I'm working on the Hopf-fibration for my undergraduate thesis, where I am giving a thoroughly geometric construction, and am stuck on finding a second trivializing map to rigorously show we have a fibration.
I'm using a geometric construction with Quaternions. The Hopf-map is given by $$h: S^3\subset\mathbb{H}\rightarrow S^2: r\mapsto ri\bar r.$$ Using this notation, we have that a fiber over a point $(p_1, p_2, p_3)$ is given by $$-\frac{\sqrt{2(p_1+1)}}{2}\sin(t)+\frac{\sqrt{2(p_1+1)}}{2}\cos(t)i+\frac{\sqrt{2}(p_2\cos(t)+p_3\sin(t))}{2\sqrt{p_1+1}}j+\frac{\sqrt{2}(-p_2\sin(t)+p_3\cos(t))}{2\sqrt{p_1+1}}k,$$ except for the special case $p_1=-1$ for which we set $h^{-1}(-1, 0, 0)=ke^{it}.$
Now assuming that $p_1 \ne -1$, it is simple to "invert" any point on the fiber and find a homomorphism which gives the point $p_1, p_2, p_3\in S^2$ and the argument $t$ for a point in $S^1$, a locally trivializing map. Using the first two components we find $t, p_1$, and using the second two we get $p_2, p_3$.
In particular, if $p_1=-1$, we can't calculate $t$ from the first two components. This leads to my question: Is it possible to extract enough information from the last two components somehow to construct a second locally trivializing map on an open cover of $(-1, 0, 0)$? Or is it necessary to switch perspectives?
From a geometric point of view, it would make sense to look at intersections of the fibers and a half-plane or hemisphere of $S^2$ for instance, which would be unique except for one fiber. However, I don't see how we get the $t$-Argument (Position on $S^1$) using this approach, even if it is possible to get from the intersections to the Hopf-basepoints homeomorphically.