Show that if the closed orientable surface \(M_g\) of genus \(g\) is a covering space of \(M_h\), then \(g = n(h-1)+1\) for some \(n\), namely, \(n\) is the number of sheets in the covering. [Conversely, if \(g = n(h-1)+1\) then there is an \(n\) sheeted covering \(M_g \to M_h\), as we saw in Example 1.41.]
Solution
For a closed orientable surface \(M_g\) of genus \(g\), \[ \chi(M_g)=2-2g. \] Suppose \[ p:M_g\to M_h \] is an \(n\)-sheeted covering. By Exercise 22, \[ \chi(M_g)=n\chi(M_h). \] Hence \[ 2-2g=n(2-2h). \] Dividing by \(-2\) gives \[ g-1=n(h-1), \] so \[ \boxed{g=n(h-1)+1}. \] The integer \(n\) is exactly the number of sheets of the covering.