How are linked rings homeomorphic to seperated links? I'm currently reading "Geometry, Topology and Physics" by Mikio Nakahara. In his book there the following exercise:

Show that two figure in figure 2.109(b) [see below] are homeomorphic to each other. Find how to unlink the right figure in $ \mathbb{R}^4$.

The figure in question is:

I have no idea how to tackle this problem. I guess I need to show that every point in space is in the same equivalence class, but I don't know how one would do that.
Any hint for me to get started on this problem are much appreciated.
 A: Note that these are two questions.
One is to show that the two figures are homeomorphic. This is trivial: just take a homeomorphism from one ring on the left to one ring on the right and an homeomorphism from the other ring on the left to the other ring on the right. Combine those two homeomorphism to a map from the two rings on the left to the two rings on the right; that's an homeomorphism as well. It is important to realize that the ambient space (the ${\mathbb R}^3$) doesn't play a role!
The other question is to argue that there is a homotopy $[0,1] \times (S^1 \sqcup S^1) \to {\mathbb R}^4$ such that (the image of) $(0,-,-) \colon (S^1 \sqcup S^1) \to {\mathbb R}^4$ is the left rings and (the image of) $(1,-,-) \colon S^1 \sqcup S^1 \to {\mathbb R}^4$ is the right rings. Assuming the left and right rings are really embedded in ${\mathbb R}^3 \times \{0\}$, use the 4th coordinate to lift one of the two rings in the figure on the left to the $(-,-,-,1)$ plane; then move it over to the location of the corresponding ring in the figure on the right; then move it back to the $(-,-,-,0)$ plane.
