# 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

## 1 Answer

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