Let \(X\) be a torus minus a hole. Explain why \(X\) is homotopy-equivalent to \(S^1\vee S^1\). Use the fundamental group to show that there does not exist a retraction \(r\colon X\to\partial X\).
Solution
1. A torus with an open disk removed deformation-retracts onto a wedge of two circles.
Use the usual square model of the torus and remove a small open disk around the unique vertex of the quotient. The remaining punctured torus has a standard spine consisting of the two generating loops \(a\) and \(b\) meeting at one point. Collapsing the complementary collar onto this spine gives a deformation retraction \[ X\simeq S^1\vee S^1. \] Thus \(\pi_1(X)\cong F(a,b)\).
2. Under this identification, the boundary loop of the puncture represents the commutator\[ [a,b]=aba^{-1}b^{-1} \]in \(\pi_1(X)\).
The boundary word of the standard fundamental polygon for the torus is \(aba^{-1}b^{-1}\). Removing a disk about the vertex turns this boundary word into the boundary component of the punctured torus.
3. There is no retraction \(r:X\to\partial X\).
Suppose such a retraction existed, and let \(i:\partial X\hookrightarrow X\) be the inclusion. Then \[ r_*\circ i_*=\operatorname{id}_{\pi_1(\partial X)}. \] Since \(\pi_1(\partial X)\cong\mathbb Z\) is abelian, every homomorphism \[ r_*:F(a,b)\to\mathbb Z \] kills the commutator subgroup, hence \(r_*([a,b])=0\). But by 2, \(i_*\) sends a generator of \(\pi_1(\partial X)\) to \([a,b]\), so \(r_*i_*\) sends that generator to \(0\), contradicting that it is the identity.