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I was wondering if the following is true:

Let $M,N$ be two manifolds such that $\dim M\leq \dim N$ and $f:M\rightarrow N$ an smooth immersion.

Assume that for any open set $U\subset M$, $f(U)$ is open in $f(M)$, does it imply that $f(M)$ is a submanifold of $N$ ?

I know that if we also ask $f$ to be injective, then it is an embedding and $f(M)$ is automatically a submanifold of $N$. But without this assumption, I am not sure that the result holds.

Being an open map on its image somehow tells us that there is no bad self-intersection in $f(M)$ but I am not sure this is enough to have a submanifold.

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  • $\begingroup$ Is $\dim M \le \dim N$ redundant please? I think it's implied by immersion. $\endgroup$ – user636532 Jul 23 '19 at 8:44
  • $\begingroup$ Actually, I notice $f: M \to N$ is actually not given as open. If it were, I think we could argue (that $f$ is a local diffeo or at least) that $f(M)$ is open. What's given is that $\tilde f: M \to f(M)$ is open, i.e. $f$ is open onto its image. Therefore, I think the title should be changed. $\endgroup$ – user636532 Jul 23 '19 at 8:47
  • $\begingroup$ Is the converse true? $\endgroup$ – user636532 Jul 24 '19 at 7:23
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Since $f$ is a local imbedding and since $\tilde f: M \to f(M)$ is an open map, there exists a coordinate patch $V \subset N$ around every point of $f(M)$ such that $V \cap N = \mathbb{R}^m$. $f(M)$ is therefore a submanifold of $N$.

More precisely:

Fix $p \in M$. Since $f$ is a local imbedding, there exist open sets $U \subset M$, $V \subset N$, $p \in U$ and charts $\phi: U \rightarrow \mathbb{R}^m, \psi: V \rightarrow \mathbb{R}^n$ such that $\psi \circ f \circ \phi^{-1}$ is the inclusion $\mathbb{R}^m \subset \mathbb{R}^n$. Now since $f(U)$ is open in $F(M)$ we may shrink $V$ (if necessary) so that $V \cap f(M) = f(U) = \psi^{-1}(\mathbb{R}^m)$. The existence of such a chart, $(V, \psi)$, at every point $f(p)$ of $f(M)$ is a necessary and sufficient condition for $f(M)$ to be an imbedded submanifold of $N$.

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    $\begingroup$ @SeleneAuckland are you asking whether a topological imbedding is an open map to its image? $\endgroup$ – Tim kinsella Jul 25 '19 at 21:08
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    $\begingroup$ btw talking about a subset of a manifold being a sub manifold can get very confusing. imo it's better to talk about maps (either injective immersions or imbeddings) as submanifolds. $\endgroup$ – Tim kinsella Jul 25 '19 at 21:14
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    $\begingroup$ then if you still want to talk about subsets, you could say "image of an injective immersion" or "image of an imbedding" $\endgroup$ – Tim kinsella Jul 25 '19 at 21:16
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    $\begingroup$ @SeleneAuckland it seems to me the answer is yes: N --> F(N) is a smooth bijection with non degenerate derivative between two manifolds. so it's a diffeomorphism. $\endgroup$ – Tim kinsella Jul 26 '19 at 3:58
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    $\begingroup$ "Since $f$ is a local imedding and an open map" Actually $\tilde f: M \to f(M)$ is the open map, not necessarily $f: M \to N$? I think you mean "Since $f$ is a local imbedding and an open map onto its mage" $\endgroup$ – user636532 Jul 27 '19 at 7:53

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