Does open map imply dimension of domain is greater than or equal to dimension of range?

Question

Update: Answer given uses Sard's Theorem. For no Sard's Theorem, see here: Does open map imply dimension of domain is greater than or equal to dimension of range? (don't use sard's theorem)

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Let $N$ be smooth $n$-manifold. Let $M$ be smooth $m$-manifold. Let $F: N \to M$ be smooth map.

What I understand:

• A. If $F$ is a submersion on $N$, then $n = \dim N \ge \dim M = m$.

• B. If $F$ is a submersion on $N$, then $F$ is open.

• C. If $F$ is smooth surjective, then $n = \dim N \ge \dim M = m$.

Questions:

1. Actually I think if $F$ is a submersion at a point $p \in N$ but not necessarily at other points in $N$, then I think $\dim N \ge \dim M$ still because $n = \dim N = \dim T_pN \ge \dim T_{F(p)}M = \dim M = m$. Is this correct?

2. Anyway, the main thing I wanted to ask is: if $F$ is open but not necessarily a submersion, then is $n = \dim N \ge \dim M = m$ still, i.e. under (B), we may relax (A) from submersion on $N$ to open?

All I know so far:

• D. The image $F(N)$ is open in $M$ (I could be wrong, but I think this condition is equivalent to saying that $F$ is open), i.e. $F(N)$ is a smooth embedded $m$-submanifold of $M$.

• E. The induced map $\tilde F: N \to F(N)$ is smooth and open.

  • E.1. $\tilde F$ is the unique map s.t. $\iota \circ \tilde F = F$, where $\iota: F(N) \to M$ is the inclusion map (which is smooth embedding if and only if $F(N)$ is a smooth embedded $k$-submanifold and so is I guess more than a smooth embedding since $k=m$. Update: I believe it's an open smooth embedding, i.e. an injective local diffeo.)

• F. In re (C) and (D), i think codimension $0$ is pretty close to surjective at least when we're thinking about dimensions. Like how an open disk/disc in $\mathbb R^2$ is diffeomorphic to the whole of $\mathbb R^2$.

Answer 1

Here are answers for the two questions:

1. You are perfectly right
2. You are also right but this is more technical. Suppose $f : N^n \to M^m$ is a smooth map and $n < m$. Then every point of $N$ is a critical point (that is, $\mathrm{d}f$ is nowhere surjective). By Sard's theorem, $f(N) \subset M$ has zero measure. It follows that it cannot be open.