# When are singular homology and singular cohomology isomorphic?

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?

• 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? Oct 12, 2021 at 2:26
• @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." Oct 12, 2021 at 2:28
• 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. Oct 12, 2021 at 4:09
• Regarding your opening paragraph, is an algebraic answer acceptable to you? Oct 12, 2021 at 11:01
• 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." Oct 14, 2021 at 0:15

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).

• 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 $H_0$. Oct 14, 2021 at 0:16
• @JohnPalmieri: thank you! If you put these comments (here and above) in an answer I would accept it. Oct 14, 2021 at 17:15