I have been trying to prove the following result for a few days now, and have made some amount of progress, but now I'm struggling. This is what I'd like to prove: Every proper knot, K has a retract from $S^3$ to K.
To approach this problem, I first solved the problem with the "cut" knot, equivalent to tying a straight line in an equivalent knot without gluing the end points together, speaking loosely, and finding a retract from $\mathbb R^3$ to this "cut" knot, which we could call $K'.$ I can solve the problem in this arena if I am permitted to speak loosely about a certain function, which I've called the "natural" homeomorphism. I'd like some help making this notion a bit more rigorous, and then working on the next step with the sphere.
Here's the argument I've got so far.
Let's begin with the $x$-axis, and wrapping it around itself into a "cut" knot $K'$, with endpoints that extend infinitely just as the $x$-axis does. Let's call this wrapping process the "natural homeomorphism" and call it $f$. Then we have $f:\mathbb R \rightarrow K'$. It's obvious that $K'$ lives in $\mathbb R^3,$ and has the subspace topology. Since $f$ is a homeomorphism, $f^{-1}$ exists, is also a homeomorphism, and has the property that $f^{-1}:K' \rightarrow \mathbb R$. However, that $K'$ is homeomorphic to $\mathbb R$ and the $x$-axis is closed in $\mathbb R^3$ means that $K'$ is closed in $\mathbb R^3$. Finally, we know that $\mathbb R^3$ is a normal space, so by the Tietze Extension Theorem, we can extend $f^{-1}$ to $(f^{-1})':\mathbb R^3 \rightarrow \mathbb R$. Now we just compose this function with the original function $f$ and we have found a retract from $\mathbb R^3 \rightarrow K'$. Note that this is a retract, since $(f^{-1})'=f^{-1}$ for any value $k$ in $K'$, and $f(f^{-1}(k))=k$.
So given this natural homeomorphism, I can do this in Euclidean space. I want to extend this result to the sphere by using one-point compactification. This will allow me to identify $\mathbb R^3 $ with $S^3$, and gluing the endpoints of the "cut" knot will allow me to identify it with a proper knot $K$.
However, I cannot come up with an analog for this same "natural" homeomorphism. I'm now thinking it doesn't exist, since obviously circles are not homeomorphic to general knots - that's why they are knots. But I'm wondering if maybe it is possible to identify the knot with the surface of some higher dimensional thing. My intuition tells me there should be an analog of this result for $\mathbb R^3$, but I'm just unclear how to find it.
A knot can't be a retract of $S^3$ except possibly if it's missing a point because the knot has nontrivial fundamental group and the inclusion homomorphism induces an injection on the fundamental group for a retract. However, $S^3$ can't retract onto a knot that's missing a point either because it isn't compact if it's missing a point. A knot that's missing a point can be a retract of $\mathbb{R}^3$ (the missing point can be chosen to be the point at infinity of $S^3$).