Removable singularity of metric? The metric of 2D plane is
\begin{equation}
ds^2=dr^2+r^2d\phi^2,
\end{equation}
which is singular at $r=0$ (is it because vanishing Jacobian?). It is said this singularity is removable if
$0<\phi<2\pi$. So how can I see this point? It seems even if $\phi \in (0,2\pi)$, $r=0$ is still a singularity.
 A: It's not quite so easy to prove the general case $\phi \in (0, \alpha)$ (the singularity can't be removed in general), but fortunately, for the case $\alpha = 2\pi$, it is fairly easy.
First let's write our metric in a more appropriate form. This spacetime can be described with two coordinate patches, $(U_1, f_1)$ and $(U_2, f_2)$, with
\begin{eqnarray}
f(U_1) &=& (0,\infty) \times (0, 2\pi)\\
f(U_2) &=& (0,\infty) \times (-\varepsilon, \varepsilon)
\end{eqnarray}
$U_1$ and $U_2$ overlap in two subsets, both $(0, \varepsilon)$ overlap, with the identity as a transition function, and $(2\pi - \varepsilon, 2\pi)$ overlaps with $(-\varepsilon, 0)$, with the transition function $\phi_2 = 2\pi + \phi_1$.
On each patch, the metric is indeed just
\begin{equation}
ds^2 = dr^2 + r^2 d\phi^2
\end{equation}
That's an entirely fine spacetime , but you'll notice that the radial curve $\gamma(l) = (R - l, 0)$ is inextendible beyond $l = R$ and has finite half-length. Therefore we have a singularity at $r = 0$. But fortunately, this spacetime is extendible, which means that there exists a larger spacetime $M'$ such that we have an inclusion map $\iota : M \to M'$ where $\iota$ is an isometry.
We don't need to look too deep for this extended spacetime, picking Cartesian coordinates will do. Then we have the inclusion map (in coordinate form)
\begin{eqnarray}
\iota(r, \phi_1) &=& (r \cos \phi_1, r \sin \phi_1)\\
\iota(r, \phi_2) &=& \left.
  \begin{cases}
    (r \cos \phi_2, r \sin \phi_2) & \text{for } \phi_2 \in (0, \varepsilon) \\
    (r \cos (2\pi + \phi_2), r \sin (2\pi + \phi_2)) & \text{for } \phi_2 \in (-\varepsilon, 0) 
  \end{cases} \right.\\
\end{eqnarray}
which is also isometric to the Euclidian metric. This will only work for $\alpha = 2\pi$, as you can notice. You can always map one of the coordinate patch to $\mathbb{R}^2$ (it's always locally extendible), but if your total angle isn't $2\pi$, the inclusion map will not work as you will not have $\cos(\phi_1) = \cos(\alpha + \phi_2)$ at the appropriate points.
