Geometrically seeing the linking of the fibres in the Hopf fibration? I'm working through the following document: https://arxiv.org/pdf/0908.1205.pdf
I can understand the construction of the Hopf map using the Riemann sphere but I struggle to understand the geometric argument in section 4.3, 'the fiber circles are linked'.
The author mentions an equitorial sphere in $S^3$, $E$,  which is of course obtained by setting $x_4 = 0$. Now we consider the fiber $F$ that lies along the equator of the equitorial sphere where $x^4 = 0$.
Now we want to show that any other fibre of the Hopf map can connect any other point in $S^3$ to antipodal points along $E$. If the point on $S^3$ is $(z_1, z_2$) I can use a complex number $\lambda$ in order to 'rotate' the fiber into the plane of  $x^4 = 0$.
I really then struggle to understand how this shows that the fibers are linked by following this geometric argument. I would appreciate it if someone could clarify the steps/logic/etc.
 A: Consider the stereographic projection of this situation. Imagine a $2$-sphere in $\mathbb{R}^3$ with a blue equator, and there is another orange circle intersecting opposite hemispheres (at red points). I think it should be relatively evident these two are linked!

For more details, you should be able to slide the red circle to be as far away from the blue one as possible: namely, the line that the blue circle is wrapped symmetrically around (this corresponds to a circle in $S^3$ through the pole we project from - in this case the two circles are in orthogonal planes). Throughout the whole sliding process (using the action of $S^3$ on itself), the red circle never intersects the blue circle, so whether or not they're linked doesn't change. Then the complement of the blue circle retracts onto the red circle (within $S^3$, or $\mathbb{R}^3$ with a point at infinity). Thus, the red circle repreents a nontrivial element of the complements' fundamental group, which isn't possible if it isn't linked.
