0

Suppose we are given spaces $X$ and $Y$. If they have the same homology groups (i.e. $H_i(X,\mathbb{Z})=H_i(Y,\mathbb{Z})$), then by universal coefficient theorem $$0\to\mathrm{Ext}^1(H_{n-1}(C_\bullet),\mathbb{Z})\to H^n(C_\bullet;\mathbb{Z})\to\mathrm{Hom}(H_n(C_\bullet),\mathbb{Z})\to0$$ tells us that all their cohomology are the same.

My question is the converse, if two spaces have the same cohomology groups, are their homology groups the same? By the same theorem, if both $X,Y$ have bounded homology groups then yes, but I cannot proceed. Any proof or disproof is welcome.

Guanyu Li
  • 530
  • As you indicate, the answer is yes if you assume that homology (and hence cohomology) is finitely generated. But there are infinitely generated counterexamples. – Jeroen van der Meer Dec 08 '21 at 18:56
  • I know for a fact that one can use Moore spaces as examples. The space $M(A,n)$ is a space defined such that the homology is $A$ in degree $n$ and $0$ elsewhere. Upon applying UCT, the cohomology in degrees $n$ and $n + 1$ are $Hom(A,\mathbb{Z})$ and $Ext^1(A,\mathbb{Z})$. The trick is that there are non-isomorphic $A$'s that produce the same $Hom(A,\mathbb{Z})$ and $Ext^1(A,\mathbb{Z})$. But I cannot recall what $A$'s one can take. Let me think about it... – Jeroen van der Meer Dec 08 '21 at 19:02
  • 1
    Ah. You can take $A = \mathbb{Q}$ and $A = \mathbb{Q} \oplus \mathbb{Q}$. In fact, there was a previous MSE question about this: https://math.stackexchange.com/questions/1268593/is-homology-determined-by-cohomology – Jeroen van der Meer Dec 08 '21 at 19:09

0 Answers0