- Show that the logarithmic spiral
\[ C = \ts{0 \times 0} \cup \ts{e^t \cos t \times e^t \sin t \mid t \in \mathbb{R}} \]
is a retract of \(\mathbb{R}^2\). Can you define a specific retraction \(r: \mathbb{R}^2 \to C\)?
- Show that the “knotted x-axis” \(K\) pictured in Figure 35.2 of the text is a retract of \(\mathbb{R}^3\).
Solution
For \(z=(x,y)\ne0\), let \[ \rho=\sqrt{x^2+y^2}. \] Define \[ r(x,y)=\rho\bigl(\cos(\log\rho),\sin(\log\rho)\bigr), \] and set \(r(0,0)=(0,0)\). The image lies in the logarithmic spiral \(C\): if \(t=\log\rho\), then \(ho=e^t\). The map is continuous away from the origin, and at the origin \[ \|r(x,y)\|=\rho\to0, \] so it is continuous there as well. If \((x,y)=e^t(\cos t,\sin t)\in C\setminus\{0\}\), then \(ho=e^t\) and \(\log\rho=t\), hence \(r(x,y)=(x,y)\). Thus \(r\) is a retraction of \(\mathbb R^2\) onto \(C\).
The knotted \(x\)-axis \(K\) in Figure 35.2 is a closed subspace of \(\mathbb R^3\) homeomorphic to \(\mathbb R\). Let \[ h:K\to\mathbb R \] be a homeomorphism. Since \(\mathbb R^3\) is normal and \(K\) is closed, the Tietze theorem extends \(h\) to a continuous function \[ H:\mathbb R^3\to\mathbb R. \] Then \[ r=h^{-1}\circ H:\mathbb R^3\to K \] is continuous and, for \(k\in K\), satisfies \(r(k)=h^{-1}(h(k))=k\). Hence \(K\) is a retract of \(\mathbb R^3\).