Are open immersions equivalent to local diffeomorphisms? (not algebraic geometry)

Question

Local diffeomorphisms are both open maps and immersions.

For the other direction:

Let $N$ and $M$ be smooth manifolds with dimensions. Let $p \in N$. Let $F: N \to M$ be a smooth map. Assume $F$ is an open map. Let $F$ be an immersion at p.

1. Because $F$ is an immersion at $p$, $\dim N \le \dim M$ at $p$ and thus at every $q \in N$ because $N$ and $M$ have dimensions.

2. Because $F$ is smooth and open, $\dim N \ge \dim M$, by this rule, from Momentum Maps and Hamiltonian Reduction By Juan-Pablo Ortega and Tudor Ratiu, which I now paraphrase:

Let $N$ and $M$ be smooth manifolds with dimensions. Let $p \in N$. Let $F: N \to M$ be a smooth map. If $F$ is open, then $\dim N \ge \dim M$.

(I know submersions are both open maps and imply $\dim N \ge \dim M$, but I didn't know openness is enough to imply $\dim N \ge \dim M$. Then again, I haven't studied this book. I just found some pages of it in a google search. I could be wrong in applying or understanding this rule since the definitions of manifold or smooth might be different)

3. By (1) and (2), $\dim N = \dim M$.

4. By (3) and this, $F$ is a local diffeomorphism at $p$ if and only if $F$ is an immersion at $p$ (if and only if $F$ is a submersion at $p$).

5. By (4), $F$ is a local diffeomorphism at $p$.

6. Therefore, by (5), open and immersion at $p$ implies local diffeomorphism at $p$.

7. I don't believe "openness" is defined pointwise. Therefore, by (6), open and immersion (everywhere) implies local diffeomorphism (everywhere)

Answer 1

Yes.

Having an open immersion guarantees that $F$ is a map between manifolds of equal dimension, and we know that immersions are equivalent to local diffeomorphisms in that context.