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I want to show the fundamental group of a topological group is abelian. In fact, the question says the topological group is path connected. I do not know where I should use path-connectedness. I think, it is still true if we do not suppose path-connectedness. Right?

I do not know homotopy. I have just learned the fundamental group.

Stefan Hamcke
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square
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    The fundamental group depends only on the path component of the base point, so the path-connectedness just says that there are no irrelevant path components, it's nothing essential. – Daniel Fischer Feb 22 '14 at 23:03
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    How can one know what the fundamental group is without knowing some basic results on homotopy? – PVAL-inactive Feb 23 '14 at 00:10
  • PVAL I know some basic results on homotopy. I have studied chapter0 and beginning of chapter1 Hatcher. I mean I have not studied chapter4. – square Mar 04 '14 at 01:53

1 Answers1

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The usual proof is to show that for all loops $\alpha,\beta \colon [0,1] \to G$ of the topological group $G$, the concatenation $\alpha \cdot \beta$ is homotop to $t \mapsto \alpha(t)\beta(t)$ and to $t \mapsto \beta(t)\alpha(t)$. It requires to exhibit formulae.

Howerver, my favourite proof of this result is the following one, from Grothendieck (I think) : the fundamental group functor $\pi_1 \colon \mathsf{pcTop} \to \mathsf{Grp}$ from the category of path-connected topological spaces to the category of groups respects products (classical lemma), so sends group objects to group objects ; the group objects of $\mathsf{pcTop}$, which are the path-connected topological groups (by definition), are send to group objects of $\mathsf{Grp}$, which are the abelian groups (easy exercise).

Pece
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    Hey, cute abstract nonsense proof! Thanks for posting it! (+1) – Bruno Stonek Feb 23 '14 at 09:41
  • @lenticcatachresis I like the cute in "cute abstract nonsense proof". You're also absolutely right. – Joachim Jan 10 '15 at 19:16
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    Actually, you should define the fundamental group functor on the category of based spaces. Otherwise you can't even define the action on morphisms. – Zhen Lin Mar 05 '15 at 16:02
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    The only proofs I can think of this "classical lemma" and this "easy exercise" use the construction you mention in your first paragraph. Is there a way to avoid the construction and use only "abstract nonsense"? :-) – André Caldas Jan 06 '19 at 19:31
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    @AndréCaldas: I cannot speak about the "classical lemma", but the "easy exercise" at the end has nothing to do with loops or topologies and is indeed cute abstract nonsense. It generalises to https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument. – Torsten Schoeneberg Nov 20 '19 at 19:14