Transverse orientability is load-bearing in the global codimension-one form

Assume the standing countable choice ${\mathrm{A}\mathrm{C}}_{\omega }$ (The countable-choice principle used in the foliation pair); the following finite quotient construction needs no further choice. On $S^2\times S^1$, with $\theta \in \mathbb{R}\mathrm{/}\mathbb{Z}$, consider $\tau (x,\theta )=(-x,-\theta )$. This involution is free, because the antipodal map on $S^2$ has no fixed point. Small disjoint neighborhoods of a point and its image give smooth quotient charts, so $X=(S^2\times S^1)\mathrm{/}⟨\tau ⟩$ is a closed connected smooth three-manifold.

The product foliation descends. For $\theta \mathrm{\ne }0,1\mathrm{/}2$ a pair of slices at $\theta ,-\theta$ has image diffeomorphic to $S^2$. At $0$ and $1\mathrm{/}2$ the slice is identified antipodally and its image is $\mathbb{R}{P}^2$. All these leaves are compact and have finite fundamental group. The leaf space is the quotient of the circle by reflection, hence a closed interval, with the two projective-plane leaves at its endpoints.

The descended foliation is not transversely orientable (Transversely oriented codimension-one foliations). Indeed a hypothetical nonzero coorientation would pull back to $a(x,\theta )d\theta$ on the connected product, where $a$ is a continuous nowhere-zero function. Invariance under $\tau$ requires $a(-x,-\theta )=-a(x,\theta )$, impossible because a continuous nowhere-zero real function on a connected space has constant sign. Thus the common-leaf and circle-fibration conclusions of global Reeb stability fail when transverse orientability is removed. Under full AC this is a counterexample to removing just that hypothesis from the global theorem, rather than an application of its proof under countable choice alone.