# Non-zero winding number on a space curve implies a linked curve in the zero set?

The following statement has been largely improved from my original post thanks to discussions with @DmitriPanov ad well as the comment from @Wojowu.

Let $$f \colon \mathbb{S}^3 \to \mathbb{R}^2$$ be real analytic. Let $$C$$ be a closed space curve in $$\mathbb{S}^3$$, which I might need to assume to be unknot. If $$f(C) \subset \mathbb{R}^2$$ has a non-zero winding number around $$0$$, then $$f^{-1}(0)$$ contains a space curve in $$\mathbb{S}^3$$ that is linked to $$C$$.

I consider this as a generalization of Kronecker's existence theorem. I sketched a plausible argument in my comment to the answer of @DmitriPanov, which I did not check with great care. The statement and argument is obviously generalizable to other dimensions, using topological degree.

I believe that this result is not new. So where do I find a reference for this and its higher dimensional versions?

• Let $f$ be the standard projection and $C$ be a flat circle. $f^{-1}(0)$ is a line going through the center of the circle and contains no curve linked to $C$. – Wojowu Apr 18 at 15:20
• @wojowu this can be fixed by compactify $\mathbb{R}^n$ to $\mathbb{S}^n$. I updated the question accordingly. – Hao Chen Apr 18 at 15:27
• What is the "winding number" for the loop in $S^2$ around 0? The sphere minus one point is contractible. You need to remove two points to have a nontrivial homotopic invariant. – Oleg Eroshkin Apr 18 at 16:07
• @OlegEroshkin You are right. As I regard $\mathbb{S}^2$ as compactification of $\mathbb{R}^2$, I mean the winding number of $f(C)$ in $\mathbb{S}^2 \setminus \infty$. It is now clarified in the question. – Hao Chen Apr 18 at 16:25

Now the statement is indeed correct. What follows below is and answer to the previous version of the question which I'll keep for the moment.

Note that the condition for $$f(C)$$ to have non-zero winding number around $$0$$ is more-less empty. Indeed, to fix the winding number of $$f(C)$$ around $$0$$ one needs to choose infinity. And for more-less any closed curve $$\eta$$ in $$S^2$$ that doesn't pass through $$0$$ one can choose $$\infty \in S^2$$ so that the winding number of $$\eta$$ in $$S^2\setminus \infty$$ around $$0$$ is non-zero.

Now, for a concrete set of counter-examples suppose that $$f: S^3\to S^2$$ is ANY map with non-zero differential at a point $$x\in S^3$$ such that $$f(x)\ne 0$$. Then take a small ball $$U$$ containing $$x$$ such that $$0\notin f(U)$$ and a curve $$\gamma\subset U$$ that projects to a small circle in $$S^2$$. Choose $$\infty\in S^2$$ such that $$f(\gamma)$$ separates $$0$$ from $$\infty$$. Then clearly the winding number of $$f(\gamma)$$ around $$0$$ is $$\pm 1$$ but $$f^{-1}(0)$$ doesn't contain a component linked to $$c$$.

I can see only one way to fix this. Ask $$f$$ not to be null-homotopic and ask $$C$$ to be a full premiage of a point $$x\in S^2$$ different from $$0$$ ...

• Thanks! Do we have a linked curve in $f^{-1}(0) \cap f^{-1}(\infty)$? Although this is not what I am intending to. – Hao Chen Apr 19 at 6:15
• Then, can it be fixed if I work not in $\mathbb{S}^n$, but in the projective space $\mathbb{R}P^n$? – Hao Chen Apr 19 at 7:59
• Concerning your first question I am just saying that $\pi_3(S^2)\cong \mathbb Z$ and if we have a map $\phi: S^3\to S^2$ whose class is equal $n\in \mathbb \pi_3(S^2)\cong \mathbb Z$ then the preimages of two generic points in $S^2$ are links in $S^3$ that have linking number $n$. Concerning your second question, changing $\mathbb S^n$ to $\mathbb RP^n$ will not make any difference, I don't see how this can be fixed – Dmitri Panov Apr 19 at 9:13
• Here is another attempt, which I think should work. If it is, I'll open another question asking for reference. Let $f$ be a continuous map from 3-ball $\mathbb{B}^3$ to $\mathbb{R}^2$, and $C$ be a closed curve in $\partial \mathbb{B}^3$. If $f(C)$ has a non-zero winding number around $0$, then the degree theory tell me that $f^{-1}(0)$ has a non-empty intersection with any disk in $\mathbb{B}^3$ bounded by $C$. If $f$ is moreover real analytic, then $f^{-1}(0)$ contains a path-connected curve that intersects every disk bounded by $C$ (is there a word for this situation?) – Hao Chen Apr 19 at 12:32
• I agree, this modification works – Dmitri Panov Apr 19 at 13:16