In regard to a retraction $r: \mathbb{R}^3 \rightarrow K$ Let $K$ be the "knotted" $x$-axis. I have been able to show that $K$ is a retract of $\mathbb{R}^3 $ using the fact that $K$ and the real line $\mathbb{R}$ are homeomorphic, $\mathbb{R}^3$ is a normal space, and then applying the Tietze Extension Theorem. But then what would be an explicit retraction $r: \mathbb{R}^3 \rightarrow K$? Any ideas?

Let $f : K \to \Bbb R$ and pick a point $x \in K$.
Pick a infinite sheet of paper on the left side of the knot and imagine pinching and pushing it inside the knot all the way right to $x$, and use this to define $g$ on the space spanned by the sheet of paper into $( - \infty ; f(x))$, such that if $y<f(x)$, $g^{-1}(\{y\})$ is homeomorphic to the sheet of paper and intersects $K$ only at $f^{-1}(y)$.
Do the same thing with another infinite sheet of paper from the right side of the knot moving all the way left to $x$.
Finally define $g(y) = f(x)$ for all $y$ in the remaining space between the two sheets of paper. 
Then $g : \Bbb R^3 \to \Bbb R$ is continuous, extends $f$, and except at $f(x)$ where the fiber is big, $g^{-1}(\{y\})$ is homeomorphic to $\Bbb R^2 $
