Nonorientable surface of genus \(g\) Obtain by removing \(g\) discs from \(S^2\) and attaching \(g\) mobius strips
Solution
1. Removing the interiors of \(g\) disjoint disks from \(S^2\) produces an orientable genus-zero surface \(S\) with \(g\) boundary circles.
Deleting one open disk gives a disk; each additional deletion adds one boundary component without adding a handle.
2. Attach a Möbius band to each boundary circle of \(S\). The resulting surface is closed and connected.
Each Möbius band has one boundary circle, so the \(g\) gluings eliminate all boundary components. Connectivity is preserved because every band is attached to the connected surface \(S\).
3. Each attached Möbius band contributes one crosscap, so the resulting surface is the connected sum\[ \boxed{N_g=\#^g\mathbb{RP}^2}. \]
A projective plane with an open disk removed is a Möbius band. The connected-sum construction removes a disk from each summand and glues boundary circles. Thus attaching a Möbius band to a boundary component is exactly adding one \(\mathbb{RP}^2\) summand.
4. Its Euler characteristic is\[ \chi(N_g)=2-g, \]and for \(g\ge1\) it is nonorientable.
The punctured sphere has Euler characteristic \(2-g\), each Möbius band has Euler characteristic \(0\), and gluing along circles (Euler characteristic \(0\)) does not change the sum. The core circle of any attached Möbius band has a nonorientable neighborhood, so the resulting surface is nonorientable.