In the surface \(M_g\) of genus \(g\), let \(C\) be a circle that separates \(M_g\) into two compact subsurfaces \(M'_h\) and \(M'_k\) obtained from the closed surfaces \(M_h\) and \(M_k\) by deleting an open disk from each. Show that \(M'_h\) does not retract onto its boundary circle \(C\), and hence \(M_g\) does not retract onto \(C\). [Hint: abelianize \(\pi_1\).] But show that \(M_g\) does retract onto the nonseparating circle \(C'\) in the figure.
Solution
1. For \(h\ge1\), the surface \(M'_h\) has fundamental group\[ \pi_1(M'_h)\cong F(a_1,b_1,\dots,a_h,b_h), \]and its boundary loop \(C\) represents\[ [a_1,b_1]\cdots[a_h,b_h]. \]
A genus-\(h\) orientable surface with one boundary component has the standard CW description with one vertex and \(2h\) one-cells \[ a_1,b_1,\dots,a_h,b_h, \] but with the interior of the usual \(2\)-cell removed. Hence its fundamental group is free on these \(2h\) generators. The boundary of the missing \(2\)-cell is the usual surface attaching word \[ [a_1,b_1]\cdots[a_h,b_h], \] so this word represents the boundary circle \(C\).
2. There is no retraction\[ r:M'_h\to C. \]
Suppose such a retraction existed, and let \[ i:C\hookrightarrow M'_h \] be the inclusion. Then \[ r_*\circ i_*=\operatorname{id}_{\pi_1(C)}. \] Identify \[ \pi_1(C)\cong\mathbb Z. \]
For \(h\ge1\), 1 says that \(i_*(1)\) is a product of commutators. Every homomorphism from a group to the abelian group \(\mathbb Z\) kills every commutator, so \[ r_*i_*(1)=0, \] contradicting \[ r_*i_*(1)=1. \]
If \(h=0\), then \(M'_0=D^2\), so \(i_*:\pi_1(S^1)\to\pi_1(D^2)\) is the zero map and cannot possess a left inverse either. Thus no retraction exists for any \(h\ge0\).
3. The closed surface \(M_g\) cannot retract onto the separating circle \(C\).
If \[ r:M_g\to C \] were a retraction, then restricting \(r\) to either subsurface \(M'_h\subset M_g\) would give a retraction \[ r|_{M'_h}:M'_h\to C, \] because \(C\subset M'_h\) and \(r\) is the identity on \(C\). This contradicts 2.
4. A nonseparating simple closed curve \(C'\subset M_g\) can be carried by a homeomorphism of \(M_g\) to the standard loop \(a_1\) in the usual one-vertex CW structure on \(M_g\).
Cutting an orientable surface along a nonseparating simple closed curve produces a connected orientable surface of genus \(g-1\) with two boundary components. The same is true for the standard generator circle \(a_1\). By the classification of compact orientable surfaces with boundary, the two cut surfaces are homeomorphic by a homeomorphism matching their two boundary circles. Regluing gives a homeomorphism of \(M_g\) carrying \(C'\) to \(a_1\). Thus it suffices to construct a retraction onto \(a_1\).
5. Define a map on the \(1\)-skeleton\[ q:M_g^1\to a_1\cong S^1 \]by mapping the edge \(a_1\) identically to itself and collapsing every other \(1\)-cell to the common vertex.
The standard \(1\)-skeleton is a wedge of the \(2g\) circles \[ a_1,b_1,\dots,a_g,b_g. \] A map from this wedge is determined continuously by its restrictions to the summand circles, and the stated restrictions agree at the wedge point.
6. The map \(q\) extends across the \(2\)-cell of \(M_g\).
The attaching map of the \(2\)-cell represents \[ [a_1,b_1]\cdots[a_g,b_g]. \] Under \(q_*\), each \(b_i\) and every \(a_i\) for \(i>1\) maps to the identity in \[ \pi_1(S^1)\cong\mathbb Z, \] while \(a_1\) maps to the generator. Therefore every commutator maps to the identity, so the entire attaching loop maps trivially in \(\pi_1(S^1)\). Hence its composite with \(q\) is null-homotopic and \(q\) extends over the \(2\)-cell.
7. The extension\[ r:M_g\to a_1 \]is a retraction, and hence \(M_g\) retracts onto \(C'\).
By construction, \(r\) restricts to the identity on the edge \(a_1\). Thus \(r\) is a retraction onto \(a_1\). Conjugating this retraction by the homeomorphism from 4 gives a retraction of \(M_g\) onto the original nonseparating curve \(C'\).