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Is it possible to prove the de Rham isomorphism between de Rham cohomology groups and singular cohomology with coefficients in the reals without resorting to any form of integration?

It seems remotely possible to me, but on the other hand, having deRham isomorphism, we can define integration by using Poincare duality $H^{n}(M, \mathbb{R}) \simeq H^{0}(M, \mathbb{R})$, because for connected manifolds we have $H^{0}(M, \mathbb{R}) \simeq \mathbb{R}$ in a canonical way.

One imagines it could be possible in principle because one can prove that any cohomology theory $E^{\bullet}$ (on, let's say, pairs of CW-complexes) that has the same coefficients as singular cohomology is necessarily isomorphic to it. We prove it by considering the cellular complex associated to the skeletal filtration

$X_{0} \subseteq \ldots \subseteq X_{n} \subseteq \ldots \subseteq X$

with respect to $E^{\bullet}$. Then because $E^{\bullet}$ has exactly the same coefficients as singular cohomology, this complex will be isomorphic to the usual cellular complex.

There are a number of obstructions to this approach for de Rham cohomology. Indeed, one should find a good replacement of the skeletal filtration of a CW-complex, something like "filtration with submanifolds". Also, it is not at all clear to me what is the right definition of de Rham cohomology on "pairs of manifolds".

I hope my question is not too trivial for MathOverflow.

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It is a corollary of the de Rham isomorphism that every function on the line has an antiderivative. How could you prove that without integration? – Tom Goodwillie Nov 21 '12 at 16:31
I think that on this site, Dmitri Pavlov has left some answers which preach an approach similar to what you are after. It is not clear to me, however, that he avoids circularity – Yemon Choi Nov 21 '12 at 17:30
I think that Tom's comment is the answer. (Checking the de Rham resolution is a resolution of the constant sheaf would require the Poincar\'e lemma, and hence integration. I don't see it can be avoided.) – Donu Arapura Nov 21 '12 at 17:49
@Yemon: I had a heated discussion with Dmitri here:…. In the end, he could cook up a quite involved argument for the Poincare lemma that avoided the explicit use of the integral. His claim, that the Lebesgue integral can be constructed by this is, however, totally unclear to me. – Johannes Ebert Nov 21 '12 at 19:55
For a purist with integral-phobia, there are some more obstructions to overcome, even if you take the Poincare lemma as god-given. It seems unclear to me how to prove that manifolds are homotopy equivalent to CW complexes or have good covers without the construction techniques of differential topology, which rely on two pillars, both deeply connected to calculus: Sards Theorem and the existence of solutions of ODEs. – Johannes Ebert Nov 21 '12 at 20:02

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