Let \(X\) be the 2-sphere, and \(A \subseteq X\) the equatorial circle in \(X\). Show that there is no retraction \(r : X \to A\).
Solution
Fix a basepoint \(x_0\in A\) and let \(i\colon A\hookrightarrow S^2\) be the inclusion. Suppose \(r\colon S^2\to A\) is a retraction.
1. \(r_*\circ i_*=\operatorname{id}_{\pi_1(A,x_0)}\).
A retraction onto \(A\) restricts to the identity on \(A\), so \(r\circ i=\operatorname{id}_A\) and \(r(x_0)=x_0\). By functoriality of \(\pi_1\), \(r_*\circ i_*=(r\circ i)_*=\operatorname{id}_{\pi_1(A,x_0)}\).
2. \(i_*\colon\pi_1(A,x_0)\to\pi_1(S^2,x_0)\) is the zero homomorphism.
Its codomain \(\pi_1(S^2,x_0)\) is trivial.
3. Q.E.D.
By step 2, \(r_*\circ i_*=0\). By step 1, the same composite is the identity of \(\pi_1(A,x_0)\cong\pi_1(S^1)\cong\ZZ\), which is nontrivial. This contradiction shows that no retraction \(S^2\to A\) exists.