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.
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 $N$.