Let $\Sigma$ be a connected (not necessarily compact) surface with or without boundary. Is it true that $\Sigma$ is homeomorphic to the sphere if it has euler characteristic $\chi(\Sigma)\geq 2$?
I could't find a definite answer if we don't request $\Sigma$ to be compact.
Best Answer
Punctures and boundary components both decrease the Euler characteristic, and the results are homotopy equivalent. For example, $\chi(S^2) = 2$. Adding a puncture gives $\chi(\mathbb{R}^2) = 1$. Adding a boundary component gives $\chi(D^2) = 1$. Puncturing the torus once gives a wedge of $2$ circles, with Euler characteristic $-1$, and puncturing it again gives a wedge of $3$ circles, with Euler characteristic $-2$.
More generally, any noncompact connected surface is homotopy equivalent to a connected $1$-dimensional CW complex, or equivalently a connected graph. These are in turn homotopy equivalent to a wedge of some number of circles, and hence have Euler characteristic at most $1$. (There may be infinitely many circles involved, in which the Euler characteristic is not even well-defined; think of $\mathbb{R}^2$ minus countably many points, for example.)