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.
 A: 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, $\operatorname{Deck}(p) \cong \pi_1(X)$.
A good reference for this material is Hatcher's Algebraic Topology.
A: 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 $p$.)
See https://en.wikipedia.org/wiki/Covering_group for an outline of the construction.
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.
