Prove that \(\mathbb R^2\) cannot be retracted to \(S^1\).
Solution
Fix a basepoint \(x_0\in S^1\subset\mathbb R^2\), and let \[ i:S^1\hookrightarrow\mathbb R^2 \] denote inclusion.
1. If a retraction\[ r:\mathbb R^2\longrightarrow S^1 \]existed, then\[ r\circ i=\id_{S^1}. \]
By definition, a retraction onto the subspace \(S^1\) restricts to the identity on \(S^1\). Thus for every \(z\in S^1\), \[ (r\circ i)(z)=r(z)=z. \] Since \(r(x_0)=x_0\), both \(i\) and \(r\) are based maps at \(x_0\).
2. Functoriality would therefore give\[ r_*\circ i_*=\id_{\pi_1(S^1,x_0)}. \]
Applying the fundamental-group functor to the identity in 1 gives \[ r_*\circ i_* =(r\circ i)_* =(\id_{S^1})_* =\id_{\pi_1(S^1,x_0)}. \]
3. On the other hand, the homomorphism\[ i_*:\pi_1(S^1,x_0)\longrightarrow\pi_1(\mathbb R^2,x_0) \]is the zero homomorphism.
The plane is contractible, hence simply connected, so \[ \pi_1(\mathbb R^2,x_0)=0. \] Therefore every homomorphism into this group, in particular \(i_*\), is zero.
4. This contradicts 2, so no retraction \(\mathbb R^2\to S^1\) exists.
By 3, \[ r_*\circ i_*=0. \] But 2 says the same composite is the identity on \[ \pi_1(S^1,x_0)\cong\mathbb Z, \] which is nontrivial. Hence the assumed retraction cannot exist.