Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ a compact and connected surface. If $S=U_1\cup U_2$, where $U_1,U_2$ are of finite character and the boundary of $U_1$ is in $U_2$.

How can I prove that S is homeomorphic to the sphere?

I'm studying an existence of triangulations proof and I'm stuck there. I need help. Thanks

Definition of finite character:

Let $\{U_i\}$ be a family of open sets in a surface S. We say $\{U_i\}$ is of finite character if the following conditions are satisfied:

(I) The family $\{U_i\}$ is locally finite, i.e, each point of S has a neighborhood which intersects only finitely many $U_i$'s

(II) The closure $\bar U_i$ of $U_i$ in S is a closed 2-cell

(III) Each $J:=∂\bar U_i$ meets at most finitely many other $J_j$'s

(IV) $J_i \cap J_j$ has finitely many connected components (which may be either arcs or points) for each i,j.

share|cite|improve this question
Do you really mean that definition of finite character you linked to? It applies to families of sets, not to (open) subsets of surfaces... – Mariano Suárez-Alvarez Nov 13 '12 at 4:33
@MarianoSuárez-Alvarez no, I will search a link with the definition I'm using. I'm using the definition in the Ahlfor's book of Riemann geometry: 1.Each family of $U_i$ is locally finite 2.the closure of $U_i$ in S is a closed 2-cell. 3.Each boundary of $U_i$ meets at most finitely many other boundaries of $U_i$ 4.intersection of boundary of $U_i$ with boundary $U_j$ has finitely many connected components(which can be either arcs or points) – user42912 Nov 13 '12 at 4:41
Please add the definition you are using in the body of the question, not in comments (and why did you link to something that has nothing to do with the question?! :P ) – Mariano Suárez-Alvarez Nov 13 '12 at 4:46
@MarianoSuárez-Alvarez I'm doing this right now. Yes, my mistake, I'm sorry – user42912 Nov 13 '12 at 4:48
Don't worry! ${}$ – Mariano Suárez-Alvarez Nov 13 '12 at 4:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.