Use the preceding problem to show that if the closed orientable surface \(M_g\) of genus \(g\) retracts onto a graph \(X \subset M_g\), then \(H_1(X)\) has rank at most \(g\). Deduce an alternative proof of Exercise 13 from this, and construct a retraction of \(M_g\) onto a wedge sum of \(k\) circles for each \(k \leq g\).
Solution
Suppose \(r:M_g\to X\) is a retraction onto a graph \(X\subset M_g\), with inclusion \(i:X\hookrightarrow M_g\). Then \[ r\circ i=\operatorname{id}_X, \] so on cohomology over a field \(F\), \[ i^*r^*=\operatorname{id}_{H^1(X;F)}. \] Hence \[ r^*:H^1(X;F)\hookrightarrow H^1(M_g;F) \] is injective.
Since \(X\) is a graph, \(H^2(X;F)=0\), so for all \(u,v\in H^1(X;F)\), \[ r^*u\smile r^*v=r^*(u\smile v)=0. \] Thus the image \(V=r^*H^1(X;F)\) is an isotropic subspace for the nonsingular skew-symmetric cup-product pairing on the \(2g\)-dimensional vector space \(H^1(M_g;F)\). By Exercise 28, \[ \dim_F V\le g. \] Taking \(F=\mathbb Q\) gives \[ \boxed{\operatorname{rank}H_1(X)\le g.} \]
For Exercise 13, a genus-\(h\) surface with one boundary component deformation retracts onto a wedge of \(2h\) circles. If there were a retraction \(M_g\to M_h'\), composing it with this spine retraction would give a retraction of \(M_g\) onto a graph of first-homology rank \(2h\). Hence \(2h\le g\), so no such retraction exists when \(h>g/2\).
Finally use the standard CW structure on \(M_g\) with one vertex, loops \[ a_1,b_1,\ldots,a_g,b_g, \] and one \(2\)-cell attached by \(\prod_i[a_i,b_i]\). For any \(k\le g\), let \(X_k\) be the wedge of the loops \(a_1,\ldots,a_k\). Define a cellular map \(M_g\to X_k\) that is the identity on these loops and sends every other \(1\)-cell to the vertex. The attaching word maps to the trivial loop, since each commutator has at least one factor sent to the vertex, so the map extends over the \(2\)-cell. It restricts to the identity on \(X_k\), hence is a retraction.