Double Covering of $SU(2) \times SU(2) \rightarrow SO(4)$ How does the map $\phi:SU(2) \times SU(2) \rightarrow SO(4)$ induce a double covering.
I understand that we can use the quaternions here by defining the action of $SU(2) \times SU(2)$ on $\Bbb V$ by $(g, h) \cdot z := g z h^* .$
I see the action is a homomorphism and has kernel { \pm 1 } But I don't see how this will induce the double covering of $\phi$.
Thanks in advance!
 A: For a group homomorphism $\phi\colon G\rightarrow H$, let $K=\{1=k_1,k_2,\dots,k_n\}$ be the kernel consisting of $n$ elements.
Then, if $h=\phi (g)$ is in its image, we can see that
$$\begin{split}
\phi^{-1}(h)&=\{g,gk_2,\dots,gk_n\}\\
&=\{g,k_2g,\dots,k_ng\},
\end{split}$$
therefore $\phi$ is exactly $n$-to-$1$ onto the image (set-theoretically).
Moreover, in your case of $\phi\colon \mathrm{SU}(2)\times \mathrm{SU}(2)\rightarrow \mathrm{SO}(4)$, the surjectivity is automatic, as follows.
Let $G'=\mathrm{Im}(\phi)$ be the image. Since $\mathrm{SU}(2)\times \mathrm{SU}(2)$ is compact, $G'$ is a closed subgroup, hence a Lie subgroup. If $\dim G'<\dim \mathrm{SO}(4)=\dim\mathrm{SU}(2)\times\mathrm{SU}(2)$, the implicit function theorem at regular points show that $\phi^{-1}(g')$ is infinity, a contradiction. Therefore $\dim G'=\dim \mathrm{SO}(4)$ and the connectedness of the latter implies $G'=\mathrm{SO}(4)$. Note that, this argument　also shows that $\phi$ is a 2-to-1 covering map topologically as well, as $\phi$ has everywhere regular points by translations.
