Q: Specific Retraction from $\mathbb{R}^2$ to the logarithmic spiral 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?

A: 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
$$
f(r, \theta) = \begin{cases}
(r, \log r) && \text{ if } r \ne 0 \\
0 && \text{ otherwise.}
\end{cases}
$$
