Is Gauss map of a free boundary convex disk a diffeomorphism? I asked this question on MSE, but obtained no answer. Maybe this is the right place to post it.
Let $D$ be a properly embedded free boundary disk in the closed unit ball $\mathbb{B}^3$ of $\mathbb{R}^3$. This means that $D$ is a smooth disk embedded in this ball, $D \cap \partial \mathbb{B}^3 = \partial D$ and this intersection is orthogonal. By orthogonality here I mean this: if $N$ is a unit normal along $D$ (its Gauss map), then $\langle N(x), x \rangle = 0$ for all $x \in \partial D$.
Assume that $D$ is strictly convex, that is to say, the principal curvatures are positive at each point of $D$ with respect to the fixed unit normal $N: D \to \mathbb{S}^2$. Does it follow that $N$ is a diffeomorphism onto its image? Equivalently, is $N$ injective?
The motivation is the following: if $S$ is a closed and connected surface in $\mathbb{R}^3$ which is also convex, then $N : S \to \mathbb{S}^2$ is a local diffeomorphism, hence a covering map. Since $\mathbb{S}^2$ is simply connected, this implies that $N$ is a global diffeomorphism.
What happens when the surface is a disk?
 A: The answer is yes. To show this one can use the fact that any topological immersion (locally one-to-one continuous map) of an n-dimensional disk into a sphere of the same dimension is an embedding (globally one-to-one) for $n\geq 2$, provided only that the map is one-to-one on the boundary of the disk. A proof may be found in
Gauss map, topology, and convexity of hypersurfaces with nonvanishing curvature,
Topology, 41 (2002) 107-117.
So it remains to show that $N$ is one-to-one on the boundary $\partial D$ of the disk $D$. To see this one can extend $D$ to a complete $\mathcal{C}^1$ convex surface by attaching to $\partial D$ all the rays which are orthogonal to $S^2$ from outside. These rays belong to a convex cone $C$ with apex at the center $o$ of $S^2$. Since $D$ has positive curvature, it follows from basic differential geometry that $\partial D$  has positive geodesic curvature in $S^2$, and hence is strictly convex, which in turn yields that $C$ is strictly convex. So $N$ will be one-to-one along $\partial D$, since $N$ is just the Gauss map of $C\setminus\{o\}$.
Incidentally, it is not necessary to assume that $D$ is convex or even embedded, but it is enough that it have positive curvature and satisfy the free boundary condition; see the following paper with Changwei Xiong
Nonnegatively curved hypersurfaces with free boundary on a sphere,
Calc. Var. Partial Differential Equations, 58 (2019), Art. 94, 20 pp.
