Let $X$ be a topological space. If $Y$ is a subspace of $X$, then $Y$ is a retract of $X$ if there exists a continuous function $r:X \rightarrow Y$ such that $r(y)=y$ for each $y\in Y$. The continuous map r is called the retraction.

So my question is how to show that the logarithmic spiral $C=\{ 0 \times 0 \} \cup \{ \mathbb{e}^t\mathbb{cos}t \times \mathbb{e}^t\mathbb{sin}t \; | \; t\in \mathbb{R}\}$ is a retract of $\mathbb{R}^2$?

And is it possible to give a specific retraction $r:\mathbb{R}^2 \rightarrow C$?

Question from Topology by J. Munkres 2nd ed., Page 223.

EDIT: Using hints in comments I have a possible retraction

Take $f: \mathbb{R}^2 \rightarrow x-axis$. Now each point (x,y) in $\mathbb{R}^2$ can be written as $(x,y)=(r\cos \theta,r\sin \theta)$. But r can be written as $e^{\log r}$. So define $f(r\cos \theta,r\sin \theta)=\left\{ \begin{array}{lr} (0,0) & : r=0\\ (\log r,0) & : otherwise\\ \end{array} \right.$

Now define $g: \{(x,0):x\ge 0\} \rightarrow C$ as $g(x,0)=\left\{ \begin{array}{lr} (0,0) & :x=0\\ (e^x\cos x,e^x\sin x) & : otherwise\\ \end{array} \right.$

The composition function $f\circ g$ takes $\{ \mathbb{e}^t\mathbb{cos}t \times \mathbb{e}^t\mathbb{sin}t \; | \; t\ge 0\}$ to itself. But what about when $t<0$?

Is this approach completely wrong? Or can this be modified somehow?

  • $\begingroup$ I edited, it is not a division sign, it is the vertical bar | which means "such that." You could also write it in polar coordinates as $r = e^\theta.$ Also, in my edition it is page 221; nothing to prevent later editions, I guess. $\endgroup$ – Will Jagy Apr 22 '14 at 0:13
  • $\begingroup$ Ah, thank you. I will add the edition I am using too. $\endgroup$ – Shalin Doctor Apr 22 '14 at 0:16
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    $\begingroup$ Start with a retract from $\Bbb R^2$ to the half-line $\{(x, 0) : x \ge 0\}$ and compose with a plane transformation that twists the line into the desired spiral. In fact, you can make this a deformation retract. $\endgroup$ – Ayman Hourieh Apr 22 '14 at 0:16
  • $\begingroup$ For readers, note that this exercise in in the section on the Urysohn Metrization Theorem, and retract was defined in an exercise of the previous section, on the Urysohn Lemma, referred to as the first deep theorem of the book. Good to know. $\endgroup$ – Will Jagy Apr 22 '14 at 0:18
  • $\begingroup$ @AymanHourieh The trick though is to make sure your first retract is injective when restricted to the spiral. $\endgroup$ – wckronholm Apr 22 '14 at 0:48

I've now had the chance to look at the book and see the question in its context. I believe you're expected to use the previous exercises.

By the Tietze extension theorem (theorem 35.1.b), $\Bbb R$ has the universal extension property (see exercise 35.5 for the definition of this property). By using a retract $\Bbb R \to [0, \infty)$, we see that $[0, \infty)$ has the universal extension property too.

By exercise 35.6.a, every space with the universal extension property is an absolute retract. Therefore $[0, \infty)$ is an absolute retract. Since $C$ is homeomorphic to $[0, \infty)$, it follows that $C$ is a retract of $\Bbb R^2$.

As for how to prove exercise 35.6.a: Suppose $Y$ has the universal extension property, $Z$ is normal and $Y_0$ is a closed subspace of $Z$ homeomorphic to $Y$. Apply the universal extension property to the homeomorphism $Y_0 \to Y$ to get a map $Z \to Y$. Now compose with the homeomorphism $Y \to Y_0$ to get the desired retract.

Of course, you can still apply my hint in the comments to construct an explicit retract. I believe the end result using polar coordinates should be $$ f(r, \theta) = \begin{cases} (r, \log r) && \text{ if } r \ne 0 \\ 0 && \text{ otherwise.} \end{cases} $$


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