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Suppose $G$ is a topological group with operation $\cdot$ and identity element $x_0$. Let $\Omega (G, x_0)$ denote the set of all loops in $G$ based at $x_0$. For $f, g\in\Omega (G, x_0)$ define $f\otimes g$ by the rule $$(f\otimes g)(s)=f(s)\cdot g(s)$$ Then $\otimes$ induces a group operation on the fundamental group $\pi_1(G, x_0)$ as follows, $$[f]\otimes[g]=[f\otimes g]$$ Is this group operation $\otimes$ same as the usual group operation on $\pi_1(G, x_0)$ ? Is $\pi_1(G, x_0)$ abelian under this new group operation ?

Stefan Hamcke
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pritam
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1 Answers1

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Yes.

Look up the «Eckmann-Hilton argument».

Mariano Suárez-Álvarez
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  • I looked in Wikipedia but I dont understand. I am new in Algebraic Topology, can we prove this directly from the definition ? – pritam Jan 12 '13 at 16:33
  • Everything can be proved directly from the definition! This is explained in pretty much every textbook on algebraic topology which deals with the fundamental group of topological groups. Try a couple. – Mariano Suárez-Álvarez Jan 12 '13 at 16:44