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.
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