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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?

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    $\begingroup$ 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 $C$. $\endgroup$ – Wojowu Apr 18 at 15:20
  • $\begingroup$ @wojowu this can be fixed by compactify $\mathbb{R}^n$ to $\mathbb{S}^n$. I updated the question accordingly. $\endgroup$ – Hao Chen Apr 18 at 15:27
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    $\begingroup$ 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. $\endgroup$ – Oleg Eroshkin Apr 18 at 16:07
  • $\begingroup$ @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. $\endgroup$ – Hao Chen Apr 18 at 16:25
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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 $0$ ...

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  • $\begingroup$ Thanks! Do we have a linked curve in $f^{-1}(0) \cap f^{-1}(\infty)$? Although this is not what I am intending to. $\endgroup$ – Hao Chen Apr 19 at 6:15
  • $\begingroup$ Then, can it be fixed if I work not in $\mathbb{S}^n$, but in the projective space $\mathbb{R}P^n$? $\endgroup$ – Hao Chen Apr 19 at 7:59
  • $\begingroup$ 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 $\endgroup$ – Dmitri Panov Apr 19 at 9:13
  • $\begingroup$ 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?) $\endgroup$ – Hao Chen Apr 19 at 12:32
  • $\begingroup$ I agree, this modification works $\endgroup$ – Dmitri Panov Apr 19 at 13:16

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