= Solution
{c}
\b[True.] Label the <pentagon> <polygon vertices> clockwise from the upper left as
$$
v_0=(0,2),\quad v_1=(2,2),\quad v_2=(2,0),\quad v_3=(1,0),\quad v_4=(0,1).
$$
Use $a$ for the upward single arrows and $b$ for the rightward double arrows in the original diagram. Their endpoint identifications are
$$
\begin{aligned}
a:&\quad v_4\sim v_2,\quad v_0\sim v_1,\\
b:&\quad v_0\sim v_3,\quad v_1\sim v_2.
\end{aligned}
$$
All five <polygon vertices> are therefore identified. The remaining diagonal <polygon edge>, from $v_3$ to $v_4$, has its endpoints identified and is the only <boundary component of a surface>.
The paired <polygon edges> join half-<disc> <neighbourhoods> into <disc> <neighbourhoods>. At the common vertex, the five corner sectors join in one chain whose two free ends lie on the unpaired <polygon edge>: its <vertex link in a polygonal schema> is a <closed interval>. Hence that point has a half-<disc> <neighbourhood>. This verifies that the <quotient space> is a <surface with boundary>, rather than merely counting its <CW cells>. The paired <polygon edges> are oppositely traversed around the <polygon>, so this <surface> is <orientable>.
Its <CW cells> comprise one vertex, three edges (the $a$ pair, the $b$ pair, and the free <polygon edge>), and one face. Therefore
$$
\chi=1-3+1=-1,\qquad b=1,\qquad 2-2g-b=-1\Longrightarrow g=1.
$$
The <classification of compact surfaces> identifies it as \b[a <torus> with an <open disc> removed].
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