 Solution

False: the result is again a torus with one open disc removed. Label the hexagon polygon vertices clockwise from the upper left asWith the same arrow labels as in part (ii),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 $v_1{v}_2$ and $v_4{v}_5$. Each joins $P$ to $Q$, so together they form 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, givingIn 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.