When are singular homology and singular cohomology isomorphic

algebraic-topologyhomology-cohomologysmooth-manifolds

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?

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.