Proof that the range of a map is determined by its behaviour on the boundary. Let f be a mapping from an open neighbourhood of the 3-dimensional unit ball to the 2-dimensional plane. Suppose that f is smooth (infinitely continuously differentiable on its domain) and regular (it's derivative, as a 2x3 matrix, has rank 2 everywhere).
I would like a proof, or informed hint, or counterexample, for the claim that the f restricted to the closed unit ball has the same range as f restricted to the unit sphere.
It is not difficult to construct a counterexample when f is allowed to be nonregular.
I have stated the problem for a mapping from 3 to 2 dimensions but there may be a similar problem going from n to n-1 dimensions. For n=2 the proof is trivial: a regular differentiable mapping from the closed unit disk to the line reaches its maximum on the boundary circle.
 A: This is a very interesting problem. Despite originally thinking the result was true and posting a flawed "proof" here, I can now show that it is false. To do this, I will make use of a published result to imply the existence of a counterexample, although a more explicit construction would be desirable.
First, a bit of notation. Regular maps as stated in the question are submersions. A knot is a circle embedded in $\mathbb{R}^3$ (we only consider smoothly embedded knots here), and a link $L\subset\mathbb{R}^3$ is a union $L=\bigcup_{i=1}^nL_i$ of a finite pairwise disjoint set of knots, $\{L_1,L_2,\ldots,L_n\}$.
Given any disjoint pair $L_i,L_j$ of knots, let $lk(L_i,L_j)$ be their linking number. I will use a result by Gilbert Hector and Daniel Peralta-Salas1. A subset $L$ of $\mathbb{R}^3$ is strongly integrable (SI) if $L=\Phi^{-1}(0)$ for some smooth submersion $\Phi\colon\mathbb{R}^3\to\mathbb{R}^2$. Now, quoting from Hector & Peralta-Salas.
Theorem 3.6.11. A link in $\mathbb{R}^3$ is SI if and only if
$$
\sum_{j\not=i}lk(L_i,L_j)=1\ mod.\ 2
$$
for all $i\in\{1,\ldots,n\}$.
For example, this condition is satisfied for the Hopf link $L=L_1\cup L_2$ for a pair of knots $L_1,L_2$ with linking number 1, so $L$ is SI. For our counterexample, we only require that there does exist a link which is SI. As links are compact, they are bounded in $\mathbb{R}^3$ so, by scaling, we can suppose that $L$ is contained in the open ball of radius $1$. Then, there is a smooth submersion $\Phi\colon\mathbb{R}^3\to\mathbb{R}^2$ such that $L=\Phi^{-1}(0)$. So, $0$ is in the image of $\Phi$ restricted to the closed unit ball, but is not in the image of $\Phi$ restricted to the unit sphere, giving the required counterexample.
1 Hector, G., Peralta-Salas, D: Integrable embeddings and foliations, American Journal of Mathematics, 134, 773-825 (2012), doi: 10.1353/ajm.2012.0018, available at arXiv:1012.4312.
