Past exam of the topology course of the University of Oxford / 2014 / 3 / b / i

Label the endpoints around the original polygon by $v_0,v_1,v_2,v_3$ so that the successive boundary arcs areThe two occurrences of $x$ have the same boundary-traversal direction, as prescribed by the word. We use cutting along a new diagonal followed by gluing the old $x$ sides. Cutting introduces two copies of the diagonal with opposite orientations, so rejoining them later exactly recovers the original quotient.

First move. Draw a diagonal $d$ from $v_0$ to $v_2$. The two pieces have boundary words $xA{d}^{-1}$ and $dxB$. Glue them along their $x$ sides, turning one piece over in the planar drawing to match the arrows. Reading the outside boundary of the joined disk givesReverse the orientation of this entire disk and change the starting point of its boundary word. The same side-paired disk then has word $dd{A}^{-1}B$. Renaming the new paired edge $d$ as $x$ yields

Second move. Instead cut along $d$ from $v_1$ to $v_3$. The pieces now have words $xdB$ and $Ax{d}^{-1}$. Gluing along the old $x$ sides gives outside word $dB{A}^{-1}d$. A cyclic change of starting point gives $A^{-1}ddB$, and renaming $d$ yieldsAll side identifications inside the strings $A$ and $B$ are carried along unchanged; reversing a piece reverses the order of its string and inverts each letter. The construction therefore gives homeomorphic quotients, not merely an equality of Euler characteristics. This is the crosscap extraction move for a polygonal word.