# Universal covering group and fundamental group of $SO(n)$

The universal cover of $SO(2)$ is $\mathbb{R}$, whilst the fundamental group is $\mathbb{Z}$. That is $$SO(2) \cong \mathrm{universal\ cover}/\pi_1$$ Likewise, I believe that the universal cover of $SO(n)$ for $n \geq 3$ satisfies $$SO(n) \cong \mathrm{universal\ cover}/\mathbb{Z}_2$$ and indeed $\pi_1(SO(n)) = \mathbb{Z}_2$. So $SO(n)$ is the quotient of its covering group by its fundamental group for all $n \geq 2$. I was wondering if this was part of a more general result? I have very little knowledge of algebraic topology so a basic answer would be appreciated. Thanks.

Yes it is. Let $$X$$ be a topological space with universal cover $$\widetilde{X}$$ and covering map $$p : \widetilde{X} \to X$$. A homeomorphism $$f : \widetilde{X} \to \widetilde{X}$$ is called a deck transformation of $$p$$ if $$p\circ f = p$$; that is, $$f$$ preserves the fibres of $$p$$ so if $$y \in p^{-1}(x)$$, $$f(y) \in p^{-1}(x)$$. The set of all deck transformations of $$p$$ is denoted $$\operatorname{Deck}(p)$$ and forms a group under composition. The quotient of $$\widetilde{X}$$ by $$\operatorname{Deck}(p)$$ is $$X$$. Moreover, $$\operatorname{Deck}(p) \cong \pi_1(X)$$.
If $H$ is a topological group which is both path-connected and locally path-connected (i.e. a connected Lie group such as $SO(n)$), then any path-connected cover of $H$ inherits a unique group structure making the covering map a group homomorphism. In fact for any such cover $p:G \to H$,we have $ker(p) \cong \pi_1(H)/p_*(\pi_1(G))$. This generalizes the statements in the question (when $\pi_1(G)=0$). (Here $p_*$ denotes the induced map on fundamental groups coming from $p$.)
For $SO(n)$, the universal covering group is called the spin group and denoted as $\operatorname{Spin}(n)$. It shows up throughout topology, geometry and physics.