# fundamental group of SO(3) through action

I want to show that the fundamental group of SO(3) is $$\mathbb{Z}/2\mathbb{Z}$$ using the following theorem.

if $$X$$ is locally path connected and simply connected and $$G$$ is groups with a properly discontinous action over $$X$$ then $$p:X\to X/G$$ is a covering and $$\Pi_1(X/G,x_0)=G$$ where $$x_0$$ is a base point.

I think that I need to use $$X=S^3$$ but I don't understand how $$SO(3)=S^3/(\mathbb{Z}/2\mathbb{Z})$$. If it's not $$S^3$$ then which topological space should I use?

Ps: for properly distoninuos action I mean that $$\forall x \in X\;\exists U(x)$$ open set such that $$g(U(x))\cap U(x)=\emptyset\;\forall g\neq e$$ given $$g\in G$$

• You might want to look into the universal covering map $SU(2)\rightarrow SO(3)$. This provides a 2-covering, which has as the Deck transformation group $\mathbb{Z}/\mathbb{Z}2$. Note $SU(2)$ is homeomorphic to the sphere $S^3$, and as a good universal covering, it is path and simply connected. Nov 11, 2021 at 15:55

Take$$SU(2)=\left\{\begin{bmatrix}a&-\overline b\\b&\overline a\end{bmatrix}\,\middle|\,a,b\in\Bbb C\wedge|a|^2+|b|^2=1\right\}\simeq S^3.$$Ang let $$G=\{\pm1\}\simeq\Bbb Z_2$$. Then $$G$$ acts on $$SU(2)$$ in a natural way: $$1$$ acts as the identity and $$-1$$ acts is $$q\mapsto-q$$. This action is properly discontinuous. So, $$\pi_1(S^3/\{\pm1\})\simeq\Bbb Z_2$$.
Now, let$$H=\left\{\begin{bmatrix}\alpha i&-\beta+\gamma i\\\beta+\gamma i&-\alpha i\end{bmatrix}\,\middle|\,\alpha,\beta,\gamma\in\Bbb R\right\}.$$It turns out that if $$q\in SU(2)$$ and $$h\in H$$, then $$qhq^{-1}\in H$$. So, if $$q\in SU(2)$$, consider$$\begin{array}{rccc}r_q\colon&H&\longrightarrow&H\\&h&\mapsto&qhq^{-1}.\end{array}$$It turns out that, if $$q,s\in SU(2)$$, then $$r_q=r_s\iff q=\pm s$$. And, if$$\left\lVert\begin{bmatrix}\alpha i&-\beta+\gamma i\\\beta+\gamma i&-\alpha i\end{bmatrix}\right\rVert=\sqrt{\alpha^2+\beta^2+\gamma^2},$$each $$r_q$$ is an orthogonal map. So, you have a map from $$SU(2)$$ onto $$SO(3,\Bbb R)$$ which happens to be a covering map. So, $$\pi_1\bigl(SO(3,\Bbb R)\bigr)\simeq\Bbb Z_2$$.