Talk:Fundamental group

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[edit] how to find a space with a given fundamental group

is there a general procedure for finding or constructing a space with a given fundamental group? especially a smooth manifold? -Lethe | Talk 22:48, August 8, 2005 (UTC)

There is such a procedure starting from a presentation for the group - see Corollary 1.28 in Hatcher's on-line book. But there's no way to get a manifold in general, because there aren't enough homeomorphism types of connected manifolds. --Zundark 11:31, 9 August 2005 (UTC)
I can't make out what Zundark is saying. There are uncountably many countable groups, true, but there are uncountably many manifolds, as well. You weren't assuming compactness, were you? I believe that it is correct to say that every finitely presented group can be obtained as the fundamental group of a compact four-manifold. Now, obtaining a general countable group seems a bit harder. But if we give up compactness it should be possible. Best, Sam nead 00:40, 11 October 2005 (UTC)
The original question didn't say anything about countable groups. A connected manifold has cardinality at most 2^{\aleph_0}, so there can't be more than 2^{2^{2^{\aleph_0}}} different connected manifolds. (In fact, I think it's known that there are only 2^{\aleph_1} connected manifolds, including non-paracompact ones). But there are an unlimited number of groups, since there is at least one of every non-zero cardinality. (Of course, the original question didn't say anything about connectedness either, but the fundamental group depends only on the path-component, so it suffices to consider connected manifolds.) --Zundark 08:26, 11 October 2005 (UTC)
I have a brief moment and would like to jump in. For finitely generated fundamental groups on CW-complexes the matter is easy. Let G be a finitely generated group on n generators. Now form the complex of n 1-cells attached to a single point, (or the one-point union or wedge of n circles, if you prefer). Label the edges to correspond to the generators. Now use 2-cells to mod out by relations, by stretching a 2-cell around each sequence of edges that correspondes to a word in the presentation. Okay, now we have a CW-complex with G as fundamental group. The next part is a bit cheap and hand-wavy, but convinces me, so I'll share, but may need help with details. We have an embedding theorem for CW-complexes into R^k, for some large k. We should also be able to find an open neighborhood of this embedding that deformation retracts to it. Then we have it! a (flat!) manifold with fundamental group G.
Oh, BTW, we're using only finitely generated groups here. These are countable. If one uses something like a hawiian ring (\cup_n\{(x,y)\in\mathbb R^2|(x-1/n)^2+y^2=1/n^2\}), one can construct CW-complexes for countably generated G, but the deformation retract idea won't work here. I haven't even addressed the case for uncountable groups here, mentioned by Zundark, above, but for finite dimensional manifolds the fundamental groups are always countable. The worst case for a manifold, i believe, is a countable direct sum (not product!!) of countable groups, which is countable. MotherFunctor 02:30, 16 May 2006 (UTC)

The classifying space construction is also relevant, but it won't give you manifolds in general. - Gauge 04:19, 15 December 2005 (UTC)

Same for Eilenberg-MacLane spaces. --Orthografer 04:36, 11 March 2006 (UTC)

Since everyone seems to be in the right mood, what about morphisms? That is, given a topological space X, a group G and a group morphism from \pi_1(X) to G is there a top. space and a continuous function that induce that situation?

hmmm, what happens if you quotient the manifold by a curve representing for each relation defining the kernal. A problem might be that the loop may be forced to intersect itself. Where is this problem from, anything specific? I don't know how to do it off hand. MotherFunctor 06:09, 15 November 2007 (UTC)
Yeah, sure -- just take a classifying space K(G,1), and realize the given homomorphism by a map X --> K(G,1). Turgidson 13:33, 15 November 2007 (UTC)

I think the symbol for the wedge sum in the third paragraph of section "Functorality" is not a wedge sum but a smash product. Sorry I couldn't change it since I didn't find the right character.. --Cheesus 16:16, 10 August 2006 (UTC)

[edit] fundamental group is abelian iff its path independent

I added this. easy to show.

128.226.164.120 (talk)jesusonfire —Preceding comment was added at 04:18, 9 December 2007 (UTC)

I have removed it because, as written, "path independent" is undefined and I have no idea what you might mean by this. Please don't use "doesn't" in main space articles. Mathsci (talk) 05:22, 9 December 2007 (UTC)