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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? – Alex Becker Dec 26 '11 at 17:03
I think in the second paragraph you meant to write "topological group" rather than "topological space". – Rudy the Reindeer Dec 26 '11 at 17:08
It cannot be a topological group, because its fundamental group is not abelian. See However, as you only require a continuous multiplication (and not also inverses), I don't know... – Manuel Araújo Dec 26 '11 at 17:09
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! – rfxoneill Dec 26 '11 at 17:19
@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. : ) – Rudy the Reindeer Dec 26 '11 at 18:01

The following lemma will be useful:

Lemma. Let $M$ be a topological space with a continuous multiplication $*$ and an identity element $e$. Then $\pi_1 (M)$ is an abelian group.

Proof. Let $u, v : [0, 1] \to M$ be continuous loops based at the identity element $e$. We construct a homotopy from $u \mathbin{.} v$ to $v \mathbin{.} u$ as follows: let $H : [0, 2] \times [0, 2] \to G$ be the map defined by

$$H(s, t) = \begin{cases} u(s) & 0 \le s \le 1, s + t \le 1 \\ v(s - 1) & 1 \le s \le 2, s - t \ge 1 \\ v(s) & 0 \le s \le 1, t - s \ge 1 \\ u(s - 1) & 1 \le s \le 2, s + t \ge 3 \\ u(\tfrac{1}{2}(s - t + 1)) * v(\tfrac{1}{2}(s + t - 1)) & \text{otherwise} \end{cases} $$

One easily verifies that $H$ is continuous (draw a picture!) and is the required homotopy.

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Thank you this is a big help! This leads me to another question: If I am given a topological space M, is it enough to show that $\pi_1(M)$ is not abelian to show that there is no continuous multiplication with an identity element? – rfxoneill Dec 26 '11 at 18:33
That is the contrapositive of the above lemma, so I'd say so. – SL2 Dec 26 '11 at 18:43

In your second paragraph, you are alluding to the idea that $S^1\vee S^1$ may be a Lie group. To be a Lie group, we require not only a continuous multiplication with identity (which we've already seen $S^1\vee S^1$ can't have), but our space must also be a manifold. However, $S^1\vee S^1$ isn't. Indeed, any open neighbourhood of the intersection point in $S^1\vee S^1$ is homeomorphic to a space that looks like an open $+$ sign, which is not homeomorphic to $\mathbb R$. We can see this because $\mathbb R\backslash \{x\}$ has two connected components for any point $x\in\mathbb{R}$, whereas if $U$ is any neighbourhood of the intersection point $*$ of $S^1\vee S^1$, then $U\backslash \{*\}$ has four connected components.

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