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?

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


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 $M$.



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 $M$.

  • 1
    $\begingroup$ Proving that submersions are open maps is a good exercise, if this is something you’re not already familiar with. $\endgroup$ – Santana Afton Oct 15 '18 at 23:19
  • $\begingroup$ About immersion implies submersion, can you help here please? $\endgroup$ – Selene Auckland Jul 20 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.