Continuity of some function related to the Hopf fibration of $S^3$ By the identification $\mathbb R^4$ with the field of quaternions $\mathbb H$ by $(x_0,x_1,x_2,x_3) \sim x_0+x_1 i+x_2 j+x_3k$ and $S^3$  with the set of unit quaternions and $S^2$ with the set of purely imaginary unit quaternions, let's consider the Hopf mapping $p: S^3 \rightarrow S^2$, $p(x)=xix^*=(x_0^2+x_1^2-x_2^2-x_3^2)i+2(x_1x_2+x_0x_3)j+2(x_1x_3-x_0x_2)k$   for $x=x_0+x_1i+x_2j+x_3k\in S^3$, where $x^*$ denotes the conjugate quaternion to $x$. For fixed $v=(v_1,v_2,v_3)\in S^2\setminus \{S\}$, where $S=(-1,0,0)\in S^2$, we have that  the preimage of $v$ is equal $p^{-1}(\{v\})=\{ \sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{1+v_1}): t\in [0,2\pi) \}$  (see here).
In order to know that the structure $(S^3, S^1, p, S^2)$ is a fiber bundle (see here), we have to know that the trivialization map $f_1: (S^2\setminus\{S\}) \times S^1 \rightarrow  p^{-1}(S^2 \setminus \{S\})$, $f_1(v,(a_1,a_2))= \sqrt{\frac{1+v_1}{2}}(a_1, a_2, \frac{v_2 a_2-v_3 a_1}{1+v_1},\frac{v_2 a_1+v_3 a_2}{1+v_1})$ for $(a_1,a_2)\in S^1$, $v=(v_1,v_2,v_3)\in S^2 \setminus \{S\}$, is the homeomorphism satisfying condition $p \circ f_1=proj_1$ and similarly for $f_2: (S^2\setminus\{N\}) \times S^1 \rightarrow  p^{-1}(S^2 \setminus \{N\})$, where $N=(1,0,0)\in S^2$.
My problem concerns continuity of the inverse function to $f_1$.
Now $p^{-1}(S^2 \setminus \{S\})=\{\sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{1+v_1}): t\in [0,2\pi), (v_1,v_2,v_3)\in S^2\setminus \{S\} \}$ 
Why the function $f_1^{-1}: p^{-1}(S^2 \setminus \{S\}) \rightarrow (S^2\setminus\{S\}) \times S^1$, $f_1^{-1}(\sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{a+v|_1}))=((v_1,v_2,v_3),(\cos t, \sin t))$
is contunuous?
 A: Just solve for the quantities $v_1,v_2,v_3,t$ in the system of equations and check continuity of the solutions:
$$(a,b,c,d)=\sqrt{\frac{1+v_1}{2}}(\cos t, \sin t, \frac{v_2 \sin t-v_3 \cos t}{1+v_1},\frac{v_2 \cos t+v_3 \sin t}{1+v_1})
$$
To start,
$$v_1 = 2(a^2+b^2)^2-1
$$
which is certainly continuous in $a,b$.
And then
$$t = \arctan(b/a)
$$
which is certainly continuous in the first quadrant of the $a,b$ plane, and you can work your way around the circle using arc tangent and arc cotangent to get the other quadrants. 
And then $v_2,v_3$ are solutions of the system of linear equations
$$(\sin t) v_2 - (\cos t) v_3 = c(1+v_1)\sqrt{\frac{1+v_1}{2}}
$$
$$(\cos t) v_2 + (\sin t) v_3 = d(1+v_1)\sqrt{\frac{1+v_1}{2}}
$$
which are certainly continuous functions of $a,b,c,d$, using the formulas from Kramer's Rule and the fact that the variables $t,v_1$ in the coefficients are already known to be continuous functions of $a,b$.
A: We can get the explicite formula on $f_1^{-1}$. From the system of equations $(a,b,c,d)=\sqrt{\frac{1+v_1}{2}}(a_1, a_2, \frac{v_2 a_2-v_3 a_1}{1+v_1},\frac{v_2 a_1+v_3 a_2}{1+v_1})$, 
we get
$v_1=2(a^2+b^2)-1$,
$v_2=2\sqrt{a^2+b^2}(c a_2+da_1)$,
$v_3=2\sqrt{a^2+b^2}(da_2-ca_1)$,
$a_1=\frac{a}{\sqrt{a^2+b^2}}$,
$a_2=\frac{b}{\sqrt{a^2+b^2}}$.
Hence we obtain
$
f_1^{-1}(a,b,c,d)=\left( \left( 2(a^2+b^2)-1, 2\sqrt{a^2+b^2}(c \frac{b}{\sqrt{a^2+b^2}}+d\frac{a}{\sqrt{a^2+b^2}}), 2\sqrt{a^2+b^2}(d\frac{b}{\sqrt{a^2+b^2}}-c\frac{a}{\sqrt{a^2+b^2}}\right), \left(\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}}\right) \right)=
$
$
\left( \left( 2(a^2+b^2)-1, 2(bc +ad),2(bd-ac)\right), \left(\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}}\right) \right).
$
Now observe that points of the form $(0,0,c,d)\in S^3$ are not in $p^{-1}(S^2\setminus \{S\})$ because $p(0,0,c,d)=(-1,0,0)=S$. Hence the formula above defines $f_1^{-1}$ on the whole set $p^{-1}(S^2\setminus \{S\})$  and proves the continuity of $f_1^{-1}$.
