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 "twocomponent 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.

2$\begingroup$ 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 $S^3$. $\endgroup$– Connor MalinSep 14, 2020 at 14:53

1$\begingroup$ I suspect you forgot to assume that $deg(f)=\pm 1$. $\endgroup$– Moishe KohanSep 14, 2020 at 15:26

$\begingroup$ 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 ? $\endgroup$– SteveSep 14, 2020 at 15:27

2$\begingroup$ Yes, I think this was added after I commented. $\endgroup$– Connor MalinSep 14, 2020 at 17:02

1$\begingroup$ Hi Connor, yes I have edited the question a little bit as suggested. $\endgroup$– SteveSep 14, 2020 at 17:05
1 Answer
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.

1$\begingroup$ just to clarify: when you say "crush the Hopf link", I think you mean individually crush each $S^1$ to a point, right? $\endgroup$ Sep 15, 2020 at 14:23

1$\begingroup$ Good point, that’s what I meant, but didn’t write it correctly. I’ll fix it. Thanks. $\endgroup$– Ian AgolSep 15, 2020 at 14:24