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.

theanswer. (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