Given Hamilton quaternions $\mathbb{H} = \{a+bi+cj+dk\mid a,b,c,d \in\mathbb{R}\}$ with subgroup $Sp(1) = \{x\in\mathbb{H}\mid \|x\| = 1\}$.
Since we can identify $\mathbb{H}$ as $\mathbb{R}^4$,now define subspace topology on $Sp(1)$.
Prove given embedding $i:Sp(1)\to \mathbb{R}^4$, $Sp(1)$ is homeomorphic to $S^3$.
we know restriction of embedding is also continuous so we have continuous bijection:$i':Sp(1)\to i(Sp(1)) = S^3$
The problem here how to show its inverse is also continuous.It seems prove $Sp(1)$ is compact is sufficient.
Is bounded since $Sp(1)$ has norm 1,and it's level set so it's closed,is my proof correct?
By the way,embedding $i':X\to i(X)$ is not homeomorphism in general,correct?
Best Answer
The same identification $I$ that maps $a + bi + cj + dk\in \Bbb H$ to $(a,b,c,d) \in \Bbb R^4$ (which is by definition a homeomorphism, as this is how you give $\Bbb H$ its topology, and which preserves norm too, so that $\textrm{Sp}(1)$ is mapped exactly to $\Bbb S^3 \subseteq \Bbb R^4$ and this means that the restriction of $I$ to $\textrm{Sp}(1)$ in the domain and $\Bbb S^3$ in the codomain, is a homeomorphism too. Nothing to with compactness: if $h: X \to Y$ is a homeomorphism between spaces, then for all $A \subseteq X$, $h\restriction_A: A \to h[A]$ is also a homeomorphism.