# Möbius Strip is no orientable

This is an exercise from Do Carmo's Riemannian Geometry book.

Let $G=\{Id,A\}$, $C= \{ (x, y, z) \in \mathbb{R}^3; x^2 + y^2 = 1, -1 < z < 1 \}$, where $A(p)=-p$. Define $\frac{C}{G}$ the Möbius band.

1. Let $G \times M \rightarrow G$ be a properly discontinuous action of a group G on a differentiable manifold $M$.

(a) Prove that the manifold $\frac{M}{G}$ is orientable if and only if there exists an orientation of $M$ that is preserved by all the diffeomorphisms of $G$.

I already did the part a). I would like to know how I use it to prove item b).

(b) Use (a) to show that the Möbius band are non-orientable.

The negation of item a) would be: The manifold $\frac{M}{G}$ is not orientable if and only for all orientation of $M$ there exist $\alpha, \ \beta$ such that $\hbox{det}(d(x_{\beta}^{-1} \circ (\varphi_g \circ x_{\alpha})))<0$ for some $g \in G$?

In the case of surfaces it is obvious because we can use normal vector fields. Now, only with the intrinsic topology of the quotient manifold I do not know how to do.

There is some diffeomorphism of this manifold with the surface " Möbius band"?