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Don't think about $SO(3)$ to start with, think about the unit quaternions, $S^3\subset \mathbb{R}^4$ where multiplication is given by quaternion multiplication and inverses are given by "complex conjugation". It might help to realize that the dot product in $\mathbb{R}^4$ is given by $q.p=Re(q\overline{p})$.

The unit quaternions act on themselves by conjugation, and this action fixes the identity, inducing an action of the unit quaternions on the tangent space of $S^3$ at $1$ which is canonically isomorphic to $\mathbb{R}^3$. Call the unit quaternions $S^3$, call this action $ad:S^3\times \mathbb{R}^3\rightarrow \mathbb{R}^3$. First prove that this action is as rigid rotations by proving it preserves the dot product on $\mathbb{R}^3$. Next prove that the kernel of the map induced by $ad$, $h:S^3\rightarrow SO(3)$ has kernel ${\pm 1}$. Finally, use a dimension count to prove the map is onto. From there you can conclude that $S^3$ is the universal cover of $SO(3)$ with group of deck transformations $\mathbb{Z}_2$.

I usually assign this as homework, though I set the kids up by giving this outline to fill in.

Another fun picture of $SO(3)$ is given by the unit tangent bundle of $S^2$. Notice that this can be described as $$T_1S^2={(\vec{u},\vec{v})\in \mathbb{R}^3\times \mathbb{R}^3|||\vec{u}||=||\vec{v}||=1, \ \vec{u}.\vec{v}=0}$$

Notice that the matrix with columns $\vec{u},\vec{v},\vec{u}\times \vec{v}$ is in $SO(3)$. This map gives a diffeomorphism between $T_1S^2$ and $SO(3)$. The projection map $p:T_1S^2\rightarrow S^2$ that sends the pair $(\vec{u},\vec{v})$ to $\vec{u}$ is a fibration, and this is the standard fibration that people use to analyze the homotopy groups of $SO(3)$.

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Don't think about $SO(3)$ to start with, think about the unit quaternions, $S^3\subset \mathbb{R}^4$ where multiplication is given by quaternion multiplication and inverses are given by "complex conjugation". It might help to realize that the dot product in $\mathbb{R}^4$ is given by $q.p=Re(q\overline{p})$.

The unit quaternions act on themselves by conjugation, and this action fixes the identity, inducing an action of the unit quaternions on the tangent space of $S^3$ at $1$ which is canonically isomorphic to $\mathbb{R}^3$. Call the unit quaternions $S^3$, call this action $ad:S^3\times \mathbb{R}^3\rightarrow \mathbb{R}^3$. First prove that this action is as rigid rotations by proving it preserves the dot product on $\mathbb{R}^3$. Next prove that the kernel of the map induced by $ad$, $h:S^3\rightarrow SO(3)$ has kernel ${\pm 1}$. Finally, use a dimension count to prove the map is onto. From there you can conclude that $S^3$ is the universal cover of $SO(3)$ with group of deck transformations $\mathbb{Z}_2$.

I usually assign this as homework, though I set the kids up by giving this outline to fill in.