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Suppose that X is a spherical complex with $H_q(X;F)=0$ for all $q>0$ where $F=\mathbb{Q}$ or $F=\mathbb{Z}/p\mathbb{Z}$, as $p$ runs over all primes. I want to prove $H_q(X;\mathbb{Z})=0$.

I know Singular Homology Theory and some of Singular Cohomology. I know that $H_q(X;\mathbb{Z})$ is a finitely generated $\mathbb Z$-module for every q, since $\mathbb{Z}$ is a Noetherian ring and $X$ is a spherical complex. But I don't know how to continue proving.

I give a definition of a spherical complex: Start with a finite discrete set of points, and successively attach cells, possibly of varying dimensions, but finite in number. The resulting Hausdorff space is called a spherical complex.

Thank you for your help!

S. Ha
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  • The term I know for what you describe is a "finite CW complex". On a separate note, do you know of the Universal Coefficient Theorem? –  Jun 10 '14 at 15:57
  • I believe you should use the universal coefficient theorem to show that $H_(X,\mathbb{Z})$ tensor with the rational and the finite fields are zero. This will imply that $H_(X,\mathbb{Z})$ has no torsion-free elements and no elemets of finte order. – Rene Schipperus Jun 10 '14 at 15:59
  • I am reading Algebraic Topology by Greenberg and Harper, and this book gives that definition. It seems that Hatcher and many references call those "finite CW complex", as you said. – S. Ha Jun 10 '14 at 16:00
  • Thank you for your comments! I will try to prove it with the universal coefficient theorem. – S. Ha Jun 10 '14 at 16:01
  • @MikeMiller I saw the latter sentence just now. Actually there is the universal coefficient theorem for singular cohomology, not for singular homology in the book. I found the homology version on the internet and it looks similar to the cohomology version. Is it possible to convert the problem to cohomology version and prove it? – S. Ha Jun 10 '14 at 16:06
  • @S.Ha I put a sketch of a sketch of the second argument below. –  Jun 10 '14 at 16:11
  • Actually the definition of a CW complex has another condition on top of what the OP is considering, namely, when a $k$-dimensional cell is attached, the image of its $k-1$-sphere boundary must be contained in the union of the cells of dimension $\le k-1$. – Lee Mosher Jun 10 '14 at 16:16

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Hint: by the universal coefficient theorem for homology, we have an isomorphism $$H_q(X, F) \cong \left(H_q(X, \mathbb Z) \otimes F\right) \oplus \text{Tor}(H_{i-1}(X,\mathbb Z), A).$$ Necessarily both terms vanish, and the fact that $H_q(X, \mathbb Z)$ is a finitely generated abelian group gives it an explicit form you can easily deal with.

EDIT: here's an approach with the UCT for cohomology. We know that $$H^i(X, F) \cong \text{Ext}_R^1(H_{i-1}(X,R),F) \oplus \text{Hom}_R(H_i(X,R),F)$$ Calculate $H^i(X,F)$ in two ways: first by using $R=F$, and then after seeing that $H^i(X,F)=0$ calculate it using $R=\mathbb Z$; again because $\text{Ext}$ and $\text{Hom}$ are well-behaved and you can explicitly write down what $H_i$ must be (by the classification of finitely generated abelian groups), the theorem you want falls out.