I was reading this post and wondered about a similar thing.
Let $\mathbb{Z}_2$ act on $S^2$ by letting the non-trivial element take $$(x,y,z)\mapsto(-x,-y,z).$$ This action is not free, so one would expect something to go "wrong" with the quotient space $S^2/\mathbb{Z}_2$.
On the surface, it seems like $S^2/\mathbb{Z}_2$ is a nice space; it is homeomorphic to $S^2$, which is a topological manifold. However, the north and south ends of $S^2/\mathbb{Z}_2$ seem somewhat crushed and "conical", instead of smooth. But I'm not sure how to match this intuition with something rigorous.
Question 1: Does $S^2/\mathbb{Z}_2$ inherit a smooth structure from $S^2$? If not, why not?
Question 2: Is there some precise sense in which the north and south ends of $S^2/\mathbb{Z}_2$ are conical? (This would seem to assume that $S^2/\mathbb{Z}_2$ has a natural Riemannian metric; why is this the case?)