Are Cartesian and spherical coordinates smoothly compatible? And is the transition map a global diffeomorphism? Consider the transition from spherical coordinates $(r, \theta, \varphi)$ to Cartesian coordinates $(x, y, z)$, given by the map 
$$F:(0,\infty) \times [0, \pi] \times [0, 2 \pi) \to \mathbb R^3,\qquad (r,\theta,\varphi)\mapsto (x,y,z)$$
where
\begin{align}
x &= r \sin \theta \cos \varphi \\
y &= r \sin \theta \sin \varphi \\
z &= r \cos \theta 
\end{align}
with the inverse relations
\begin{align}
r&=\sqrt{x^2 + y^2 + z^2} \\ 
\theta &= \arccos\frac{z}{\sqrt{x^2 + y^2 + z^2}} = \arccos\frac{z}{r} \\ 
\varphi &= \text{angle}(y,x)
\end{align}
where the function $\text{angle}:\mathbb R^2\backslash\{0\}\to [0,2\pi)$ is defined (awkwardly) as
\begin{align}\label{eq:angle_function}
\text{angle}(y,x)= \left\{ \begin{matrix}
\arctan(\frac y x) &\text{if } x > 0 \text{ and } y\geq 0\\[2px]
\arctan(\frac y x) +2\pi &\text{if } x > 0 \text{ and } y< 0\\[2px]
\arctan(\frac y x) + \pi &\text{if } x < 0 \\[2px]
+\frac{\pi}{2} &\text{if } x = 0 \text{ and } y > 0 \\[2px]
+\frac{3 \pi}{2} &\text{if } x = 0 \text{ and } y < 0
\end{matrix}\right.
\end{align}
The Jacobian matrix of $F$ is given by
\begin{align}
J_{\mathbf F}(r, \theta, \varphi) = \begin{bmatrix}
  \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \theta} & \dfrac{\partial x}{\partial \varphi} \\[10px]
  \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \theta} & \dfrac{\partial y}{\partial \varphi} \\[10px]
  \dfrac{\partial z}{\partial r} & \dfrac{\partial z}{\partial \theta} & \dfrac{\partial z}{\partial \varphi}\end{bmatrix}
= \begin{bmatrix}
      \sin \theta \cos \varphi & r \cos \theta \cos \varphi & - r \sin \theta \sin \varphi \\
      \sin \theta \sin \varphi & r \cos \theta \sin \varphi & r \sin \theta \cos \varphi \\
      \cos \theta & - r \sin \theta & 0 \end{bmatrix}
\end{align}
which has determinant $\det J_{\mathbf F}(r, \theta, \varphi) = r^2\sin\theta$.
I have some questions about this map:
Is $F$ a diffeomorphism between its domain and its image $\mathbb R\backslash\{0\}$? If so, can someone show how to prove this? References to proofs are also appreciated. At least I see that $F$ is bijective and smooth, but I am not so sure about the inverse, especially the $\varphi$ part, although intuitively it seems clear.
I found that the map is discussed in John Lee's Introduction to Smooth Manifolds on page 167, where he proves that $F$ (but restricted to $0<\theta<\pi$) is a local diffeomorphism. Then he states that $F$ is also a diffeomorphism whenever the domain is restricted to an open set in $\mathbb R^3$, but I don't see how he comes to that conclusion. Could someone explain this? Moreover, (as I already asked above) does this result als extend to 'my' domain of $F$?
And finally, I imagine that the fact that $F$ is an (at least local) diffeomorphism implies that spherical coordinates are smoothly compatible with the standard differentiable structure on $\mathbb R$, so that a map between manifolds with domain or codomain $\mathbb R$ is smooth w.r.t. Cartesian coordinates if and only if it is smooth w.r.t. spherical coordinates, am I right about this?
Am I right about this? 
Thanks!
(Note that in Lee's textbook the roles of $\theta$ and $\phi$ are interchanged.)

What you can use here is
1) A smooth map whose differential at $p$ is an isomorphism is a diffeomorphism in a neighbourhood of $p$.
2) An bijective local diffeomorphism is a diffeomorphism.
So in particular from the Jacobian determinant you see that $F$ is a local diffeomorphism at any point for which $r\neq 0$ and $\theta\neq 0, \theta\neq\pi$. Therefore with domain $(0, \infty)\times(0,\pi)\times(0,2\pi)$ the map $F$ is a diffeomorphism. To cover the whole sphere you need more than one chart.
