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My intuition tells me that any two topological embeddings of the closed 2-ball (aka unit disk) into $R^3$ are isotopic in $R^3$. Is this correct? Maybe easy to prove?

It seems like it should be easy because the disk is compact and contractible, but I sill don't know what to do about it. I am working on problems in mathematical origami and I'm hoping to use this fact as a starting point in a much bigger proof.

This is not my area of expertise, so any tips on what to search or how to go about proving (or disproving) this would be helpful. I have searched different ways of wording this question and haven't found anything helpful.

I suspect this may generalize to closed $n$-balls embedded in $R^{n+1}$ or even an $(n+1)$-manifold, but I only need the case $n=2$.

Update: After researching some more based on ARupinski's comment, I realized that I am concerned specifically with piecewise-linear embeddings of the disk into $R^3$ with finitely many pieces. I don't think you can make a piecewise-linear Alexander Horned Shere with only finitely many pieces. So my question would be, if $f$ is a piecewise-linear embedding of the closed 2-ball $B^2$ into $R^3$ with finitely many pieces, and $g$ is another embedding of $B^2$ into $R^3$ such that $g[B^2]$ is a closed triangle (with interior), is there an isotopy between $f$ and $g$ using piecewise-linear embeddings along the way? This sounds a bit easier to go about proving... if it's true... :( This is somewhat related to things like the Carpenter's Rule Theorem for chain linkages in the plane.

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    $\begingroup$ If one of your embeddings consists of an Alexander Horned Sphere with an open disk removed then it seems that should not be isotopic to the embedding of a disk as a filled unit circle in some plane. Perhaps someone with more of a topology background can confirm if I this is indeed true, or conversely correct me if my intuition about puncturing a Horned Sphere is in fact wrong. $\endgroup$
    – ARupinski
    Commented Feb 12, 2016 at 3:46
  • $\begingroup$ Very good observation! Simple topological questions always seem to result in some pathological counterexample that redefines our intuition. Since the complement of the usual disk is simply connected while the complement of the Alexander Horned Sphere (AHS) is not, the next question would be, is the homotopy type of the complement of an embedding an isotopy invariant? It appears so. This only gets more and more confusing/exciting. $\endgroup$
    – Jaeba
    Commented Feb 12, 2016 at 4:39
  • $\begingroup$ I'm not sure your question is simple. You are talking about topological embeddings, not smooth embeddings. Your idea about the disc being contractible, if you follow it through, is a smooth argument. I think your question is complicated because you have overlooked how difficult it is to construct all possible topological embeddings. $\endgroup$
    – Ryan Budney
    Commented Feb 12, 2016 at 6:48
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    $\begingroup$ It's not clear to me that an embedding mapping the sphere to a horned sphere can't be isotopic (= homotopic where each step is an embedding) to a standard embedding. Isn't the horned sphere defined as a uniform limit of smooth embeddings? $\endgroup$
    – YCor
    Commented Feb 12, 2016 at 9:07
  • $\begingroup$ There exist so-called wild disks in $R^3$ (cf. J. R. Stallings, Uncountably many wild disks, Ann. of Math. (2) 71 (1960), 185— 186.) Now the definition of a wild disk is a disk which is not equivalent to a PL disk, and the definition of "equivalent" is the following : $D_1$ is equivalent to $D_2$ if there is a self-homeomorphism of $R^3$ which takes $D_1$ to $D_2$. However, I don't see any reason why two inequivalent disks couldn't be isotopic. $\endgroup$
    – Sylvain Maillot
    Commented Feb 12, 2016 at 15:35

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The answer to your updated question is yes. In fact more is true: any two PL embeddings $f,g$ of the 2-disk in any connected 3-manifold $M$ are isotopic.

One way to see this is to use the following statement, which follows from Guggenheim's theorem: any two PL embedded $n$-balls in a connected PL $n$-manifold are isotopic. (Guggenheim's theorem is slightly stronger in that it gives an ambient isotopy.)

The deduction is as follows: let $D_1$ (resp. $D_2$) be the image of $f$ (resp. $g$). For $i\in \{1,2\}$ by thickening $D_i$ you can construct an embedded 3-ball $B_i$ containing $D_i$ such that $\partial D_i=\partial B_i\cap D_i$ and $D_i$ is isotopic rel boundary to any of the two 2-disks contained in $\partial B_i$ and bounded by $\partial D_i$. By Guggenheim's theorem in dimension 3, we may assume that $B_1=B_2$. By the same theorem in dimension 2 we may assume that $\partial D_1=\partial D_2$, hence $D_1$ is isotopic to $D_2$.

You are asking for something stronger: namely $f,g$ should be isotopic, not just their images. By the previous argument we can assume that $D_1=D_2$. Now use the fact that any orientation-preserving self-homeomorphism of the disk is isotopic to the identity (if the orientations are wrong, just flip one of the disks; this can be achieved by an isotopy.)

A good book on elementary PL topology is Introduction to piecewise linear Topology by Rourke and Sanderson.

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