Why the teardrop is a bad orbifold? I found this The teardrop and the spindle are bad orbifolds, but I do not totally understand the explanation. Could you help me, please?
 A: There are two ways to solve this problem. A better solution is to use the van Kampen's theorem for orbifold fundamental groups. For the teardrop $S^2(n)$ this yields the presentation $\langle a| a^n=1, a=1\rangle$ which obviously gives the trivial group. For the double teardrop $S^2(n,m)$, this yields the presentation $\langle a| a^n=1, a^m=1\rangle$, which is the trivial group as well (assuming $n, m$ are coprime). Notice that the double teardrop $S^2(N,M)$ (with $N\ne M$) has a finite covering by $S^2(n,m)$ with $n, m$ coprime, so it suffices to consider only double teardrops $S^2(n,m)$ with coprime $n, m$. From this, we conclude that the orbifold is not a manifold but has trivial fundamental group, hence, is bad. 
Here is an alternative direct proof. Consider the teardrop orbifold $O=S^2(n)$ and suppose that $p: S\to S^2(n)$ is the (universal) manifold covering. The complement $C$ to the singular locus of $O$ is homeomorphic to $R^2$ which is simply connected. Hence, the preimage $p^{-1}(C)$ is a disjoint union of copies of $R^2$. But the singular locus of $O$ is zero-dimensional, hence, its preimage in $S$ cannot separate $S$. Thus, $p^{-1}(C)$ is connected and the restriction of $p$ to it is a homeomorphism. Now, take a small neighborhood $B$ of the singular point $x\in O$ ($B$ is a topological disk if we ignore the orbifold structure). Then $p^{-1}(B)$ is a disjoint union of  disks $B_i$ such that $p|_{B_i}$ is a finite cyclic orbifold covering). By the previous observation, the preimage $p^{-1}(B)$ is connected, equal to $B_1$, and, moreover, the restriction of $p$ to $p^{-1}(C)\cap B_1$ is 1-1. But this contradicts the assumption that $x$ is the singular point of the orbifold. 
The proof for the double teardrop (the spindle) is similar. Take $C$ to be the complement to the singular locus of $O=S^2(n,m)$. It is a topological cylinder. The preimage $p^{-1}(C)$ is connected and is a finite covering of the cylinder $C$, hence,   is again a cylinder. From this we conclude that the covering group $G$ of $p: S\to O$ is cyclic. It has to have the order $n$ as well as the order $m$ (this you observe by looking at the restriction to $p$ to preimages of small disk neighborhoods of the singular points $x, y\in O$). This implies that $n=m$. 
