# Is a manifold $N$ smoothly embedded in a manifold $M$ of the same dimension open in $M$?

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?

• What do you mean by "less regularity"? – Paul Frost Oct 15 '18 at 23:12
• For example, continuous embeddings, or worse, topological manifolds. – W. Rether Oct 15 '18 at 23:16

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 $$M$$.

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 $$M$$.

• Proving 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
• About immersion implies submersion, can you help here please? – Selene Auckland Jul 20 at 7:54