Is every based embedding from $S^n$ to $\mathbb{R}^{n+1} \setminus \{0\}$ either the unit or a generator of $\pi_n(\mathbb{R}^{n+1} \setminus \{0\})$?

Question

In the case $n = 1$, by complex analysis we can prove (Index of a Jordan curve) that for any smooth Jordan curve the winding number around a point (defined by integral $\frac{1}{2\pi i} \oint \frac{\mathrm{d}z}{z - z_{0}}$) is either $0$, $-1$ or $1$. Complex analysis also concludes that two smooth closed curves with a homotopy in $\mathbb{C} \setminus \{z\}$ have the same winding number around $z$, thus there’s a definition of winding number around a point of any closed curve (that doesn’t go through the point), namely $\operatorname{wind}(\gamma; z) := \operatorname{wind}(\gamma'; z)$, where $\gamma'$ is homotopic to $\gamma$ in $\mathbb{C} \setminus \{z\}$ and smooth. By these, we know the winding number of a closed curve around $0$ is actually the element it represents in $\pi_{1}(\mathbb{C} \setminus \{0\}) \cong \mathbb{Z}$ while the circle $e^{2\pi i \,\cdot} \colon [0, 1] \rightarrow \mathbb{C}$ represents $1$ in $\mathbb{Z}$. (It doesn’t matter if the closed curve is a based map.)

So by complex analysis we know when $n = 1$ and the embedding $S^1 \rightarrow \mathbb{C} \setminus \{0\}$ is smooth that the proposition is true. I guess the winding number of any Jordan curve is either $0$, $-1$ or $1$, i.e., the case $n = 1$ without the smoothness.

Furthermore, I want to know whether the proposition is true when $n > 1$.

Answer 1

This answer started out as only a partial solution. Additional details provided by the commenters, which completed the solution, are given after the Edit break below.

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Let $S \subset \mathbb R^{n+1} \setminus \{0\}$ be the image of an embedding of $S^{n}$. By the Jordan separation theorem, $\mathbb R - S$ has two components, a bounded component $B$ and an unbounded component $U$. There are two cases depending on whether $0 \in B$ or $U$.

Case 1: $0 \in U$. I can prove that $S$ is homotopic to a constant in $\mathbb R^{n+1} \setminus \{0\}$ and therefore represents the identity element of $\pi_n(\mathbb R^{n+1} \setminus \{0\})$. Choose a homotopy $H' : S \times [0,1] \to \mathbb R^{n+1}$ from the identity map on $S$ to a constant map, let $K$ be the image of this homotopy, and let $R$ be so large that $K$ (including $S$ itself) is contained in the ball of radius $R$ around $0$. Since $U$ is path connected, we may choose a path $\gamma : [0,1]$ in $U$ from $\gamma(0)=0$ to a point $\gamma(1)$ whose distance to $0$ is $>4R$.

Now do a homotopy of $S$ in two steps.

In the first step use the homotopy $H_1(x,t) = x - \gamma(t)$, and notice that $H_1(x,t) \ne 0$ for all $x \in S$, $t \in [0,1]$. At the end of this homotopy, $S$ has been translated to $S - \gamma(1)$ which lies outside of the ball of radius $3R$ around $0$. The diameter of $K$ is $\le 2R$, and so the diameter of $K-\gamma(1)$ is $\le 2R$, and hence $K-\gamma(1)$ is contained outside the ball of radius $R$ around $0$. We may therefore homotopy $S-\gamma(1)$ to a point through $K-\gamma(1)$, using $H_2 = H' - \gamma_1$, and so $H_2(x,t) \ne 0$ for all $x \in S$, $t \in [0,1]$. Concatenating the homotopy $H_1$ with the homotopy $H_2$, the result is a homotopy from the identity map on $S$ to a constant map such that the entire homotopy stays away from $0$.

Case 2: $0 \in B$. If we could assume that $S$ was tame then we could use the Brown-Masur Theorem (aka the generalized Schönflies theorem) to conclude that there is a homeomorphism of $\mathbb R^{n+1}$ that fixes $0$ and takes $S$ to $S^{n+1}$ and we would be done.

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Edit: For a wild sphere in Case 2, there's one more step; thanks to the commenters for pushing this issue forward.

If $S \subset \mathbb R^{n+1}$ is a wildly embedded $n$-sphere then by work of Ancel linked in the comments below, the inclusion map $S \hookrightarrow \mathbb R^{n+1}$ may be approximated by a tame map $h : S \to \mathbb R^{n+1}$, meaning that $h(S)$ is a tame embedding and $d(x,h(x)) < \epsilon$, where $\epsilon>0$ may be chosen arbitrarily. By choosing $\epsilon$ equal to the distance from $S$ to $0$, the homotopy from $S$ to $h(S)$ misses $0$. One can then apply the earlier solution to $h(S)$. It follows that $S$ satisfies the desired conclusion.

Answer 2

Another way to do it is via homological degree, which generalizes the winding number argument. The Schonflies Theorem is a rather deep theorem, but I wanted to show that the question in the OP does not depend on it. You can use standard homological arguments, although you are still relying on some heavy machinery. It does give a nice flavor for some geometric topology though. (I do use the fact that the $n$ sphere separates $\mathbb R^{n+1}$ into two components, but that is a far easier theorem to prove than Schonflies.)

Consider the composition of your embedding with radial projection to the $n$ sphere: $$S^n \overset{g}{\hookrightarrow} \mathbb R^{n+1}\setminus \{0\} \overset{\pi}{\twoheadrightarrow} S^n$$

As in Lee Mosher's argument, perturb $g$ slightly so that it is a smooth embedding. Then the degree of the composition $\pi\circ g$ is well defined, and can be calculated by considering the preimage of a regular value $p$. The preimage of a small disk around $p$ is a disjoint union of disks, one for each point in the preimage, and we can assign $\pm 1$ to each preimage depending on whether the map preserves or reverses orientation. The degree is then the sum of these local degrees. The degree tells you the induced map on $H_n(S^n)\cong \pi_n(S^n)$.

So all we need to do is show that the degree is $\pm 1$ or $0$. This follows by the fact that the sphere separates $\mathbb R^n$ into two components. Look at the preimages $\pi^{-1}(p)$ in $\mathbb R^{n+1}$. By definition of $\pi$, they lie along a radial line out from the origin, and we can calculate the sign by whether the ray is entering or exiting the interior of the embedded sphere. Thus the degree is the algebraic intersection number of the ray and the sphere. If the sphere contains the origin in its interior, then this intersection number is $\pm 1$ since it starts inside the sphere and ends outside the sphere, while if the origin is not in the sphere's interior, then the intersection number is 0.

This completes the proof.