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I've come up with some examples to apply the Hurewicz theorem to compute $H_1(X)$.

This is only interesting if $\pi_1(X)$ is not abelian. The only examples of $X$ such that $\pi_1(X)$ not abelian I can come up with are $\vee_i S^1$ and $\Sigma_g$ the surface of genus $g$ for $g > 1$.

Does anyone know any other examples, preferably easy ones? Many thanks for your help!

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    Very easy examples would be products of the ones you know already :)2011-07-31
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    your first comment: I was looking for something more interesting, yet not too difficult. : )2011-07-31
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    @gary: the Hurewicz theorem doesn't say that...2011-07-31
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    More seriously (I temporarily confused myself): Take $S^3$ identified with the unit quaternions. The quaternions $\pm 1, \pm i, \pm j, \pm k$ form the quaternion group $Q$. The quotient $S^3/Q$ has fundamental group $Q$. Interesting examples also arise from knot complements (try to compute the fundamental group of the complement of the trefoil knot, for example).2011-07-31
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    @Matt: are you aware that every group is the fundamental group of some space? It's not particularly hard to find a finite $2$-complex whose fundamental group is any finitely presented group, either (start with a wedge of circles, then glue in $2$-cells corresponding to the relations), and their homology isn't difficult to compute.2011-07-31
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    @Qiaochu: No, I wasn't aware of that, thanks.2011-07-31
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    Would it be appropriate to make this a community wiki question?2011-07-31
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    @Matt: ah. Then see http://math.stackexchange.com/questions/36775/is-the-group-of-rational-numbers-the-fundamental-group-of-some-space .2011-07-31
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    @Qiaochu's answer [here](http://math.stackexchange.com/questions/36775/) explains his comment in somewhat more detail and lhf gives the precise reference to Hatcher.2011-07-31

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One more example: complement to any non-trivial knot in $S^3$ has non-abelian $\pi_1$.

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    Thank you, this is my favourite answer!!2011-08-03
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I am late to the party here, but take any finitely presented group, say $G=\langle x_1,\ldots , x_n: r_1,\ldots r_m \rangle$. Take a bouquet of circles embedded in 4-space and a tubular neighborhood thereof. This is a space with fund. group free of rank $n$. Now carve out a tubular neighborhood of the words represented by the relations, and sew in $D^2 \times S^2$s in their place. There is plenty of room to do this in 4-space. So you can construct a $4$-manifold with $\pi_1$ being any finitely presented group that you like.

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    Late to the party or not, at least you gave the first reasonable answer in this thread. In case you don't know it, there's a beautiful and (sort of) related paper by Martin Bridson, *[The geometry of the word problem](http://people.maths.ox.ac.uk/bridson/papers/bfs/bfs.pdf)*.2011-08-01
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Consider the figure 8. Using Van Kampen's theorem, you know that its fundamental group is the free product of the additive group integers with itself. Said group is nastily non-abelian.

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    Matt already gave the bouquets of circles in his question :)2011-07-31
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Try the cube with a twist

Take the quotient space of the cube $I^3$ obtained by identifying each square face with opposite square via the right handed screw motion consisting of a translation by 1 unit perpendicular to the face, combined with a one-quarter twist of its face about it's center point.

I think it is a problem in Hatcher's book somewhere

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    I believe the fundamental group of this quotient ends up being precisely the $8$-element quaternion group.2013-03-29