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Is there a simple proof of the fact that:

If $A\subset S^3$ is homeomorphic to $S^1$, then there is a circle $B$ embedded into $S^3\setminus A$ that such that the circles $A$ and $B$ are linked with the linking number $1$?

If $A$ is smoothly embedded, than it is easy. The problem is that the general topological embedding can be very complicated and I do not see a simple geometric explanation of this fact. I know that $H_1(S^3\setminus A)=\mathbb Z$ (Corollary 1.29 in Vick's Homology Theory) so the generator of the homology group should be a circle.

Edit 1: This is actually true as John Klein pointed out in his answer below, but I do not know the answer to the higher dimensional version of the problem stated below. Only a partial answer is given in comments below.

The question can be then generalized to higher dimensional spheres.

If $A\subset S^n$ is homeomorphic to $S^k$, there should be a topological sphere in $S^n\setminus A$ of dimension $n-k-1$ that links $A$ with the linking number $1$.

Edit 2. A counterexample is given below in my answer.

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Let $C = S^3 \setminus A$. Alexander duality says that $$ H_1(C) \cong H^1(A) \cong \Bbb Z\, . $$ Let $\alpha: S^1 \to C$ be any map representing a generator of $H_1(C)$ (every first homology class is spherical by the Hurewicz theorem). By homotopical approximation we can assume $\alpha$ is a smooth embedding. We can also assume without loss in generality that the image of $\alpha$ misses the north pole of $S^3$. Identify the complement of the north pole with $\Bbb R^3$. Set $B= \alpha(S^1)$. Then $A\amalg B\subset \Bbb R^3$. The degree of the map $\require{AMScd}$ \begin{CD} \ell: A \times B @>>> S^2 \end{CD} given by $(x,y) \mapsto (x - y)/|x - \alpha(y)|$ is the linking number of $A$ with $B$ by definition. On the other hand, this map has degree one (after choosing appropriate homology generators).

The point is that the pushforward of a generator \begin{CD} H_1(S^1) @>\alpha_\ast >> H_1(C) \end{CD} coincides with the degree of $\ell$.

How can we check this? Well, assuming $A$ has a nice regular neighborhood, we could redefine $C$ as the complement of that neighborhood. Then $\ell$ can be redefined as the degree of the map $$ A \times C \to S^2 $$ again given by the same formula, where we are assuming our new $C$ misses the north pole of $S^3$. Alexander duality says that the induced slant product pairing $$ H_1(A) \otimes H_1(C) \to H_2(S^2) = \Bbb Z $$ is non-singular, so the degree is $\pm 1$.

Even if $A$ fails to have a nice regular neighborhood, we can assume it misses the north pole $x$ and that $C:= S^n \setminus A$ misses another point $y$ of $S^n$. Identify $S^n \setminus x \cong \Bbb R^n \cong S^n \setminus y$. Then, similarly, we obtain a map $$ A\times C \to S^2 $$ and there a linking map $A\times B\to S^2$. With these changes, the argument proceeds as before.

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  • $\begingroup$ You are right. But how about the higher dimensional case? Then, as far as I understand, the generator of homology, in general, need not be represented by a map from a sphere, but I still believe that there should be an $n-k-1$ embedded sphere that has linking number $1$ with $A$. $\endgroup$ – Piotr Hajlasz Mar 11 '18 at 20:47
  • $\begingroup$ It works the same way at least if the embedding of $S^k\subset S^n$ has a nice regular neighborhood of the form $S^k \times \Bbb R^{n-k}$ (this is always true in the smooth case). Then the inclusion $\ast \subset S^{n-k} \subset S^n$ has linking number +1 with $S^k$. $\endgroup$ – John Klein Mar 11 '18 at 21:15
  • $\begingroup$ The problem is that I am interested in wild embedding of $S^k$ so there is no information about neighborhood of $S^k$. $\endgroup$ – Piotr Hajlasz Mar 11 '18 at 21:26
  • $\begingroup$ Suppose $f: S^k \to S^n$ is a topological embedding. I think what is generally true is that there is a $j > 0$ such that the composition $$S^k \to S^n \subset S^{n+j}$$ has the property that there is an embedded sphere $S^{j+n-k-1} \subset S^{n+j}$ having linking number +1 with $f(S^k)$. If you want me to, I can sketch the proof. $\endgroup$ – John Klein Mar 11 '18 at 22:04
  • $\begingroup$ I think I have an idea how to prove it using Klee's trick, see Proposition 3 in Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857, but I haven't checked the details. Of course, I am interested in your argument. $\endgroup$ – Piotr Hajlasz Mar 11 '18 at 22:23
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The question can be then generalized to higher dimensional spheres. If $A\subset S^n$ is homeomorphic to $S^k$, there should be a topological sphere in $S^n\setminus A$ of dimension $n-k-1$ that links $A$ with the linking number $1$.

