# Is the Gauss map $N : \partial \Omega \rightarrow \mathcal{S}^{d-1}$ an open map?

Let $$\Omega \subset \mathbb{R}^{d}$$ be open, bounded and of class $$C^{1}$$. The Gauss map $$N : \partial \Omega \rightarrow \mathcal{S}^{d-1}$$ is defined as the unit outward normal map to $$\Omega$$ at each $$x\in\partial\Omega$$. I want to prove (or to find a reference) that $$N$$ is surjective and open.

For the surjective part, I have been able to find this result for an even dimension in terms of the Brouwer degree of $$N$$ and the Euler characteristic, and other results such as (Proposition 4.33, Differential Geometry of Curves and Surfaces (Tapp, K.; 2016)); but I am sure there exists a global reference independent of the dimension. For example, from (Theorem, Chapter 6, Elementary Topics in Differential Geometry (Thorpe, J. A.; 1979)) is independent of the dimension and shows a local result similar to what I need, but not quite the one. The introduction to the theorem suggests that the result can be made globally.

For the openness of the Gauss map I have not been able to find a satisfying reference, which I am quite sure there exists since some authors mention that $$N$$ is open in $$\mathbb{R}^{2}$$ and $$\mathbb{R}^{3}$$ and it seems a logical result.

Any help is welcome, thank you.

• You asked if the map $N:\partial \Omega \to \mathbb R^d$ is surjective and open. Then map is not surjective, since it is sending each point $x$ to a unit vector, so the image is contained in $S^{d-1}$ Commented Aug 1 at 9:41
• @FedericoFallucca Thanks for noticing, I corrected the mistake! Commented Aug 1 at 9:44
• I am not sure about the Gauss map being open. There seems to be a simple counterexample for domains in $\mathbb R^2$: Just take the unit square, round off the corners by small circles and look at the interior of the resulting closed curve. This should be a $C^1$-domain and the Gauss map is just the unit normal to the closed curve. But of coures this is constant on open parts of the curve whose image under the Gauss map therefore is not open. I don't really see why something similar should not be possible in higher dimensions. Commented Aug 2 at 8:54
• I agree with @Andreas. Moreover, you want $\Omega$ to be a $C^1$ $d$-dimensional manifold with boundary. Saying an open subset is $C^1$ is tautological. Commented Aug 3 at 22:05