Local embedding and disk in domain perturbation

Question

Consider say $M=(\mathbb{S}^1\times\dotsb\times \mathbb{S}^1)-q$ ($n$-times). Assume that $B$ is an $n$ disk in $M$ (for instance, thinking of $\mathbb{S}^1$ as gluing $-1$ and $1$, the cube $B=[-\frac{1}{2},\frac{1}{2}]\times \dotsb\times [-\frac{1}{2},\frac{1}{2}]$ is in $M$ provided $q$ is not contained in it). Can we always find an (local) embedding $f:M\rightarrow \mathbb{R}^N$ for some $N$ AND $f\rvert_{B}$ is the identity?

Remarks:

• I think local embedding always exists from some abstract theory for $N=n$ but I don't see why the extra condition holds for $N=n$ (maybe true for large $n$?).
• Somehow I think the idea might be to find a smaller disk in $B$ such that for some embedding it is the identity there and then somehow stretch this to all $B$. I don't know how to make this rigorous.

More Questions:

I found a reference [essays of top manifolds] which I think sais above is true because two orientation preserving embeddings of $B^n$ in to $\mathbb{R}^n$ are equivalent by homeo of $\mathbb{R}^n$. I don't know why this is true? and why it implies it?

Is this question trivial? Comments suggest it is.

Answer 1

I believe it is sensible to ask that $B$ is collared (e.g. smooth). I don't know what happens otherwise.

The key fact is the consequence of the annulus theorem that if $M$ is a connected $n$-manifold, any two collared embeddings $B^k\rightarrow M$ are ambiently isotopic as long as $k<n$ or $k=n$ and either $M$ is non-orientable or $M$ is orientable and the embeddings are equioriented. This is explained by Lee Mosher in this answer. The analogous claim in the smooth category is much easier and known as Palais' disk theorem.

Now, if $M$ is any $n$-manifold, $B\subseteq M$ a collared $n$-disk and $f\colon M\rightarrow\mathbb{R}^N$ a local embedding, then you can find a disk $B^{\prime}\subseteq B$ s.t. there is a homeomorphism $h\colon M\rightarrow M$ taking $B$ to $B^{\prime}$ and $f\vert_{B^{\prime}}$ is an embedding (you can find a $B^{\prime}$ since $f$ is a local embedding and then shrink $B$ to $B^{\prime}$ within a slightly larger ball). Replace $f$ by $fh$ and assume w.lo.g. that $f\vert_B$ is an embedding. (If $f$ was already an embedding, this step is trivial.) If $M$ is oriented and $N=n$, we compose with a reflection if necessary and assume $f\vert_B$ is orientation-preserving. In either case, the initially mentioned result ensures us the existence of a homeomorphism $k\colon\mathbb{R}^N\rightarrow\mathbb{R}^N$ s.t. $kf\vert_B$ is your preferred orientation-preserving embedding of $B$ into $\mathbb{R}^N$, hence $kf$ is the desired local embedding $M\rightarrow\mathbb{R}^N$. In particular, your question comes down entirely to how we may (locally) embed $M$ into $\mathbb{R}^N$ as everything else is then automatic.

Now, for any manifold $M$, there are embeddings $M\hookrightarrow\mathbb{R}^N$ for $N$ large enough, so that isn't hard to achieve. For your specific example $M=(S^1)^n-\{\ast\}$, the punctured $n$-torus, there is a local embedding (in fact, a smooth immersion) $M\looparrowright\mathbb{R}^n$ to which we can apply these observations. This does follow from Hirsch-Smale theory, but there also exist multiple explicit constructions. A detailed discussion is in Chapter 16 of these Lectures Notes.