# Dimension of domain is greater than/less than/equal to dimension of range for a smooth surjection/injection/submersion/immersion?

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 $$\dim N \le \dim M$$.

• 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). – Mindlack Jul 24 '19 at 9:30
• @Mindlack Thanks! – user636532 Jul 24 '19 at 9:32
• @Mindlack Sard's Theorem is related to 1A or 1B? 1A looks like an injective version for Sard's Theorem. – user636532 Jul 24 '19 at 9:39
• 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). – Mindlack Jul 24 '19 at 10:01