= Solution
{c}
The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a <circle>; the apex becomes its <cone> point. Thus this <quotient> triangle is a \b[<disc>], whose boundary is the doubly marked edge.
In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa b^{-1}$, where $b$ is the remaining doubly marked edge. This is a \b[<Möbius band>] with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.
The double markings glue the <disc> boundary to the boundary of that <Möbius band>. Capping a <Möbius band> by a <disc> gives
$$
\boxed{\mathbb{RP}^2,\text{ the real projective plane}.}
$$
This description also verifies that the <quotient> is a <closed surface>, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The <Euler characteristic> is $2-3+2=1$, consistent with the <projective plane>. The <Möbius band> exhibits nonorientability directly.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2006-options-projective-plane.png]
{title=The identified triangles become a <disc> and a <Möbius band> joined along their boundary}
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