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Let $a$ be in a point in $\mathbb{R}^m$.

$f$ is a partial function from $\mathbb{R}^m$ to $\mathbb{R}^n$, where $m\geq n$, that is differentiable in a neighbourhood $V$ of $a$ and continuously differentiable at $a$.

I'm trying to prove that if $J_{f_a}$ has full rank then $f(V)$ is a neighbourhood of $f(a)$.

I figured that if $m=n$ then this means that $D_{f_a}$ is invertible and the proof is obtained via the inverse function theorem. Assuming this is correct, then to handle the case where $m>n$ I would have to form some other function from $f$ so that I can again use the inverse function theorem somehow. The problem is I simply cannot see what kind of function I can form.

I even tried considering a concrete example with $(x,y,z) \mapsto (x+y,z)$ to see which kind of function would be helpful here (maybe $(x+y,z,x)$?), but I'm struggling to generalise this.

I also thought of somehow using the implicit function theorem because of the similar hypotheses and that it allows for dimensionality reduction, but I'm not really seeing how it could be of help.

What am I missing?

Anon
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  • Does https://math.stackexchange.com/questions/1607049/show-that-a-submersion-is-open help ? – NaNoS Mar 02 '24 at 11:31
  • @NaNoS Thanks, but I'm not familiar with half the terminology there so it's a bit too advanced for me to see the relation. – Anon Mar 02 '24 at 11:34
  • Being a submersion is the same thing as $J_{f_a}$ having full rank. Unfortunately I will be unavailable until tonight, I can give a bibliographic reference of the result if you are interested and if you can access it: An Introduction to Manifolds, Loring W.Tu. You can read from page 343, it does not take too much preliminary material. If you need help, I’ll be back tonight. – NaNoS Mar 02 '24 at 13:43
  • @NaNoS Wow, thanks a lot! Will try to read about this and see how far I can get. – Anon Mar 02 '24 at 14:10

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