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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.

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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

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