From: Johannes Ebert Date: Tue, 27 Feb 2007 23:01:16 +0100 (CET) Hello, here is a partial answer to your question on fundamental groups of topological group. If G is a Lie group, then your seqeunce is exact: Because the fundamental group of a topological group is abelian and because the multiplication in pi_1 (G) is induced by the multiplication in G, the map G \times G \to [G;G] induces the zero map on \pi_1. So you need to know that [G,G] \to G is injective on \pi_1. If you look at the fibration sequence [G,G] \to G \to G^ab, you see that this is true because \pi_2 (G^ab)=0 (here you use that G is a Lie group). In general, it might be the case that \pi_2 (G^ab) is nontrivial, and I do not know an argument in that case. Kind regards, Johannes Ebert >> >> Subject: Question about fundamental groups of topological groups >> From: Masoud >> Date: Mon, 26 Feb 2007 23:07:00 -0600 >> >> Let G be a connected topological group. Is the following complex exact? >> >> \pi_1(G\times G) ----> \pi_1([G,G])---->\pi_1(G) >> >> Here all the fundamental groups have as their base points the identity >> elements of the respective groups. The first map is induced by the >> commutator map >> >> (g,h) \mapsto ghg^{-1}h^{-1}, >> >> the second map is induced by the natural inclusion. >> >> Warning: The commutator map is not a group morphism. _________________________________________________________________________