I am a bit stuck on a homework assignment and I'm hoping someone can push me in the right direction.
Consider the 3-dimensional sphere: $$S^3=\{(z_1,z_2)\in\mathbb{C}:|z_1|^2+|z_2|^2=1\}$$ and the circle $S^1=\{z\in\mathbb{C}:|z|=1\}$, viewed as a group with respect to the multiplication of complex numbers. We consider the group action of $S^1$ on $S^3$ given by $$z\cdot(z_1,z_2):=(zz_1,\bar{z}z_2).$$ Let $X:=S^3/S^1$. Consider $\tilde{f}:S^3\to\mathbb{R}$ given by $$\tilde{f}(z_1,z_2):=z_1z_2+\bar{z}_1\bar{z}_2.$$ Show that $\tilde{f}$ induces a function $f:X\to\mathbb{R}$.
My approach is to show that $\tilde{f}(x)=\tilde{f}(y)\iff\exists\gamma\in S^1:\gamma\cdot x=y\;\forall x,y\in S^3$, for then $\tilde{f}=f\circ\pi$, where $\pi$ is the surjection $\pi:S^3\to X$ that produces the quotient $S^3/S^1$.
The direct implication is simple: consider $(z_1,z_2),(w_1,w_2)\in S^3$ and let $\gamma\in S^1$ such that $w_1=\gamma z_1$ and $w_2=\bar{\gamma}z_2$. Then we have
$$\begin{aligned}
\tilde{f}(w_1,w_2)&=w_1w_2+\bar{w}_1\bar{w}_2\\
&=\gamma z_1\bar{\gamma}z_2+\overline{\gamma z_1}\overline{\bar{\gamma}z_2}\\
&=|\gamma|^2z_1z_2+|\gamma|^2\bar{z}_1\bar{z}_2\\
&=z_1z_2+\bar{z}_1\bar{z}_2.
\end{aligned}$$
Hence the required equality, however the converse got me stuck. I don't really see a way to acquire such a $\gamma$ from the equality $\tilde{f}(z_1,z_2)=\tilde{f}(w_1,w_2)$ and the few other properties. I've also tried assuming no such $\gamma$ exists and get to a contradiction, but I couldn't make that work either and got stuck pretty quickly.
Could anyone please give me a push in the right direction on how this can be shown? I'd appreciate that a lot! Thanks in advance!