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).
Best Answer
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)