The fundamental group of SO(3)

Idea. A connected, simply connected cover of $X$ with deck group $G$ has $\pi_1(X)\cong G$.

$\Phi$ is a covering map. $\Phi$ identifies $SO(3)$ with the quotient $S^3/\{\pm1\}$, where $\{\pm1\}$ acts by $q\mapsto-q$. This action is free (no fixed points: $-q\ne q$) and properly discontinuous, so the quotient map $S^3\to S^3/\{\pm1\}$ is a covering map of degree $2$. Via the homeomorphism $\bar\Phi$ of the previous problem, $\Phi:S^3\to SO(3)$ is a $2$-sheeted covering.

$S^3$ is simply connected. The sphere $S^m$ is simply connected for $m\ge2$ (e.g. by the Seifert-van Kampen theorem, covering $S^3$ by two open sets homeomorphic to $\mathbb{R}^3$ with connected intersection $\simeq S^2$). So $\Phi$ is the universal cover of $SO(3)$.

Deck group. The deck transformations of the covering $S^3\to S^3/\{\pm1\}$ are $\mathrm{id}$ and $q\mapsto-q$, a group of order $2$. For the universal cover of a path-connected, locally path-connected space $X$, the deck group is isomorphic to $\pi_1(X)$. Hence
$$\pi_1(SO(3))\cong\{\pm1\}\cong\mathbb{Z}/2\mathbb{Z}.$$

Alternative viewpoint (lifting). Take a loop $\gamma$ in $SO(3)$ based at $I$ and lift it to $\tilde\gamma$ in $S^3$ starting at $1$. The endpoint is $1$ or $-1$; the class of $\gamma$ is determined by this endpoint. Loops with endpoint $-1$ are the non-trivial class.

Remark. Consequence: the group $SU(2)\cong S^3$, the *spin group* $\mathrm{Spin}(3)$, is the universal cover of $SO(3)$. This is why electrons (spinors) change sign under a $2\pi$ rotation but not under a $4\pi$ rotation (next problem).