Let \(M_h' \subset M_g\) be a compact subsurface of genus \(h\) with one boundary circle, so \(M_h'\) is homeomorphic to \(M_h\) with an open disk removed. Show there is no retraction \(M_g \to M_h'\) if \(h > g/2\).
Solution
Suppose \[ r:M_g\longrightarrow M_h' \] is a retraction, and let \[ i:M_h'\hookrightarrow M_g \] be the inclusion. Since \(r\circ i=\operatorname{id}\), on cohomology over a field \(F\) one has \[ i^*r^*=\operatorname{id}_{H^1(M_h';F)}. \] Hence \[ r^*:H^1(M_h';F)\hookrightarrow H^1(M_g;F) \] is injective.
Now \(M_h'\) is a compact genus-\(h\) surface with one boundary component, so \[ \dim_F H^1(M_h';F)=2h, \qquad H^2(M_h';F)=0. \] Therefore for all \(u,v\in H^1(M_h';F)\), \[ r^*u\smile r^*v=r^*(u\smile v)=0. \] Thus the \(2h\)-dimensional subspace \[ V=r^*H^1(M_h';F)\subset H^1(M_g;F) \] is isotropic for the cup-product pairing \[ H^1(M_g;F)\times H^1(M_g;F)\longrightarrow H^2(M_g;F)\cong F. \] This pairing is nonsingular and skew-symmetric. Hence an isotropic subspace has dimension at most half of \[ \dim_F H^1(M_g;F)=2g. \] Consequently \[ 2h\le g. \] In particular, if \(h>g/2\), no retraction \(M_g\to M_h'\) can exist.