# When does the topological boundary of an embedded manifold equal its manifold boundary?

Suppose I embed a manifold-with-boundary $M$ in some $\mathbb{R}^n$. Are there conditions (necessary, sufficient, or both) that can help determine when the topological boundary of $M$ is equal to the manifold boundary?

By "topological boundary," I'm referring to $\text{Bd } M$, which is the closure minus the interior (relative to $\mathbb{R}^n$).

By "manifold boundary," I mean the boundary $\partial M$ that is specified in the definition of "manifold-with-boundary."

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## 1 Answer

If you embed a smooth $m$-manifold $M$ smoothly in $\mathbb{R}^n$ then the "topological boundary" of $M$ is the closure of $M$. As locally each point of $M$ has a neighbourhood in $\mathbb{R}^n$ where $M$ looks like $\mathbb{R}^m$, then no point of $M$ is interior.

When $m=n$ and $M$ is compact then, yes (at least in the smooth case) the topological and manifold boundaries coincide.

I expect the above hold for topological embeddings but won't swear to it; they need not be locally flat (see nasties like the Alexander horned sphere),

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@Rasmus: Why might that be? It seems correct to me... –  Jesse Madnick Sep 30 '10 at 6:53
@Jesse Madnick: You are right, thanks. –  Rasmus Sep 30 '10 at 8:51