I was reading differential geometry from a lecture note, there I found the following theorem,
If $S$ is a compact connected oriented $n-$ surface in $\mathbb{R}^{n+1}.$ Then the gauss map is surjective.
Now in the proof , the picture of the surface $S$ sketched as if $S$ is homeomorphic to $S^{n-1}$ or$D^{n}.$
I am wondering that, is it true?
Is there any characterisation of $n-$ surfaces?
Definition: An $n-$ surface $S$ is a level curve of a smooth function whose gradient is non-zero on $S.$
No. Consider the following function: $$ f(x,y,z)=(x^2+y^2-1)^2+z^2 $$ Note that $f^{-1}(\{0\})$ is the unit circle in the $xy$-plane, while $f^{-1}(\epsilon)$ is a torus around that circle for sufficiently small $\epsilon>0$. For instance, the $\epsilon=1/2$ case looks like this:
One can in fact obtain any oriented closed surface as the regular level set of a smooth function (see here for some examples).
As an aside, the unit disc (open or closed) is not applicable to this statement (since it concerns compact surfaces without boundary), and indeed the Gauss map of a disc need not be surjective.