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Define the maps $F_\pm:B^2\rightarrow S^2$, $(x,y)\rightarrow(x,y,\pm\sqrt{1-x^2-y^2})$. Prove that for any $\omega\in\Omega^2(S^2)$: $$\int_{S^2}\omega=\int_{B^2}F_+^*\omega-\int_{B_2}F_-^*\omega$$

I guess I have to prove first, that $S^2$ is an orientable manifold. I know the characterization that $S^2$ is orientable if there exists an atlas $\mathcal{A}$ such that for any $\phi,\psi\in\mathcal{A}$, $\det D(\psi\circ\phi^{-1})>0$. For example I could define the open sets $$U_A^\pm:=\{(x,y,z)\in S^2~|~x \gtrless 0\}\\ U_B^\pm:=\{(x,y,z)\in S^2~|~y \gtrless 0\}\\ U_C^\pm:=\{(x,y,z)\in S^2~|~z \gtrless 0\}$$ together with the maps $$ \phi_A^+:U_A^+\rightarrow B^2,~(x,y,z)\rightarrow (y,z)\\ \phi_A^-:U_A^-\rightarrow B^2,~(x,y,z)\rightarrow (z,y)\\ \phi_B^+:U_B^+\rightarrow B^2,~(x,y,z)\rightarrow (x,z)\\ \phi_B^-:U_B^-\rightarrow B^2,~(x,y,z)\rightarrow (z,x)\\ \phi_C^+:U_C^+\rightarrow B^2,~(x,y,z)\rightarrow (x,y)\\ \phi_C^-:U_C^-\rightarrow B^2,~(x,y,z)\rightarrow (y,x) $$

I guess these maps form an oriented atlas. For example $$(\phi_A^+\circ(\phi_B^-)^{-1})(z,x)=\phi_A^+(x,\sqrt{1-x^2-z^2},z)=(\sqrt{1-x^2-z^2},z)$$ satisfies $$D(\phi_A^+\circ(\phi_B^-)^{-1})=\begin{pmatrix} \frac{-z}{\sqrt{1-x^2-z^2}} & \frac{-x}{\sqrt{1-x^2-z^2}}\\ 1 & 0 \end{pmatrix}$$ with positive determinant. Therefore $S^2$ is an orientable manifold with fixed orientation induced by the maps. It is quite obvious that $\int_{S^2}\omega=\int_{U_C^+}\omega+\int_{U_C^-}\omega$ (although I'm not sure how to justify this). Furthermore $F_+$ is orientation-preserving and $F_-$ orientation-reversing and the claim follows.

Three questions about this:

  1. Is the argument correct?
  2. Can this be done more directly? Is there a quick argument to prove that this atlas is oriented?
  3. How can I justify the splitting of the integral?
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