Question about the proof that the Gauss map on a compact surface is surjective

Question

I'm reading through Kristopher Tapp's Differential Geometry of Curves and Surfaces, and it has a proof (first suggested in one of the exercises, then later written out in the proof of one of the propositions) that the Gauss map $N$ on a compact surface is surjective. The proof given involves choosing an arbitrary unit vector $v$, taking a plane orthogonal to that vector far away from the surface, then moving it toward the surface until it just touches the surface at a point $p$ (i.e. taking the maximum of $H_v(p) = \langle p, v\rangle$). We will then have $N(p) = v$. The same proof appears in several of the questions here on StackExchange, but neither the book nor those questions fully address the possibility that $N(p) = -v$ (except, in some, by way of mentioning the Jordan-Brouwer theorem, which the book does not cover). One questioner did ask whether the minimum would produce the necessary point if $N(p) = -v$, but never had their concern addressed. Thus, how do you guarantee you get $N(p) = v$ and not just $N(p) = \pm v$?

Answer 1

Given a unit vector $v$ and the boundedness of the surface $S$, you can pick two hyperplanes perpendicular to $v$, $\langle v,x\rangle=a$ and $\langle v,x\rangle=b$ with $a<b$ far away from the surface in such a way that your surface is in between, $a<\langle v,x\rangle<b$ for all $x\in S$. They have normal vectors $v$ and $-v$ pointing towards the outside (not towards the region where $S$ is located. Then you start moving the one with normal vector $v$. Eventually you touch the surface at a point with normal vector $v$ (not $-v$).