TITLE: unlinking when relaxing the homeomorphism condition
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: 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.