I'm learning about de Rham's theorem, but from a very analytical point of view: my algebra is very weak.

My understanding is that de Rham's theorem tells us that for a smooth manifold $M$, the de Rham cohomology groups $H_{DR}^k(M)$ are isomorphic to the singular cohomology groups $H^k(M;\mathbb{R})$ with real coefficients.

From what I gather by looking at simple examples (sphere, torus) and also informal discussions of topological "holes," these real singular cohomology groups are also isomorphic to the corresponding real singular homology groups, $H_k(M;\mathbb{R})$. But I can't find a straightforward discussion or statement of this fact.

So I have two questions:

  1. Is it true that for a smooth manifold $M$ (or possibly any topological space $M$), $H^k(M;\mathbb{R})\cong H_k(M;\mathbb{R})$? If so, how is this proven?

Reading the page on cohomology on Wikipedia, I find the following sentence after a remark about the universal coefficient theorem: "A related statement is that for a field $F$, $H^i(X,F)$ is precisely the dual space of the vector space $H_i(X,F)$." If I could prove this stament, I could then answer my question in the affirmative: since $H_i(X,F)$ and $H^i(X,F)$ are finite-dimensional vector spaces and dual, they must be isomorphic (although not canonically so). But I don't know how to prove the quoted statement, and I'm not sure of its precise relationship to the universal coefficient theorem; Wikipedia merely says this fact is "related," not that it is a consequence of UCT.

  1. Does this fact descend to integral coefficients too, at least for smooth manifolds manifolds? That is, is it true that $H^k(M;\mathbb{Z})\cong H_k(M;\mathbb{Z})$? If not, what is a counterexample?

Edit. Regarding question 2, I've just noticed that real projective spaces provide counterexamples for the case of integral coefficients; thus the answer is no. But does the result hold if none of the groups have torsion?

  • $\begingroup$ It’s not so much whether they’re isomorphic or not, because that could be coincidence. It’s if they are, where did the iso come from, and what does it mean? $\endgroup$
    – Randall
    Oct 12, 2021 at 2:26
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    $\begingroup$ @Randall: I'm not sure what your point is. I would like yes/no answers to these questions. In case of "yes," I would like to know why too, of course: hence my request "how is this proven." $\endgroup$
    – curious21
    Oct 12, 2021 at 2:28
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    $\begingroup$ For fields of characteristic $0$ and nice spaces (like manifolds or cell complexes with finitely many cells in each dimension) homology and cohomology are dual vector spaces. This is indeed a consequence of UCT. $\endgroup$ Oct 12, 2021 at 4:09
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    $\begingroup$ Regarding your opening paragraph, is an algebraic answer acceptable to you? $\endgroup$
    – Lee Mosher
    Oct 12, 2021 at 11:01
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    $\begingroup$ With coefficients in the integers or a field, as long as the homology groups are free (automatic over a field) and finitely generated over the coefficient ring, then the answers are "yes; prove using the universal coefficient theorem." $\endgroup$ Oct 14, 2021 at 0:15

2 Answers 2


When the coefficient ring $R$ is the integers or a field, as long as the homology groups of $X$ are free (automatic over a field) and finitely generated over the coefficient ring, then $H_k(X; R) \cong H^k(X; R)$, by the Universal Coefficient Theorem. Furthermore, if $X$ has the structure of a finite CW complex, for example if $X$ is a compact manifold, then the homology groups will be finitely generated. There is no nice topological description that ensures that the integral homology groups will be torsion-free; spaces with cells in only even dimensions have this property, for example, but that's not a very broad class of spaces.


I believe I have found a partial answer to (1) in the negative.

Let $M\subset\mathbb{R}^2$ be the complement of $\left\{(n,0)\,|\,n\in\mathbb{N}\right\}$ in the plane, construed as a smooth manifold in the natural way. In other words, we delete a countably infinite number of isolated points from the plane.

Then $H^1(M;\mathbb{R})$ and $H_1(M;\mathbb{R})$ are infinite-dimensional. But if, as the sentence I quoted from Wikipedia claims, $H^1$ and $H_1$ must be dual vector spaces (because we are working with real coefficients), then $H^1$ and $H_1$ cannot be isomorphic: as shown by Arturo Magidin in his answer here, the dual of an infinite-dimensional vector space has strictly larger dimension and thus cannot be isomorphic.

This leads me to wonder what restrictions we might put on (1) to prevent this kind of example:

Does assuming $M$ is compact suffice to make (1) true?

It would if compactness implies that real homology or cohomology must be finite-dimensional (and if the statement from Wikipedia is correct).

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    $\begingroup$ Being compact along with having the structure of a CW complex (or less generally, a simplicial complex) will be enough, so as long as you stick to manifolds, it should be fine. If you allow non-manifolds, then a compact space with infinitely many path-connected components (like the Cantor set) will have an infinite-dimensional $H_0$. $\endgroup$ Oct 14, 2021 at 0:16
  • $\begingroup$ @JohnPalmieri: thank you! If you put these comments (here and above) in an answer I would accept it. $\endgroup$
    – curious21
    Oct 14, 2021 at 17:15

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