@LeakyNun the formula you gave for $S\to S^3$ seems to be some kind of augmentation of the identification $S^1\star S^1\simeq S^3$ (where $\star$ is
join), which would have just been $(\varphi,\theta,\omega)\mapsto(\sin\varphi\exp(i\omega),\cos\varphi\exp(i\theta))$. not sure how to interpret the extra "twisting" factor you have geometrically or how you derived it.