Do unit normal vectors of extremal points on a compact hypersurface have opposite directions? Problem
To be definite, by a hypersurface of $\mathbb R^{n+1}$, I mean a connected submanifold of $\mathbb R^{n+1}$ of dimension $n$ without boundary.
Suppose $M$ is a compact smooth orientable hypersurface of $\mathbb R^{n+1}$ with the Gauss map $N\colon M\to S^n$ and $0\neq v\in\mathbb R^{n+1}$. Let $f_v(x)=\langle x,v\rangle$ be a so-called height function. Since $M$ is compact, we can take a maximal point $p\in M$ and a minimal point $q\in M$ of $f_v$ restricted on $M$. It's not difficult to show that $N(p),N(q),v$ are collinear. Is it true that $N(p)+N(q)=0$, i.e., $N(p),N(q)$ are of opposite direction?
Discussion
If $M$ is a level hypersurface, i.e., the zero set of a submersion $F\colon\mathbb R^{n+1}\to\mathbb R$, then the statement is true. The idea of the proof is depicted as follows. Take a sufficient large sphere $S\subseteq\mathbb R^{n+1}$, and then we can use it to construct a continuous path $\alpha\colon[a,b]\to\mathbb R^{n+1}$, differentiable at $p$ and $q$, such that $\alpha(a)=p,\alpha(b)=q,\alpha'(a)=\alpha'(b)=v$, but $\alpha(t)\not\in M$ for $a<t<b$. For more details, see J. A. Thorpe's Elementary Topics in Differential Geometry, Chapter 6, page 33.
If the compactness is removed, even if $f_v$ attains its maximum and minimum, the statement is false. Consider $n=1$ and $M$ to be the graph of $y=\sin x$ and take $v=(0,1)$.
I heard that there's a magic theorem that any compact (smooth) hypersurface in $\mathbb R^{n+1}$ is a level hypersurface and therefore orientable. I don't want to appeal to such a theorem. I want a direct proof for this.
Any idea? Thanks!
 A: Discussion:
What you're trying to show is global in nature. The example of the sinusoidal curve you describe in $\mathbb{R}^2$ or the twice-twisted annulus Behaviour describes in $\mathbb{R}^3$ shows this. You'll need techniques that respect this global nature of your result.
The result you want, being global in nature, seems to me just as hard to prove as the statement that any closed, connected, dimension $n$ submanifold $M$ of $\mathbb{R}^{n+1}$ separates $\mathbb{R}^{n+1}$ into an unbounded piece and a bounded piece (i.e. $\mathbb{R}^{n+1}-M$ has two components, one unbounded and one bounded) such that, at $p \in M$, one (local) side of $M$ is in the unbounded component, and the other (local) side of $M$ is in the bounded component. 
This fact actually proves that $M$ is oriented, where the orientation is given by a normal vector pointing into the unbounded component, say. (The other choice would be pointing into the bounded component.)
It also proves your result: There are only two orientations on a connected, oriented manifold, so your orientation must be one of these two. Then, at the max the unbounded component is necessarily up and at the min the unbounded component is necessarily down, giving one orientation satisfying what you want; the other orientation swapping both of these also does what you want.
You need some techniques to give a global result like this. Transversality is the standard tool to do this in differential topology. There are also algebraic topology techniques that would prove this. I'll describe the transversality techniques and how to prove this below.

General Transversality Counts:
A nice reference for the following would be Differential Topology by Guillemin and Pollack.
The mod $2$ transversality intersection count of two submanifolds $A$ and $B$ of dimension $n$ and $k$ inside an $n+k$ dimensional manifold is a count of the number of times they intersect modulo $2$, in the case when they intersect transversally (i.e. "go straight across" each other; the technical condition is that $TA_p + TB_P = TM_p$). When they don't, it's defined by homotoping them slightly so they do intersect transversally and then counting (a general theorem states that such a homotopy is always available and can be made very small in any appropriate sense).
This count is invariant under homotopies in general, which is a way of saying it is getting at something global in nature. The fact that our counts were mod $2$ was essential for this, and the intuitive idea is that if we introduce new intersections by a homotopy, they always come in pairs: for example, think of passing a curve across a line, generating two intersection points. (In the case when everything has an orientation, we can also define these counts with signs, but that won't come into play here.)
This is defined even in the case where $A$ and/or $B$ has boundary, as long as we specify that the boundaries are disjoint and are required to remain so under homotopy. It's also defined when one of the manifolds is non-compact, so long as it's proper (intuitively, runs off to infinity, so that after some point it doesn't interact with the other; formally this means that any compact subset of the ambient manifold intersects it in a compact set).

The transversality counts for this situation:
Let $M$ be a connected, closed, dimension $n$ sub manifold of $\mathbb{R}^{n+1}$.
Given a point $x \in \mathbb{R}^{n+1}-M$, define $m_x$ as the mod $2$ transverse count of intersections of any half-infinite ray starting at $x$ and $M$.
I claim this is independent of the ray chosen. To see this, consider two rays. I claim they have the same count mod $2$. Stick the two rays together. We must show this $V$ shape intersects the manifold zero times mod $2$. To see this, close it up by a circular arc at some large distance (out past some finite ball that $M$ is contained in). Then we have a closed circle in $\mathbb{R}^{n+1}$ we are intersecting with $M$ with the same mod $2$ intersection count as our $V$ shape. But any circle in $\mathbb{R}^{n+1}$ is homotopic any other; in particular it's homotopic to a tiny circle that misses $M$, so the count is zero mod $2$.

Using the counts:
We are to show that there are precisely two components to $\mathbb{R}^{n+1} - M$, an unbounded component and a bounded component.
Let $C_0 = \{x \in \mathbb{R}^{n+1} - M : m_x = 0\}$ and $C_1 = \{x \in \mathbb{R}^{n+1} - M : m_x = 1\}$.
These are disjoint and are also open (by the fact the counts remain the same under homotopy where the boundary point of the ray doesn't hit $M$).
Additionally, near any point in $M$, by the local form of an immersion, points on opposite sides locally have counts that differ by $1 \mod 2$.
Finally, every point outside a large ball containing $M$ must be in $C_0$ (take a ray heading outward, away from the ball), so that's the unbounded one and $C_1$ is bounded.
This finishes the claim up in the discussion section, which then proves your result as discussed there.

It isn't necessary to actually show each of $C_0$ and $C_1$ is connected, but here's a summary of that:
To see each is connected, I claim that a tubular neighborhood $T$ of $M$ (see Guillemin & Pollack e.g. for discussion of this if it isn't familiar) in $\mathbb{R}^{n+1}$ such that $T-M$ has at most two components. The at most two components part can be seen by local form for an immersion plus the fact that any two points in $M$ can be connected by a compact arc.
Now, from any point $x \in C_0$, draw a ray starting at $x$. This ray has a first time it hits $M$. Just prior to that it was inside $T-M$. This shows that any two points in $C_0$ are in the same connected component of $C_0$. Similarly for $C_1$.
