PeterDonis said:
It does not meet the requirements for a global coordinate chart, since an open interval of ##\varphi## either does not cover the entire circle, or covers at least one point more than once. To cover every point on the circle exactly once, you need a half-closed interval of ##\varphi##.
A similar issue arises with any compact manifold. Many physics texts ignore or gloss over this technical issue, often because problems of interest can be analyzed without having to deal with it. But that doesn't mean it isn't there.
Exactly that's the important point! Ironically that's precisely what's behind the here discussed issue with the non-integrable phase factor describing the AB effect. The apperently simple case of the infinitely long DC-current carrying solenoid (radius ##a##) seems not that trivial after all. In this case you have the vector potential (in Coulomb gauge and standard cylinder coordinates, ##(R,\varphi,z)##):
$$\vec{A}=\begin{cases} \frac{I \mu_0 N R}{2L} \vec{e}_{\varphi} & \text{for} \quad R<a, \\ \frac{I \mu_0 N a^2}{2R} \vec{e}_{\varphi} & \text{for} \quad R>a.\end{cases}$$
For the B-field we get, of course,
$$\vec{B}=\vec{\nabla} \times \vec{A}=\begin{cases} \frac{I \mu_0 N}{L} \vec{e}_z &\text{for} \quad R<a, \\ \vec{0} & \text{for} \quad R>a. \end{cases}$$
For ##R>a## since ##\vec{\nabla} \times \vec{A}=0##, there is locally a potential everywhere, but since this region is not simply connected there's no unique global potential. Indeed for ##R>a##
$$\vec{A}=-\vec{\nabla} \Phi, \quad \Phi=-\frac{\mu_0 I N a^2}{2L} \varphi \quad \text{for} \quad R>a.$$
Now it's clear that you can choose an arbitrary open (!) interval of length ##2 \pi## for ##\varphi##, i.e., with an arbitrary ##\varphi_0## you can make ##\varphi \in (\varphi_0,\varphi_0+2 \pi)##. The potential ##\Phi## jumps along the half-plane ##\varphi=\varphi_0##, and this gives the non-zero phase factor for any closed path encircling the solenoid. The value is, of course, independent of the shape of the path, always the magnetic flux through the solenoid. Using a circle with radius ##b>a## parallel to the ##x_1x_2## plane, indeed leads to
$$\oint_{\mathcal{C}} \mathrm{d} \vec{r} \cdot \vec{A} =\frac{\mu_0 I N a^2}{2L} 2 \pi=\pi a^2 |\vec{B}|,$$
where ##\vec{B}## is the homogeneous magnetic field inside the solenoid.