In general the answer is in the negative. Here is an example. The double suspension of the Poincare homological sphere $M^3$ is homeomorphic to $\mathbb{S}^5$. This is a well known Cannon-Edwards theorem [1,2,3]. The suspension circle is an embedding of $\mathbb{S}^1$ to $\mathbb{S}^5$. Thus Poincare 3-sphere $M^3$ is a deformation retract of $\mathbb{S}^5\setminus\mathbb{S}^1$. Since $|\pi_1(M^3)|=120$ and the degree of a map $f:\mathbb{S}^3\to M^3$ is a multiple of $|\pi_1(M^3)|=120$, It follows that the linking number between the suspension circle $\mathbb{S}^1\subset\mathbb{S}^5$ and any embedding of $\mathbb{S}^3$ to $\mathbb{S}^5\setminus\mathbb{S}^1$ must be a multiple of $120$.

[1] https://en.m.wikipedia.org/wiki/Double_suspension_theorem

[2] J. W. Cannon, Shrinking cell-like decompositions of manifolds. Codimension three. Ann. of Math. 110 (1979), 83-112.

[3] R. D. Edwards, The topology of manifolds and cell-like maps. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 111–127, Acad. Sci. Fennica, Helsinki, 1980.

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    $\begingroup$ Even though I’m well aware of Cannon-Edwards’ result, this answer still surprised me. Smoothly embedded circles in 4 dimensions and higher are unknotted, so this example is quite counterintuitive, in some sense it can’t be well-approximated by smooth circles. Using other homology 3-spheres with infinite $π_1$, you get examples where any embedded 3-sphere must have linking number 0, since a 3-sphere can have a finite degree map only to a manifold with finite fundamental group. $\endgroup$ – Ian Agol Jan 20 '19 at 2:40
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    $\begingroup$ @IanAgol Thank you for your nice comment. Where can I find an example of a homology 3-sphere with infinite $\pi_1$? (I am not a topologist, but an analyst who is using topology in their research; topology is a shaky ground for me). $\endgroup$ – Piotr Hajlasz Jan 20 '19 at 15:26
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    $\begingroup$ see en.wikipedia.org/wiki/Homology_sphere?wprov=sfti1 The Brieskorn spheres might be the easiest to describe. When $(p,q,r)$ satisfy $1/p+1/q+1/r \leq 1$, $\Sigma(p,q,r)$ will have infinite fundamental group. $\endgroup$ – Ian Agol Jan 20 '19 at 15:49
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Let me add another answer. This answer stems from the fruitful discussion on MathOverflow.

Theorem. There is a topological embedding $\iota:\mathbb{S}^1\to\mathbb{S}^5$ such that $\pi_3(\mathbb{S}^5\setminus\iota(\mathbb{S}^1))=0$. Therefore, no $3$-sphere can be linked with $\iota(\mathbb{S}^1)$.

Proof. It is well known that there are $3$-dimensional integer homology spheres whose universal cover is $\mathbb{R}^3$. For example, there are particular constructions in [1,2,5] of hyperbolic integer homology spheres. Note that the universal cover of a hyperbolic $3$-manifold is the hyperbolic space that is homeomorphic to $\mathbb{R}^3$. Other examples are listed in Homology sphere with $\mathbb{R}^3$ as the universal cover. Let $\mathcal M$ be such an integer homology sphere. Since the universal cover of $\mathcal M$ is contractible, $\pi_3(\mathcal M)=0$. According to the celebrated theorem of Cannon and Edwards [3,4], the double suspension of an integer homology sphere is homeomorphic to a topological sphere. Let $h:S^2{\mathcal M}\to\mathbb{S}^5$ be such a homeomorphism. $\mathcal M$ is a deformation retract of the complement of the vertices of the suspension $S\mathcal M$. Therefore, $\mathcal M$ is also a deformation retract of the complement of the suspension of the vertices in $S^2\mathcal M$. Denote the suspension of the vertices by $X$, so $\mathcal M$ is a deformation retract of $S^2{\mathcal M}\setminus X$ and hence $\pi_3(S^2\mathcal{M}\setminus X)=0$. $X$ is homeomorphic to $\mathbb{S}^1$. If $g:\mathbb{S}^1\to X$ is a homeomorphism, then $\iota=h\circ g:\mathbb{S}^1\to\mathbb{S}^5$ is a topological embedding and clearly $\pi_3(\mathbb{S}^5\setminus\iota(\mathbb{S}^1))=\pi_3(\mathbb{S}^5\setminus h(X))=\pi_3(S^2{\mathcal M}\setminus X)=0$. The proof is complete. $\Box$

[1] J. Baldwin, J., S. Sivek, Stein fillings and $SU(2)$ representations. Geom. Topol. 22 (2018), 4307-4380.

[2] J. Brock, N. M. Dunfield, Injectivity radii of hyperbolic integer homology $3$-spheres. Geom. Topol. 19 (2015), 497-523.

[3] J. W. Cannon, Shrinking cell-like decompositions of manifolds. Codimension three. Ann. of Math. 110 (1979), 83-112.

[4] R. D. Edwards, The topology of manifolds and cell-like maps. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 111–127, Acad. Sci. Fennica, Helsinki, 1980.

[5] J. Hom, T. Lidman, A note on surgery obstructions and hyperbolic integer homology spheres. Proc. Amer. Math. Soc. 146 (2018), 1363-1365.

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