TITLE: Exercise in Do Carmo's "Riemannian Geometry": the Möbius band is nonorientable.
QUESTION [11 upvotes]: Of course there are many ways to prove this.  However, I came across the following exercise (Ch. 0 #3).

Prove that:
  (a) a regular surface
  $S\subset \mathbb{R}^3$ is an
  orientable manifold if and only if
  there exists a differentiable mapping
  of $N:S\rightarrow \mathbb{R}^3$ with
  $N(p)\perp T_p(S)$ and $|N(p)|=1$, for
  all $p\in S$. (b) the Möbius band
  (Example 4.9 (b)) is non-orientable.

In Example 4.9 (b), he constructs the Möbius band as the quotient by the antipodal map of the cylinder $C=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2=1,|z|<1\}$.  The problem, of course, is that this isn't given as a surface in $\mathbb{R}^3$!  I was thinking for a second that maybe I should try and construct a map $C\rightarrow S^2$ with the right properties and check that it doesn't descend to a map on $M$, but that's stupid because if I were to embed $M\subset \mathbb{R}^3$, I'm pretty sure it couldn't possibly have those tangent planes anyways.
Does anyone have any insight?  Presumably the solution to (b) should use the fact given in (a).

REPLY [2 votes]: What doCarmo might have had in mind: 
I didn't check, but I think the quotient map 
$$ \pi: C \to M $$
is precisely the orientation covering of $M$. And one can prove that this covering is connected iff $M$ is not orientable (see Lee "Intro to Smooth Manifolds", p. 331 for example).
If the above is not useful: Maybe you could try to pull back the orientation of $M$ to $C$ and get a contradiction or so. (although this really has not much to do with part a) of the exercise, so it might be that doCarmo wants you to imbed the Möbius strip)