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A necessary condition for a manifold with non-empty manifold boundary embedded in $\mathbb{R}^n$ to have its topological boundary coincide with its manifold boundary is that its manifold interior is an open subset of $\mathbb{R}^n$.

Questions:

  1. Is this also a sufficient condition? (Note that I am assuming a prior that the manifold boundary is non-empty -- this does not hold otherwise.)

  2. What conditions are necessary for a sort of "converse" to this to hold? I.e., given a proper open subset of $\mathbb{R}^n$, which is always a smooth embedded submanifold, when is its topological closure an embedded smooth submanifold with (non-empty) boundary?

Attempt: The second question is not as straightforward as I first thought: consider the open hypercube, which is clearly a proper open subset of $\mathbb{R}^n$ -- however, unless $n=1$, its topological closure is a manifold with corners, but not a manifold with boundary.

Since the open hypercube is homeomorphic (diffeomorphic too, right?) to any open ball, whatever conditions need to be placed on such a proper open subset of $\mathbb{R}^n$ for this "converse" to hold must be "more than topological" (I don't know how to phrase that precisely).

Note: this is not a duplicate of this question, it is a follow-up to it.


Context: This answer established that, if an $m$-dimensional manifold with boundary $M$ can be embedded into a Euclidean space $\mathbb{R}^n$ such that its topological and manifold boundaries coincide, then necessarily one has that $m=n$.

This implies in particular that for $n$-dimensional manifolds with or without boundary which cannot be embedded into $\mathbb{R}^n$, the topological and manifold boundaries never coincide when embedded in any Euclidean space. (If the manifold with boundary has empty manifold boundary then of course the topological and manifold boundaries coincide when it is embedded in itself.)

The interior of any such embedded manifold with boundary is always an embedded manifold (without boundary). We also have that the embedded submanifolds of codimension $0$ of any manifold are exactly the open subsets of that manifold (e.g. Proposition 5.1. p.99 of Lee's Introduction to Smooth Manifolds).

Thus we have a necessary condition for the topological and manifold boundaries to coincide, at least when the embedding is inside of $\mathbb{R}^n$ (although I suspect that this holds for embeddings in an arbitrary $n$-dimensional manifold): the manifold interior of $M$ is an open subset of $\mathbb{R}^n$.

However, this is can not be a sufficient condition: if the manifold boundary is empty then this fails because the topological boundary is non-empty (unless the manifold is all of $\mathbb{R}^n$ itself -- e.g. see this answer). It is unclear to me whether this is a sufficient condition if we assume a priori that the manifold boundary is non-empty.

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    $\begingroup$ Actually a closed cube is a manifold with boundary. The definition says that each boundary point has to have a neighbourhood homeomorphic to "half" ball $\{(x_1,\ldots, x_n)\ |\ \sum x_{i}^{2} < 1\ \mbox{and}\ x_j \geq 0\}$. Note that point $(0, \ldots, 0)$ is a corner of "half" ball. Corners of a cube can definitely be mapped to $(0,\ldots, 0)$. $\endgroup$ – freakish Jan 11 '17 at 16:24
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    $\begingroup$ It looks like it is enough if the topological boundary is a manifold. I'm not able to prove it though. $\endgroup$ – freakish Jan 11 '17 at 16:27
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A sufficient condition for a codimension-$0$ manifold with non-empty manifold boundary embedded in $\mathbb{R}^n$ to have its manifold and topological boundary coincide may be that it is properly embedded.

Consider the following excerpt from Lee's Introduction to Smooth Manifolds (p. 120, 2nd edition):

If $M$ is a smooth manifold with or without boundary, a regular domain in $M$ is a properly embedded codimension-$0$ submanifold with boundary. Familiar examples are the closed upper half space $\mathbb{H}^n \subseteq \mathbb{R}^n$, the closed unit ball $\bar{\mathbb{B}}^n \subseteq \mathbb{R}^n$, and the closed upper hemisphere in $\mathbb{S}^n$.

Proposition 5.46 Suppose $M$ is a smooth manifold without boundary and $D \subseteq M$ is a regular domain. The topological interior and boundary of $D$ are equal to its manifold interior and domain, respectively.

The definition of proper embedded is found on p. 100 of the same:

An embedded submanifold $S \subseteq M$ is said to be properly embedded if the inclusion $S \hookrightarrow M$ is a proper map [preimages of compact sets are compact].

Since $\mathbb{R}^n$ is connected, a necessary condition for a submanifold with or without boundary of codimension $0$ to be properly embedded is that its manifold boundary is non-empty:

Proposition 5.49.(c) Suppose $M$ is a smooth manifold with or without boundary and $S \subseteq M$ is an embedded submanifold with boundary. Then $S$ is properly embedded if and only if it is a closed subset of $M$.

(See also this related question on Math.SE) This is because a codimension $0$ embedded submanifold with empty manifold boundary must be an open subset of $\mathbb{R}^n$, and if it is a proper subset, then it can not also be a closed subset of $\mathbb{R}^n$, since $\mathbb{R}^n$ is connected.

This doesn't say anything about my second question though, since Proposition 5.49(c) only applies to manifolds with boundary, and not proper embeddings of more general subsets.

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A partial answer to the second question: if $W$ is an open set in an affine space and $H=\overline W\setminus W$ is a smooth hypersurface, then $\overline W$ is a smooth manifold, with or without boundary (think of $W=\mathbb R^2\setminus\{x=0\}$). Notice that if $\overline W$ is a manifold, its boundary is a hypersurface contained in $H$. In fact the argument below works to see that that boundary must be a union of connected components of $H$.

The argument. For points $x\in W$ there is nothing to do. If $H$ is a hypersurface, every point $x\in H$ has an open neighborhood $U$ such that $U\setminus H$ has two connected components $A,B$, and $\overline A=U\setminus B$ is a manifold with boundary $U\cap H$, and viceversa. Since $$D:=W\cap(U\setminus H)=\overline W\cap(U\setminus H),$$ the non-empty set $D$ is open and closed in $W\cap(U\setminus H)$. It follows either $D=A$, or $D=B$, or $D=A\cup B$. Hence either $\overline D\cap U= U\setminus B$ or $\overline D\cap U= U\setminus A$, which are both manifolds with boundary $U\cap H$, or $\overline D\cap U$, which is a manifold without boundary.

Hope this is not redundant!

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