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The sphere $S^2$ can be covered by the following $6$ subsets (hemispheres) $$ O_i = \{(x^1, x^2, x^3) \in \mathbb{R}^3 | x^i > 0, i = 1, 2, 3\}$$

Each of these subsets can be mapped by the unit open disk $D$ or $\mathbb{R}^2$ via the projection: $$ f_i : O_i \rightarrow D_i$$

For example: $$ f_1 : O_1 \rightarrow D_1$$ $$ (x^1, x^2, x^3) \rightarrow (x^2, x^3)$$

Check that the mappings $f_i o (f_j)^{-1}$ are $C^{\infty}$.

I have just started studying Manifolds and I already feel lost and cannot find a starting point to show this. I do not want a solution, but some insight on how to start reasoning a proof out would be greatly appreciated.

mesllo
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  • Your definition of $O_i$ seems ambiguous, in that you have placed the letter $i$ inside the description after the $|$. If it's moved out, it seems thare are only 3 subsets. – coffeemath Mar 03 '15 at 17:30
  • I forgot to mention that $x^i<0$ is also considered. Am I correct? – mesllo Mar 03 '15 at 17:41
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    jablesauce-- Yes that would give six but the part $i=1,2,3$ should be moved outside, e.g. $O_i^+={(x^1,x^2,x^3) \in \mathbb{R}^3 | x^i>0}$, for $i=1,2,3$ and similarly for $O_i^-$ with the $>$ replaced by $<.$ – coffeemath Mar 03 '15 at 18:08
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    Another thing: instead of $\mathbb{R}^3$ you should have $S^2$ (the subset of $\mathbb{R}^3$ for which the squares of coordinates add to 1). – coffeemath Mar 03 '15 at 18:11

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I'll use $(x_1,x_2,x_3)$ for a typical point in $S^2,$ so that we have $x_1^2+x_2^2+x_3^2=1.$ I'll also use $p_i^>$ or $p_i^<$ for the projections from the intersections of the sphere with the open half spaces $x_i>0,x_i<0$ onto the open unit disc. To find $p_1^> \circ (p_2^<)^{-1},$ we start from the point $(x_1,x_3)$ in the open unit disc with the assumed $x_2<0,$ and then $x_2=-\sqrt{1-x_1^2-x_3^2}.$ Here note that since $x_2<0$ it follows that $x_1^2+x_3^2<1,$ so that under the radical we have a positive quantity, making the radical $C^\infty$ at that point. So we now have

$$(x_1,-\sqrt{1-x_1^2-x_3^2},x_3)$$

as the point on $S^2$ reached by applying the inverse of the map $p_2^<.$

Note that if this point is to be in the domain of the map $p_1>,$ we must assume at this point that $x_1>0.$

To finish we now apply $p_1^>$ to the above point of $S_2$ and arrive at $(-\sqrt{1-x_1^2-x_3^2},x_3)$ of the unit disc.

coffeemath
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  • So using the same reasoning as in the first part, it follows that the point outputted by ${p_1}^{>}$ is also $C^{\infty}$. I think I understand, the reason why I was finding it hard was because I put too much thought in applying an Epsilon-delta approach and induction as well. But this seems to suffice, right? And similarly the same idea can be applied to the other cases. – mesllo Mar 05 '15 at 15:52
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    @jablesauce Yes all the cases are similar (there are 6 choices for the "up" maps from 2-disc into sphere, and then 5 choices for the "back down" maps, so 30 in all). One just has to keep track of positive versus negative radicals, and the bit about domains so radicals are not applied at zero. – coffeemath Mar 05 '15 at 16:42