The conjugation action $\mathbb{H}^*\times \mathbb{H} \rightarrow \mathbb{H}$ restricted to unit-quaternions yields an orthogonal representation Consider the action $\mathbb{H}^*\times\mathbb{H} \rightarrow \mathbb{H}, (h,h')\mapsto hh'h^{-1}$.
Show that it preserves the orthogonal-decomposition $\mathbb{R}\bigoplus $Im$\mathbb{H}$, and therefore when restricted to unit quaternions $S^{3} = \{h \in \mathbb{H} | \overline{h}h = 1\}$ yields an orthogonal representation
\[S^3 \xrightarrow{\phi}\text{SO(Im}\mathbb{H})\cong SO(3)\]
with kernel $\{\pm 1\}$. Conclude that $\phi$ is a local diffeomorphism and hence a covering map.
My ideas so far:
$S^3$ has dimension $3$ as well as $SO(3)$  since dim $SO(n) = \frac{n\cdot(n-1)}{2}$.
Therefore by Prop.$4.8$ in Lee it suffices to show that $\phi$ is either a submersion or an immersion. Since $S^3$ and $SO$(Im($\mathbb{H})$ are embedded submanifolds, ($SO$(Im($\mathbb{H})$ is a linear subspace) the map $\phi$ should at least be smooth.
But I don't quite understand why the action preserves the orthogonal decomposition.
I would be very grateful for any help.
 A: Quick exercise-review of how to write linear operators as matrices with respect to bases. Consider the vector space of real-coefficient polynomials with degree $\le 2$. The derivative $D$ is a linear operator. If we pick the basis $\{1,x,x^2\}$, we may apply $D$ to the basis elements and express the results as linear combinations:
$$\begin{array}{ccccc}
D1 & = & 0(1) & + & 0(x) & + & 0(x^2) \\
Dx & = & 1(1) & + & 0(x) & + & 0(x^2) \\
Dx^2 & = & 0(1) & + & 2(x) & + & 0(x^2)
\end{array}$$
Therefore we can represent $D$ by the $3\times3$ matrix
$$ \begin{bmatrix}
0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0
\end{bmatrix} $$
So for example, if you think of $1,x,x^2$ as $e_1,e_2,e_3$ respectively, then $De_1=0$, $De_2=e_1$, $De_3=2e_2$.

Pick the basis $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ for $\mathrm{Im}\,\mathbb{H}=\mathbb{R}^3$.
To turn a unit quaternion $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ into a $3\times3$ matrix, check how conjugating by it affects the basis vectors. For example, if we let $p=\mathbf{i}$ then we can compute
$$\begin{array}{lcr}
\mathbf{iii}^{-1} & = & \mathbf{i} \\
\mathbf{iji}^{-1} & = & -\mathbf{j} \\
\mathbf{iki}^{-1} & = & -\mathbf{k}
\end{array}$$
and therefore $\mathbf{i}\in S^3$ corresponds to the matrix
$$ \mathbf{i}\longleftrightarrow
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix} .$$
Can you do the same for $p=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$? Note the reciprocal of a unit quaternion is its quaternion conjugate so there's no need for rational functions. You should then be able to define $S^3\to SO(3)$ as an explicit function.
