Why is the Möbius strip not a product manifold despite being parameterizable as $(\theta, h)$?

Question

I was reading this^† lecture note and came across to the section on fibre bundles. On page 2 they mention that there is no unambiguous and continuous way to write a point $m$ on a Möbius strip $M_0$ as a Cartesian pair $(s,t) \in S^1 \times L$, and therefore conclude that the Möbius strip is not a product manifold globally. However, in this Quora answer an explicit parameterization of the Möbius strip in terms of $(\theta, h)$ is shown where $\theta \in [0, 2\pi]$ and $h\in [-1, 1]$ i.e.,

$$\displaystyle(X(\theta,h),Y(\theta,h),Z(\theta,h)) = (\cos(\theta)\left(R+h\cos(\frac{\theta}{2})\right),\sin(\theta)\left(R+h\cos(\frac{\theta}{2})\right),h\sin(\frac{\theta}{2}))$$

Could someone clarify this apparent contradiction?

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^{†: Topology of Fibre bundles and Global Aspects of Gauge Theories (Andres Collinucci & Alexander Wijns, 2006) [arXiv:hep-th/0611201]}

Answer 1

The explicit parameterization you've given has different values when $\theta = 0$ and $\theta = 2\pi$. Thus, while's it's a continuous map from $[0, 2\pi] \times L$ to the Möbius strip, it doesn't descend to a continuous map from $S^1 \times L$ to the Möbius strip.