Universal cover $\mathbb{S}^3 \rightarrow SO(3)$ through Quaternions. 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 $v \bot \mu$.
 A: You dont need this ; your map is an homomomorphism between two Lie groups, and it is differentiable. The  rank of its derivative is constant (as this is an homomomorphism). As your map is surjective the rank must be maximal, and your map is a local diffeomorphism.
A: 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.
