Let \(Z\) be a topological space. If \(Y\) is a subspace of \(Z\), we say that \(Y\) is a retract of \(Z\) if there is a continuous map \(r: Z \to Y\) such that \(r(y) = y\) for each \(y \in Y\).
Show that if \(Z\) is Hausdorff and \(Y\) is a retract of \(Z\), then \(Y\) is closed in \(Z\).
Let \(A\) be a two-point set in \(\mathbb{R}^2\). Show that \(A\) is not a retract of \(\mathbb{R}^2\).
Let \(S^1\) be the unit circle in \(\mathbb{R}^2\); show that \(S^1\) is a retract of \(\mathbb{R}^2 - \ts{\mathbf{0}}\), where \(\mathbf{0}\) is the origin. Can you conjecture whether or not \(S^1\) is a retract of \(\mathbb{R}^2\)?
Solution
Let \(i:Y\hookrightarrow Z\) be inclusion and \(r:Z\to Y\) a retraction. Then \[ R=i\circ r:Z\to Z \] is continuous, and \[ Y=\{z\in Z:R(z)=z\}. \] Since \(Z\) is Hausdorff, the equalizer of \(R\) and the identity map is closed. Hence \(Y\) is closed in \(Z\).
If a two-point set \(A\subset\mathbb R^2\) were a retract, the retraction \(r:\mathbb R^2\to A\) would be a continuous surjection. But \(\mathbb R^2\) is connected, so its continuous image \(A\) would be connected, whereas a two-point subspace of the Hausdorff plane is disconnected. Contradiction.
The radial map \[ r:\mathbb R^2\setminus\{0\}\to S^1, \qquad r(x)=\frac{x}{\|x\|}, \] is continuous and restricts to the identity on \(S^1\), so \(S^1\) is a retract of the punctured plane.
One should conjecture that \(S^1\) is not a retract of the whole plane. This is true; it is the classical no-retraction theorem for the disk/circle, proved later by algebraic-topological methods.