Prove that the Euler characteristic of a compact surface with boundary which has \(k\) boundary components is less than or equal to \(2 - k\).
Solution
1. Use the compact-surface convention that a surface is connected, and let \(S\) be the given compact surface with boundary components\[ C_1,\dots,C_k. \]Cap each boundary component by a disk to obtain a closed connected surface\[ \widehat S = S\cup_{C_1}D_1^2\cup\cdots\cup_{C_k}D_k^2. \]
Each boundary component of a compact surface is a circle. Gluing a copy of \(D^2\) to each boundary circle removes all boundary components and does not disconnect the surface, so \(\widehat S\) is a closed connected compact surface.
The connectedness convention matters here: if disconnected \(2\)-manifolds were allowed under the word ``surface,’’ the stated inequality would not hold in general; for example, the disjoint union of two disks has \(k=2\) and Euler characteristic \(2\).
2. Capping the \(k\) boundary circles increases Euler characteristic by \(k\):\[ \chi(\widehat S)=\chi(S)+k. \]
Compact surfaces admit finite CW structures for which the boundary circles are subcomplexes. For finite CW complexes, Euler characteristic satisfies \[ \chi(A\cup B) = \chi(A)+\chi(B)-\chi(A\cap B) \] when the intersection is a subcomplex.
Attaching one capping disk gives \[ \chi(S\cup_{S^1}D^2) = \chi(S)+\chi(D^2)-\chi(S^1) = \chi(S)+1-0 = \chi(S)+1. \] Applying this successively to the \(k\) boundary components yields \[ \chi(\widehat S)=\chi(S)+k. \]
3. Every closed connected compact surface has Euler characteristic at most \(2\).
By the classification theorem for compact surfaces, a closed connected surface is either
orientable of genus \(g\ge0\), with \[ \chi=2-2g\le2, \] or
nonorientable of genus \(h\ge1\), with \[ \chi=2-h\le1<2. \]
Therefore \[ \chi(\widehat S)\le2. \]
4. Hence\[ \boxed{\chi(S)\le2-k}. \]
By 2, \[ \chi(S)=\chi(\widehat S)-k. \] Using 3 gives \[ \chi(S) \le 2-k, \] as required.