Are any two $n$-balls in $\mathbb{R}^n$ isotopic?

Question

I have two related questions in the topological category. Let $B^n$ denote the closed unit $n$-ball, and let $S^{n-1}$ denote its boundary sphere.

1. Is it true that there are exactly two isotopy classes of embeddings $B^n\to\mathbb{R}^n$?

Here "isotopy" means isotopy in the class of embeddings, not ambient isotopy in $\mathbb{R}^n$. The two isotopy classes that I have in mind are the orientation-preserving embeddings and the orientation-reversing embeddings.

2. Is it true that there are exactly two isotopy classes of embeddings $S^{n-1}\to\mathbb{R}^n$?

I haven't been able to track down answers to these questions. The "obvious" counterexample would seem to be the Alexander horned sphere embedding $S^2 \to \mathbb{R}^3$, but it seems to me that this embedding is isotopic the standard one by shrinking in along radial lines. That is, if $f\colon B^3\to \mathbb{R}^3$ is an embedding that maps $S^2$ to the Alexander horned sphere, then the embedding $g_t\colon S^2\to \mathbb{R}^3$ defined by $$ g_t(p) = f(tp) $$ is wild for $t=1$ but ought to be tame for $0<t<1$.

Incidentally, I'm interested in these questions because they would give an entirely elementary definition for orientability of topological manifolds. Specifically, if the first statement is true then a connected topological manifold $M^n$ is orientable if and only if there exist two distinct isotopy classes of embeddings $B^n\to M^n$.

Edit:

The first question has now been resolved (see the answer below). The second question is still unanswered, but is equivalent to either of the following:

• Is every embedding $S^{n-1}\to\mathbb{R}^n$ isotopic to one whose image bounds a ball?

• Is every embedding $S^{n-1}\to\mathbb{R}^n$ isotopic to one whose image is bicollared?

Answer 1

Answer to the First Question

Okay, as suggested by Moishe Kohan's comment, the argument by Lee Mosher in this MathOverflow post is sufficient to answer the first question. It is based on a couple of major theorems in geometric topology.

For $r>0$, let $B^n(r)$ be the closed ball in $\mathbb{R}^n$ centered at the origin with radius $r$, and let $S^{n-1}(r)$ be its boundary sphere. Also, let $$ B^n = B^n(1) \qquad\text{and}\qquad S^{n-1}=S^{n-1}(1). $$

Collared Balls and Bicollared Spheres

Recall the following definitions:

• A topological $n$-ball $B\subset \mathbb{R}^n$ is collared if there exists an $\epsilon>0$ and an embedding $$ f\colon B^n(1+\epsilon) \to \mathbb{R}^n $$ such that $B=f(B^n)$.
• A topological $(n-1)$-sphere $S\subset \mathbb{R}^n$ is bicollared if there exists an $\epsilon > 0$ and an embedding $$ f\colon S^{n-1}\times [-\epsilon,\epsilon]\to\mathbb{R}^n $$ such that $f(S^{n-1}\times\{0\}) = S$.

(Note: Since we are doing topology, we could just use $\epsilon=1$ in each case.)

For example, the Alexander horned sphere is not bicollared, and it follows that the closed ball bounded by the Alexander horned sphere is not collared. (Indeed, it is easy to see that a topological $n$-ball in $\mathbb{R}^n$ is collared if and only if its boundary sphere is bicollared.)

Some Big Theorems

Next we use two big theorems in geometric topology. The first was proven by Morton Brown (1960) and independently by Barry Mazur and Marston Morse (1960).

Generalized Schoenflies Theorem.
Let $S\subset\mathbb{R}^n$ be a topological $(n-1)$-sphere. If $S$ is bicollared, then there exists a homeomorphism $h\colon \mathbb{R}^n\to\mathbb{R}^n$ such that $h(S)=S^{n-1}$.

Since the boundary of a collared ball is always a bicollared sphere, it follows that every collared $n$-ball can be mapped to $B^n$ by a homeomorphism of $\mathbb{R}^n$.

Next we need the following theorem. According to another MathOverflow post, this can be proven using the Stable Homeomorphism Conjecture (now a theorem), which was proven by Robion Kirby in 1969 for all $n\ne 4$ and by Frank Quinn in 1982 for $n=4$.

Isotopy Theorem for Spheres.
For $n\geq 1$, every orientation-preserving homeomorphism of $S^n$ is isotopic to the identity.

This theorem has two important consequences:

• Since we can always arrange for an isotopy to fix a point, it follows that every orientation-preserving homeomorphism of $\mathbb{R}^n$ is isotopic to the identity.

• Also, we can combine this theorem with the Alexander trick to deduce that every orientation-preserving homeomorphism of $B^n$ is isotopic to the identity.

Conclusions

Combining all this together gives us the following theorem:

Theorem.

• Every orientation-preserving embedding $f\colon S^{n-1}\to\mathbb{R}^n$ whose image is bicollared is isotopic to the inclusion map.

• Every orientation-preserving embedding $f\colon B^n\to\mathbb{R}^n$ whose image is collared is isotopic to the inclusion map.

In both cases the proof is to first use the Schoenflies theorem to find a homeomorphism $h\colon\mathbb{R}^n\to\mathbb{R}^n$ such that $(h\circ f)(S^{n-1})=S^{n-1}$ or $(h\circ f)(B^n)=B^n$. Composing with a reflection if necessary, we can assume that $h$ is orientation-preserving. By the isotopy theorem, $h$ is isotopic to the identity, which means that $h\circ f$ is isotopic to $f$. But $h\circ f$ maps $S^{n-1}$ or $B^n$ to itself by some orientation-preserving homeomorphism, and is therefore isotopic to the identity by the isotopy theorem.

All that remains of the first question is the following observation:

Observation.
Every embedding $B^n\to \mathbb{R}^n$ is isotopic to an embedding whose image is collared.

To see this, let $f\colon B^n\to\mathbb{R}^n$ be an embedding, and define an embedding $g\colon B^n\to\mathbb{R}^n$ by $$ g(p) = f(p/2). $$ Then $f$ is isotopic to $g$ via $(p,t)\mapsto f\bigl((1-t/2)p)$, and $g(B^n)$ is collared by the embedding $S^{n-1}\times [-1/4,1/4] \to \mathbb{R}^n$ defined by $(p,\delta) \mapsto f\bigl((\tfrac{1}{2}+\delta)p)$.

This doesn't answer the second question, but it does reduce it to the following form:

Question.
Is every embedding $S^{n-1}\to\mathbb{R}^n$ isotopic to an embedding whose image is bicollared?

If the answer is yes then every orientation-preserving embedding $S^{n-1}\to \mathbb{R}^n$ is isotopic to the inclusion, by the arguments given above. Alternatively, it would suffice to answer the following question:

Question.
Is every embedding $S^{n-1}\to\mathbb{R}^n$ isotopic to an embedding whose image bounds a ball?