I'm trying to prove that the map given by \begin{equation*} \begin{gathered} \varpi: \mathbb{S}^3 \rightarrow SO(3)\\ \mu \mapsto M(f_\mu) \end{gathered} \end{equation*} where $f_\mu: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ ($a \mapsto \mu a \mu^{-1}$), and $M(f)$ stands for the matrix of $f$, is a covering (in fact, the universal cover).
(In the sense that $a \in \mathbb{R}^3 = Im(\mathbb{H})$, $\mu \in \mathbb{S}^3 = \{q \in \mathbb{H} : |q| = 1\}$ and $\mu a \mu^{-1}$ denotes the quaternion product).
I've already proved that is well defined and is surjective using the relations between the quaternions and the rotations in $\mathbb{R}^3$. https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation (Obviously, it is also continuous).
However I don't know how to prove that is a local homeomorphism.
In the book $\textit{Conformal Geometry of Surfaces in $\mathbb{S}^4$ and Quaternions}$ by F. E. BURSTALL, D. FERUS, K. LESCHKE, F. PEDIT and U. PINKALL the differential of the mapping is computed and it is proved that is a local isomorphism (and then $\varpi$ is a local diffeomorphism).
It is written that for $\mu \in \mathbb{S}^3, v\in T_{\mu}\mathbb{S}^3 = (\mathbb{R}_{\mu})^{\bot}$ the differential is \begin{equation*} d_{\mu}\varpi(v)(a) = va\mu^{-1} - \mu a \mu^{-1}v\mu^{-1} =\mu(\mu^{-1}va - a\mu^{-1}v )\mu^{-1}. \end{equation*} But I don't know where this formula come from. ¿How I should calculate this?
From there, it is easy to show that the differential is a local isomorphism because $\mu^{-1}v$ commutes with all $a\in Im(\mathbb{H})$ if and only if $v = r\mu$ for some real $r$. But then $v = 0$ since $v \bot \mu$.
It's easier than I though.
Fixed $\mu \in \mathbb{S}^3$, the differential of $\varpi$ at $\mu$ is \begin{equation} \begin{gathered} d\varpi(\mu): T_{\mu}\mathbb{S}^3 \rightarrow T_{\varpi{(\mu)}}SO(3)\\ \hspace{2cm}v \mapsto d\varpi(\mu)(v) \end{gathered} \end{equation} The space $T_{\varpi{(\mu)}}SO(3)$ is a matrix space. In fact, is exactly the Lie Algebra $\mathfrak{so}(3)$ of skew-symmetric 3x3 matrices. https://en.wikipedia.org/wiki/Rotation_group_SO(3)
(Although we don't need this).
Therefore, the notation $d_{\mu}\varpi(v)(a)$ makes sense since it is the vector of $\mathbb{R}^3$ given by the multiplication of the matrix $d\varpi(\mu)(v)$ and the vector $a\in\mathbb{R}^3$.
Also, the formula comes from the direction derivative: \begin{equation} \begin{gathered} d_{\mu}\varpi(v)(a) = d\varpi(\mu)(v)(a) = \frac{d}{dt}\mid_{t = 0} \varpi(\mu + tv)a = \frac{d}{dt}\mid_{t = 0}(\mu + tv)a(\mu + tv)^{-1} = (va + (\mu + tv)\cdot 0)(\mu + tv)^{-1} + (\mu + tv)a \frac{-v}{(\mu + tv)^2}\mid_{t = 0} = va\mu^{-1} - \mu a \mu^{-1}v\mu^{-1}. \end{gathered} \end{equation} As we wanted.