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Jul 9, 2011 at 20:47 comment added J.C. Ottem Some of these answers have $H_1=0$ also.
Jul 9, 2011 at 20:18 comment added user16335 Ops, I have just seen the mistake in the title. Sorry, in fact I mean H_1 and not H^1... . I would like to thank you all for your quick answers and useful examples.
Jul 8, 2011 at 9:48 comment added Tom Mrowka To explain Joel Fine's answer. By transversality you can make a loop in the regular neighborhood miss the 2-complex and hence a smaller regular neighborhood. Then the loop is homotopic into the boundary. If a loop in the boundary is null homotopic in the regular neighborhood the disk it bounds in the regular neighbor can similarly be made to miss the 2-complex and hence is homotopic into the boundary. Thus the two complex and the boundary of the regular neighborhood have the same fundamental group. – Tom Mrowka 11 hours ago
Jul 7, 2011 at 19:19 comment added Neil Strickland @Joel: why is the fundamental group of the boundary of a regular neighbourhood the same as that of the original complex?
Jul 7, 2011 at 14:27 comment added Joel Fine You've almost answered your own question. You can realise any finitely presented group as the fundamental group of a closed 4-manifold. You take your 2-dimensional CW complex, embed it in R^5 (dim 5 is necessary so you can separate all the 2-cells) and then take the boundary of a tubular neighbourhood (and perhaps smooth off any corners). The result is a closed 4-manifold with the same fundamental group as the CW complex, unless I'm mistaken.
Jul 7, 2011 at 14:12 answer added J.C. Ottem timeline score: 9
Jul 7, 2011 at 12:26 answer added Neil Strickland timeline score: 5
Jul 7, 2011 at 12:16 answer added Francesco Polizzi timeline score: 4
Jul 7, 2011 at 12:08 history edited agt
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Jul 7, 2011 at 12:06 answer added Matthew Kahle timeline score: 12
Jul 7, 2011 at 11:39 history asked Luciano Mari CC BY-SA 3.0