I have the following problem. I need to compute homology of $\mathbb{C}^n$ with hyperplanes of the form $z_i=z_j$ and $z_i=-z_j$ thrown out. By the tools of Orlik-Solomon algebras I know the integral cohomology of this:
- All cohomology groups are free;
- The Poincaré polynomial is given by $$\prod_{i=0}^{n-2}(1+(2i+1)t)(1+(n-1)t)$$
Thus I know my cohomology ring. How can I pass, if that is possible, to integral homology?
One way may be to use UCT, but in that case I have to prove that homology groups are free. Other way may be to use some sort of duality, but then I have to compactify my space - so I am not sure how the cohomology of one-point compactification behave.
Use the universal coefficient theorem and the fact that the exact sequence splits. The module $Ext(H_{r-1}(X),\mathbb{Z})$ is torsion, and since it appears as a direct summand of the free module $H^r(X)$ it must be zero. You can use the structure theorem for abelian groups, the distributivity of $Ext(-,\mathbb{Z})$ over direct sums, and the other elementary properties of this functor to conclude that $H_{r-1}(X)$ is free.