# Submanifold given by an open immersion

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.

• Is $\dim M \le \dim N$ redundant please? I think it's implied by immersion. – user636532 Jul 23 '19 at 8:44
• 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. – user636532 Jul 23 '19 at 8:47
• Is the converse true? – user636532 Jul 24 '19 at 7:23

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 $$N$$.
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 $$N$$.
• "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" – user636532 Jul 27 '19 at 7:53