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i have a question. I think that is not difficult, but i can't find a solution.7 I want to show the following:

Cutting a sphere along a curve always results in two discs.

Therefore i want to use the classification theorem and the Euler characteristic. Here my own ideas: Notice that we have a sphere, thus every curve seperates the surface and $\chi(F)=2$ (in addition these two facts are equivalent). But then, cutting along a curve must give: $$2=\chi(F_1)+\chi(F_2)$$ But $\chi(F_i)\leq2$ because the Euler-characteristic is always lower of equal to $2$. But then we have two possbilities: $2=2+0$ or $2=1+1$. The first can not be true because every curve seperates the sphere. Thus it must be the second one. But then we may conclude that we have two discs.

My Problem: I don't know if i can make such a argumentation. I think i miss something, true?!

Thank you for help, hints and solutions.

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Isn't this exactly Jordan curve theorem? a classical non-trivial result. – KotelKanim Mar 19 '13 at 10:41
Your argument makes perfect sense by summation formula as stated by Damien. But this only works when the curve is a simple closed curve. This condition will exclude all the weird behaviors including the case $2=0+2$ in you question. – Hesky Cee Apr 26 '13 at 13:04

True, the Euler Characteristic is a homotopical measure, so $\chi ( \mathrm S^2) = \chi(\mathrm F_1) + \chi(\mathrm F_2) - \chi(\mathrm S^1)$ and as $\chi(\mathrm S^1) = 0$, you can find your conclusion !

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Thank you but why does $\chi(S^2)=\chi(F_1)+\chi(F_2)−\chi(S^1)$ holds? is this not the fact we want to prove? – Adrian57 Mar 19 '13 at 10:33
$\chi(\rm A \cup B) = \chi(A) + \chi(B) - \chi(A \cap B)$ is true for very general $\rm A$ and $\rm B$. – Damien L Mar 19 '13 at 10:35
Oh yeah thanks :) – Adrian57 Mar 19 '13 at 10:42
Is it true that $\chi(F)=1 \iff$ $F$ is a disc? – Adrian57 Mar 19 '13 at 10:49
If by $\rm F$ you mean a topological manifold of dimension 1 with boundary $\rm S^1$ yes. – Damien L Mar 19 '13 at 11:03

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