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My book is An Introduction to Manifolds by Loring W. Tu.

Let $N$ and $M$ be smooth manifolds with dimensions. Let $p \in N$. Let $F: N \to M$ be a smooth map.

Question 1. Are these correct?

A. If $F$ is injective, then $\dim N \le \dim M$, by this (from Momentum Maps and Hamiltonian Reduction by Juan-Pablo Ortega, Tudor Ratiu)

B. If $F$ is open, then $\dim N \ge \dim M$, by this (from Momentum Maps and Hamiltonian Reduction by Juan-Pablo Ortega, Tudor Ratiu).

C. If $F$ is a submersion, then $F$ is open and so $\dim N \ge \dim M$. Alternatively, we may use this.

D. If $F$ is an immersion, then $\dim N \le \dim M$, by this.

Question 2. Given that injections, immersions and submersions imply either $\dim N \le \dim M$ or $\ge$, I guess surjections imply one of those too. Which if any does surjection imply, and why?

  • I think it's the same as submersion (and open): $\dim N \ge \dim M$.

  • An example would be retraction $r(x) = \frac{x}{||x||}, r: \mathbb R^2 \setminus 0 \to S^1$ (I recall this is smooth. Not sure). At the very least, I think the example (if correct) proves that surjections definitely do not imply $\dim N \le \dim M$.

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    $\begingroup$ 1C and 1D follow just by definition of submersion/immersion, considering one single tangent space. 1A and 1B are right (I’d prefer a complete argument based on constant-rank theorem, but okay). For 2, if $F:N \rightarrow M$ is smooth surjective, then $\dim\,N \geq \dim\,M$ (there is an issue with Sard’s theorem otherwise). $\endgroup$ – Mindlack Jul 24 '19 at 9:30
  • $\begingroup$ @Mindlack Thanks! $\endgroup$ – user636532 Jul 24 '19 at 9:32
  • $\begingroup$ @Mindlack Sard's Theorem is related to 1A or 1B? 1A looks like an injective version for Sard's Theorem. $\endgroup$ – user636532 Jul 24 '19 at 9:39
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    $\begingroup$ Sard is for your question 2. I suppose it also works for question 1B (if the dimension of the source is smaller, the only regular values are the values not reached by the function, which is a full-measure closed subset, thus the entire manifold, a contradiction). $\endgroup$ – Mindlack Jul 24 '19 at 10:01

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