[proofplan] We use the standard quaternionic double cover from $SU(2)$ to $SO(3)$, recorded as the smooth surjective homomorphism whose kernel is $\{I_2,-I_2\}$. The finite kernel makes the quotient map $SU(2)\to SU(2)/\{I_2,-I_2\}$ a two-sheeted covering, and the induced Lie group isomorphism with $SO(3)$ transports this covering to $SO(3)$. Finally, $SU(2)$ is simply connected, so this covering is universal. [/proofplan] [step:Take the standard double covering homomorphism from $SU(2)$ to $SO(3)$] By [citetheorem:8775], there exists a smooth surjective Lie [group homomorphism](/page/Group%20Homomorphism) \begin{align*} p:SU(2)\to SO(3) \end{align*} such that \begin{align*} \ker p=\{I_2,-I_2\}. \end{align*} The same result identifies the induced homomorphism \begin{align*} \overline p:SU(2)/\{I_2,-I_2\}\to SO(3) \end{align*} as an isomorphism of Lie groups. Thus the only remaining point is to check that the quotient projection by the finite subgroup $\{I_2,-I_2\}$ is a covering map, and then to use simple connectedness of $SU(2)$. [/step] [step:Show the quotient by $\{I_2,-I_2\}$ is a two-sheeted covering] Let \begin{align*} K:=\{I_2,-I_2\}\le SU(2) \end{align*} and let \begin{align*} q:SU(2)\to SU(2)/K \end{align*} be the quotient map. Since $K$ is finite and acts on $SU(2)$ by right multiplication, \begin{align*} R_k:SU(2)\to SU(2),\qquad U\mapsto Uk \end{align*} is a smooth diffeomorphism for each $k\in K$. For each $U\in SU(2)$, the two points $U$ and $-U$ are distinct. Since $SU(2)$ is Hausdorff, choose disjoint open neighbourhoods $O_U\subset SU(2)$ of $U$ and $O_{-U}\subset SU(2)$ of $-U$. The map $R_{-I_2}:SU(2)\to SU(2)$ is a homeomorphism, so \begin{align*} W_U:=O_U\cap R_{-I_2}^{-1}(O_{-U}) \end{align*} is an open neighbourhood of $U$. With \begin{align*} -W_U:=\{-V:V\in W_U\}=R_{-I_2}(W_U), \end{align*} we have $W_U\subset O_U$ and $-W_U\subset O_{-U}$, hence \begin{align*} W_U\cap (-W_U)=\varnothing. \end{align*} Then \begin{align*} q^{-1}(q(W_U))=W_U\cup (-W_U), \end{align*} and the two sets $W_U$ and $-W_U$ are disjoint open subsets of $SU(2)$. Moreover, the restrictions \begin{align*} q|_{W_U}:W_U\to q(W_U) \end{align*} and \begin{align*} q|_{-W_U}:-W_U\to q(W_U) \end{align*} are bijections, because two points in either $W_U$ or $-W_U$ have the same $q$-image exactly when they differ by multiplication by an element of $K$, and the non-identity element sends $W_U$ onto the disjoint set $-W_U$. The quotient map $q$ is open: for every [open set](/page/Open%20Set) $O\subset SU(2)$, \begin{align*} q^{-1}(q(O))=O\cup (-O) \end{align*} is open, so $q(O)$ is open by the [quotient topology](/page/Quotient%20Topology). Hence $q(W_U)$ is open, and the inverse of $q|_{W_U}$ sends each open set $q(O)\subset q(W_U)$ with $O\subset W_U$ open to $O$; the same argument applies on $-W_U$. Therefore both restrictions are homeomorphisms. Therefore $q$ is a two-sheeted covering map. [guided] We need to show that the finite quotient projection \begin{align*} q:SU(2)\to SU(2)/K,\qquad K=\{I_2,-I_2\}, \end{align*} is locally a disjoint union of copies of the base. The subgroup $K$ acts on $SU(2)$ by right multiplication. The identity element $I_2\in K$ fixes every point, while the non-identity element $-I_2\in K$ sends a matrix $U\in SU(2)$ to $-U$. Fix $U\in SU(2)$. Since $U\neq -U$ and $SU(2)$ is Hausdorff, choose disjoint open neighbourhoods $O_U\subset SU(2)$ of $U$ and $O_{-U}\subset SU(2)$ of $-U$. The right multiplication map $R_{-I_2}:SU(2)\to SU(2)$ is a homeomorphism, so \begin{align*} W_U:=O_U\cap R_{-I_2}^{-1}(O_{-U}) \end{align*} is an open neighbourhood of $U$. Its translate by the non-identity element of $K$ is \begin{align*} -W_U=\{-V:V\in W_U\}=R_{-I_2}(W_U), \end{align*} and this set is contained in $O_{-U}$. Since $W_U\subset O_U$ and $O_U\cap O_{-U}=\varnothing$, we obtain \begin{align*} W_U\cap (-W_U)=\varnothing. \end{align*} This is the key local separation: it prevents a point of $W_U$ from being identified with another point of $W_U$ by the nontrivial element of the kernel. Now compute the full preimage of $q(W_U)$. A point $Z\in SU(2)$ lies in $q^{-1}(q(W_U))$ exactly when $q(Z)=q(V)$ for some $V\in W_U$. By the definition of the quotient by right multiplication of $K$, this means \begin{align*} Z=Vk \end{align*} for some $V\in W_U$ and some $k\in K$. If $k=I_2$, then $Z\in W_U$. If $k=-I_2$, then $Z=-V\in -W_U$. Hence \begin{align*} q^{-1}(q(W_U))=W_U\cup (-W_U). \end{align*} The union is disjoint by the choice of $W_U$. It remains to check that each sheet maps homeomorphically onto $q(W_U)$. The map $q|_{W_U}$ is surjective onto $q(W_U)$ by definition. It is injective because if $V_1,V_2\in W_U$ and $q(V_1)=q(V_2)$, then $V_2=V_1k$ for some $k\in K$. For $k=I_2$ this gives $V_2=V_1$. For $k=-I_2$ it gives $V_2=-V_1\in -W_U$, contradicting $V_2\in W_U$ and $W_U\cap(-W_U)=\varnothing$. Thus $q|_{W_U}$ is bijective. We also need the inverse map to be continuous. The quotient map $q$ is open: if $O\subset SU(2)$ is open, then \begin{align*} q^{-1}(q(O))=O\cup (-O) \end{align*} is open, and therefore $q(O)$ is open in the quotient topology. In particular $q(W_U)$ is open. If $O\subset W_U$ is open, then $q(O)$ is open in $q(W_U)$, and the inverse of $q|_{W_U}$ sends $q(O)$ back to $O$. Hence $(q|_{W_U})^{-1}$ is continuous, so $q|_{W_U}$ is a homeomorphism. The same argument applies to $q|_{-W_U}$. Therefore every point of $SU(2)/K$ has an evenly covered neighbourhood, and $q$ is a two-sheeted covering map. [/guided] [/step] [step:Transfer the covering map to $SO(3)$] By the first step, the induced map \begin{align*} \overline p:SU(2)/K\to SO(3) \end{align*} is a Lie group isomorphism. In particular, its underlying map of topological spaces is a homeomorphism. Since \begin{align*} p=\overline p\circ q, \end{align*} and $q$ is a covering map, it follows that $p$ is a covering map. Because $q$ has two sheets, the covering $p:SU(2)\to SO(3)$ is two-sheeted. [/step] [step:Use simple connectedness of $SU(2)$ to identify the universal cover] By [citetheorem:8816], $SU(n)$ is simply connected for every $n\geq 2$. Applying this with $n=2$, the Lie group $SU(2)$ is simply connected. The map \begin{align*} p:SU(2)\to SO(3) \end{align*} is a covering map by the previous step, and its total space $SU(2)$ is simply connected. Hence $p$ is the universal covering map of $SO(3)$. This proves that $SU(2)$ is the universal cover of $SO(3)$, with kernel $\{I_2,-I_2\}$ and quotient \begin{align*} SO(3)\cong SU(2)/\{I_2,-I_2\}. \end{align*} [/step]