That singular and de Rham cohomologies of a smooth manifold are isomorphic has two proofs that I know of. The classical one uses Stokes' theorem to give the isomorphism explicitly. The second proof that I know uses the machinery of derived categories. Namely, instead of thinking of singular and de Rham cohomologies as the homologies of complexes of abelian groups, one thinks of them as being the hypercohomologies of complexes of sheaves (whose global sections are the abelian groups that appear in the classic definition of singular and de Rham cohomologies). The isomorphism then follows from the following two steps:
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Showing that both complexes of sheaves are exact, implying that they are both quasi-isomorphic to $..\rightarrow 0\rightarrow \mathbb{R}\rightarrow 0\rightarrow …$.
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Showing that the sheaves that appear in both complexes are $\Gamma$-acyclic, implying that the hypercohomology is simply the homology of the complex after taking global sections. (In fact, showing that the sheaves are "flabby".)
I find this second proof to be very comforting, and a good way in general to show that various definitions of cohomology are isomorphic. Indeed, the same argument (mutatis mutandis) would follow for simplicial cohomology. One can also make a similar argument for algebraic de Rham in the affine case. (In the non-affine case the sheaves are not $\Gamma$-acyclic in general; in the affine case this is the algebraic version of Cartan's theorems A and B.)
Sadly, this second proof does not apply to cellular cohomology for the simple reason that the complex of abelian groups for which cellular cohomology is the homology, is not the global sections of a complex of sheaves in any natural way.
In Hatcher the proof of the isomorphism between cellular cohomology and singular cohomology goes through a rather unintuitive argument using relative cohomology. Is there a more satisfying proof? Perhaps, though not necessarily, one that uses derived categories? Alternatively, is there a proof that uses the Grothendieck group over the category of CW-complexes, or some other variation on motivic arguments?
Best Answer
Here is how I like best to think about this, although I'll point to a few other proofs. Consider a general space $X$ (say compactly generated). There is a natural weak homotopy equivalence $\epsilon\colon |SX|\to X$ from the geometric realization of the singular simplicial set of $X$ to $X$. Moreover, $|SX|$ is a CW complex whose cellular chains (and cochains) are isomorphic to the (normalized) singular chains and cochains of $X$. If $X$ itself is a CW complex, then $\epsilon$ is a homotopy equivalence and is homotopic to a cellular homotopy equivalence. The induced map of cellular chains is a chain homotopy equivalence between the singular chains of $X$ and the cellular chains of $X$.
Another proof just checks the Eilenberg-Steenrod axioms for both. Relative homology is irrelevant since there is a version of the axioms for the reduced homology of spaces that is equivalent to the usual axioms for pairs of spaces. There is also a version of the axioms just on CW complexes, where the excision axiom reduces to the tautology that if a CW complex $X$ is the union of subcomplexes $A$ and $B$, then $A/A\cap B$ and $X/B$ are isomorphic CW complexes. The general excision axiom reduces to this version by a purely topological argument (no use of homology in any form needed). Here is a concise reference for these statements: http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf.