The following quoted material is from The Geometry of Celestial Mechanics by Hansjörg Geiges.
Remark 1.4: The transformation from polar to Cartesian coordinates
is described by the smooth map
\begin{align*}
p:\mathbb{R}^{+}\times\mathbb{R} & \to\mathbb{R}^{2}\backslash\left\{ \mathbf{0}\right\} \\
\left(r,\theta\right) & \mapsto\left(r\cos\theta,r\sin\theta\right).
\end{align*}
The Jacobian determinant of this map is
\begin{align*}
\det J_{p,\left(r,\theta\right)} & =\begin{vmatrix}\cos\theta & -r\sin\theta\\
\sin\theta & r\cos\theta
\end{vmatrix}=r\ne0,
\end{align*}
so $p$ is a local diffeomorphism by the inverse function theorem.
Moreover, $p$ is covering map, which means the following. For
any point in $\mathbb{R}^{2}\backslash\left\{ \mathbf{0}\right\} $
one can find an open, path-connected neighborhood $U$ whose preimage
$p^{-1}\left(U\right)$ is a non-empty, disjoint union of sets $U_{\lambda},\lambda\in\Lambda,$such
that $p\vert_{U_{\lambda}}:U_{\lambda}\to U$ is a homeomorphism for
each $\lambda$ in the relevant index set $\Lambda.$ (You are asked
to verify this property in Exercise 1.3)
Exercise 1.3: Verify the map $p$ in Remark 1.4 is a covering map
in the sense described there. As an index set one can take $\Lambda=\mathbb{Z}.$
How does one have to chose the neighborhood $U$ of a given point
in $\mathbb{R}^{2}\backslash\left\{ \mathbf{0}\right\} ?$
I have multiple questions about this, but I will only pose one in this post. What is meant by "a non-empty, disjoint union of sets"?
The non-empty part is easy. I take disjoint to mean that no point in $p^{-1}\left(U\right)$ is in more than one $U_\lambda.$ The path-connected property of $U$ says that there is a continuous path connecting every pair of points in $U.$ If $\mathbf{p}_1$ lies in $p\left(U_1\right)$ and $\mathbf{p}_2$ lies in $p\left(U_2\right)$, apparently the preimage of a path connecting $\mathbf{p}_1$ and $\mathbf{p}_2$ is a path in $p^{-1}\left(U\right)$ connecting a point of $U_1$ and a point of $U_2,$ even though they are disjoint.
Intuitively, that seems contrary to the assertion that $U_1$ and $U_2$ are disjoint.
Best Answer
$U$ itself is path-connected, but in general, $p^{-1}(U)$ need not be path-connected. This is where your confusion lies.
If we have $a_1 \in U_1$ and $a_2 \in U_2$, there is a path connecting $p(a_1)$ and $p(a_2)$. But there is no reason to think this path can be used to get a path between $a_1$ and $a_2$.
For a very simple example of this, consider $p : \{0, 1\} \times \mathbb{R} \to \mathbb{R}$ defined by $p(x, y) = y$. Then $p$ is a covering map; in particular, $p^{-1}(\mathbb{R})$ is the disjoint union of $\{0\} \times \mathbb{R}$ and $\{1\} \times \mathbb{R}$, which are both copies of $\mathbb{R}$. Let $a_0 = (0, 0)$, and let $a_1 = (1, 1)$. Then there is clearly a path from $p(a_0)$ to $p(a_1)$ in $\mathbb{R}$. But this path cannot lift to a path from $a_0$ to $a_1$. We can, however, lift the path to get a path from $(0, 0)$ to $(0, 1)$, and $p(0, 1) = p(1,1)$.