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I am trying to determine whether the rose with two petals ($S^1 \vee S^1$ or the figure-eight) has a continuous multiplication with identity element. I know that this is true for the unit circle $S^1$ in the complex plane, where $S^1 = \{ z \in \mathbb{C} \mid |{z}| = 1 \}$.

I also know that $S^1$ is a Lie Group, and I believe that because of the intersection point of the figure eight, this space does not have a continuous multiplication with identity element and is therefore not a topological group. Would someone mind pointing me in the right direction for how to approach this idea?

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    How do you define multiplication on $S^1\wedge S^1$? If you don't specify this you haven't specified the group. Or are you asking if there is *any* group structure on this set in which multiplication is a continuous map?2011-12-26
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    I think in the second paragraph you meant to write "topological group" rather than "topological space".2011-12-26
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    It cannot be a topological group, because its fundamental group is not abelian. See http://en.wikipedia.org/wiki/Topological_group#Properties However, as you only require a continuous multiplication (and not also inverses), I don't know...2011-12-26
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    Sorry, I meant to say topological group in the second paragraph. Also, you are correct, I am asking if there is any group structure on this set in which multiplication is a continuous map. Thanks for the help!2011-12-26
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    @ManuelAraújo He asks for a topological group so it has to have inverses, too. So the previous version of your comment without the "However..." answers the question. : )2011-12-26

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