A stereographic manipulation (part 2)

First, do have a look at the first part .

Continuing where we left off, we found our two stereographic projections, $f$ and $f'$ which cover the whole of our circle $S^1$ We can think of $f$ as a mappings of the form : $f_{1} : U_1 \rightarrow \overline{U}_1$ where $U_1 \subseteq S^1 \backslash (0,1) \subseteq \mathbb{R}$ and $\overline{U}_1 \subseteq \mathbb{R}$

and similarly for $f'$ : $f_{2} : U_2 \rightarrow \overline{U}_2$ where $U_2 \subseteq S^1 \backslash (0,-1) \subseteq \mathbb{R}$ and $\overline{U}_2 \subseteq \mathbb{R}$

First, convince yourself that these two functions are bijections!

Now, let $0 \neq \overline{x} \in \overline{U}_2$ . How does $\overline{x}$ look like in $\overline{U}_1$ ,i.e. to what element in $\overline{U}_1$ does $\overline{x}$ corresponds to?

Hint : Find the point $(x,y) \in S^1 \backslash \{ (0,1) ; (0,-1) \}$ such that $f_2(x,y) = \overline{x}$ and then project this point in $\overline{U}_1$ via the function $f_1$