In this question, the OP asks whether it is possible to find a CW complex with prescribed homology groups and fundamental group. In my (partial) answer, I point out that by taking the wedge sum of Moore spaces, the homology condition can be achieved. However, the resulting space need not have the prescribed fundamental group. In order to rectify this, it is sufficient to have a positive answer to the following question:

Let $G$ be a group. Does there exist a CW complex $X$ with $\pi_1(X) = G$ and $H_i(X; \mathbb{Z}) = 0$ for $i \geq 2$?

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    $\begingroup$ Sure, just attach whatever cells you need to kill the homology. Cells in degree 3 and higher (which is what you'll use) can't change the fundamental group. $\endgroup$ – user98602 Mar 11 '17 at 18:04
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    $\begingroup$ @MikeMiller: This approach had occurred to me. However, in the case of a torus $S^1\times S^1$, the 2-cell corresponds to a non-zero element in second homology. How can you glue on a 3-cell to kill this? Every map $S^2 \to S^1\times S^1$ is nullhomotopic so the resulting space just becomes $S^1\times S^1\vee S^3$. $\endgroup$ – Michael Albanese Mar 11 '17 at 20:58
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    $\begingroup$ Good point. Indeed the answer is no. A space with fundamental group $\Bbb Z^2$ necessarily has nontrivial second cohomology by looking at the map on cohomology to $K(G,1)$. $\endgroup$ – user98602 Mar 11 '17 at 21:00
  • $\begingroup$ @MikeMiller: Would you mind expanding this into an answer (at some point)? $\endgroup$ – Michael Albanese Mar 11 '17 at 21:01
  • $\begingroup$ Yeah, I'm trying to work out the generality the observation works in. $\endgroup$ – user98602 Mar 11 '17 at 21:02

Let $G$ be a group, and $X$ a space with $\pi_1 = G$. Then there is a map $f: X \to K(G,1)$ (unique up to homotopy) that is the identity on fundamental groups. Then $f$ is surjective on second homology, and the kernel of $f$ is precisely the homology classes representable by 2-spheres. This is because you can build a model of $K(G,1)$ out of $X$ - attach 3-cells and higher to kill off homotopy groups - and can extend the map $f$ over the new $K(G,1)$ so that it's still the identity on $\pi_1$. So the map $f$ can be considered as the inclusion into the model of $K(G,1)$ we built, and all we did was kill off the 2-spheres in second homology.

On the other hand, we can build a space $X$ with $\pi_1 X = G$, $H_2(X) = H_2(G)$, and $H_k(X) = 0$ for all $k \geq 3$. Just write down a 2-complex whose fundamental group is $G$ (for instance, a presentation complex), and attach 3-cells to kill off any homology classes represented by spheres. These don't contribute in third homology because their boundary is nonzero in the cellular homology chain complex. (You could possibly still have third homology, but not with $\Bbb Z$ coefficients.)

  • $\begingroup$ I don't know anything about homology of groups. Is $H_2(G) = H_2(K(G, 1); \mathbb{Z})$? $\endgroup$ – Michael Albanese Mar 19 '17 at 13:07
  • $\begingroup$ @MichaelAlbanese Yeah - for the purpose of this answer, take that as a fact. $\endgroup$ – user98602 Mar 19 '17 at 15:49
  • $\begingroup$ Sorry, I meant to say "as a definition"! $\endgroup$ – user98602 Mar 20 '17 at 3:59
  • $\begingroup$ For the record, the question and its answer are related to exercise 4.2.24 of Hatcher's Algebraic Topology. $\endgroup$ – Michael Albanese Aug 27 '17 at 21:04

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