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$i.$ $S^1 \vee S^1$ can be embedded in a topological group

$ii.$ $S^1 \vee S^1$ can be covered by a topological group

I think $i.$ is true since we can embed the wedge sum into $\mathbb{R}^2$, which is a topological group under addition.

Not sure about $ii$ though. I know a covering map will induce an injective homomorphism on $\pi_1$ and that $\pi_1$ is abelian for topological groups. It appears that all of $S^1 \vee S^1$'s covering spaces have nonabelian fundamental groups, except for the universal cover, but I'm not seeing an obvious way of putting a group structure on that one.

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    $\begingroup$ I just wanted to note that there are topological groups which cover things with non-abelian $\pi_1$. Probably the simplest example is that $S^3$, thought of as the unit quaternions, covers the Poincare dodecahedral space. $\endgroup$ – Jason DeVito Jul 30 '14 at 2:41
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ii is impossible. A topological group is locally homogeneous: for any two points $p,q$ there exist neighborhoods $U,V$ such that $U$ is homeomorphic to $V$. But no covering space of $S^1 \vee S^1$ has that property, because any covering space is a graph having a vertex of valence $4$, and the only locally homogeneous graph is a 1-manifold.

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  • $\begingroup$ I should of thought of that! Thanks:) $\endgroup$ – Ashley Jul 30 '14 at 2:23

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