Prove that the Möbius band is not orientable.
I know that in the Möbius band the central circle is orientable. If I let $Y$ be the Möbius band and $Z$ be a compact submanifold of $Y$ with $\dim(Z)=\frac 12 \dim(Y)$. From one of my exercise I have show that if $Z$ is globally defined by independent function then $I_2(Z,Z)=0$. The central circle is orientable but it is not definable by an independent function, so there is no smooth choice of orientation for all the tangent space $T_y(Y)$. If $Y$ can't be given an orientation, it's not orientable.
I'm not sure if my explaining is understandable. Please help me improve it.
The key fact here is that a hypersurface $Z$ in an orientable manifold $Y$ is orientable if and only if its normal bundle $N(Z,Y)$ is trivial. Here we have $Z$ the central circle in the Möbius strip $M$, and $I_2(Z,Z)=1$, so, by the Tubular Neighborhood Theorem, $N(Z,Y)$ cannot be trivial.