= Normal reversal around the Möbius core
{title2=$n(2\pi)=-n(0)$}
For the model $F$ of the <Möbius band>, the <tangent vectors> along its <core circle of a Möbius band> are
$$
F_\theta(\theta,0)=(-R\sin\theta,R\cos\theta,0),\qquad F_w(\theta,0)=(\cos(\theta/2)\cos\theta,\cos(\theta/2)\sin\theta,\sin(\theta/2)).
$$
Their <cross product>, divided by its <norm> $R$, gives the continuous <unit normal>
$$
n(\theta)=\bigl(\cos\theta\sin(\theta/2),\ \sin\theta\sin(\theta/2),\ -\cos(\theta/2)\bigr).
$$
Although $F(2\pi,0)=F(0,0)$, we have $n(2\pi)=-n(0)$. Any other continuous choice of <unit normal> along this parametrized interval is either $n$ everywhere or $-n$ everywhere: its relative sign is a continuous map into the <discrete space> $\{+1,-1\}$, and the interval is <connected>. Thus no choice can agree at the identified endpoints. This short calculation proves that the <Möbius band> has no global continuous <unit normal> and is not an <orientable surface>.
The plotted quantity is $n(\theta)\cdot n(0)=\cos(\theta/2)$. It compares ambient <vectors> at generally different points; after one circuit it is $-1$ at the starting point, and after two circuits it is $+1$ there. Only these endpoint comparisons are needed for the <orientability> argument.
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