My apologies if this is fairly basic. I'm trying to understand the proof that $Sp(1)$ and $SO(3)$ are locally isomorphic but are not isomorphic but have run into some trouble. Here's what my understanding is so far:
Take $x\in Im\mathbb{H}$. Each $q\in Sp(1)$ defines a map:
$\rho(q):Im\mathbb{H} \rightarrow Im\mathbb{H}, \rho(q)(x)=qx\bar{q}$
We note that if $x\in Im\mathbb{H}$,then we write $x=bi+cj+dk$, and then $\bar{x}=-bi-cj-dk=-x$. So for purely imaginary quaternions, $\bar{x}=-x$. So $\rho(q)(x)\in Im\mathbb{H}$ if $x \in Im\mathbb{H}$.
For $q\in Sp(1)$, $|\rho(q)(x)|=|qx\bar{q}|=|q||x||\bar{q}|=|x|$ since $|q|=|\bar{q}|=1$. So $\rho(q) \in O(3)$ is orthogonal.
We can see that we have a homomorphism $\rho:Sp(1) \rightarrow O(3)$.
However, now the notes that I'm following say that this is actually a homomorphism $\rho:Sp(1) \rightarrow SO(3)$ and I can't quite see why. Is it because $Sp(1)$ is connected, and the connected component of the identity in $O(3)$ is $SO(3)$ or is it something else?
Many thanks for any responses.
Yes you are right. $Sp(1)$ can be identified with $S^3$, so it is connected. Your map $\rho$ is clearly continuous, so it maps $Sp(1)$ into a connected component of $O(3)$. But being a homomorphism of groups it has to map the identity of $Sp(1)$ to the identity of $O(3)$, therefor $\rho (Sp(1))$ will be contained in the identity component of $O(3)$, which is just $SO(3)$ as you observed.