= Solution
Number the <polygon vertices> $v_0,\ldots,v_9$ 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 give
$$
\begin{array}{c|cc}
\text{label}&\text{first endpoints}&\text{second endpoints}\\
a_1&v_0\sim v_4&v_1\sim v_3\\
a_2&v_2\sim v_7&v_3\sim v_6\\
a_3&v_5\sim v_9&v_6\sim v_8\\
a_4&v_2\sim v_9&v_1\sim v_0\\
a_5&v_5\sim v_7&v_4\sim v_8
\end{array}
$$
Consequently there are two vertex <equivalence classes>,
$$
\{v_0,v_1,v_3,v_4,v_6,v_8\},\qquad\{v_2,v_5,v_7,v_9\}.
$$
The <quotient space> has one face, five edges and two vertices, so the <polygonal-schema Euler count> gives
$$
\boxed{\chi(X)=2-5+1=-2.}
$$
Each 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 \b[a closed <orientable surface> of <genus> two], equivalently the <connected sum of surfaces> of two <tori>.
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