Is homology determined by cohomology?

Question

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.

Thanks!

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).

Answer 2

The only purpose of this answer is to exhibit a less well known universal coefficient theorem and specialise it to give a(n) (alternative?) verification of the facts offered in Qiaochu's answer.

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By one of the many variants of the universal coefficient theorem, namely Spanier's Algebraic Topology Theorem $6.5.12$, we know that for any PID $R$ and bounded complex $C$ of free $R$-modules with homology of finite type (finitely generated in every degree) and any $R$-module $M$ there are natural short exact sequences $\forall n$: $$0\to\mathsf{Ext}(H^{n+1}(C; R),M)\to H_n(C; M)\to\mathsf{Hom}_R(H^n(C; R), M)\to0$$Where cohomology of $C$ is a shorthand for the cohomology of its dual complex $\mathsf{Hom}_R(C;R)$. This sequence splits naturally (! yes, really) in the coefficient module $M$ (but unnaturally in the complex $C$). Unfortunately all the hypotheses in this theorem are very necessary for the proof, including the boundedness hypothesis (for many of the other UCTs the boundedness is not necessary).

The maps in the sequence are only well defined up to isomorphism, since the proof works by replacing $C$ with another complex $L$. Modulo such isomorphisms however, the final map in the sequence is induced by the following operation: take a pure tensor $c\otimes m$ of $C\otimes_R M$ and assign to it the map which takes a module homomorphism $f:C\to R$ to $f(c)\cdot m$.

Anyway, if $X$ is any topological space then we care about the singular complex $SX$, which is free and nonnegative - certainly bounded - and because $\mathsf{Hom}_R(SX\otimes_{\Bbb Z}R, R)\cong\mathsf{Hom}_{\Bbb Z}(SX,R)$ (as $R$-modules) we can compute its cohomology with coefficients in $R$ by computing the cohomology of the complex of $R$-modules $C:=SX\otimes_{\Bbb Z}R$ which is also free and nonnegative.

The homology of $C$ is of finite type (as an $R$-module) by one of the other universal coefficient theorems if the homology of $SX$ is of finite type (as an Abelian group). In other words, if $X$ is a space of finite type, then the hypotheses of the theorem hold and we know there is that natural short exact sequence which splits. But we could do better, technically, since I believe the converse direction does not hold; if you happen to know your space $X$ is of finite type when considering $R$-coefficients, I don't think that forces $X$ to be of finite type in the usual sense, so we possibly have an even more general domain of discourse. That doesn't seem likely to occur in practice however.

Thus if $X,Y$ are spaces of finite type and have isomorphic cohomology with coefficients in $R$ then (take $M=R)$ they in particular have (unnaturally) isomorphic homology with coefficients in $R$ (by the splitting) and if the continuous function $f:X\to Y$ induces an isomorphism on cohomology with coefficients in $R$ then $f$ induces an isomorphism on homology with coefficients in $R$ by the $5$-lemma. Moreover naturality of the splitting ensures a very nice understanding of how $H_\ast(X;-)$ varies as the coefficient module varies, using only the cohomology data.

Simple examples of spaces with finite type:

• Compact manifolds
• "Levelwise" finite CW and simplicial complexes (to steal the language of Qiaochu's answer)