The aim of this note is to compute the fundamental group of $SO(3)$ using quaternions. More precisely, an isomorphism $SU(2)/ \mathbb{Z}_2 \simeq SO(3)$ will be found leading to:

Property 1: $SU(2)$ is the univeral covering of $SO(3)$ and $\pi_1(SO(3)) \simeq \mathbb{Z}_2$.

Before introducing quaternions, recall the construction of $\mathbb{C}$. Usually, $\mathbb{C}$ is defined as the splitting field of the polynomial $X^2+1$ over $\mathbb{R}$, namely $\mathbb{R}[X]/(X^2+1)$. Taking the companion matrix $A= \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right)$ of $X^2+1$, we get an isomorphism $\mathbb{C} \simeq \mathbb{R}[A] = \left\{ \left( \begin{matrix} a & -b \\ b & a \end{matrix} \right) \mid a,b \in \mathbb{R} \right\}$ with a subalgebra of $M_2(\mathbb{R})$. Thus, a possible construction of the quaternions $\mathbb{H}$ is to replace real coefficients with complex ones, namely $\mathbb{H} = \left\{ \left( \begin{matrix} a & -\overline{b} \\ b & \overline{a} \end{matrix} \right) \mid a,b \in \mathbb{C} \right\}$. Notice that if $1= \left( \begin{matrix} 1&0 \\ 0&1 \end{matrix} \right)$, $I= \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right)$, $J= \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right)$ and $K= \left( \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right)$, then every element of $\mathbb{H}$ can be uniquely written as a $\mathbb{R}$-linear combination of $1$, $I$, $J$ and $K$.

Formally, $\mathbb{H}$ is a four-dimensional real algebra with a basis $(1,i,j,k)$ satisfying $i^2=j^2=k^2=ijk=-1$; in particular, $ij=-ijk^2=-(ijk)k=k$ and $ji=i^2j(ij)j=i(ijk)j=-ij=-k$, so $\mathbb{H}$ is not commutative. In fact, it can be shown that $\mathbb{H}$ is a noncommutative division algebra.

Viewing $\mathbb{H}$ as a subalgebra of $M_2(\mathbb{C})$, there are a natural conjugaison $\overline{ a+bi+cj+dk}= a-bi-cj-dk$ and a natural norm $|a+bi+cj+dk|=a^2+b^2+c^2+d^2$ corresponding respectively to the conjugaison and the determinant in $M_2(\mathbb{C})$. As a real vector space, the norm $|\cdot |$ define a definite positive quadratic form on $\mathbb{H}$, whose bilinear form is $\langle h_1,h_2 \rangle= \frac{1}{2}(h_1\overline{h_2} + \overline{h_1}h_2 )$. In particular, notice that $\mathbb{R}1$ and $\mathbb{I}= i \mathbb{R}+j \mathbb{R}+k \mathbb{R}$ are orthogonal.

Now let $SU(2)$ act on $\mathbb{H}$ by conjugaison, $\varphi : \left\{ \begin{array}{ccc} SU(2) & \to & \text{Aut}(\mathbb{H}) \\ h & \mapsto & (\varphi_h : k \mapsto hkh^{-1}) \end{array} \right.$ ($\text{Aut}(\mathbb{H})$ is the group of automorphisms of $\mathbb{H}$ viewed as a real vector space). Since $\text{Aut}(\mathbb{H}) \simeq GL_4(\mathbb{C})$ and each $\varphi_h$ preserve the quadratic form $\det$, taking the restrictions of the $\varphi_h$‘s to $\mathbb{I}$ we can suppose that $\text{Im}(\varphi) \subset O(3) \simeq \mathbb{I}$. Moreover, $SU(2)$ is connected and $\varphi$ is continuous so $\text{Im}(\varphi) \subset O_0(3)=SO(3)$.

To show that $\varphi : SU(2) \to SO(3)$ is onto, notice that $\varphi$ is $C^1$ and  that $d \varphi(I_2) : H \mapsto (K \mapsto HK-KH)$ induces a linear map between the Lie algebra $\mathfrak{su}(2)=\mathfrak{u}(2)= \{ M \in M_2(\mathbb{C}) \mid M^*+M=0 \}$ and $\mathfrak{so}(3)=\mathfrak{o}(3) = \{ M \in M_3(\mathbb{R}) \mid M^t +M=0 \}$; by connectedness of $SO(3)$ and using the inverse function theorem, it is sufficient to show that $d\phi(I_2)$ is invertible. But $\text{ker}(d \varphi(I_2))= \{ H \in \mathfrak{su}(2) \mid \forall K \in \mathbb{I}, HK=KH\}= \{I_2\}$ and $\dim(\mathfrak{su}(2))=\dim(\mathfrak{so}(3))= 3$. Therefore, $\varphi : SU(2) \to SO(3)$ is onto.

Noticing finally that $\text{ker}(\varphi)= \{\pm I_2\}$, we get the isomorphism $SU(2)/\mathbb{Z}_2 \simeq SU(2) / \{ \pm I_2\} \simeq SO(3)$. But the isomorphism is also a homeomorphis since the coordinate functions of $\varphi$ are just rational functions, and identifying $SU(2)$ with the element of $\mathbb{H} \simeq \mathbb{R}^4$ of norm $1$, $SU(2)$ is homeomorphic to the sphere $\mathbb{S}^3$; in particulier, $SU(2)$ is simply connected. Moreover, the action of $\mathbb{Z}_2$ on $SU(2)$ by left multiplication is properly discontinuously, so we can deduce property 1.

In fact, the isomorphism $SU(2)/ \mathbb{Z}_2 \simeq SO(3)$ has a topological interpretation:

Property 2: $SO(3)$ is homeomorphic to the projective space $P^3\mathbb{R}$.

By definition, $P^3\mathbb{R}$ is the quotient of $\mathbb{R}^4 \backslash \{0\}$ by the equivalence relation $(x,y,z,t) \sim (x',y',z',t') \Leftrightarrow \exists \lambda \in \mathbb{R}^*, (x',y',z',t')= \lambda (x,y,z,t)$.  In fact, all points of $\mathbb{R}^4 \backslash \{0\}$ are not needed, and $P^3 \mathbb{R}$ is also homeomorphic to the quotient of $X:=\{ (x,y,z,t) \in \mathbb{R}^4 \backslash \{0\} \mid x^2+y^2+z^2+t^2=1, t \geq 0 \}$ by $\sim$; since the projection $\text{pr}_{\mathbb{R}^3 \times \{0\}}$ induces a bijection from $X$ to $B^3:= \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2+z^2 \leq 1 \}$, we get that $P^3\mathbb{R}$ is also homeomorphic to $B^3$ quotiented by the symmetry with respect to $(0,0,0)$.

Now let $\psi : \left\{ \begin{array}{ccc} B^3 & \to & SO(3) \\ x & \mapsto & r(x,||x||\pi) \end{array} \right.$ where $r(x,\theta)$ denotes the rotation of axis $x$ and angle $\theta$. Then $\psi$ is a continuous surjection such that for all $x,x' \in B^3$, $\psi(x)=\psi(x')$ is equivalent to $x= \pm x'$. Therefore, $\psi$ induces a continuous bijection $\tilde{\psi} : P^3\mathbb{R} \to SO(3)$; since $P^3\mathbb{R}$ is compact we deduce that $\tilde{\psi}$ is in fact a homeomorphism.