0\sim v_2,\\
b:&\quad v_0\sim v_4,\quad v_1\sim v_3.
\end{aligned}
$$
There are two vertex <equivalence classes>, $P=\{v_0,v_2,v_4\}$ and $Q=\{v_1,v_3,v_5\}$. The free diagonal <polygon edges> are $v_1v_2$ and $v_4v_5$. Each joins $P$ to $Q$, so together they form \b[one <boundary component of a surface>, not two]. At each vertex <equivalence class> the three corner sectors join in a single chain between these two free <polygon edges>. Thus each <vertex link in a polygonal schema> is a <closed interval>, and the <quotient space> is an <orientable> <surface with boundary>.
Its <CW complex> has <CW cells> comprising two vertices, four edges and one face, giving
$$
\chi=2-4+1=-1,\qquad b=1,\qquad g=1.
$$
In contrast, a <torus> with two disjoint <open discs> removed has two <boundary components of a surface> and <Euler characteristic> $2-2-2=-2$. Either <topological invariant> rules out the proposed <homeomorphism>.
\Image[/past-exam-of-the-topology-course-of-the-university-of-oxford/2021-polygon-boundaries.png]
{title=After identifying equally labelled arrows, both polygons have one boundary circle; in the hexagon the two free edges join the two vertex classes}
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