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Let G be a topological group and let $s_1$ and $s_2$ be loops in G (both loops are based at the identity e of G). Is it true that the loop $s_1s_2$ (where the multiplication is the one of the group structure of G) is equal, in $\pi_1(G,e)$, to the loop $s_1*s_2$ where this product is given by first going around $s_1$ and then $s_2$ (i.e., do we have $[s_1s_2] = [s_1*s_2]$)? If yes, what is the proof for this?

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up vote 10 down vote accepted

Yes and this is known as the Eckmann-Hilton argument. (Wikipedia)

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Thank you (and thanks to Sam Lichtenstein too)! – Down Sep 23 '10 at 15:32

Instead of using the Eckmann-Hilton argument abstractly, you can make it explicit, as follows:

We can reparameterize the loop $s_1$ to be constant (and equal to the neutral element) on $[1/2,1]$ and $s_2$ to be constant on $[0,1/2]$. Then $[s_1 \ast s_2] = [s_1 s_2]$ and since $s_1s_2 = s_2s_1$ we also get that the fundamental group is abelian.

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Well, that's just E-H argument for this situation written down explicitly, isn't it?.. – Grigory M Nov 20 '11 at 20:39
Yes, but I still think that it's shorter and more intuitive this way. (Maybe I shouldn't have said Instead...) – t.b. Nov 20 '11 at 20:44
I agree. (But something like more explicitly would seem more... precise, indeed.) – Grigory M Nov 20 '11 at 20:50
Fixed, I hope. Thanks for pointing it out! – t.b. Nov 20 '11 at 21:00

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