Prove that each $R \in SO(3)$ has exactly 2 preimages with respect to the covering map $\Phi_U : SU(2) \rightarrow SO(3)$
I tried in this way:
Let's consider $U \in SU(2)$ and $X \in V = i su(2)$
\begin{array}{ccc} && \text{U =} & \begin{bmatrix} \alpha & \beta\\ -\beta^{*} & \alpha^{*} \end{bmatrix} & \text{$\alpha$, $\beta$ $\in \mathbb{C}$, $|\alpha|^2 + |\beta|^2 = 1$} & && \end{array}
\begin{array}{ccc} & & \\ & \\ \text{X = } & \begin{bmatrix} x_1 & x_2+ix_3 \\ x_2 - ix_3 & -x_1\\ \end{bmatrix} \end{array}
where $x_1, x_2, x_3$ are the coordinates of a vector in $\mathbb{R}^3$.
$$\Phi : SU(2) \rightarrow SO(3)$$ is given by $\Phi_U(X) = U X U^{-1}$ and
\begin{array}{ccc} \text{$\Phi_U$ = } & \begin{bmatrix} |\alpha|^2 - |\beta|^2 & -2Re(\alpha \beta) & 2 Im( \alpha \beta) \\ 2Re(\alpha \beta) & Re(\alpha^2 - \beta^2) & Im(\beta^2 - \alpha^2) \\ 2Im(\alpha \beta) & Im(\alpha^2 + \beta^2) & Re(\alpha^2 + \beta^2) \end{bmatrix} & \end{array}
$\Phi$ (i.e. the map $U \rightarrow \Phi_U$) is a Lie group homomorphism of $SU(2)$ into $SO(3)$ which is surjective. Moreover its kernel is: $$ \text{Ker} \Phi =: K = \{ -\text{Id}, \text{Id} \}$$ Indeed $\Phi_U = \text{Id}$ means $\Phi_U(X) = X = U X U^{-1}$, i.e. $UX = XU$ for all $U$, so $U$ should be a multiple of the identity matrix, and since $\text{det}(Id)=1$ and $U \in SU(2)$, this implies $U= \pm Id$.
So by a theorem related to universal covering maps between group, we know that, since $SO(3)$ is a connected Lie group, then $SU(2)$ is the unique simply connected Lie group and $\Phi$ is the unique Lie group homomorphism that is surjective, such that: $$SO(3) \cong SU(2)/ K$$
As $\Phi$ is surjective, every $R \in SO(3)$ has a preimage. Then, by definition, if $\Phi_U = R$, $\Phi_{-U} = R$ too. Cannot be more than two because $SO(3) \cong SU(2)/K$ means that $$SO(3) = \text{Id} \hspace{2mm} SU(2) \cup (-\text{Id}) \hspace{2mm} SU(2)$$
This proves the statement. Is my proof correct?
The transformation in question is for $g=\begin{bmatrix}z &-\overline{w}\\ w & \overline{z}\end{bmatrix}$ with $|z|^2+|w|^2=1$ and $$ \operatorname{Ad}\, : \,\operatorname{SU}(2,\mathbb{C})\longrightarrow \operatorname{SO}\left(\mathfrak{su}_\mathbb{R}(2,\mathbb{C})\right)=\operatorname{SO}(3,\mathbb{R}) $$ given by $\operatorname{Ad}(g)(X)=gXg^{-1}$ \begin{equation}\operatorname{Ad}(g) = \begin{bmatrix} \mathfrak{Re}(z^2-w^2)&\mathfrak{Im}(z^2-w^2)&2\; \mathfrak{Re}(w \cdot \bar{z})\\[6pt] -\, \mathfrak{Im}(z^2+w^2)&\mathfrak{Re}(z^2+w^2)&2\; \mathfrak{Im}(w \cdot \bar{z})\\[6pt] -2\; \mathfrak{Re}(w \cdot z)&-2\; \mathfrak{Im}(w \cdot z)& |z|^2-|w|^2 \end{bmatrix} \end{equation} according to the basis \begin{equation}\label{S-II} \sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\, , \, \sigma_2 = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\, , \, \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \end{equation}