I am aware of the universal coefficients theorem for cohomology which implies that the homology groups completely determine the cohomology groups. I am wondering if cohomology determines homology in a similar sense? If two spaces have the same cohomology groups do they necessarily have the same homology groups? Can we compute the homology groups from the cohomology groups?

I am aware of Poincare duality - this only applies to "nice" spaces - seeing as homology determines cohomology in the above sense for general spaces, I was wondering if such a fact still holds switching the roles of homology and cohomology.

The universal coefficients theorem for homology tells us about how the homology group with arbitrary coefficients relates to the integer homology groups - so I suppose that this is not the result I am looking for.


  • $\begingroup$ It's a standard fact in algebraic topology that for closed oriented $n$-manifolds $M$, $H_{n - k}(M) \cong H^k(M)$. Look up Poincare duality. $\endgroup$ May 5, 2015 at 17:51
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    $\begingroup$ @BalarkaSen How is that relevant? The post doesn't even contain the word "manifold." $\endgroup$ May 5, 2015 at 17:54
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    $\begingroup$ @MattSamuel It's relevant in the sense that one at least knows that for (good) manifolds, two spaces with same homology has the same cohomology and vice versa, which is what the question asks. $\endgroup$ May 5, 2015 at 17:55

1 Answer 1


Suppose that $X$ is a space such that all of its homology groups are finitely generated. This holds, for example, if $X$ is a "levelwise finite" CW complex (finitely many cells in each dimension), which is a very broad class of spaces. Then universal coefficients implies that all of the cohomology groups are also finitely generated. Moreover, the isomorphism class of each homology and cohomology group is determined by rank and torsion subgroup, and universal coefficients implies that

  • $H^k(X, \mathbb{Z})$ and $H_k(X, \mathbb{Z})$ have the same rank, and
  • The torsion subgroup of $H^k(X, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H_{k-1}(X, \mathbb{Z})$.

Hence the two sequences of isomorphism classes of groups determine each other in this case. They are nearly the same, except that the torsion subgroups are shifted one degree.

Without the finitely generated hypothesis this is no longer true. For every abelian group $A$ and positive integer $n \ge 1$ it is possible to construct a Moore space $X = M(A, n)$, which is a space whose homology vanishes except in degree $n$, where it is isomorphic to $A$, and in degree $0$, where it is $\mathbb{Z}$. By universal coefficients, the cohomology of a Moore space is

  • $H^0(X, \mathbb{Z}) \cong \mathbb{Z}$
  • $H^n(X, \mathbb{Z}) \cong \text{Hom}(A, \mathbb{Z})$
  • $H^{n+1}(X, \mathbb{Z}) \cong \text{Ext}^1(A, \mathbb{Z})$

and all other cohomology vanishes. So the question is whether an abelian group $A$ is determined up to isomorphism by the isomorphism class of $\text{Hom}(A, \mathbb{Z})$ and $\text{Ext}^1(A, \mathbb{Z})$, and the answer is no. Counterexamples cannot be finitely generated: the first one that comes to mind is the following. We have

$$\text{Hom}(\mathbb{Q}, \mathbb{Z}) = 0$$

(straightforward) and

$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \mathbb{R}$$

(nontrivial), and $\text{Ext}^1(-, -)$ preserves direct sums in the first factor, so it follows that the groups $\mathbb{Q}$ and $\mathbb{Q} \oplus \mathbb{Q}$ cannot be distinguished this way, since the two groups are not isomorphic, but $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{R}$ (as abelian groups).

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    $\begingroup$ Do you know an example (Moore space or otherwise) which has trivial cohomology but non-trivial homology? $\endgroup$ Apr 7, 2016 at 1:02
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    $\begingroup$ @Michael: by the universal coefficient theorem this is equivalent to finding an abelian group $A$ such that $\text{Hom}(A, \mathbb{Z})$ and $\text{Ext}^1(A, \mathbb{Z})$ are both trivial (so an example exists iff an example exists which is a Moore space), assuming that you mean reduced cohomology. I don't know an example off the top of my head and am not sure if one exists. You could probably ask this as a new question. $\endgroup$ Apr 7, 2016 at 3:39
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    $\begingroup$ For those interested, I asked this as a new question. $\endgroup$ Apr 9, 2016 at 19:38
  • $\begingroup$ Is the answer unchanged if we allow the coefficient group to be arbitrary, rather than just $\mathbb{Z}$? $\endgroup$
    – Yly
    Dec 8, 2021 at 21:10

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