# The teardrop and the spindle are bad orbifolds

I should prove that the teardop $S^2(p)$ (the orbifold with underlying surface $S^2$ and a single cone point of order $p>1$) and the spindle $S^2(p,q)$ (the orbifold with underlying surface $S^2$ and two cone points of orders $p,q>1$, $p\neq q$) are bad orbifolds, that is they are not orbifold covered by any surface. I worked out the teardrop case using an argument involving the Euler characteristic, but the same idea fails with the spindle. I guess I should find a geometric proof for the spindle, but I tried without any success. Could you help me with that? Thank you.

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The central idea is that orbifolds have universal covers, much like manifolds. So an orbifold, $X$, is covered by a unique simply connected cover. This space also covers every connected cover of $X$. Now an orbifold that is not a manifold can never cover one that is. (A covering map cannot destroy all the singular points, since the image of a singular point must be singular.) Therefore, the universal cover of every good orbifold is a manifold.
What is the universal cover of the tear-drop? What of the $p \neq q$ spindle?