A free action need not have a manifold quotient

Fix an irrational number $\alpha$ and let $\mathbb{R}$ act on ${\mathbb{T}}^2$ by $t\cdot (z,w)=(e^{2\pi it}z,e^{2\pi i\alpha t}w)\mathrm{.}$ By [F1], this is a smooth free action.

Its orbit quotient ${\mathbb{T}}^2\mathrm{/}\mathbb{R}$ is not Hausdorff. Indeed, the identity orbit is proper because its intersection with $\{1\}\times S^1$ is the countable set $\{(1,e^{2\pi i\alpha n}):n\in \mathbb{Z}\}$ rather than the whole circle, while it is dense by [F1]. If the quotient were Hausdorff, its singleton orbit classes would be closed, and continuity of the quotient map would make every orbit closed, a contradiction.

By [F2], a non-Hausdorff space is not a topological manifold and therefore cannot be a smooth manifold. Hence the free action in step 1.1 has no manifold orbit space, refuting the claim. The contrast with [F3] identifies properness, not freeness, as the missing hypothesis. No choice principle is used.