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From these:

Does open map imply dimension of domain is greater than or equal to dimension of range? (please don't use sard's theorem)

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

I learned that...

Let $N,M$ be smooth manifolds with respective dimensions $n,m$ and $f:N \rightarrow M$ smooth.

  1. Assume that the image $f(N)$ has nonempty interior. Then $n \geq m$.

    • 1.1. Note: Not sure if this holds if $f(N)$ isn't a smooth embedded/regular $m$-submanifold of $M$. (I asked for clarification in comment in answer in question above.)

  2. If $f$ satisfies any of the ff conditions, then $f$ also satisfies the condition 'The image $f(N)$ has nonempty interior' (and $f(N)$ is a smooth embedded/regular $m$-submanifold of $M$).

    • 'open'

    • 'surjective'

    • 'submersion'

    • 'submersion at some point $p \in N$'

  3. If $f$ satisfies any of the ff conditions, then $n \leq m$.

    • 'injective'

    • 'immersion'

    • 'immersion at some point $p \in N$'

Question 1: Is there some weaker condition that the conditions in (3) have in common analogous to how the conditions in (2) have the common weaker condition 'The image $f(N)$ has nonempty interior'? (Hopefully, the word 'analogous' here rules out the trivial answer of '$n \leq m$'.)

Question 2: Btw, very briefly (or say more if you want), what if anything does closed map imply?

  • (2A. I just figure: since I'm asking all the above, then, hey, maybe 'closed' fits somewhere above. But maybe not since I guess immersion doesn't imply closed the way submersion implies open.)
BCLC
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    If $M=]-\pi/2,\pi/2[$, then $f(x) = x \in M$, $g(x) = (x,\tan{x}) \in \mathbb{R}^2$ and $h(x) = (x,0)\in \mathbb{R}^2$ respectively have open, closed and nor closed nor open image, and are all injective immersions. Thus, I don't think we can have any topological property on the image $f(M)$ derived from an injective / immersion asumption – Didier May 24 '21 at 11:12
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    Moreover, for the question "what do immersion and injective functions have in common" I can only give this answer: If $u : M \to N$ is an injective $C^1$ map, then $\dim M \leqslant \dim N$ and there exists a dense open subset $W \subset M$ such that $u : W \to N$ is an immersion (this is a clever application of the constant rank Theorem). – Didier May 24 '21 at 11:16
  • thanks @Didier ! – BCLC May 24 '21 at 11:42

1 Answers1

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For Q1 the common property is "nowhere open" meaning there is no nonempty open subset whose image is open. You do not even need smoothness of the map, continuity will suffice. Maps on your list in part 3 all have this property and nowhere open implies that the image has empty interior. However, it does not imply that $n\le m$.

For Q2: Nothing of interest to you.

Moishe Kohan
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