Prove that if \(p: M_g \to M_h\) is an \(n\)-sheeted covering space between connected, closed, orientable surfaces of genus \(g\) and \(h\), then: \[g = n(h - 1) + 1.\]
Solution
1. For a closed orientable surface of genus \(r\),\[ \chi(M_r)=2-2r. \]
2. Euler characteristic multiplies by the number of sheets in a finite covering:\[ \chi(M_g)=n\,\chi(M_h). \]
Give \(M_h\) a finite CW structure. Each open cell is evenly covered, and an \(n\)-sheeted covering has exactly \(n\) lifts of each cell. Thus every cell count is multiplied by \(n\), hence so is the alternating sum defining Euler characteristic.
3. Therefore\[ 2-2g=n(2-2h). \]Dividing by \(2\) and rearranging gives\[ g-1=n(h-1), \]so\[ \boxed{g=n(h-1)+1}. \]