# unlinking when relaxing the homeomorphism condition

Say that we have two knots $$K_1$$ and $$K_2$$ in $$S^3$$ linked together in $$S^3$$ and forming the Hopf link. Usually, we can prove that we cannot unlink them by using a link invariant that shows that the "two-component unlink" that consists of two separate circles in $$S^3$$ have a different value (with respect to the invariant) in comparison to its value on the Hopf link. This effectively shows that there is no homomorphism from $$S^3$$ to itself that separates the two links. I want to relax the condition of homomorphism a little bit and ask: is there a continuous function that separates the images of the two links? in other words, is there a continuous function $$f:S^3\to S^3$$ with $$deg(f)=\pm 1$$ such that $$f(K_1)$$ is contained in a closed disk $$D_1$$ and $$f(K_2)$$ is contained in another closed disk $$D_2$$ and $$D_1$$ and $$D_2$$ are disjoint? Any pointer is appreciated.

• If I’m understanding your question correctly, the Tietze extension theorem tells you that since your links are closed subsets, it is possible to separate them with a function to the interval. After that you may just embed the interval into $S^3$. Sep 14, 2020 at 14:53
• I suspect you forgot to assume that $deg(f)=\pm 1$. Sep 14, 2020 at 15:26
• Thank you for your answer. Actually, I don't want my function to "factor through" another continues function because in that case I know we can separate them even with a homeomorphism ( we send them to a higher space and the unlink them then send them back to $S^3$). In that case can we find such a map ? Sep 14, 2020 at 15:27
• Yes, I think this was added after I commented. Sep 14, 2020 at 17:02
• Hi Connor, yes I have edited the question a little bit as suggested. Sep 14, 2020 at 17:05

It's possible to give such a map with degree $$1$$. The complement of a Hopf link $$H=H_1\cup H_2$$ is $$T^2\times I$$. So if we take $$S^3$$ and crush each component of the Hopf link to a point, we get a map to the suspension of the torus $$T^2$$, $$S^3 \to S^3/H_1/H_2 \cong ST^2$$. Moreover, if we had such a degree 1 map $$f:S^3\to S^3$$ with $$f(K_i) \subset D_i$$, we could get a map factoring through the suspension, since we can homotope the image $$f(K_i)$$ to a point in $$D_i$$, and then extend by homotopy extension to a map factoring through $$S^3/H_1/H_2$$.
Now we take a degree 1 map from $$T^2$$ to the sphere $$S^2$$, e.g. by crushing a wedge of circles in $$T^2$$ whose complement is a disk.
This degree one map suspends to a degree 1 map $$S^3 /H _1/H_2 \cong ST^2 \to S S^2$$. The composition of these maps has the desired property.
• just to clarify: when you say "crush the Hopf link", I think you mean individually crush each $S^1$ to a point, right? Sep 15, 2020 at 14:23