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)
Let $N$ be smooth $n$-manifold. Let $M$ be smooth $m$-manifold. Let $F: N \to M$ be smooth map.
What I understand:
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A. If $F$ is a submersion on $N$, then $n = \dim N \ge \dim M = m$.
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B. If $F$ is a submersion on $N$, then $F$ is open.
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C. If $F$ is smooth surjective, then $n = \dim N \ge \dim M = m$.
Questions:
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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?
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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:
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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 $M$.
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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.)
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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 $\mathbb R^2$.
Best Answer
Here are answers for the two questions: