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.
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.