I study mathematics as a hobby and therefore do not have access to a professor to check if my work is correct, so I am coming here for help. Can you please check if my answer is correct?
The exercise is as follows:
If $f: U \to \mathbb{R}^3, U \subset \mathbb{R}^4$ an open set, is $\mathcal{C}^1$ in $U$ and has rank $3$ everywhere in this set, then $|f(x)|$ does not have a maximum for $x \in U$.
My proof:
The conditions given guarantee $f$ is a submersion. Let $\mathbb{R}^4=\mathbb{R}^1\oplus\mathbb{R}^3$ be a direct sum decomposition wherein $f'(x)|\mathbb{R}^3\to\mathbb{R}^3$ is an isomorphism for all $x \in U$. From the local form of submersions, we know there is a homeomorphism $h:V\times W\to Z; x\in V \subset\mathbb{R}^1, f(x) \in W\subset\mathbb{R}^3, (x,f(x))\in Z \subset\mathbb{R}^4$, V, W and Z all open, such that $$fh(x,w)=w\;\forall(x,w)\in V \times W$$ Write $$h^{-1}:Z\to V \times W; h^{-1}(a)=(a_1, a_2)$$ Now, $f(a)=fhh^{-1}(a)=fh(a_1,a_2)=a_2$. But $f$ is a submersion, hence it is an open mapping, meaning $f(U)$ is open, where then $|f(a)|=|a_2|$ cannot attain a maximum.