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?
Is a manifold $N$ smoothly embedded in a manifold $M$ of the same dimension open in $M$
differential-geometrygeneral-topologymanifoldssmooth-manifolds
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Best Answer
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 $M$.