Consider a manifold smooth manifold $N$ smoothly embedded in another manifold $M$ of the same dimension. Is it true that $N$ is open in $M$? I think this is true, due to the open mapping theorem.
If so, is this also true with less regularity?
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What do you mean by "less regularity"? – Paul Frost Oct 15 '18 at 23:12
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For example, continuous embeddings, or worse, topological manifolds. – W. Rether Oct 15 '18 at 23:16
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It is true for topological manifolds (without boundary). This is a consequence of the invariance of domain. See for example
Version of Invariance of Domain for n-manifolds
https://en.wikipedia.org/wiki/Invariance_of_domain
Given an embedding $f : N \to M$, each $x \in N$ has an open chart neighborhood $U$ which is mapped homeomorphically into an open chart neighborhood $V$ of $f(x)$. Both $U,V$ are homeomorphic to open subsets of $\mathbb{R}^n$, hence invariance of domain implies that $f(U)$ is open in $V$ and therefore open in $M$. Since $f(U) \subset f(N)$, we see that $f(N)$ is open in $M$.
Paul Frost
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Yes.
A smooth immersion of one manifold into another of the same dimension is also a submersion. Submersions are open maps. Thus, the image of $N$ is open in $M$.
Santana Afton
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1Proving that submersions are open maps is a good exercise, if this is something you’re not already familiar with. – Santana Afton Oct 15 '18 at 23:19
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