Free and finitely generated (co)homology If the (integral) cohomology of a space $X$ is free and finitely generated, will the (integral) homology also be free and finitely generated?
In Hatcher's book there's a Proposition 3F.12, that tells me that the integral homology will be finitely generated, but will it also be free?
In my case I'm working with $X=BU(n)$, the classifying space of the unitary group, in which case $H^*(X,\mathbb{Z}) \cong \mathbb{Z}[x_1, \ldots, x_n]$.
I'm trying to prove that $H^*(X,k) \cong k[x_1, \ldots, x_n]$, for a field $k$. I tried using the Universal Coefficient Theorem, but then I think I will need that the homology should be free. I have seen that in general $H^p(X,k)≃Hom_k(H_p(X,k),k)$
, but I'm stuck from here.
Any help regarding the general case or my specific case is greatly appreciated.
Edit: To add to my question, I have managed to solve this for $k=\mathbb{Q}$, but here I use only the finitely generated part, and the fact that $\mathbb{Z}/m \otimes \mathbb{Q} = 0$. If I could say the homology is free in each degree, then a similar argument would apply, without having to use these facts.
 A: Let $X$ be a topological space with finitely generated free $\mathbb{Z}$-cohomology. Assume that the $\mathbb{Z}$-homology is finitely generated as well. Let $T_i$ be the torsion subgroup of $H_i(X,\mathbb{Z})$ (it is finite). All we need to show is that $T_i=0$.
By the UCT, $E_i=\mathrm{Ext}^1(T_i,\mathbb{Z})$ embeds into $H^{i+1}(T,\mathbb{Z})$ so is free of finite rank. But $|T_i|E_i=0$, so that $E_i=0$.
We have an exact sequence $0 \rightarrow A \rightarrow B \rightarrow T_i \rightarrow 0$ with $A,B$ free of finite rank. Then $\mathrm{Hom}(B,\mathbb{Z}) \rightarrow \mathrm{Hom}(A,\mathbb{Z})$ is onto. By the “adapted basis” theorem, it follows that $B/A \cong T_i$ is torsion-free, so $T_i=0$.
A: If $X$ has finitely generated homology and for some $n$, $H_n(X)$ is not free, then it is of the form $(\text{free}) \oplus F$ where $F$ is finite and nonzero. By the Universal Coefficient Theorem, $H^{n+1}(X)$ has $\text{Ext}^1(F, \mathbb{Z}) \cong F$ as a summand.
So by the contrapositive, if each cohomology group is free, the same must be true for the homology groups.
