# “Abstract nonsense” proof that the fundamental group of a topological group is abelian

I seem to remember reading somewhere an "abstract nonsense" proof of the well-known fact that the fundamental group of a connected topological group is abelian. By "abstract nonsense" I mean that the proof used little more than the fact that topological groups are the group objects in the category of topological spaces and the fact that $\pi_1$ is a homotopy functor. Does anybody remember how this works? References are fine.

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Google the "Eckman-Hilton argument". – Ryan Budney Dec 8 '10 at 1:43
@Ryan: I'm sure you're right. Would you perhaps consider leaving this as answer (possibly with a little more information and/or a link to a nice reference)? – Pete L. Clark Dec 8 '10 at 2:11

He proves that if $X$ is an $H$-cogroup, then $[X,Y]$ is a group, and that if $Y$ is an $H$-group, then $[X,Y]$ is also a group, so when $X$ and $Y$ are an $H$-cogroups and an $H$-group, respectively, then $[X,Y]$ is a group in two ways. And so on.
I just found this old question and happen to know a very nice "abstract nonsense" argument of a completely different flavor. Let $G$ be a topological group and consider the fundamental groupoid $\Pi(G)$ of $G$, which becomes a monoidal category under the group operation. The unit object is of of course the identity. But it is well known that the endomorphism monoid of the unit object, which in this case is the fundamental group at the identity, embeds into the Bernstein center (the monoid of endomorphisms of the identity functor on $\Pi(G)$), which is always commutative.