The zero section $C=\{t=0\}\subset M$ is a circle. Its <normal bundle> is the real line bundle obtained from
$$
(x,v)\sim(x+2\pi,-v),
$$
so its clutching map reverses sign once around $C$. Its <mod-two Euler class of a real line bundle>, equivalently its first <Stiefel–Whitney class>, therefore satisfies
$$
\langle w_1(\nu_C),[C]_{\mathbb F_2}\rangle=1.
$$
If $M$ were orientable, the splitting
$$
TM|_C\cong TC\oplus\nu_C
$$
and the orientation of the circle would orient $\nu_C$, forcing this mod-two Euler number to vanish. This contradiction proves that the <Möbius band> is nonorientable.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 / 1 / b / Solution

{title2=$e_2(L)=w_1(L)$}

For a real line bundle, the mod-two Euler class is its top <Stiefel–Whitney class>, namely $w_1(L)\in H^1(X;\mathbb F_2)$. Its evaluation on a closed curve records whether parallel transport reverses the fiber orientation.

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