Is the quotient space obtained by identifying the poles of a sphere homeomorphic to a closed surface?

Question

I'm interested in the quotient space of $S^2$ obtained by identifying the poles, and in particular whether it is homeomorphic to a closed surface. I'm pretty sure its homotopic to one, just by squeezing together or extending the class of the poles, however the homeomorphic part is more troublesome to me.

Here a closed surface means compact, hausdorff space in which each point has a neighborhood homeomorphic to the plane.

Intuitively, I would guess no. I don't think it is possible to find a neighborhood of the pole equivalence class homeomorphic to the plane. This is easy for a "small" neighborhood since removing the poles disconnects it but not the plane. But I guess I should show this for a large neighborhood as well?

I am currently studying at the level of Armstrong's basic topology so I would appreciate it if any advice is kept at a relatively beginner-friendly level (fundamental groups and simplicial complexes are fine, but no higher homotopies or CW complexes).

Thanks in advance.

Answer 1

You've shown that there is a connected neighborhood $U$ of the point $\ast$ of interest in the quotient space such that $U − \{ \ast \}$ has two components, but you know that no point on the plane admits such a neighborhood; therefore, the quotient space cannot be a closed surface.